Important Questions For You!

Mathematics is one of the crucial and most scoring subjects in ICSE Class 10. TopperLearning presents study materials for ICSE Class 10 Mathematics which will enable students to score well in the board examination. The syllabus includes certain challenging concepts like Value Added Tax, Ratio and Proportion, Quadric Equations among others which require effective study materials. Our study materials for ICSE Class 10 consist of video lessons, question banks, sample papers, revision notes and past year papers which improve the quality of learning.

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Chapters 1: GST [Goods and Services Tax]

Q 1. M/s Ram Traders, Delhi, provided the following services to M/s Geeta Trading Company in Agra (UP). Find the amount of bill.                [3M]

Number of services

8

12

10

16

Cost of each service (in Rs.)

680

320

260

420

GST %

5

12

18

12


Q 2.
 National Trading Company, Meerut (UP) made the supply of the following goods/services to Samarth Traders, Noida (UP). Find the total amount of bill if the rate of GST = 12%                                                                                                      [4M]

Quantity (no. of pieces)

20

30

12

40

MRP (in Rs. per piece)

225

320

300

250

Discount %

40

30

50

40


Q 3.
Mr. Malik went on a tour to Goa. He took a room in a hotel for two days at the rate of Rs. 5000 per day. On the same day, his friend John also joined him. Hotel provided an extra bed charging Rs. 1000 per day for the bed. How much GST, at the rate of 28% is charged by the hotel in the bill to Mr. Malik for both the days?                                                                              [3M]


Q 4.
A is a dealer in Meerut (U.P.). He supplies goods/services, worth Rs. 15,000 to a dealer B in Ratlam (M.P.). Dealer B, in turn, supplies the same goods/services to dealer C in Jabalpur (M.P.) at a profit of Rs. 3000. If rate of tax (under GST system) is 18%, find                                                                                                                                                                           [4M]

  1. The cost of goods/services to the dealer C in Jabalpur.
  2. Net tax payable by dealer B.

Q 5.
For a dealer A, the list price of an article is Rs. 9000, which he sells to dealer B at some lower price. Further, dealer B sells the same article to a customer at its list price. If the rate of GST is 18% and dealer B paid a tax, under GST, equal to Rs. 324 to the government, find the amount (inclusive of GST) paid by dealer B.                                                                    [4M]


Q 6.
A is a manufacturer of T.V. sets in Delhi. He manufacturers a particular brand of T.V. set and marks it at Rs. 75,000. He then sells this T.V. set to a wholesaler B in Punjab at a discount of 30%. The wholesaler B raises the marked price of the T.V. set bought by 30% and then sells it to dealer C in Delhi. If the rate of GST = 5% find tax (under GST) paid by wholesaler B to the government.    [4M]  


Chapter 2: Banking (Recurring Deposit Account)

Q 1. Renu deposited Rs. 500 per month in a bank for 15 months under a recurring deposit account scheme. What will be the maturity value of her deposits, if the rate of interest is 6% per annum and the interest is calculated at the end of every month? [3M]


Q 2.
Mr A opened a recurring deposit account in a certain bank and deposited Rs. 2520 per month for 5 years. Find the maturity value of this account if the bank pays interest at the rate of 9% per year.                                                                [3M]   


Q 3.
Arun has a recurring deposit account for 2 years at 5% p.a. He receives Rs. 1250 as interest on maturity.           [4M]

  1. Find the monthly instalment amount
  2. Find the maturity amount


Q 4. 
Mrs Singh deposits Rs. 1750 per month in her recurring bank account for a period of 2 years. At the time of maturity, she would get Rs. 47250. Find the                                                                                                                       [4M] 

  1. Rate of interest p.a.
  2. Total interest earned by Mrs Singh


Chapter 3: Linear Inequations (in one variable)

Q 1. Solve the given inequation and graph the solution on the number line.              [3M]

6y – 2 < y + 8 ≤ 3y + 6; y ε  R                                                                                                                         


Q 2.
If A = {x : 9x – 3 > 2x + 4, x ε R} and B = {x : 12x – 7 ≥ 5 + 6x, x ε R}, find the range of set A ∩ B and represent it on the number line.                                                                                                                                                     [4M] 


Q 3.
Solve the inequation and represent the solution set on the number line.                                                            [4M]

begin mathsize 12px style 7 plus straight x less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus 2 less or equal than 9 over 2 2 straight x comma space straight x element of straight I end style

Chapter 4: Quadratic Equations

Q 1. Without solving the quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.    [3M] 

x2 + 2(m – 3)x + (m + 3) = 0


Q 2.
 Solve: begin mathsize 12px style fraction numerator x over denominator x minus 2 end fraction minus fraction numerator x minus 2 over denominator x end fraction equals 1 1 half end style                                                                                                                           [3M]


Q 3.
 begin mathsize 12px style open parentheses 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses fraction numerator 1 over denominator straight x minus 1 end fraction equals 7 over 8. end style Give your answer correct to two significant decimal places.                                            [3M]


Q 4.
Solve for x: begin mathsize 12px style 4 open parentheses straight x minus 1 over straight x close parentheses squared minus 8 open parentheses straight x plus 1 over straight x close parentheses equals 29 semicolon space straight x not equal to 0 end style                                                                                               [4M]


Q 5.
The sum of a square of a certain whole number and 45 times it is 900. Find the number.                                         [3M]


Q 6. The product of a boy’s age two years ago with his age 5 years later is 60. Find his present age.                               [3M]


Q 7. By increasing the speed of a car by 15 km/hr, the time of the journey for a distance of 60 km is reduced by 20 minutes. Find the original speed of the car.                                                                                                                                       [4M]


Q 8. The area of a big room is 375 m2. If the length is decreased by 10 m and the breadth is increased by 10 m, then the area would remain unaltered. Find the length of the room.                                                                                      [4M]


Q 9. A motor boat, whose speed is 6 km/h in still water, goes 8 km downstream and returns in a total time of 3 hours. Find the speed of the stream.                                                                                                                                [3M]


Q 10. Rs. 600 is divided equally among ‘x’ children. If the number of children were 10 more, then each would have got Rs. 10 less. Find ‘x’.                                                                                                                                                                [3M]

 

Chapter 5: Ratio and Proportion

Q 1. If begin mathsize 12px style fraction numerator square root of straight a plus square root of straight b over denominator square root of straight a minus square root of straight b end fraction equals 5 over 2 end style, find the value of  begin mathsize 12px style fraction numerator straight a squared plus straight b squared over denominator straight a squared minus straight b squared end fraction end style.                                                                                                    [3M]


Q 2. 
If a, b, c and d are in proportion, prove that  begin mathsize 12px style fraction numerator ma squared plus nb squared over denominator mc squared plus nd squared end fraction equals fraction numerator ma squared minus nb squared over denominator mc squared minus nd squared end fraction end style .                                                             [4M]


Q 3.
If begin mathsize 12px style fraction numerator straight x cubed plus 3 xy squared over denominator straight y cubed plus 3 straight x squared straight y end fraction equals 63 over 62 end style, then find x:y.                                                                                                                     [4M]


Q 4.
Find the numbers such that their mean proportion is 14 and third proportion is 112.                                                 [3M] 

 

Chapter 6:  Factorisation of Polynomials

Q 1. x – 2 is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by x – 3, the remainder is 3. Find the values of a and b.                                                                                                                                          [3M]


Q 2.
Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence, factorise the given expression completely using the factor theorem.           [3M]


Q 3.
The polynomials 3x3 – 9x2 + 5x – 13 and (a + 1)x2 – 7x + 5 leaves the same remainder when divided by x – 3. Find the value of a.                                                                                                                                                [3M]

   

Chapter 7: Matrices

Q 1. Find the value of x given that A2 = B                                                                                                                    [3M]

begin mathsize 12px style straight A equals open square brackets table row cell negative 4 end cell 2 row 3 1 end table close square brackets comma space straight B equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets end style


Q 2.
If begin mathsize 12px style straight A equals open square brackets table row 2 3 row cell negative 2 end cell cell negative 1 end cell end table close square brackets end style and begin mathsize 12px style straight B equals open square brackets table row 1 4 row cell negative 3 end cell 0 end table close square brackets end style; find the matrix X such that AX = B.                                                              [4M]


Q 3.
Given begin mathsize 12px style straight A equals open square brackets table row 2 cell negative 1 end cell row 2 0 end table close square brackets end style, begin mathsize 12px style straight B equals open square brackets table row cell negative 3 end cell 2 row 4 0 end table close square brackets end style and begin mathsize 12px style straight C equals open square brackets table row 1 0 row 0 2 end table close square brackets end style, find the matrix X such that  A + X = 2B + C.                           [3M]


Q 4.
Given begin mathsize 12px style straight A equals open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets end style and  begin mathsize 12px style straight I equals open square brackets table row 1 0 row 0 1 end table close square brackets end style,  and A2 = 9A + mI. Find m.                                                                          [4M]


Q 5.
Solve for x and y : begin mathsize 12px style open square brackets table row cell straight x minus straight y space space end cell cell straight x plus straight y end cell end table close square brackets open square brackets table row 3 cell negative 4 end cell row 5 cell negative 1 end cell end table close square brackets equals open square brackets table row 24 2 end table close square brackets end style                                                                                         [3M]


Q 6.
Evaluate: begin mathsize 12px style open square brackets table row cell 4 space sin space 30 degree end cell cell 2 space cos space 60 degree end cell row cell sin space 90 degree end cell cell 2 space cos space 0 degree end cell end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets end style                                                                                                             [3M]

 

Chapters 8: Arithmetic Progression

Q 1. The 11th  term of an AP is 80 and the 16th term is 110. Find the 31st term.                                                                [3M]


Q 2.
If the third term of an AP is 18 and the seventh term is 30, then find the series.                                                         [3M]


Q 3.
If the 3rd and the 9th terms of an Arithmetic Progression are 4 and –8 respectively, then find:  S20                              [4M]


Q 4.
Split 180 into three parts such that these parts are in AP and the product of the two extreme terms is 3500.                [4M]

 

Chapters 9: Coordinate Geometry

Q 1. Using a graph paper, plot the points A(3, 5) and (0, 5).                                                                                            [4M]

  1. Reflect A and B in the origin to get the images A’ and B’.
  2. Write the co-ordinates of A’ and B’.
  3. State the geometrical name for the figure ABA’B’.
  4. Find its perimeter.


Q 2.
Use a graph paper for this question. A(1, 1), B(5, 1), C(4, 2) and D(2, 2) are the vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C and D are reflected in the origin A’, B’, C’ and D’, respectively. Locate them on the graph sheet and write their co-ordinates.                                                                                                                                  [4M]


Q 3.
Use a graph paper for this question:                                                                                                                     [6M]

(Take 2cm = 1 unit on both x and y axes)

i.  Plot the following points:
A(0,4), B(2,3), C(1,1) and D(2,0).
ii. Reflect points B, C, D about the y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
iii. Join the points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation to the line about which if this closed figure obtained is folded, the two parts of the figure exactly coincide.


Q 4.
The mid-point of the line segment joining (2a, 4) and (−2, 2b) is (1, 2a + 1). Find the values of a and b.                  [3M]


Q 5.
In the given figure, line ACB meets the y-axis at point B and the x-axis at point A. C is the point (−6, 6) and AC:CB = 2:3. Find the coordinates of A and B.                                                                          [3M]


Q 6.
Show that P(2, m – 2) is a point of trisection of the line segment joining the points A(4, −2) and B(1, 4). Hence, find the value of m.                                                                                                                                                       [3M]


Q 7.
The equation of the line is 4x – 5y = 9. Find the                                                                                                        [4M]

  1. Slope of the line
  2. Equation of the passing through the intersection of the lines x + y = 1 and 2x + y = 2 with slope begin mathsize 12px style fraction numerator negative 5 over denominator 4 end fraction end style.


Q 8.
Find the equation of a line                                                                                                                      [3M]

  1. Whose inclination is 30˚ and y-intercept is −2
  2. With inclination 60˚ and passing through (−4, 2)
  3. Passing through the points (3, −4) and (7, 1)


Q 9.
Write the equation of the line whose gradient is begin mathsize 12px style 3 over 2 end style and which passes through P, where P divides the line segment joining A(−2, 6) and B(3, −4) in the ratio 2:3.                                                                                                                   [3M]


Q 10.
If the slope of a line joining P(6, k) and Q(1 – 3k, 3) is ½, find                                                                                [3M]

  1. k
  2. the mid-point of PQ using the value of k found in (i)


Chapter 10: Similarity

Q 1. In ΔABC, DE is drawn parallel to BC. If AB = 16 cm, AD = 4 cm, and AC = 24 cm, find                                             [4M]

  1. EC
  2. begin mathsize 12px style fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses DABC close parentheses end fraction end style
  3. begin mathsize 12px style fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses trapezium space BCED close parentheses end fraction end style

Q 2.
The scale of the map is 1:200000. A plot of land of area 20 km2 is to be represented on the map. Find the                  [3M]
  1. Number of kilometres on the ground which is represented by 1 cm on the map
  2. Area in sq km which can be represented by 1 cm2
  3. Area of the map which represents the plot of land


Q 3.
A model of a ship is made to a scale 1: 300.                                                                                                           [3M]

  1. The length of the model of the ship is 2 m. Calculate the length of the ship.
  2. The area of the deck of the ship is 180,000 m2. Calculate the area of the deck of the model.
  3. The volume of the model is 6.5 m3. Calculate the volume of the ship.

 

Chapters 11: Circles

Q 1. In the figure, AD = BC and AB || DC, begin mathsize 12px style angle end styleBAC = 40˚ and begin mathsize 12px style angle end styleCBD = 60˚. Find                                                              [4M]

  1. begin mathsize 12px style angle end styleBCD
  2. begin mathsize 12px style angle end styleBCA
  3. begin mathsize 12px style angle end styleABC
  4. begin mathsize 12px style angle end styleADB


Q 2.
AB and CD are two chords on the same side. AB = 12 cm and CD = 24 cm. Distance between two parallel chords is 4 cm. Find the radius of the circle.                                                                [4M]


Q 3.
 In the given figure, ABCD is a cyclic quadrilateral. Find the values of x and y.                                                            [3M]


Q 4. 
In the given figure, O is the centre of the circle. AB is a tangent to it at point B. begin mathsize 12px style angle end styleBDC = 60˚. Find begin mathsize 12px style angle end styleBAO.          [4M]


Q 5. 
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If begin mathsize 12px style angle end styleACO = 25˚, find                   [3M]

  1. begin mathsize 12px style angle end styleBCO
  2. begin mathsize 12px style angle end styleAOB
  3. begin mathsize 12px style angle end styleAPB


Q 6. 
AB is a line segment and M is its mid-point. Semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle C(O, r) is drawn so that it touches all the three semi-circles. Prove that begin mathsize 12px style straight r equals 1 over 6 AB end style                           [4M] 

 

Chapter 12: Area and Volume of Solids

Q 1. A conical tent is to accommodate 11 people. Each person must have 4 sq. metres of the space on the ground and 20 cubic metres of air to breathe. Find the height of the cone.                                                                                                        [4M]


Q 2. 
A steel wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the mass of the wire, assuming the density of steel to be 8.88 g per cm3.                      [3M]


Q 3. 
The area of the base of a right circular cone is 28.26 sq. cm. If its height is 4 cm, find its volume and the curved surface area. (use π = 3.14)                                                                                                                                                   [3M]


Q 4.
The volume of a right circular cone is 660 cm3 and the diameter of its base is 12 cm. Calculate                              [4M]

  1. the height of the cone,
  2. the slant height of the cone,
  3. the total surface area of the cone. 

 

Chapters 13: Trigonometrical Identities

Q 1. Prove that: begin mathsize 12px style fraction numerator cosA over denominator 1 plus sinA end fraction plus tanA equals secA end style                                                                                                                [3M]


Q 2.
 Evaluate: begin mathsize 12px style 2 open parentheses fraction numerator tan 38 degree over denominator cot 52 degree end fraction close parentheses squared plus open parentheses fraction numerator cot 67 degree over denominator tan 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator sec 83 degree over denominator cosec 7 degree end fraction close parentheses end style                                                                                      [3M]


Q 3.
If sin A + tan A = p and tan A - sin A = q, prove that begin mathsize 12px style straight p squared minus straight q squared equals 4 square root of pq end style.                                                              [4M]


Q 4.
Prove that: sin4A - cos4A = 1 - 2cos4A                                                                                                                [3M]


Q 5.
If x = a sinθ and y = b tanθ, show that begin mathsize 12px style straight a squared over straight x squared minus straight b squared over straight y squared equals 1 end style.                                                                                         [3M]

Chapters 14: Heights and Distances

Q 1. A guard observes a boat from a tower at a height of 150 m above sea level to be at an angle of depression of 25° (Take tan25° = 0.4663).                                                                                                         [4M]                   

  1. Calculate the distance of the boat from the foot of the tower to the nearest metre.
  2. After some time, it is observed that the boat is 180 m from the foot of the tower. Calculate the new angle of depression.


Q 2.
A vertical pole and a vertical tower are at the same ground level. From the top of the pole, the angle of elevation at the top of the tower is 30° and the angle of depression of the foot of the tower is 60°. Find the height of the pole if the height of the tower is 100 m.                                                                                                          [4M]

 

Chapters 15: Graphical representation

Q 1. Draw a histogram for the following discontinuous distribution:                                         [6M]

Class interval

11–20

21–30

31–40

41–50

51–60

Frequency

25

10

27

18

20

 

Q 2. Draw a frequency polygon for the following distribution (continuous data):                 [4M]

Age group

0-8

8-16

16-24

24-32

32-40

40-48

48-56

No. of girls

5

16

9

29

7

10

2


Q 3. Draw a frequency polygon for the following distribution (discontinuous data):           [6M]

Marks

11–20

21–30

31–40

41–50

51–60

61–70

No. of boys

10

18

36

58

72

80

 

Chapters 16: Measures of Central Tendency

Q 1. The weight of 40 children in a class was recorded in kg as follows:                                   [4M]

Weight (in kg)

45

47

49

51

53

55

No. of children

8

6

7

12

5

2

Calculate the following for the given distribution:

  1. Median
  2. Mode


Q 2. 
The height of 25 students of a class is given in the following table:                                    [4M]

Height (in cm)

140

150

160

170

180

No. of students

6

8

4

5

2

Find the mean height using the short-cut method.


Q 3.
From the given frequency distribution table, find the following using a graph:             [6M]

  1. Lower quartile  
  2. Upper quartile
  3. Inter – quartile range

Class interval

5–10

10–15

15–20

20–25

25–30

30–35

Frequency

4

8

5

13

11

9

 

Q 4. The following numbers are written in the descending order of their values:                    [3M]

70, 62, 50, a – 2, a – 6, a – 8, 28, 22, 16 and 10

If their median is 35, find the value of ‘a’.                                                                                  


Q 5.
Calculate the mean marks of the following distribution using the step-deviation method.               [4M]

Class interval

25–30

30–35

35–40

40–45

45–50

50–55

Frequency

8

15

25

17

14

11


Q 6.
Using a graph, draw an ogive for the following distribution which shows the marks obtained in the Mathematics paper by 150 students.                                                                                                                                        [6M]

Marks

0–10

10–20

20–30

30–40

40–50

50–60

60–70

70–80

No. of students

5

8

20

34

26

31

12

14

Use the ogive to estimate

i.  Median    ii. Number of students who score more than 55

 

Chapter 17: Probability

Q 1. Twenty identical cards are numbered from 1 to 20. A card is drawn randomly from those 20 cards. Find the probability that the number on the card drawn is                                                                                                                          [4M]

  1. divisible by both 2 and 5
  2. greater than 20


Q 2. A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting.                                          [3M]

  1. an ace
  2. a queen
  3. not a jack.

Q 3. 
A single letter is selected at random from the word ‘STATISTICS’. Find the probability that it is a                         [3M]
  1. vowel  
  2. consonant


Q 4.
A face of a die is marked with a number 1, 2, 3, -1, -2 and -3, respectively. The die is thrown once. Find the probability of getting                                                                                                  [4M]

  1. a negative integer
  2. the smallest positive integer


Q 5.
Three identical coins are tossed together. Find the probability of getting                       [4M]

  1. exactly two tails        
  2. at least one tail

Chapters 1: GST [Goods and Services Tax]

Q 1. M/s Ram Traders, Delhi, provided the following services to M/s Geeta Trading Company in Agra (UP). Find the amount of bill.   [3M]

Number of services

8

12

10

16

Cost of each service (in Rs.)

680

320

260

420

GST %

5

12

18

12

Solution: 

Number of services

Cost of each service (in Rs.)

GST %

MRP

IGST 

8

680

5

5440

272

12

320

12

3840

460.8

10

260

18

2600

468

16

420

12

6720

806.4

 

 

 

18,600

2007.2

Amount of bill = Selling price + IGST = 18,600 + 2007.2 =Rs. 20,607.2

 

Q 2. National Trading Company, Meerut (UP) made the supply of the following goods/services to Samarth Traders, Noida (UP). Find the total amount of bill if the rate of GST = 12%                                                                                                      [4M]

Quantity (no. of pieces)

20

30

12

40

MRP (in Rs. per piece)

225

320

300

250

Discount %

40

30

50

40

Solution: 

MRP (in Rs. per piece)

Quantity (no. of pieces)

Discount %

MRP

Selling price

SGST @ 6%

CGST @ 6%

225

20

40

4500

2700

162

162

320

30

30

9600

6720

403.2

403.2

300

12

50

3600

1800

108

108

250

40

40

10,000

6000

360

360

 

 

 

 

17,220

1033.2

1033.2

Amount of bill = Selling price + SGST + CGST = 17,220 + 2066.4 = Rs. 19,286.4


Q 3. Mr. Malik went on a tour to Goa. He took a room in a hotel for two days at the rate of Rs. 5000 per day. On the same day, his friend John also joined him. Hotel provided an extra bed charging Rs. 1000 per day for the bed. How much GST, at the rate of 28% is charged by the hotel in the bill to Mr. Malik for both the days?                                                                              [3M]

Solution: According to the question,

The amount of bill = 5000 × 2 + 1000 + 1000

                           = 10,000 + 2000 = Rs. 12,000

GST = 28% of 12,000 = begin mathsize 12px style 28 over 100 cross times 12 comma 000 equals 3360 end style

GST charged by Mr. Malik Rs. 3360.


Q 4. A is a dealer in Meerut (U.P.). He supplies goods/services, worth Rs. 15,000 to a dealer B in Ratlam (M.P.). Dealer B, in turn, supplies the same goods/services to dealer C in Jabalpur (M.P.) at a profit of Rs. 3000. If rate of tax (under GST system) is 18%, find                                                                                                                              [4M]

  1. The cost of goods/services to the dealer C in Jabalpur.
  2. Net tax payable by dealer B.
Solution: For A (case of inter-state transaction)

S.P. in Meerut = Rs. 15,000

For B (case of inter-state transaction)

C.P.= Rs. 15,000
IGST = 18% of 15,000 = begin mathsize 12px style 18 over 100 cross times 15 comma 000 end style= Rs. 2700

Input tax for B = Rs. 2,700

C.P. in Ratlam = Rs. 2700 and profit = Rs. 3000
S.P. in Ratlam = 15,000 + 3000 = Rs. 18,000

For C (case of intra-state transaction)

C.P = Rs. 18,000
CGST = 9% of 18,000 = begin mathsize 12px style 9 over 100 cross times 18 comma 000 end style = Rs. 1620
SGST = begin mathsize 12px style 9 over 100 cross times 18 comma 000 end style = Rs. 1620

Output tax for B = Rs. 3240

i. C.P. for the dealer C in Jabalpur

= S.P. for the dealer in Ratlam + GST

= 18,000 + 1620 + 1620

= Rs. 21,240

ii. Net GST payable by the dealer B

= Output tax – Input tax

= 1620 + 1620 – 2700

= Rs. 540

 

Q 5. For a dealer A, the list price of an article is Rs. 9000, which he sells to dealer B at some lower price. Further, dealer B sells the same article to a customer at its list price. If the rate of GST is 18% and dealer B paid a tax, under GST, equal to Rs. 324 to the government, find the amount (inclusive of GST) paid by dealer B.                                                          [4M]

Solution: Let A sells to dealer B at Rs. x lower price.

According to the question,

Net Tax paid by dealer B is

⇒ Out put tax – Input Tax = Rs. 324

⇒ 18% of 9000 – 18% of (9000 – x) = 324

⇒ 1620 – 1620 + 18% of x = 324

⇒ 18% of x = 324

⇒ x = 1800

Hence, selling price of B = 9000 – 1800 = Rs. 7200

The amount (inclusive of GST) paid by dealer B

= 7200 + 18% of 7200 = 7200 + 1296 = Rs. 8496


Q 6. A is a manufacturer of T.V. sets in Delhi. He manufacturers a particular brand of T.V. set and marks it at Rs. 75,000. He then sells this T.V. set to a wholesaler B in Punjab at a discount of 30%. The wholesaler B raises the marked price of the T.V. set bought by 30% and then sells it to dealer C in Delhi. If the rate of GST = 5% find tax (under GST) paid by wholesaler B to the government.                                                                                                                  [4M]  

Solution: Initial marked price by manufacturer A is Rs. 75,000

B bought the T.V. at a discount of 30%.

Cost price of B = 70% of 75,000 = Rs. 52,500 ….(i)

GST paid by B for purchase = 5% of 52,500 = Rs. 2625 ….(ii)

B sells T.V. by increasing marked price by 30%.

Selling price for B = 75,000 + 30% of 75,000 = Rs. 97,500 …(iii)

GST charged by B on selling of T.V. = 5% of 97,500  = Rs. 4875 … (iv)

GST paid by B to the government

= GST charged on selling price – GST paid against purchase price

= 4875 – 2625

= Rs. 2250

 

Chapter 2: Banking (Recurring Deposit Account)

Q 1. Renu deposited Rs. 500 per month in a bank for 15 months under a recurring deposit account scheme. What will be the maturity value of her deposits, if the rate of interest is 6% per annum and the interest is calculated at the end of every month?                            [3M]

Solution: According to the question,

P = Rs. 500, n = 15 months, r = 6% per year

begin mathsize 12px style Interest equals straight P cross times fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 cross times 12 end fraction cross times straight r over 100 equals 500 cross times fraction numerator 15 cross times 16 over denominator 2 cross times 12 end fraction cross times 6 over 100 equals Rs. space 300 end style

Maturity value = P × n + interest = 500 × 15 + 300 = Rs. 7800

 

Q 2. Mr A opened a recurring deposit account in a certain bank and deposited Rs. 2520 per month for 5 years. Find the maturity value of this account if the bank pays interest at the rate of 9% per year.                                       [3M]   

Solution: According to the question,

P = Rs. 2520, n = 5 years = 5 × 12 months = 60 months, r = 9% per year

begin mathsize 12px style Interest equals straight P cross times fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 cross times 12 end fraction cross times straight r over 100 equals 2520 cross times fraction numerator 60 cross times 61 over denominator 2 cross times 12 end fraction cross times 9 over 100 equals Rs. space 34 comma 587 end style

Maturity value = P × n + interest

                              = 2520 × 60 + 34587

                              = 151200 + 34587

                              = Rs. 1,85,787

Hence, Mr A will get Rs. 185787 as maturity value.                                                                                                                                                                

Q 3. Arun has a recurring deposit account for 2 years at 5% p.a. He receives Rs. 1250 as interest on maturity.                      [4M]

  1. Find the monthly instalment amount
  2. Find the maturity amount

Solution: According to the question,

n = 2 years = 24 months, r = 5% p.a., Interest = Rs. 1250

i. Interest = begin mathsize 12px style straight P cross times fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 cross times 12 end fraction cross times straight r over 100 end style

begin mathsize 12px style rightwards double arrow 1250 equals straight P cross times fraction numerator 24 cross times 25 over denominator 2 cross times 12 end fraction cross times 5 over 100 end style

⇒ 1250 = 1.25P

⇒ P = Rs. 1000

ii. Maturity amount = P × n + interest

                           = 1000 × 24 + 1250

                           = 24000 + 1250

                           = Rs. 25250

 

Q 4. Mrs Singh deposits Rs. 1750 per month in her recurring bank account for a period of 2 years. At the time of maturity, she would get Rs. 47250. Find the                                                                                                                      [4M] 

  1. Rate of interest p.a.
  2. Total interest earned by Mrs Singh

Solution: According to the question,

i. Maturity value = Rs. 47250, P = Rs. 1750, n = 2 years = 24 months

 Maturity value = begin mathsize 12px style straight P cross times straight n plus straight P cross times fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 cross times 12 end fraction cross times straight r over 100 end style

begin mathsize 12px style rightwards double arrow 47250 equals 1750 cross times 24 plus 1750 cross times fraction numerator 24 cross times 25 over denominator 2 cross times 12 end fraction cross times straight r over 100 end style

⇒ 47250 = 42000 + 437.5r

⇒ 437.5r = 5250

⇒ r = 12% p.a.

ii. Interest =  begin mathsize 12px style straight P cross times fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 cross times 12 end fraction cross times straight r over 100 equals 1750 cross times fraction numerator 24 cross times 25 over denominator 2 cross times 12 end fraction cross times 12 over 100 equals Rs. space 5250 end style


Chapter 3: Linear Inequations (in one variable)

Q 1. Solve the given inequation and graph the solution on the number line.                                       [3M]

6y – 2 < y + 8 ≤ 3y + 6; y ε  R                                                                                                                      

Solution:  6y – 2 < y + 8 ≤ 3y + 6; y ε R

⇒ 6y – 2 < y + 8 and y + 8 ≤ 3y + 6
6y – y < 8 + 2 ⇒ 5y < 10  ⇒ y < 2
Also,
y + 8 ≤ 3y + 6 ⇒ 8 – 6 ≤ 3y – y ⇒ 2 ≤ 2y ⇒ 1 ≤ y ⇒ y ≥ 1
The solution set is {y : y ε R, 1 ≤ y < 2}.
This can be represented on the number line as shown below:


Q 2. If A = {x : 9x – 3 > 2x + 4, x ε R} and B = {x : 12x – 7 ≥ 5 + 6x, x ε R}, find the range of set A ∩ B and represent it on the number line.                                                                                                                                                     [4M]     

Solution: 

9x – 3 > 2x + 4 ⇒ 9x – 2x > 4 + 3 ⇒ 7x > 7 ⇒ x > 1

A = {x : 9x – 3 > 2x + 4, x ε R} = {x : x > 1, x ε R}

Also, 12x – 7 ≥ 5 + 6x ⇒ 12x – 6x ≥ 5 + 7 ⇒ 6x ≥ 12 ⇒ x ≥ 2

B = {x : 12x – 7 ≥ 5 + 6x, x ε R} = {x : x ≥ 2, x ε R}

A ∩ B = {x : x ≥ 2 x ε R}

The solution set should be x greater than or equal to 2.

This can be represented on the number line as shown below:


Q 3. Solve the inequation and represent the solution set on the number line.                                                            [4M]

begin mathsize 12px style 7 plus straight x less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus 2 less or equal than 9 over 2 2 straight x comma space straight x element of straight I end style

Solution: 

begin mathsize 12px style 7 plus straight x less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus 2 less or equal than 9 over 2 plus 2 straight x comma space straight x element of straight I
7 plus straight x less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus 2 space and space fraction numerator 5 straight x over denominator 2 end fraction minus 2 less or equal than 9 over 2 plus 2 straight x
Now comma space 7 plus straight x less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus 2
rightwards double arrow 7 plus 2 less or equal than fraction numerator 5 straight x over denominator 2 end fraction minus straight x rightwards double arrow 9 less or equal than fraction numerator 5 straight x minus 2 straight x over denominator 2 end fraction rightwards double arrow 9 less or equal than fraction numerator 3 straight x over denominator 2 end fraction rightwards double arrow 6 less or equal than straight x
Also comma
fraction numerator 5 straight x over denominator 2 end fraction minus 2 less or equal than 9 over 2 plus 2 straight x rightwards double arrow fraction numerator 5 straight x over denominator 2 end fraction minus 2 straight x less or equal than 9 over 2 plus 2 rightwards double arrow fraction numerator 5 straight x minus 4 straight x over denominator 2 end fraction less or equal than fraction numerator 9 plus 4 over denominator 2 end fraction rightwards double arrow straight x over 2 less or equal than 13 over 2 rightwards double arrow straight x less or equal than 13 end style

The solution set is {x : X ε I, 6 ≤ x ≤ 13} = {6, 7, 8, 9, 10, 11, 12, 13}.
This can be represented on the number line as shown below:


Chapter 4: Quadratic Equations

Q 1. Without solving the quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.    [3M] 

x2 + 2(m – 3)x + (m + 3) = 0

Solution: Comparing x2 + 2(m – 3)x + (m + 3) = 0 with ax2 + bx + c = 0

a = 1, b = 2(m – 3) and c = m + 3
According to the question,
b2 – 4ac = 0
[2(m – 3)]2 – 4 × 1 × (m + 3) = 0
⟹ 4(m – 3)2 – 4(m + 3) = 0
⟹ 4(m2 – 6m + 9 – m – 3) = 0
⟹ 4(m2 – 7m + 6) = 0
⟹ m2 – 7m + 6 = 0
⟹ m2 – 6m – m + 6 = 0
⟹ m(m – 6) – (m – 6) = 0
(m – 6)(m – 1) = 0
⟹ m = 6 or m = 1


Q 2. Solve: begin mathsize 12px style fraction numerator x over denominator x minus 2 end fraction minus fraction numerator x minus 2 over denominator x end fraction equals 1 1 half end style                                                                                                                           [3M]

Solution:

begin mathsize 12px style fraction numerator straight x squared minus open parentheses straight x minus 2 close parentheses squared over denominator straight x open parentheses straight x minus 2 close parentheses end fraction equals 3 over 2 end style

⟹ 2[x2 - (x - 2)2] = 3x (x - 2)

⟹ 2[x2 - x2 + 4x - 4] = 3x2 - 6x

⟹ 2(4x - 4) = 3x2 - 6x

⟹ 8x - 8 = 3x2 - 6x

⟹ 3x2 - 6x - 8x + 8 = 0

⟹ 3x2 - 14x + 8 = 0

⟹ 3x2 - 12x - 2x + 8 = 0

⟹ 3x(x - 4) -2(x - 4) = 0

⟹ (x - 4)(3x - 2) = 0

⟹ x = 4 or x = begin mathsize 12px style 2 over 3 end style


Q 3. begin mathsize 12px style open parentheses 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses fraction numerator 1 over denominator straight x minus 1 end fraction equals 7 over 8. end style Give your answer correct to two significant decimal places.                                            [3M]

Solution:

begin mathsize 12px style open parentheses 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses fraction numerator 1 over denominator straight x minus 1 end fraction equals 7 over 8
Let space fraction numerator 1 over denominator straight x minus 1 end fraction equals straight m
rightwards double arrow open parentheses 1 plus straight m close parentheses straight m equals 7 over 8 end style

⟹ 8(m + m2) = 7

⟹ 8m2 + 8m - 7 = 0

Comparing 8m2 + 8m - 7 = 0 with ax2 + bx + c = 0

a = 8, b = 8 and c = -7

begin mathsize 12px style straight x equals fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction
rightwards double arrow straight x equals fraction numerator negative 8 plus-or-minus square root of 8 squared minus 4 cross times 8 cross times open parentheses negative 7 close parentheses end root over denominator 2 cross times 8 end fraction
rightwards double arrow straight x equals fraction numerator negative 8 plus-or-minus square root of 288 over denominator 16 end fraction
straight x equals fraction numerator negative 8 plus-or-minus 12 square root of 2 over denominator 16 end fraction equals fraction numerator negative 2 plus-or-minus 3 square root of 2 over denominator 4 end fraction end style


Q 4. Solve for x: begin mathsize 12px style 4 open parentheses straight x minus 1 over straight x close parentheses squared minus 8 open parentheses straight x plus 1 over straight x close parentheses equals 29 semicolon space straight x not equal to 0 end style                                                                                               [4M]

Solution:

begin mathsize 12px style 4 open parentheses straight x minus 1 over straight x close parentheses squared minus 8 open parentheses straight x plus 1 over straight x close parentheses equals 29 semicolon space straight x not equal to 0
Let space straight x plus 1 over straight x equals straight y
We space know space that
open parentheses straight x plus 1 over straight x close parentheses squared minus open parentheses straight x minus 1 half close parentheses squared equals 4
rightwards double arrow straight y squared minus open parentheses straight x minus 1 over straight x close parentheses squared equals 4
rightwards double arrow 4 open parentheses straight x minus 1 over straight x close parentheses squared minus 8 open parentheses straight x plus 1 over straight x close parentheses equals 29 semicolon space straight x not equal to 0 end style

⟹ 4(y2 - 4) - 8y = 29

⟹ 4y2 - 16 - 8y = 29

⟹ 4y2 - 8y - 45 = 0

⟹ 4y2 - 18y + 10y - 45 = 0

⟹ 2y(2y - 9) + 5(2y - 9) = 0

⟹ (2y + 5)(2y - 9) = 0

begin mathsize 12px style rightwards double arrow straight y equals negative 5 over 2 or space straight y equals 9 over 2
rightwards double arrow straight x plus 1 over straight x equals negative 5 over 2 or space straight x plus 1 over straight x equals 9 over 2
rightwards double arrow straight x plus 1 over straight x equals negative 5 over 2
rightwards double arrow fraction numerator straight x squared plus 1 over denominator straight x end fraction equals negative 5 over 2 end style

⟹ 2(x2 + 1) = -5x

⟹ 2x2 + 5x + 2 = 0

⟹ 2x2 + 4x + x + 2 = 0

⟹ 2x(x + 2) + (x + 2) = 0

⟹ (2x + 1)(x + 2) = 0

begin mathsize 12px style rightwards double arrow straight x equals negative 1 half or space straight x equals negative 2
Also comma
straight x plus 1 over straight x equals 9 over 2
rightwards double arrow fraction numerator straight x squared plus 1 over denominator straight x end fraction equals 9 over 2 end style

⟹ 2(x2 + 1) = 9x

⟹ 2x2 - 9x + 2 = 0

Comparing 2x2 - 9x + 2 = 0 With ax2 + bx + c = 0,

a = 2, b = -9 and c = 2

begin mathsize 12px style straight x equals fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction
rightwards double arrow straight x equals fraction numerator 9 plus-or-minus square root of open parentheses negative 9 close parentheses squared minus 4 cross times 2 cross times 2 end root over denominator 2 cross times 2 end fraction
rightwards double arrow straight x equals fraction numerator 9 plus-or-minus square root of 81 minus 16 end root over denominator 4 end fraction
rightwards double arrow straight x equals fraction numerator 9 plus-or-minus square root of 65 over denominator 4 end fraction
Hence comma space straight x equals fraction numerator 9 plus-or-minus square root of 65 over denominator 4 end fraction comma space minus 2 comma space minus 1 half
end style

Q 5. The sum of a square of a certain whole number and 45 times it is 900. Find the number.                                         [3M]

Solution: Let the number be x.

According to the question,
x2 + 45x = 900
⟹ x2 + 45x – 900 = 0
⟹ x2 + 60x – 15x – 900 = 0
⟹ x(x + 60) – 15(x + 60) = 0
(x + 60)(x – 15) = 0
x = – 60 or x = 15
As x should be a whole number.
Hence, the required number is 15.


Q 6. The product of a boy’s age two years ago with his age 5 years later is 60. Find his present age.                               [3M]

Solution: Let the boy’s present age be x years.

His age two years ago was (x – 2) years.
His age five years later will be (x + 5) years.
According to the question,
(x – 2)(x + 5) = 60
x2 + 3x – 10 = 60
x2 + 3x – 70 = 0
(x  + 10)(x – 7) = 0
x = -10 or x = 7
x cannot be negative, so x = 7.
Hence, the boy’s present age is 7 years.


Q 7. By increasing the speed of a car by 15 km/hr, the time of the journey for a distance of 60 km is reduced by 20 minutes. Find the original speed of the car.                                                                                                                                       [4M]

Solution: Let x km/hr be the original speed of the car.

Time taken to cover 60 km = begin mathsize 12px style 60 over straight x end style hour
Time taken to cover 60 km when the speed is increased by 15 km/hr = begin mathsize 12px style fraction numerator 60 over denominator straight x plus 15 end fraction end style hour
It is given that the time to cover 60 km reduced by 20 minutes =  begin mathsize 12px style 20 over 60 end style hour
According to the question,

begin mathsize 12px style 60 over straight x minus fraction numerator 60 over denominator straight x plus 15 end fraction equals 20 over 60
60 open parentheses 1 over straight x minus fraction numerator 1 over denominator straight x plus 15 end fraction close parentheses equals 1 third
60 open parentheses fraction numerator straight x plus 15 minus straight x over denominator straight x open parentheses straight x plus 15 close parentheses end fraction close parentheses equals 1 third
fraction numerator 60 cross times 15 over denominator straight x open parentheses straight x plus 15 close parentheses end fraction equals 1 third end style

x(x + 15) = 2700

x2 +  15x - 2700 = 0

x2 + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x + 60)(x - 45) = 0

x = -60 or x = 45

Since x cannot be negative.

Hence, x = 45

Thus, the original speed of the car is 45 km/hr.


Q 8. The area of a big room is 375 m2. If the length is decreased by 10 m and the breadth is increased by 10 m, then the area would remain unaltered. Find the length of the room.                                                                               [4M]

Solution: Let the length and breadth of the room be x m and y m, respectively.

Area of the room = length × breadth
375 = xy

begin mathsize 12px style straight y equals 375 over straight x end style

Length when decreased by 10 m = (x – 10) m
Breadth when increased by 10 m = (y + 10) m = begin mathsize 12px style open parentheses 375 over straight x plus 10 close parentheses straight m end style
According to the question,

begin mathsize 12px style open parentheses straight x minus 10 close parentheses open parentheses 375 over straight x plus 10 close parentheses equals 375
open parentheses straight x minus 10 close parentheses open parentheses fraction numerator 375 plus 10 straight x over denominator straight x end fraction close parentheses equals 375 end style

(x - 10)(375 + 10X) = 375

375x + 10x2 - 3750 - 100x = 375x

10x2 - 100x - 3750 = 0

x2 - 10x - 375 = 0

(x + 15)(x - 25) = 0

As x cannot be negative, x = 25.
Thus, the length of the room is 25 m.


Q 9. A motor boat, whose speed is 6 km/h in still water, goes 8 km downstream and returns in a total time of 3 hours. Find the speed of the stream.                                                                                                                                      [3M]

Solution:

Let the speed of the stream = x km/hr
Speed of the boat downstream = (6 + x) km/hr
Speed of the boat upstream = (6 – x) km/hr
Also, time taken to go 8 km downstream = begin mathsize 12px style fraction numerator 8 over denominator 6 plus straight x end fraction hr end style
Time taken to return =  begin mathsize 12px style fraction numerator 8 over denominator 6 minus straight x end fraction hr end style
According to the question,

begin mathsize 12px style fraction numerator 8 over denominator 6 plus straight x end fraction plus fraction numerator 8 over denominator 6 minus straight x end fraction equals 3
8 open parentheses fraction numerator 6 minus straight x plus 6 plus straight x over denominator open parentheses 6 plus straight x close parentheses open parentheses 6 minus straight x close parentheses end fraction close parentheses equals 3 end style

96 = 3(6 + x)(6 - x)

32 = 36 - x2

x2 = 4

x = ±2

x cannot be negative.
Hence, x = 2.
Thus, the speed of the stream is 2 km/hr.


Q 10. Rs. 600 is divided equally among ‘x’ children. If the number of children were 10 more, then each would have got Rs. 10 less. Find ‘x’.                                                                                                                                         [3M]

Solution: When Rs. 600 is divided equally among ‘x’ children.

Each child gets = Rs. begin mathsize 12px style 600 over straight x end style

When the number of children were 10 more than the original number of children, each child gets = Rs. begin mathsize 12px style fraction numerator 600 over denominator straight x plus 10 end fraction end style

According to the question,
begin mathsize 12px style 600 over straight x minus fraction numerator 600 over denominator straight x plus 10 end fraction equals 10
60 open parentheses 1 over straight x minus fraction numerator 1 over denominator straight x plus 10 end fraction close parentheses equals 1 end style

60 × 10 = x(x + 10)

x2 + 10x - 600 = 0

x2 + 30x - 20x - 600 = 0

x(x + 30) -20 (x + 30) = 0

(x + 30) (x - 20) = 0

x = -30 or x = 20

But x cannot be negative.

Hence, x = 20.

 

Chapter 5: Ratio and Proportion

Q 1. If begin mathsize 12px style fraction numerator square root of straight a plus square root of straight b over denominator square root of straight a minus square root of straight b end fraction equals 5 over 2 end style, find the value of  begin mathsize 12px style fraction numerator straight a squared plus straight b squared over denominator straight a squared minus straight b squared end fraction end style.                                                                                                    [3M]

Solution:

begin mathsize 12px style fraction numerator square root of straight a plus square root of straight b over denominator square root of straight a minus square root of straight b end fraction equals 5 over 2
Using space componendo space and space dividendo comma
rightwards double arrow fraction numerator square root of straight a plus square root of straight b plus square root of straight a minus square root of straight b over denominator square root of straight a plus square root of straight b minus left parenthesis square root of straight a minus square root of straight b right parenthesis end fraction equals fraction numerator 5 plus 2 over denominator 5 minus 2 end fraction
rightwards double arrow fraction numerator 2 square root of straight a over denominator 2 square root of straight b end fraction equals 7 over 3
rightwards double arrow fraction numerator square root of straight a over denominator square root of straight b end fraction equals 7 over 3
rightwards double arrow straight a over straight b equals 49 over 9
rightwards double arrow straight a squared over straight b squared equals 2401 over 81
Using space componendo space and space dividendo comma
rightwards double arrow fraction numerator straight a squared plus straight b squared over denominator straight a squared minus straight b squared end fraction equals fraction numerator 2401 plus 81 over denominator 2401 minus 81 end fraction
rightwards double arrow fraction numerator straight a squared plus straight b squared over denominator straight a squared minus straight b squared end fraction equals 2482 over 2320 end style


Q 2. If a, b, c and d are in proportion, prove that  begin mathsize 12px style fraction numerator ma squared plus nb squared over denominator mc squared plus nd squared end fraction equals fraction numerator ma squared minus nb squared over denominator mc squared minus nd squared end fraction end style .                                                             [4M]

Solution: 

begin mathsize 12px style straight a comma space straight b comma space straight c space and space straight d space are space in space proportion.
rightwards double arrow straight a over straight b equals straight c over straight d equals straight k
rightwards double arrow straight a equals bk space and space straight c equals dk
LHS equals fraction numerator ma squared plus nb squared over denominator mc squared plus nd squared end fraction equals fraction numerator straight m open parentheses bk close parentheses squared plus nb squared over denominator straight m left parenthesis dk right parenthesis squared plus nd squared end fraction equals fraction numerator mb squared straight k squared plus nb squared over denominator md squared straight k squared plus nd squared end fraction equals fraction numerator straight b squared open parentheses mk squared plus straight n close parentheses over denominator straight d squared open parentheses mk squared plus straight n close parentheses end fraction equals straight b squared over straight d squared
rightwards double arrow fraction numerator ma squared plus nb squared over denominator mc squared plus nd squared end fraction equals straight b squared over straight d squared space space space space space... open parentheses straight i close parentheses
RHS equals fraction numerator ma squared minus nb squared over denominator mc squared minus nd squared end fraction equals fraction numerator straight m open parentheses bk close parentheses squared minus nb squared over denominator straight m left parenthesis dk right parenthesis squared minus nd squared end fraction equals fraction numerator mb squared straight k squared minus nb squared over denominator md squared straight k squared minus nd squared end fraction equals fraction numerator straight b squared open parentheses mk squared minus straight n close parentheses over denominator straight d squared open parentheses mk squared minus straight n close parentheses end fraction equals straight b squared over straight d squared
rightwards double arrow fraction numerator ma squared minus nb squared over denominator mc squared minus nd squared end fraction equals straight b squared over straight d squared space space space space space... open parentheses ii close parentheses
From space open parentheses straight i close parentheses space and space open parentheses ii close parentheses
rightwards double arrow fraction numerator ma squared plus nb squared over denominator mc squared plus nd squared end fraction equals fraction numerator ma squared minus nb squared over denominator mc squared minus nd squared end fraction end style


Q 3. If begin mathsize 12px style fraction numerator straight x cubed plus 3 xy squared over denominator straight y cubed plus 3 straight x squared straight y end fraction equals 63 over 62 end style, then find x:y.                                                                                                                     [4M]

Solution:

begin mathsize 12px style fraction numerator straight x cubed plus 3 xy squared over denominator straight y cubed plus 3 straight x squared straight y end fraction equals 63 over 62
Using space componendo space and space dividendo comma
fraction numerator straight x cubed plus 3 xy squared plus straight y cubed plus 3 straight x squared straight y over denominator straight x cubed plus 3 xy squared minus open parentheses straight y cubed plus 3 straight x squared straight y close parentheses end fraction equals fraction numerator 63 plus 62 over denominator 63 minus 62 end fraction
fraction numerator straight x cubed plus straight y cubed plus 3 xy open parentheses straight x plus straight y close parentheses over denominator straight x cubed minus straight y cubed minus 3 xy open parentheses straight x minus straight y close parentheses end fraction equals 125 over 1
fraction numerator open parentheses straight x plus straight y close parentheses open parentheses straight x squared minus xy plus straight y squared close parentheses plus 3 xy open parentheses straight x plus straight y close parentheses over denominator open parentheses straight x minus straight y close parentheses open parentheses straight x squared plus xy plus straight y squared close parentheses minus 3 xy open parentheses straight x minus straight y close parentheses end fraction equals 125 over 1
fraction numerator open parentheses straight x plus straight y close parentheses open square brackets open parentheses straight x squared minus xy plus straight y squared close parentheses plus 3 xy close square brackets over denominator open parentheses straight x minus straight y close parentheses open square brackets open parentheses straight x squared plus xy plus straight y squared close parentheses minus 3 xy close square brackets end fraction equals 125 over 1
fraction numerator open parentheses straight x plus straight y close parentheses open parentheses straight x squared plus 2 xy plus straight y squared close parentheses over denominator open parentheses straight x minus straight y close parentheses open parentheses straight x squared minus 2 xy plus straight y squared close parentheses end fraction equals 125 over 1
fraction numerator open parentheses straight x plus straight y close parentheses open parentheses straight x plus straight y close parentheses squared over denominator open parentheses straight x minus straight y close parentheses open parentheses straight x minus straight y close parentheses squared end fraction equals 125 over 1
open parentheses straight x plus straight y close parentheses cubed over open parentheses straight x minus straight y close parentheses cubed equals 125 over 1
fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction equals 5 over 1
Using space componendo space and space dividendo comma
fraction numerator straight x plus straight y plus straight x minus straight y over denominator straight x plus straight y minus open parentheses straight x minus straight y close parentheses end fraction equals fraction numerator 5 plus 1 over denominator 5 minus 1 end fraction
fraction numerator 2 straight x over denominator 2 straight y end fraction equals 6 over 4
straight x over straight y equals 3 over 2 end style


Q 4. Find the numbers such that their mean proportion is 14 and third proportion is 112.                                                 [3M] 

Solution:

Let the numbers be x and y.

xy = 142 

begin mathsize 12px style rightwards double arrow xy equals 196
rightwards double arrow straight x equals 196 over straight y end style

And 112x = y2

Put x = begin mathsize 12px style 196 over straight y end style in 112x = y2, we get

begin mathsize 12px style rightwards double arrow fraction numerator 112 cross times 196 over denominator straight y end fraction equals straight y squared
rightwards double arrow straight y equals 28
And space straight x equals 196 over 28 equals 7 end style

So, numbers are = 7, 14 and 28.

 

Chapter 6:  Factorisation of Polynomials

Q 1. x – 2 is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by x – 3, the remainder is 3. Find the values of a and b.                                                                                                                                           [3M]

Solution: Let f(x) = x3 + ax2 + bx + 6

x – 2 is a factor of f(x)

⇒ f(2) = 0

⇒ 23 + a × 22 + b × 2 + 6 = 0

⇒ 8 + 4a + 2b + 6 = 0

⇒ 4a + 2b = −14

⇒ 2a + b = −7 … (i)

⇒ f(x) is divided by x – 3, it leaves the remainder 3.

⇒ f(3) = 3

⇒ 33 + a × 32 + b × 3 + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b = −30

⇒ 3a + b = −10 … (ii)

⇒ Solving (i) and (ii), we get a = −3 and b = −1.

 

Q 2. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence, factorise the given expression completely using the factor theorem.           [3M]

Solution: Let p(x) = 2x3 + 5x2 – 11x – 14

To prove that 2x + 7 is a factor of p(x).

Hence, we find

begin mathsize 12px style straight p open parentheses fraction numerator negative 7 over denominator 2 end fraction close parentheses equals 2 cross times open parentheses fraction numerator negative 7 over denominator 2 end fraction close parentheses cubed plus 5 open parentheses fraction numerator negative 7 over denominator 2 end fraction close parentheses squared minus 11 cross times open parentheses fraction numerator negative 7 over denominator 2 end fraction close parentheses minus 14
equals 2 cross times open parentheses fraction numerator negative 343 over denominator 8 end fraction close parentheses plus 5 open parentheses 49 over 4 close parentheses plus 77 over 2 minus 14
equals fraction numerator negative 343 over denominator 4 end fraction plus 245 over 4 plus 77 over 2 minus 14 equals 0 end style

Hence, 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14.

p(x) = 2x3 + 5x2 – 11x – 14
⇒ p(x) = (2x + 7)(x2 – x – 2)
⇒ p(x) = (2x + 7)(x2 – 2x + x – 2)
⇒ p(x) = (2x + 7)[x(x – 2) + (x – 2)]
⇒ p(x) = (2x + 7)(x – 2)(x + 1)

 

Q 3. The polynomials 3x3 – 9x2 + 5x – 13 and (a + 1)x2 – 7x + 5 leaves the same remainder when divided by x – 3. Find the value of a.                                                                                                                                                           [3M]

Solution: Let p(x)=3x3 – 9x2 + 5x – 13 and q(x)=(a + 1)x2 – 7x + 5

According to the question,

⇒ p(3) = q(3)

⇒ 3 × 33 – 9 × 32 + 5 × 3 – 13 = (a + 1) × 32 – 7 × 3 + 5

⇒ 81 – 81 + 15 – 13 = 9a + 9 – 21 + 5

⇒ 2 = 9a – 7 9a = 9 a = 1                                           


Chapter 7: Matrices

Q 1. Find the value of x given that A2 = B                                                                                                                    [3M]

begin mathsize 12px style straight A equals open square brackets table row cell negative 4 end cell 2 row 3 1 end table close square brackets comma space straight B equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets end style

Solution:

begin mathsize 12px style straight A equals open square brackets table row cell negative 4 end cell 2 row 3 1 end table close square brackets comma space straight B equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets
straight A squared equals straight B
rightwards double arrow open square brackets table row cell negative 4 end cell 2 row 3 1 end table close square brackets open square brackets table row cell negative 4 end cell 2 row 3 1 end table close square brackets equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets
rightwards double arrow open square brackets table row cell negative 4 cross times open parentheses negative 4 close parentheses plus 2 cross times 3 end cell cell open parentheses negative 4 close parentheses cross times 2 plus 3 end cell row cell 3 cross times open parentheses negative 4 close parentheses plus 3 end cell cell 3 cross times 2 plus 1 end cell end table close square brackets equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets
rightwards double arrow open square brackets table row 22 cell negative 6 end cell row cell negative 9 end cell 7 end table close square brackets equals open square brackets table row 22 cell negative 6 end cell row straight x 7 end table close square brackets
rightwards double arrow Comparing space both space sides comma space we space get space straight x equals negative 9 end style


Q 2. If begin mathsize 12px style straight A equals open square brackets table row 2 3 row cell negative 2 end cell cell negative 1 end cell end table close square brackets end style and begin mathsize 12px style straight B equals open square brackets table row 1 4 row cell negative 3 end cell 0 end table close square brackets end style; find the matrix X such that AX = B.                                                              [4M]

Solution:

begin mathsize 12px style AX equals straight B
rightwards double arrow open square brackets table row 2 3 row cell negative 2 end cell cell negative 1 end cell end table close square brackets straight X equals open square brackets table row 1 4 row cell negative 3 end cell 0 end table close square brackets
Let space straight X equals open square brackets table row straight a straight b row straight c straight d end table close square brackets
rightwards double arrow open square brackets table row 2 3 row cell negative 2 end cell cell negative 1 end cell end table close square brackets open square brackets table row straight a straight b row straight c straight d end table close square brackets equals open square brackets table row 1 4 row cell negative 3 end cell 0 end table close square brackets
rightwards double arrow open square brackets table row cell 2 straight a plus 3 straight c end cell cell 2 straight b plus 3 straight d end cell row cell negative 2 straight a minus straight c end cell cell negative 2 straight b minus straight d end cell end table close square brackets equals open square brackets table row 1 4 row cell negative 3 end cell 0 end table close square brackets end style

Comparing both sides, we get
2a + 3c = 1  … (i)
−2a – c = −3 … (ii)
2b + 3d = 4 … (iii)
−2b – d = 0 … (iv)
Solving (i) and (ii), we get c = −1 and a = 2.

Solving (iii) and (iv), we get d = 2 and b = −1.

begin mathsize 12px style Hence comma space straight X equals open square brackets table row straight a straight b row straight c straight d end table close square brackets equals open square brackets table row 2 cell negative 1 end cell row cell negative 1 end cell 2 end table close square brackets end style

Q 3. Given begin mathsize 12px style straight A equals open square brackets table row 2 cell negative 1 end cell row 2 0 end table close square brackets end stylebegin mathsize 12px style straight B equals open square brackets table row cell negative 3 end cell 2 row 4 0 end table close square brackets end style and begin mathsize 12px style straight C equals open square brackets table row 1 0 row 0 2 end table close square brackets end style, find the matrix X such that  A + X = 2B + C.                           [3M]

Solution:

A + X = 2B + C
⇒ −A + A + X = −A + 2B + C
⇒ X = −A + 2B + C

begin mathsize 12px style rightwards double arrow straight X equals negative open square brackets table row 2 cell negative 1 end cell row 2 0 end table close square brackets plus 2 open square brackets table row cell negative 3 end cell 2 row 4 0 end table close square brackets plus open square brackets table row 1 0 row 0 2 end table close square brackets
rightwards double arrow straight X equals open square brackets table row cell negative 2 end cell 1 row cell negative 2 end cell 0 end table close square brackets plus open square brackets table row cell negative 6 end cell 4 row 8 0 end table close square brackets plus open square brackets table row 1 0 row 0 2 end table close square brackets
rightwards double arrow straight X equals open square brackets table row cell negative 2 minus 6 plus 1 end cell cell 1 plus 4 end cell row cell negative 2 plus 8 end cell 2 end table close square brackets equals open square brackets table row cell negative 7 end cell 5 row 6 2 end table close square brackets end style


Q 4. Given begin mathsize 12px style straight A equals open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets end style and  begin mathsize 12px style straight I equals open square brackets table row 1 0 row 0 1 end table close square brackets end style,  and A2 = 9A + mI. Find m.                                                                          [4M]

Solution: 

begin mathsize 12px style straight A equals open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets space and space straight I equals open square brackets table row 1 0 row 0 1 end table close square brackets
rightwards double arrow straight A squared equals open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets equals open square brackets table row 4 0 row cell negative 9 end cell 49 end table close square brackets
rightwards double arrow straight A squared equals 9 straight A plus mI
rightwards double arrow open square brackets table row 4 0 row cell negative 9 end cell 49 end table close square brackets equals 9 open square brackets table row 2 0 row cell negative 1 end cell 7 end table close square brackets plus straight m open square brackets table row 1 0 row 0 1 end table close square brackets
rightwards double arrow open square brackets table row 4 0 row cell negative 9 end cell 49 end table close square brackets equals open square brackets table row 18 0 row cell negative 9 end cell 63 end table close square brackets plus open square brackets table row straight m 0 row 0 straight m end table close square brackets
rightwards double arrow open square brackets table row 4 0 row cell negative 9 end cell 49 end table close square brackets equals open square brackets table row cell 18 plus straight m end cell 0 row cell negative 9 end cell cell 63 plus straight m end cell end table close square brackets end style

Comparing both sides, we get
18 + m = 4
m = −14

Q 5.
 Solve for x and y : begin mathsize 12px style open square brackets table row cell straight x minus straight y space space end cell cell straight x plus straight y end cell end table close square brackets open square brackets table row 3 cell negative 4 end cell row 5 cell negative 1 end cell end table close square brackets equals open square brackets table row 24 2 end table close square brackets end style                                                                                         [3M]

Solution:

begin mathsize 12px style open square brackets table row cell straight x minus straight y end cell cell straight x plus straight y end cell end table close square brackets open square brackets table row 3 cell negative 4 end cell row 5 cell negative 1 end cell end table close square brackets equals open square brackets table row 24 2 end table close square brackets
rightwards double arrow open square brackets table row cell 3 open parentheses straight x minus straight y close parentheses plus 5 open parentheses straight x plus straight y close parentheses end cell cell negative 4 open parentheses straight x minus straight y close parentheses minus open parentheses straight x plus straight y close parentheses end cell end table close square brackets equals open square brackets table row 24 2 end table close square brackets
rightwards double arrow open square brackets table row cell 8 straight x plus 2 straight y end cell cell negative 5 straight x plus 3 straight y end cell end table close square brackets equals open square brackets table row 24 2 end table close square brackets end style

Comparing both sides, we get
8x + 2y = 24 … (i)
−5x + 3y = 2 … (ii)

Solving (i) and (ii), we get x = 2 and y = 4.


Q 6. Evaluate: begin mathsize 12px style open square brackets table row cell 4 space sin space 30 degree end cell cell 2 space cos space 60 degree end cell row cell sin space 90 degree end cell cell 2 space cos space 0 degree end cell end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets end style                                                                                                             [3M]

Solution:

begin mathsize 12px style open square brackets table row cell 4 sin 30 degree end cell cell 2 cos 60 degree end cell row cell sin 90 degree end cell cell 2 cos 0 degree end cell end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets
equals open square brackets table row cell 4 cross times 1 half end cell cell 2 cross times 1 half end cell row 1 cell 2 cross times 1 end cell end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets
equals open square brackets table row 2 1 row 1 2 end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets
equals open square brackets table row cell 8 plus 5 end cell cell 10 plus 4 end cell row cell 4 plus 10 end cell cell 5 plus 8 end cell end table close square brackets
equals open square brackets table row 13 14 row 14 13 end table close square brackets
rightwards double arrow open square brackets table row cell 4 sin 30 degree end cell cell 2 cos 60 degree end cell row cell sin 90 degree end cell cell 2 cos 0 degree end cell end table close square brackets open square brackets table row 4 5 row 5 4 end table close square brackets equals open square brackets table row 13 14 row 14 13 end table close square brackets end style

 

Chapters 8: Arithmetic Progression

Q 1. The 11th  term of an AP is 80 and the 16th term is 110. Find the 31st term.                                                                [3M]

Solution: 

We know that, nth term of an AP = an = a + (n - 1)d

⇒ a11 = a + (11 - 1)d = a + 10d = 80 ...(i) and

a16 = a + (16 - 1)d = a + 15d = 110 ...(ii)

Subtracting (i) from (ii), we get

5d = 30 ⟹ d = 6

Substituting d = 6 in (i), we get

a + 10(6) = 80 ⟹ a = 20

begin mathsize 12px style therefore end style a31 = a + (31 - 1)d = 20 + 30 × 6 = 200


Q 2. If the third term of an AP is 18 and the seventh term is 30, then find the series.                                                         [3M]

Solution: The third term of an AP is 18.

⇒ t3 = 18
⇒ a + 2d = 18 … (i)
The seventh term of an AP is 30.
⇒ a + 6d = 30 … (ii)
⇒ Solving (i) and (ii), we get d = 3 and a = 12
⇒ tn = a + (n – 1)d
⇒ tn = 12 + (n – 1) × 3
⇒ a = 12, t2 = 15, t3 = 18, t4 = 21,…
⇒ The required series is 12, 15, 18, 21,…


Q 3. If the 3rd and the 9th terms of an Arithmetic Progression are 4 and –8 respectively, then find:  S20                              [4M]

Solution: The 3rd term of an AP is 18.

⇒ t3 = 4
⇒ a + 2d = 4 … (i)
The 9th term of an AP is –8 .
⇒ a + 8d = –8 … (ii)

Subtracting (i) from (ii), we get

6d = –12 ⇒ d = –2

Substitute d = –2 in equation (i), we get

a + 2(–2) = 4 ⇒ a = 8

We know that,

begin mathsize 12px style straight S subscript straight n equals straight n over 2 open square brackets 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close square brackets
rightwards double arrow straight S subscript 20 equals 20 over 2 open square brackets 2 cross times 8 plus open parentheses 20 minus 1 close parentheses straight x minus 2 close square brackets equals 10 open parentheses 16 minus 38 close parentheses equals 10 straight x minus 22 equals negative 220 end style

 

Q 4. Split 180 into three parts such that these parts are in AP and the product of the two extreme terms is 3500.                [4M]

Solution: Let the three parts be a – d, a, a + d.

According to the question,

⇒ a – d + a + a + d = 180

⇒ 3a = 180

⇒ a = 60

Hence, the numbers are 60 – d, 60, 60 + d

Also,

⇒ (60 – d)(60 + d) = 3500

⇒ 3600 – d2 = 3500

⇒ d2 = 100

⇒ d = ±10

⇒ If a = 60 and d = 10, then the parts are 50, 60, 70.

⇒ If a = 60, d = −10, then the parts are 70, 60, 50.


Chapters 9: Coordinate Geometry

Q 1. Using a graph paper, plot the points A(3, 5) and (0, 5).                                                                                            [4M]

  1. Reflect A and B in the origin to get the images A’ and B’.
  2. Write the co-ordinates of A’ and B’.
  3. State the geometrical name for the figure ABA’B’.
  4. Find its perimeter.

Solution: 

i. Choose the co-ordinate axes as shown in the graph paper.
Take 1 cm = 1 unit on both axes.
Plot the points A(3, 5) and B(0, 5) on the graph paper.
Reflect the points A and B in the origin onto the points A’ and B’, respectively.

ii. The co-ordinates of these points are A’ (-3, -5) and B’ (0, -5).

iii. ABA’B’ is a parallelogram.

iv. From the figure,
AB = 3 units

begin mathsize 12px style AB apostrophe equals square root of open parentheses 3 minus 0 close parentheses squared plus open parentheses 5 plus 5 close parentheses squared end root equals square root of 109
Perimeter space of space the space parallelogram space ABA ’ straight B ’
equals 2 open parentheses AB plus AB apostrophe close parentheses equals 2 open parentheses 3 plus square root of 109 close parentheses units equals 6 plus 2 square root of 109 units end style


Q 2. Use a graph paper for this question. A(1, 1), B(5, 1), C(4, 2) and D(2, 2) are the vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C and D are reflected in the origin A’, B’, C’ and D’, respectively. Locate them on the graph sheet and write their co-ordinates.                                                                                                                                [4M]

Solution:


Q 3. Use a graph paper for this question:                                                                                                                     [6M]

(Take 2cm = 1 unit on both x and y axes)

i.  Plot the following points:
A(0,4), B(2,3), C(1,1) and D(2,0).
ii. Reflect points B, C, D about the y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
iii. Join the points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation to the line about which if this closed figure obtained is folded, the two parts of the figure exactly coincide.

Solution:

i. Plotting A(0,4), B(2,3), C(1,1) and D(2,0). 

ii. Reflected points B’(-2,3), C’(-1,1) and D’(-2,0).

iii. The figure is symmetrical about x = 0

 

Q 4. The mid-point of the line segment joining (2a, 4) and (−2, 2b) is (1, 2a + 1). Find the values of a and b.                  [3M]

Solution:

Let P be the mid-point of (2a, 4) and (−2, 2b).
Hence,

begin mathsize 12px style rightwards double arrow open parentheses fraction numerator 2 straight a minus 2 over denominator 2 end fraction comma space fraction numerator 4 plus 2 straight b over denominator 2 end fraction close parentheses equals open parentheses 1 comma space 2 straight a plus 1 close parentheses
rightwards double arrow fraction numerator 2 straight a minus 2 over denominator 2 end fraction equals 1 comma space fraction numerator 4 plus 2 straight b over denominator 2 end fraction equals 2 straight a plus 1
rightwards double arrow straight a equals 2 space put space it space in space fraction numerator 4 plus 2 straight b over denominator 2 end fraction equals 2 straight a plus 1
rightwards double arrow fraction numerator 4 plus 2 straight b over denominator 2 end fraction equals 2 apostrophe space 2 plus 1
rightwards double arrow 4 plus 2 straight b equals 10 rightwards double arrow 2 straight b equals 6 rightwards double arrow straight b equals 3 end style

 

Q 5. In the given figure, line ACB meets the y-axis at point B and the x-axis at point A. C is the point (−6, 6) and AC:CB = 2:3. Find the coordinates of A and B.                                                                                                                                [3M]

Solution: From the diagram, point A is on the x-axis; hence, the coordinates of point B are (0, y) and those of point A are (x, 0).

C divides AB in the ratio 2:3.

begin mathsize 12px style rightwards double arrow open parentheses fraction numerator 2 cross times 0 plus 3 straight x over denominator 5 end fraction comma space fraction numerator 2 straight y plus 3 cross times 0 over denominator 5 end fraction close parentheses equals open parentheses negative 6 comma 6 close parentheses
rightwards double arrow fraction numerator 3 straight x over denominator 5 end fraction equals negative 6 space and space fraction numerator 3 straight y over denominator 5 end fraction equals 6
rightwards double arrow straight x equals negative 10 space and space straight y equals 10 end style

Hence, point A is (−10, 0) and point B is (0, 10).


Q 6. Show that P(2, m – 2) is a point of trisection of the line segment joining the points A(4, −2) and B(1, 4). Hence, find the value of m.                                                                                                                                                      [3M]

Solution: P will be a point of trisection of AB if it divides AB in the ratio 1:2 or 2:1.
begin mathsize 12px style rightwards double arrow straight x equals fraction numerator straight m subscript 1 straight x subscript 2 plus straight m subscript 2 straight x subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction
rightwards double arrow fraction numerator straight m subscript 1 cross times 1 plus straight m subscript 2 cross times 4 over denominator straight m subscript 1 plus straight m subscript 2 end fraction equals 2
rightwards double arrow straight m subscript 1 plus 4 straight m subscript 2 equals 2 open parentheses straight m subscript 1 plus straight m subscript 2 close parentheses
rightwards double arrow straight m subscript 1 plus 4 straight m subscript 2 equals 2 straight m subscript 1 plus 2 straight m subscript 2
rightwards double arrow 2 straight m subscript 2 equals straight m subscript 1
rightwards double arrow straight m subscript 1 over straight m subscript 2 equals 2 over 1
Hence comma space straight P space is space straight a space point space of space trisection space of space AB.
rightwards double arrow straight y equals fraction numerator straight m subscript 1 straight y subscript 2 plus straight m subscript 2 straight y subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction equals fraction numerator 2 cross times 4 plus 1 cross times open parentheses negative 2 close parentheses over denominator 2 plus 1 end fraction equals 2
rightwards double arrow straight m minus 2 equals 2
rightwards double arrow straight m equals 4 end style


Q 7. The equation of the line is 4x – 5y = 9. Find the                                                                                                        [4M]

  1. Slope of the line
  2. Equation of the passing through the intersection of the lines x + y = 1 and 2x + y = 2 with slope begin mathsize 12px style fraction numerator negative 5 over denominator 4 end fraction end style.
Solution: 
i. Given equation of the line is 4x – 5y = 9
begin mathsize 12px style therefore space 5 straight y equals 4 straight x minus 9 rightwards double arrow straight y equals fraction numerator 4 straight x over denominator 5 end fraction minus 9 over 5 end style
Comparing it with y = mx + c, we get begin mathsize 12px style straight m equals 4 over 5 end style, which is the required slope.
ii. To find the point of intersection of the two lines

x + y = 1 … (i)  and 2x + y = 2 … (ii)
Subtracting (i) from (ii),
We get x = 1

Putting it in (i), we get y = 0
Thus, the intersection point of the given two lines is (1, 0). Now, the equation of the required line is passing through (1, 0) with slope begin mathsize 12px style fraction numerator negative 5 over denominator 4 end fraction end style is given by
∴ y – y1 = m(x – x1)
∴ y – 0 = begin mathsize 12px style fraction numerator negative 5 over denominator 4 end fraction end style(x – 1)
∴ 4y = –5x + 5
∴ 5x + 4y = 5
 
Q 8. Find the equation of a line                                                                                                                                      [3M]
  1. Whose inclination is 30˚ and y-intercept is −2
  2. With inclination 60˚ and passing through (−4, 2)
  3. Passing through the points (3, −4) and (7, 1)
Solution: 
i. Inclination is θ = 30˚ and y-intercept c = −2
begin mathsize 12px style therefore space Slope open parentheses straight m close parentheses equals tanθ equals tan 30 degree equals fraction numerator 1 over denominator square root of 3 end fraction
The space equation space of space the space line space is space straight y space equals fraction numerator 1 over denominator square root of 3 end fraction straight x minus 2
therefore space straight x minus square root of 3 straight y minus 2 square root of 3 equals 0 end style
 
ii. Inclination is θ = 60˚ and passing through (−4, 2)
slope (m) = tan θ = tan 60˚ = begin mathsize 12px style square root of 3 end style
The equation of the line is
∴ y – y1 = m(x – x1)
∴ y – 2 = begin mathsize 12px style square root of 3 end style(x + 4)
∴ begin mathsize 12px style square root of 3 straight x minus straight y plus 2 plus 4 square root of 3 equals 0 end style
iii. To find the equation of the line passing through the points (3, −4) and (7, 1) is
begin mathsize 12px style therefore space Slope space of space the space line equals straight m fraction numerator 1 minus open parentheses negative 4 close parentheses over denominator 7 minus 3 end fraction equals 5 over 4 end style
The equation of the line passing through (3, −4) with slope = begin mathsize 12px style 5 over 4 end style is given by
begin mathsize 12px style therefore space straight y minus open parentheses negative 4 close parentheses equals 5 over 4 open parentheses straight x minus 3 close parentheses
therefore space 4 straight y plus 16 equals 5 straight x minus 15
therefore 5 straight x minus 4 straight y minus 31 equals 0 end style
 

Q 9. Write the equation of the line whose gradient is begin mathsize 12px style 3 over 2 end style and which passes through P, where P divides the line segment joining A(−2, 6) and B(3, −4) in the ratio 2:3.                                                                                                                         [3M]

Solution: P divides the line segment joining A(−2, 6) and B(3, −4) in the ratio 2:3.

begin mathsize 12px style The space co minus ordinates space of space straight P space are space open parentheses fraction numerator 2 cross times 3 plus 3 cross times open parentheses negative 2 close parentheses over denominator 5 end fraction comma fraction numerator 6 cross times 3 plus open parentheses negative 4 close parentheses cross times 2 over denominator 5 end fraction close parentheses equals open parentheses 0 comma space 2 close parentheses end style

The equation of the line whose gradient is begin mathsize 12px style 3 over 2 end style and passing through (0, 2) is

begin mathsize 12px style therefore space straight y minus 2 equals 3 over 2 open parentheses straight x minus 0 close parentheses rightwards double arrow 2 straight y minus 4 equals 3 straight x rightwards double arrow 3 straight x minus 2 straight y plus 4 equals 0 end style

 

Q 10. If the slope of a line joining P(6, k) and Q(1 – 3k, 3) is ½, find                                                                                [3M]

  1. k
  2. the mid-point of PQ using the value of k found in (i)

Solution: 

i. The slope of a line joining P(6, k) and Q(1 – 3k, 3) is ½.

begin mathsize 12px style therefore space fraction numerator 3 minus straight k over denominator 1 minus 3 straight k minus 6 end fraction equals 1 half
therefore space 6 minus 2 straight k equals negative 3 straight k minus 5
therefore space straight k equals negative 11 end style

ii. Hence, point Q becomes (1 - 3 × (-11), 3) = (34, 3)

∴ Mid-point of P (6, -11)  and Q (34, 3) is begin mathsize 12px style open parentheses fraction numerator 6 plus 34 over denominator 2 end fraction comma space fraction numerator negative 11 plus 3 over denominator 2 end fraction close parentheses equals open parentheses 20 comma space minus 4 close parentheses end style

Chapter 10: Similarity

Q 1. In ΔABC, DE is drawn parallel to BC. If AB = 16 cm, AD = 4 cm, and AC = 24 cm, find                                             [4M]

  1. EC
  2. begin mathsize 12px style fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses DABC close parentheses end fraction end style
  3. begin mathsize 12px style fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses trapezium space BCED close parentheses end fraction end style
Solution:
i. AB = 16 cm, AD = 4 cm
∴  BD = AB – AD = 16 – 4 = 12 cm
Let EC = x cm, then AE = (24 – x) cm
In ΔABC, DE || BC
begin mathsize 12px style therefore AD over DB equals AE over EC space space space space space space space space space space space space space space space because Basic space Proportionality space Theorem
therefore 4 over 12 equals fraction numerator 24 minus straight x over denominator straight x end fraction
therefore 1 third equals fraction numerator 24 minus straight x over denominator straight x end fraction end style
 
∴ 72 – 3x = x

∴ 4x = 72
∴ x = 18 cm
∴ EC = 18 cm

ii. In ΔADE and ΔABC,
begin mathsize 12px style angle end styleADE = begin mathsize 12px style angle end styleABC                     ∵ corresponding angles
begin mathsize 12px style angle end styleAED = begin mathsize 12px style angle end styleACB                     ∵ corresponding angles
∴ ΔADE ~ ΔABC                        ∵ by AA similarity

begin mathsize 12px style therefore fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses DABC close parentheses end fraction equals AD squared over AB squared equals 4 squared over 16 squared equals 1 over 16 end style

iii. 

begin mathsize 12px style fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses trapezium space BCED close parentheses end fraction equals fraction numerator ar open parentheses DADE close parentheses over denominator ar open parentheses DABC close parentheses minus ar open parentheses DADE close parentheses end fraction equals fraction numerator 1 over denominator 16 minus 1 end fraction
therefore space fraction numerator ar left parenthesis DADE right parenthesis over denominator ar left parenthesis trapezium space BCED right parenthesis end fraction equals space 1 over 15 end style

 

Q 2. The scale of the map is 1:200000. A plot of land of area 20 km2 is to be represented on the map. Find the                  [3M]
  1. Number of kilometres on the ground which is represented by 1 cm on the map
  2. Area in sq km which can be represented by 1 cm2
  3. Area of the map which represents the plot of land
Solution:
begin mathsize 12px style Scale space factor space straight k equals 1 over 200000 space and space 1 space cm equals 1 over 200000 space Lenght space of space the space ground
straight i. space Length space of space the space ground equals 200000 space cm equals fraction numerator 200000 over denominator 100 cross times 1000 end fraction km equals 2 km
ii. space Area space of space map equals straight k squared space times space the space corresponding space area space of space the space ground
therefore space 1 space cm squared equals open parentheses 1 over 200000 close parentheses squared space area space of space ground
therefore space Area space of space the space ground equals 200000 cross times 200000 space cm squared equals 4 space km squared
iii. space Total space area space of space the space map equals straight k squared space times space the space corresponding space area space of space the space plot
equals open parentheses 1 over 200000 close parentheses squared cross times 20 space km squared
equals open parentheses 1 over 40000000000 close parentheses cross times 20 cross times open parentheses 100 cross times 1000 close parentheses squared space cm squared equals 5 space km squared end style


Q 3.
 A model of a ship is made to a scale 1: 300.                                                                                                           [3M]

  1. The length of the model of the ship is 2 m. Calculate the length of the ship.
  2. The area of the deck of the ship is 180,000 m2. Calculate the area of the deck of the model.
  3. The volume of the model is 6.5 m3. Calculate the volume of the ship.

 Solution:

i. Scale factor k = begin mathsize 12px style 1 over 300 end style

Length of the model of the ship = k × Length of the ship

⇒ 2 = begin mathsize 12px style 1 over 300 end style× Length of the ship

⇒ Length of the ship = 600 m

ii. Area of the deck of the model = k2 × Area of the deck of the ship

begin mathsize 12px style rightwards double arrow Area space of space the space deck space of space the space model space equals open parentheses 1 over 300 close parentheses squared cross times 180 comma 000
equals 1 over 90000 cross times 180 comma 000
equals 2 space straight m squared end style

iii. Volume of the model = k3 × Volume of the ship

begin mathsize 12px style rightwards double arrow 6.5 equals open parentheses 1 over 300 close parentheses cubed cross times Volume space of space the space ship end style

⇒ Volume of the ship = 6.5 × 27000000 = 17,55,00,000 m3


Chapters 11: Circles

Q 1. In the figure, AD = BC and AB || DC, begin mathsize 12px style angle end styleBAC = 40˚ and begin mathsize 12px style angle end styleCBD = 60˚. Find                                                              [4M]

  1. begin mathsize 12px style angle end styleBCD
  2. begin mathsize 12px style angle end styleBCA
  3. begin mathsize 12px style angle end styleABC
  4. begin mathsize 12px style angle end styleADB
Solution:

ABCD is a cyclic quadrilateral.
AC and BD are the diagonals of a cyclic quadrilateral.
∴ begin mathsize 12px style angle end styleBAC = 40˚ and begin mathsize 12px style angle end styleCBD = 60˚
begin mathsize 12px style angle end styleCAD = begin mathsize 12px style angle end styleCBD = 60˚                 ∵ angle in the same segment
Also, begin mathsize 12px style angle end styleBAC = begin mathsize 12px style angle end styleBDC = 40˚
begin mathsize 12px style angle end styleBAD = begin mathsize 12px style angle end styleBAC + begin mathsize 12px style angle end styleCAD
= 40˚ + 60˚
= 100˚

i. begin mathsize 12px style angle end styleBCD + begin mathsize 12px style angle end styleBAD = 180˚     ∵ opposite angles of a cyclic quadrilateral

begin mathsize 12px style angle end styleBCD + 100˚ = 180˚
begin mathsize 12px style angle end styleBCD  = 80˚

 ii. AD = BC

ABCD is an isosceles trapezium and AB || DC
begin mathsize 12px style angle end styleBAC = begin mathsize 12px style angle end styleDCA                 ∵ alternate angles
begin mathsize 12px style angle end styleDCA = 40˚
begin mathsize 12px style angle end styleABD = begin mathsize 12px style angle end styleACD = 40˚      ∵ angles in the same segment
begin mathsize 12px style angle end styleBCA = begin mathsize 12px style angle end styleBCD – begin mathsize 12px style angle end styleACD
= 80˚ – 40˚
= 40˚
 
iii. begin mathsize 12px style angle end styleABC = begin mathsize 12px style angle end styleABD + begin mathsize 12px style angle end styleCBD
= 40˚ + 60˚
= 100˚
 
iv. begin mathsize 12px style angle end styleADB = begin mathsize 12px style angle end styleBCA = 40˚    ∵ angles in the same segment


Q 2.
 AB and CD are two chords on the same side. AB = 12 cm and CD = 24 cm. Distance between two parallel chords is 4 cm. Find the radius of the circle.                                                                                                                             [4M]

Solution:

AB = 12 cm, CD = 24 cm and LM = 4 cm
Let OL = x cm
Then, OM = (x + 4) cm
Join OA and OC.
Then, OA = OC = r
Since the perpendicular from the centre to a chord bisects the chord,
∴ AM = MB = 6 cm and CL = LD = 12 cm
In triangles OAM and OCL,
∴ OA2 = OM2 + AM2 and OC2 = OL2 + CL2
∴ r2 = (x + 4)2 + 62 and r2 = x2 + 122
∴ (x + 4)2 + 62 = x2 + 122
∴ x2 + 8x + 16 + 36 = x2 + 144
∴ 8x = 92
∴ x = 11.5 cm
∴ r2 = 11.52 + 122
∴ r = 16.62 cm (approx.)


Q 3.
 In the given figure, ABCD is a cyclic quadrilateral. Find the values of x and y.                                                            [3M]

Solution:

ABCD is a cyclic quadrilateral.

Sum of opposite angles of a cyclic quadrilateral is 180o

x + 10 + 5y + 5 = 180

x + 5y = 165 … (1)

4y – 4 + 2x + 4 = 180

2x + 4y = 180 … (2)

Solving (1) and (2) simultaneously

x = 40 and y = 25


Q 4. In the given figure, O is the centre of the circle. AB is a tangent to it at point B. begin mathsize 12px style angle end styleBDC = 60˚. Find begin mathsize 12px style angle end styleBAO.          [4M]

Solution:

In ΔBDC,
begin mathsize 12px style angle end styleDBC = 90˚                      ∵ tangent begin mathsize 12px style perpendicular end styleradius
begin mathsize 12px style angle end styleBDC = 60˚       …(i)
In ΔBDC,

begin mathsize 12px style angle end styleDCB + begin mathsize 12px style angle end styleDBC + begin mathsize 12px style angle end styleBDC = 180˚         ∵ angle sum property in the triangle
begin mathsize 12px style angle end styleDCB + 90˚ + 60˚ = 180˚   
begin mathsize 12px style angle end styleDCB = 30˚
In ΔOEC,
OE = OC                            ∵ radius
begin mathsize 12px style angle end styleOEC = begin mathsize 12px style angle end styleOCE              ∵ isosceles triangle property
begin mathsize 12px style angle end styleOEC = begin mathsize 12px style angle end styleBCD = 30˚
In ΔAEB,
begin mathsize 12px style angle end styleADE + begin mathsize 12px style angle end styleBDE = 180˚            ∵ straight line property

begin mathsize 12px style angle end styleADE + 60˚  = 180˚            ∵ begin mathsize 12px style angle end styleBDE = begin mathsize 12px style angle end styleBDC from (i)
begin mathsize 12px style angle end styleADE = 120˚
begin mathsize 12px style angle end styleDEA = begin mathsize 12px style angle end styleOEC = 30˚…vertically opposite angle
begin mathsize 12px style angle end styleDAE + begin mathsize 12px style angle end styleADE + begin mathsize 12px style angle end styleDEA = 180˚    ∵ angle sum property in the triangle
begin mathsize 12px style angle end styleDAE + 120˚ + 30˚ = 180˚
begin mathsize 12px style angle end styleDAE = 30˚
begin mathsize 12px style angle end styleBAO = begin mathsize 12px style angle end styleDAE = 30˚


Q 5. 
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If begin mathsize 12px style angle end styleACO = 25˚, find                   [3M]

  1. begin mathsize 12px style angle end styleBCO
  2. begin mathsize 12px style angle end styleAOB
  3. begin mathsize 12px style angle end styleAPB

Solution:

i. In ΔOAC and ΔOBC,
OA = OB                          radii of the same circle
CA = CB                         ∵ tangents from an external point to a circle are equal
OC = OC                         ∵ common side
ΔOAC ≅ ΔOBC             ∵ by SSS congruence
begin mathsize 12px style angle end styleBCO = begin mathsize 12px style angle end styleACO = 25˚

ii. In ΔOAC,
begin mathsize 12px style angle end styleAOC = 180˚  –  begin mathsize 12px style angle end styleOAC  – begin mathsize 12px style angle end styleACO
              = 180˚ – 90˚ – 25˚
              = 65˚
ΔOAC ≅ ΔOBC
begin mathsize 12px style angle end styleBOC = begin mathsize 12px style angle end styleAOC = 65˚
begin mathsize 12px style angle end styleAOB = begin mathsize 12px style angle end styleAOC + begin mathsize 12px style angle end styleBOC
              = 65˚ + 65˚
              = 130˚
ii. begin mathsize 12px style angle end styleAPB = ½ begin mathsize 12px style angle end styleAOB    ∵ angle subtended at the centre = 2 times angle on the circle
begin mathsize 12px style angle end styleAPB = ½ × 130 = 65˚


Q 6. AB is a line segment and M is its mid-point. Semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle C(O, r) is drawn so that it touches all the three semi-circles. Prove that begin mathsize 12px style straight r equals 1 over 6 AB end style                       [4M] 

Solution:

Since M is the mid-point of AB,
∴ AM = MB = begin mathsize 12px style AB over 2 end style
Since P and S are the mid-points of AM and MB,
∴ AP = PM = MS = SB = begin mathsize 12px style AB over 4 end style
Radius of a circle = r
∴ CP = r + PQ = r +  begin mathsize 12px style AB over 4 end style
∴ CM = DM – DC = begin mathsize 12px style AB over 2 end style – r

In Δ CMP,
begin mathsize 12px style angle end styleCMP = 90˚
∴ CM2 + MP2 = CP2
begin mathsize 12px style therefore space open parentheses AB over 2 minus straight r close parentheses squared plus open parentheses AB over 4 close parentheses squared equals open parentheses straight r plus AB over 4 close parentheses squared
therefore space AB squared over 4 minus rAB plus straight r squared plus AB squared over 16 equals straight r squared plus AB squared over 16 plus fraction numerator 2 rAB over denominator 4 end fraction
therefore AB squared over 4 equals fraction numerator 2 rAB over denominator 4 end fraction plus rAB
therefore AB over 4 equals fraction numerator 2 straight r over denominator 4 end fraction plus straight r
therefore AB over 4 equals fraction numerator 6 straight r over denominator 4 end fraction
therefore AB over 6 equals straight r
therefore space straight r equals AB over 6 end style

 

Chapter 12: Area and Volume of Solids

Q 1. A conical tent is to accommodate 11 people. Each person must have 4 sq. metres of the space on the ground and 20 cubic metres of air to breathe. Find the height of the cone.                                                                                                        [4M]

Solution:

Let the height of the conical tent be h metres and r be the radius in metres.
Number of people the tent can accommodate = 11
Space required by each person on the ground = 4 m2
Volume of air required by each person = 20 m3
⇒ Area of the base = 11 × 4 = 44 m2
⇒ πr2 = 44  …(i)
⇒ Volume of the cone = 11 × 20 = 220 m3

begin mathsize 12px style text ⇒ end text 1 third πr squared straight h equals 220 space space... open parentheses ii close parentheses
Dividing space open parentheses ii close parentheses space by space open parentheses straight i close parentheses comma
rightwards double arrow fraction numerator begin display style 1 third end style πr squared straight h over denominator πr squared end fraction equals 220 over 44
rightwards double arrow straight h over 3 equals 5
rightwards double arrow straight h equals 15 space straight m
therefore space The space height space of space the space cone space is space 15 space straight m. end style


Q 2. 
A steel wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the mass of the wire, assuming the density of steel to be 8.88 g per cm3.                      [3M]

Solution:

Length of the wire used in one round = 3 mm = 0.3 cm

Number of rounds required to cover 12 cm of length = begin mathsize 12px style fraction numerator 12 over denominator 0.3 end fraction end style = 40

Diameter of the cylinder = 10 cm ⇒ Radius of the cylinder = 5 cm

Length of wire required for one round = 2∏r = 2∏ × 5 cm = 10∏ cm

begin mathsize 12px style therefore end style Length of wire required for 40 rounds = 10∏ × 40 = 400∏ cm = 400 × 3.14 = 1256 cm

Now, radius of the wire = begin mathsize 12px style 3 over 2 mm equals 3 over 20 cm end style

Volume of the wire = ∏r2h = begin mathsize 12px style straight pi open parentheses 3 over 20 close parentheses squared cross times 1256 space cm squared end style

Weight of the wire = Volume of the wire × 8.88 gm = begin mathsize 12px style 3.14 cross times open parentheses 3 over 20 close parentheses squared cross times 1256 space cross times space 8.88 equals 787.98 space gm end style

 
Q 3. The area of the base of a right circular cone is 28.26 sq. cm. If its height is 4 cm, find its volume and the curved surface area. (use π = 3.14)                                                                                                                                                   [3M]

Solution:

 begin mathsize 12px style Area space of space base space of space cone equals 28.26 space cm squared
rightwards double arrow πr squared equals 28.26 rightwards double arrow straight r squared equals fraction numerator 28.26 over denominator 3.14 end fraction equals 9 rightwards double arrow straight r equals 3 space cm
Volume space of space the space cone equals 1 third πr squared straight h equals 37.68 space cm cubed
Curved space surface space area space of space the space cone space equals πr calligraphic l equals πr square root of straight r squared plus straight h squared end root equals 3.14 cross times 3 cross times square root of 3 squared plus 4 squared end root equals 3.14 cross times 3 cross times 5 equals 47.10 space cm squared end style


Q 4. The volume of a right circular cone is 660 cm3 and the diameter of its base is 12 cm. Calculate                              [4M]

  1. the height of the cone,
  2. the slant height of the cone,
  3. the total surface area of the cone. 

Solution:

Diameter of a cone, d = 12 cm

⇒ Radius, r = 6 cm

Volume of cone = 660 cm3

begin mathsize 12px style straight V equals 1 third πr squared straight h
rightwards double arrow 660 equals 1 third cross times 22 over 7 cross times 6 cross times 6 cross times straight h
rightwards double arrow straight h equals fraction numerator 660 cross times 7 over denominator 22 cross times 12 end fraction cm
therefore space straight h equals 17.5 space cm end style

i. Hence, height of the cone = 17.5 cm

begin mathsize 12px style ii. space calligraphic l squared equals straight r squared plus straight h squared
rightwards double arrow calligraphic l squared equals open parentheses 6 close parentheses squared plus open parentheses 35 over 2 close parentheses squared
rightwards double arrow calligraphic l squared equals 36 plus 1225 over 4
rightwards double arrow calligraphic l squared equals fraction numerator 144 plus 1225 over denominator 4 end fraction
rightwards double arrow calligraphic l equals square root of 1369 over 4 end root
rightwards double arrow calligraphic l equals 37 over 2 cm
rightwards double arrow calligraphic l equals 18.5 space cm
therefore space Slant space height equals 18.5 space cm end style

begin mathsize 12px style iii. space Total space surface space area equals πr open parentheses straight r plus calligraphic l close parentheses
equals 22 over 7 cross times 6 open parentheses 6 plus 18.5 close parentheses space cm squared equals 22 over 7 cross times 6 cross times 24.5 space cm squared equals 462 space cm squared
therefore space Total space surface space area space equals 462 space cm squared end style

 

Chapters 13: Trigonometrical Identities

Q 1. Prove that: begin mathsize 12px style fraction numerator cosA over denominator 1 plus sinA end fraction plus tanA equals secA end style                                                                                                                [3M]

Solution:

begin mathsize 12px style fraction numerator cos space straight A over denominator 1 plus sin space straight A end fraction plus tan space straight A equals sec space straight A
straight L. straight H. straight S equals fraction numerator cos space straight A over denominator 1 plus sin space straight A end fraction plus tan space straight A equals fraction numerator cos space straight A over denominator 1 plus sin space straight A end fraction plus fraction numerator sin space straight A over denominator cos space straight A end fraction equals fraction numerator cos squared straight A plus sin space straight A plus sin squared straight A over denominator cos space straight A open parentheses 1 plus sin space straight A close parentheses end fraction
equals fraction numerator 1 plus sin space straight A over denominator cos space straight A open parentheses 1 plus sin space straight A close parentheses end fraction equals fraction numerator 1 over denominator cos space straight A end fraction equals sec space straight A equals straight R. straight H. straight S
Hence space proved. end style


Q 2.
 Evaluate: begin mathsize 12px style 2 open parentheses fraction numerator tan space 38 degree over denominator cot space 52 degree end fraction close parentheses squared plus open parentheses fraction numerator cot space 67 degree over denominator tan space 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator sec space 83 degree over denominator cosec space 7 degree end fraction close parentheses end style                                                                                [3M]

Solution:

 begin mathsize 12px style Consider
2 open parentheses fraction numerator tan space 38 degree over denominator cot space 52 degree end fraction close parentheses squared plus open parentheses fraction numerator cot space 67 degree over denominator tan space 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator sec space 83 degree over denominator cosec space 7 degree end fraction close parentheses
equals 2 open parentheses fraction numerator tan open parentheses 90 degree minus 52 degree close parentheses over denominator cot space 52 degree end fraction close parentheses squared plus open parentheses fraction numerator cot open parentheses 90 degree minus 23 degree close parentheses over denominator tan space 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator sec open parentheses 90 degree minus 7 degree close parentheses over denominator cosec space 7 degree end fraction close parentheses
equals 2 open parentheses fraction numerator cot space 52 degree over denominator cot space 52 degree end fraction close parentheses plus open parentheses fraction numerator tan space 23 degree over denominator tan space 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator cosec space 7 degree over denominator cosec space 7 degree end fraction close parentheses
equals 2 open parentheses 1 close parentheses plus 1 minus 3 open parentheses 1 close parentheses equals 3 minus 3 equals 0
Hence comma space 2 open parentheses fraction numerator tan space 38 degree over denominator cot space 52 degree end fraction close parentheses squared plus open parentheses fraction numerator cot space 67 degree over denominator tan space 23 degree end fraction close parentheses minus 3 open parentheses fraction numerator sec space 83 degree over denominator cosec space 7 degree end fraction close parentheses equals 0 end style


Q 3. If sin A + tan A = p and tan A - sin A = q, prove that begin mathsize 12px style straight p squared minus straight q squared equals 4 square root of pq end style.                                                              [4M]

Solution: 

Given: sin A + tan A = p and tan A - sin A = q

Consider,

L.H.S = p2 - q2

= (p - q) (p + q)

= [sin A + tan A - tan A + sin A]

[sin A + tan A + tan A - sin A]

= (2 sin A) (2 tan A)

= 4 sin A tan A   ...(I)

begin mathsize 12px style straight R. straight H. straight S equals 4 square root of pq equals 4 square root of open parentheses sin space straight A plus tan space straight A close parentheses open parentheses tan space straight A minus sin space straight A close parentheses end root
equals 4 square root of tan squared straight A minus sin squared straight A end root
equals 4 square root of fraction numerator sin squared straight A over denominator cos squared straight A end fraction minus sin squared straight A end root
equals 4 square root of sin squared straight A open parentheses fraction numerator 1 over denominator cos squared straight A end fraction minus 1 close parentheses end root
equals 4 sin space straight A square root of sec squared straight A minus 1 end root
equals 4 sin space straight A square root of tan squared straight A end root
equals 4 sin space straight A space tan space straight A space... open parentheses II close parentheses
USing space open parentheses straight I close parentheses space and space open parentheses II close parentheses
straight p squared minus straight q squared equals 4 square root of pq
Hence space Proved. end style


Q 4.
 Prove that: sin4A - cos4A = 1 - 2cos4A                                                                                                                [3M]

Solution:

To prove: sin4 A - cos4 A = 1 - 2cos2 A

Consider LHS,

= sin4 A - cos4 A

= (sin2 A)2 - (cos2 A)2

= (sin2 A - cos2 A) (sin2 A + cos2 A)

= (sin2 A - cos2 A)(1)

= sin2 A - cos2 A

= 1 - cos2 A - cos2 A

= 1 - cos2 A

Hence proved.


Q 5. 
If x = a sinθ and y = b tanθ, show that begin mathsize 12px style straight a squared over straight x squared minus straight b squared over straight y squared equals 1 end style.                                                                                         [3M]

Solution:

Given: x = a sinθ and y = b tanθ

Consider

begin mathsize 12px style straight a squared over straight x squared minus straight b squared over straight y squared
equals straight a squared over open parentheses straight a space sinθ close parentheses squared minus straight b squared over open parentheses straight b space tanθ close parentheses squared
equals fraction numerator straight a squared over denominator straight a squared space sin squared straight theta end fraction minus fraction numerator straight b squared over denominator straight b squared space tan squared straight theta end fraction
equals fraction numerator 1 over denominator sin squared straight theta end fraction minus fraction numerator 1 over denominator tan squared straight theta end fraction
equals fraction numerator 1 over denominator sin squared straight theta end fraction minus fraction numerator cos squared straight theta over denominator sin squared straight theta end fraction equals fraction numerator 1 minus cos squared straight theta over denominator sin squared straight theta end fraction equals fraction numerator sin squared straight theta over denominator sin squared straight theta end fraction equals 1
Hence space proved. end style


Chapters 14: Heights and Distances

Q 1. A guard observes a boat from a tower at a height of 150 m above sea level to be at an angle of depression of 25° (Take tan25° = 0.4663).                                                                                                                            [4M]                   

  1. Calculate the distance of the boat from the foot of the tower to the nearest metre.
  2. After some time, it is observed that the boat is 180 m from the foot of the tower. Calculate the new angle of depression.

Solution:

i. 

begin mathsize 12px style Let space straight A space be space the space boat space and space PQ space be space the space tower.
In space triangle space APQ comma
tan space angle PAQ equals PQ over PA
tan space 25 degree equals 150 over PA
rightwards double arrow PA equals fraction numerator 150 over denominator tan space 25 degree end fraction equals fraction numerator 150 over denominator 0.4663 end fraction equals 321.68 space straight m end style

ii. 

Let B be the new position of the boat and θ be the new angle of depression.

begin mathsize 12px style rightwards double arrow tanθ equals PQ over PB equals 150 over 180 equals 0.8333...
rightwards double arrow straight theta equals tan to the power of negative 1 end exponent open parentheses 0.83333... close parentheses equals 39.80 degree end style

Q 2. A vertical pole and a vertical tower are at the same ground level. From the top of the pole, the angle of elevation at the top of the tower is 30° and the angle of depression of the foot of the tower is 60°. Find the height of the pole if the height of the tower is 100 m.    [4M]

Solution:

According to the statements given, the figure will be as shown below:
Here, BC is the pole and AE is the tower.
Clearly, it is given that AE = 100 m, begin mathsize 12px style angle end styleABD = 30° and begin mathsize 12px style angle end styleEBD = 60°.


begin mathsize 12px style Let space BC equals straight y equals DE space and space AD equals straight x comma space so space AE equals straight x plus straight y
Now space in space increment ABD comma
tan space 30 degree equals AD over BD rightwards double arrow fraction numerator 1 over denominator square root of 3 end fraction equals straight x over BD rightwards double arrow BD equals straight x square root of 3 space space... open parentheses straight i close parentheses
Also comma space in space increment BDE comma
tan space 60 degree equals DE over BD rightwards double arrow square root of 3 equals straight y over BD rightwards double arrow BD equals fraction numerator straight y over denominator square root of 3 end fraction space... open parentheses ii close parentheses
Using space open parentheses straight i close parentheses space and space open parentheses ii close parentheses comma space we space have
straight x square root of 3 equals fraction numerator straight y over denominator square root of 3 end fraction rightwards double arrow 3 straight x equals straight y rightwards double arrow 3 straight x minus straight y equals 0 space... open parentheses iii close parentheses end style

Also, AE = x + y, i.e.
x + y = 100 m … (iv)
Solving equations (iii) and (iv), we get
4x = 100 m
⇒ x = 25 m = AD
Substituting in (iv), we get
y = 75 m = DE = BC
Hence, the height of the pole is 75 m.


Chapters 15: Graphical representation

Q 1. Draw a histogram for the following discontinuous distribution:                                         [6M]

Class interval

11–20

21–30

31–40

41–50

51–60

Frequency

25

10

27

18

20

Solution:

In this case, the class intervals are in the inclusive form (discontinuous).
So, first of all, we have to convert them to the exclusive form (continuous).
We know that the adjustment factor

begin mathsize 11px style equals 1 half space left parenthesis difference space between space the space upper space limit space of space straight a space class space and space the space lower space limit space of space the space next space class right parenthesis space end style
begin mathsize 11px style equals 1 half open parentheses 21 minus 20 close parentheses equals 0.5 end style
Therefore, to convert the class intervals to the exclusive form, subtract the adjustment factor from all the lower limits and add it to all the upper limits.
The adjusted class intervals would then be as follows:

Class Interval
(Inclusive form)

Class Interval
(Exclusive form)

Frequency

11–20

21–30

31–40

41–50

51–60

10.5–20.5

20.5–30.5

30.5–40.5

40.5–50.5

50.5–60.5

25

10

27

18

20

The required histogram will be as shown:

Q 2. Draw a frequency polygon for the following distribution (continuous data):                 [4M]

Age group

0-8

8-16

16-24

24-32

32-40

40-48

48-56

No. of girls

5

16

9

29

7

10

2

Solution:
Let us prepare the frequency table as follows:

Class Interval

Midpoints

Frequency

0–8

8–16

16–24

24–32

32–40

40–48

48–56

56–64

4

12

20

28

36

44

52

60

5

16

9

29

7

10

2

0

Mark the class intervals along the x-axis and frequencies along the y-axis.
Take the imagined class 56–64 at the end with frequency 0.
Plot the points (4, 5), (12, 16), (20, 9), (28, 29), (36, 7), (44, 10), (52, 2) and (60, 0) on the graph.
Draw line segments joining each consecutive point as shown to obtain the polygon.


Q 3.
 Draw a frequency polygon for the following distribution (discontinuous data):           [6M]

Marks

11–20

21–30

31–40

41–50

51–60

61–70

No. of boys

10

18

36

58

72

80

 Solution:

In this case, the class intervals are in the inclusive form (discontinuous).
So, first of all, we have to convert them to the exclusive form (continuous).
We know that the adjustment factor

begin mathsize 11px style equals 1 half left parenthesis difference space between space the space upper space limit space of space straight a space class space and space the space lower space limit space of space the space next space class right parenthesis end style

begin mathsize 11px style equals 1 half open parentheses 21 minus 20 close parentheses equals 0.5 end style
Therefore, to convert the given class intervals to the exclusive form, subtract the adjustment factor from all the lower limits and add it to all the upper limits.

The adjusted class intervals would then be as follows:

Class Interval
(Inclusive form)

Class Interval
(Exclusive form)

Midpoints

Frequency

1–10

11–20

21–30

31–40

41–50

51–60

61–70

71–80

0.5–10.5

10.5–20.5

20.5–30.5

30.5–40.5

40.5–50.5

50.5–60.5

60.5–70.5

70.5–80.5

5.5

15.5

25.5

35.5

45.5

55.5

65.5

75.5

0

10

18

36

58

72

80

0

Mark the class intervals along the x-axis and frequencies along the y-axis.
Take the imagined classes 1–10 at the beginning and 71–80 at the end each with frequency 0.
Plot the points (5.5, 0), (15.5, 10), (25.5, 18), (35.5, 36), (45.5, 58), (55.5, 72), (65.5, 80) and (75.5, 0) on the graph.
Draw line segments joining each consecutive point as shown to obtain the polygon.


Chapters 16: Measures of Central Tendency

Q 1. The weight of 40 children in a class was recorded in kg as follows:                                   [4M]

Weight (in kg)

45

47

49

51

53

55

No. of children

8

6

7

12

5

2

Calculate the following for the given distribution:

  1. Median
  2. Mode

Solution:

First we will prepare the cumulative frequency table as follows:
i. Median

Weight(in kg)
(x)

No. of children
(f)

Cumulative frequency
(c.f.)

45

47

49

51

53

55

8

6

7

12

5

2

8

14

21

33

38

40

 

Sum = 40

 

The total number of children = 40
begin mathsize 11px style therefore Median equals 1 half open square brackets open parentheses 40 over 2 close parentheses to the power of th term thin space plus open parentheses 40 over 2 plus 1 close parentheses to the power of th term close square brackets equals 1 half cross times open square brackets weight thin space of thin space open parentheses 20 to the power of th thin space plus thin space 21 to the power of st thin space child close parentheses close square brackets end style
From the table above, it can be observed that the weight of each child from 15th to 21st is 49 kg. So, the weight of 20th and 21st child is 49 kg each.
begin mathsize 11px style therefore Median equals 1 half open square brackets 49 plus 49 close square brackets equals 49 end style

ii. Mode:
In the given data, the frequency of 51 is maximum i. e. 12.
∴ Mode = 51


Q 2. 
The height of 25 students of a class is given in the following table:                                    [4M]

Height (in cm)

140

150

160

170

180

No. of students

6

8

4

5

2

Find the mean height using the short-cut method.
Solution:

Let the assumed mean A = 160.
Then the frequency table can be obtained as follows:

Height (x)

(in cm)

No. of students (f)

d = x – A

= x − 160

fd

140

150

A = 160

170

180

6

8

4

5

2

−20

−10

0

10

20

−120

−80

0

50

40

 

 begin mathsize 11px style sum straight f equals 25 end style

 

 Error converting from MathML to accessible text.

begin mathsize 11px style therefore Mean equals straight A plus fraction numerator sum fd over denominator sum straight f end fraction equals 160 plus fraction numerator negative 110 over denominator 25 end fraction equals 160 minus 4.4 equals 155.6 end style

Q 3. From the given frequency distribution table, find the following using a graph:             [6M]

  1. Lower quartile  
  2. Upper quartile
  3. Inter – quartile range

Class interval

5–10

10–15

15–20

20–25

25–30

30–35

Frequency

4

8

5

13

11

9

Solution: Let us first prepare the cumulative frequency table as follows:

Class Interval

Frequency

Cumulative frequency
(c.f.)

5–10

10–15

15–20

20–25

25–30

30–35

4

8

5

13

11

9

4

12

17

30

41

50

 

 begin mathsize 11px style straight n equals sum straight f equals 50 end style

 

Plot the points (10, 4), (15, 12), (20, 17), (25, 30), (30, 41) and (35, 50) on a graph paper.
Draw an ogive curve as shown:


begin mathsize 11px style straight i. Lower space quartile space open parentheses straight Q subscript 1 close parentheses equals open parentheses 50 over 4 close parentheses to the power of th thin space term thin space equals thin space 12.5 to the power of th thin space term thin space equals thin space 16

ii. space Upper space quartile thin space open parentheses straight Q subscript 3 close parentheses equals open parentheses fraction numerator 3 cross times 50 over denominator 4 end fraction close parentheses to the power of th thin space term thin space equals thin space 37.5 to the power of th thin space term thin space equals thin space 28.75
iii. space Inter minus quartile space range space equals straight Q subscript 3 minus straight Q subscript 1 space 28.75 space – space 16 space equals space 12.75 end style


Q 4. The following numbers are written in the descending order of their values:                         [3M]

70, 62, 50, a – 2, a – 6, a – 8, 28, 22, 16 and 10

If their median is 35, find the value of ‘a’.
Solution
Here, n = number of terms = 10, which is even.
begin mathsize 11px style therefore Median space equals space 1 half open square brackets open parentheses 10 over 2 close parentheses to the power of th thin space thin space term thin space plus thin space open parentheses 10 over 2 plus 1 close parentheses to the power of th thin space thin space term close square brackets
rightwards double arrow 35 equals 1 half open square brackets 5 to the power of th thin space term thin space plus thin space 6 to the power of th thin space term close square brackets
rightwards double arrow 70 space equals space straight a space minus space 6 space plus space straight a space minus space 8
rightwards double arrow 2 straight a space equals space 84
rightwards double arrow straight a space equals space 42 end style
                                                                                   

Q 5. Calculate the mean marks of the following distribution using the step-deviation method.               [4M]

Class interval

25–30

30–35

35–40

40–45

45–50

50–55

Frequency

8

15

25

17

14

11

Solution
Let us prepare the frequency table using the direct method as follows:

C.I.

f

Class mark (x)

Assumed mean A = 37.5

∴ d = x - A 

 begin mathsize 11px style straight t equals fraction numerator straight x minus straight A over denominator straight i end fraction end style

ft

25–30

30–35

35–40

40–45

45–50

50–55

8

15

25

17

14

11

27.5

32.5

37.5

42.5

47.5

52.5

−10

−5

0

5

10

15

−2

−1

0

1

2

3

−16

−15

0

17

28

33

 

 begin mathsize 11px style sum straight f equals 90 end style

 

 

 

 begin mathsize 11px style sum ft equals 47 end style

begin mathsize 11px style text ∴Mean=A+ end text fraction numerator sum f t over denominator n end fraction cross times i
rightwards double arrow Mean equals 37.5 plus 47 over 90 cross times 5 equals 37.5 plus 2.61 equals 40.11 end style


Q 6.
 Using a graph, draw an ogive for the following distribution which shows the marks obtained in the Mathematics paper by 150 students.                                                                                                                                        [6M]

Marks

0–10

10–20

20–30

30–40

40–50

50–60

60–70

70–80

No. of students

5

8

20

34

26

31

12

14

Use the ogive to estimate

i.  Median    ii. Number of students who score more than 55

Solution

Let us prepare the frequency table as follows:

Class Interval

Frequency

Cumulative Frequency

0–10

10–20

20–30

30–40

40–50

50–60

60–70

70–80

5

8

20

34

26

31

12

14

5

13

33

67

93

124

136

150

 

Mark the class intervals along the x-axis and cumulative frequencies along the y-axis.
Taking the upper class limits along the x-axis and corresponding cumulative frequencies (less than) along the y-axis, mark the points (10, 5), (20, 13), (30, 33), (40, 67), (50, 93), (60, 124), (70,136) and (80, 150).
Join the points marked by a free-hand curve (as shown below):


i. Here n = 150 ⇒ n/2 = 150/2 = 75

Draw a horizontal line from 75 which is on the y-axis parallel to the x-axis and meets the curve. Now, draw a vertical line which cuts the x-axis.
Thus, the median = 43
ii. Number of students who score more than 55 = 150 – 110 = 40


Chapter 17: 
Probability

Q 1. Twenty identical cards are numbered from 1 to 20. A card is drawn randomly from those 20 cards. Find the probability that the number on the card drawn is                                                                                                                          [4M]

  1. divisible by both 2 and 5
  2. greater than 20

Solution

There are a total of 20 cards numbered from 1 to 20 and a card is drawn at random.
Let S denote the sample space of this experiment, then all the possible outcomes are
i.
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
So, the number of all possible outcomes is n(S) = 20.

i. Let A be an event of getting a number which is divisible by both 2 and 5.
Now, since lcm of 2 and 5 is 10.
So, 10 and 20 are the only numbers which are divisible by both 2 and 5.
⇒ All favourable outcomes are 10 and 20, i.e. A = {10, 20}
⇒ Number of favourable outcomes is n(A) = 2
∴ Probability of getting a number which is divisible by 2 and 5 is
begin mathsize 11px style straight P open parentheses straight A close parentheses equals fraction numerator straight n open parentheses straight A close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 2 over 20 equals 1 over 10 end style
ii.

Let B be an event of getting a number greater than 20.
The cards are numbered from 1 to 20.
⇒ There are no favourable outcomes, i.e. B = {}
⇒ Number of favourable outcomes = n(B) = 0

begin mathsize 11px style therefore Probability space of space getting space straight a space number space greater space than space 20 space equals space straight P left parenthesis straight B right parenthesis equals fraction numerator straight n open parentheses straight B close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 0 over 20 equals 0 end style

Q 2. A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting.                                          [3M]

  1. an ace
  2. a queen
  3. not a jack
Solution

Let S denote the sample space of this experiment.
Since only one card is drawn from a deck of 52 cards, the total number of possible outcomes = n(S) = 52.

i.
Number of ace cards in a deck of 52 cards ace cards
Let A be an event getting an ace card.
⇒ The number of possible outcomes = n(A) = 4

begin mathsize 11px style rightwards double arrow Probability space of space getting space an space ace space card space equals space straight P left parenthesis straight A right parenthesis space fraction numerator straight n open parentheses straight A close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 4 over 52 equals 1 over 13
straight i. straight e. space straight P left parenthesis getting space an space ace space card right parenthesis space equals space 1 over 13 end style

ii.

Number of queens in a deck of 52 cards queens
Let B be an event getting a queen card
⇒ The number of possible outcomes = n(B) = 4

begin mathsize 11px style rightwards double arrow Probability space of space getting space an space queen space equals space straight P left parenthesis straight B right parenthesis space fraction numerator straight n open parentheses straight B close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 4 over 52 equals 1 over 13
rightwards double arrow straight P left parenthesis getting space a space queen right parenthesis space equals space 1 over 13 end style

iii.

Number of jacks in a deck of 52 cards jacks
Let C be an event getting a jack card
⇒ The number of possible outcomes = n(C) = 4

begin mathsize 11px style rightwards double arrow Probability space of space getting space straight a space jack space equals space straight P left parenthesis straight C right parenthesis space fraction numerator straight n open parentheses straight C close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 4 over 52 equals 1 over 13
rightwards double arrow straight P left parenthesis getting space straight a space jack right parenthesis space equals space 1 over 13
So comma space straight P left parenthesis card space drawn space is space not space straight a space jack right parenthesis space space equals space 1 space minus space straight P left parenthesis getting space straight a space jack right parenthesis
thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 minus 1 over 13 equals 12 over 13
Thus comma space straight P left parenthesis card space drawn space is space not space straight a space jack right parenthesis space equals 12 over 13 end style



Q 3. 
A single letter is selected at random from the word ‘STATISTICS’. Find the probability that it is a                         [3M]
  1. vowel  
  2. consonant

Solution

A single letter is selected at random from the word ‘STATISTICS’.
There are a total of 10 letters.
⇒ The total number of possible outcomes is 10.
The vowels are A, I, I and the consonants are S, T, T, S, T, C, S.
So, there are 7 consonants and 3 vowels.
i.
Let A be an event of getting a vowel.

⇒ The total number of favourable outcomes is 3.

begin mathsize 11px style rightwards double arrow straight P left parenthesis straight a space vowel right parenthesis space equals space fraction numerator Number space of space favourable space out space comes over denominator Number space of space all space possible space out space comes end fraction equals 3 over 10 end style
ii.

Let B be an event of getting a consonant.
⇒ The total number of favourable outcomes is 7.

begin mathsize 11px style rightwards double arrow straight P left parenthesis straight a space consonant right parenthesis space equals space fraction numerator Number space of space favourable space out space comes over denominator Number space of space all space possible space out space comes end fraction equals 7 over 10 end style

Q 4.
 A face of a die is marked with a number 1, 2, 3, -1, -2 and -3, respectively. The die is thrown once. Find the probability of getting    [4M]

  1. a negative integer
  2. the smallest positive integer

Solution

A die is thrown once, so all the possible outcomes are 1, 2, 3, -1, -2 and -3.
Then, the total number of outcomes is 6.
i.
Let A be an event of getting a negative integer.
So, all the favourable outcomes are A = {-1, -2, -3}
Total number of favourable outcomes is n(A) = 3

Error converting from MathML to accessible text.
ii.

Let B be an event of getting the smallest positive integer.
So, all the favourable outcomes are B = {1}
Total number of favourable outcomes is n(B) = 1
Error converting from MathML to accessible text.


Q 5.
 Three identical coins are tossed together. Find the probability of getting                       [4M]

  1. exactly two tails
  2. at least one tail

Solution

Let S denote the sample space of this experiment.
Three identical coins are tossed simultaneously, so all the possible outcomes are
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The number of all possible outcomes = n(S) = 8
i.

Let A be an event of getting exactly two tails.

⇒ All favourable outcomes are A = {HTT, THT, TTH}
⇒ Number of all favourable outcomes = n(A) = 3

begin mathsize 11px style therefore space Probability left parenthesis exactly space two space tails right parenthesis space equals space straight P left parenthesis straight A right parenthesis space equals fraction numerator straight n open parentheses straight A close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 3 over 8 end style

ii.
Let B be an event of getting at least one tail.

⇒ All favourable outcomes are B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
⇒ Number of all favourable outcomes = n(B) = 7

begin mathsize 11px style therefore Probability left parenthesis at space least space one space tail right parenthesis space equals space P left parenthesis B right parenthesis space equals fraction numerator straight n open parentheses straight B close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 7 over 8 end style

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