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.

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]

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:                                                                                                                            [3M]

Q 3.

Q 4.
Solve for x:                                                                                                [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 , find the value of  .                                                                                                    [3M]

Q 2.
If a, b, c and d are in proportion, prove that   .                                                             [4M]

Q 3.
If , 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]

Q 2.
If  and ; find the matrix X such that AX = B.                                                              [4M]

Q 3.
Given ,  and , find the matrix X such that  A + X = 2B + C.                           [3M]

Q 4.
Given  and  ,  and A2 = 9A + mI. Find m.                                                                          [4M]

Q 5.
Solve for x and y :                                                                                          [3M]

Q 6.
Evaluate:                                                                                                              [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 .

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  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

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, BAC = 40˚ and CBD = 60˚. Find                                                              [4M]

1. BCD
2. BCA
3. ABC

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. BDC = 60˚. Find BAO.          [4M]

Q 5.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ACO = 25˚, find                   [3M]

1. BCO
2. AOB
3. APB

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                            [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:                                                                                                                 [3M]

Q 2.
Evaluate:                                                                                       [3M]

Q 3.
If sin A + tan A = p and tan A - sin A = q, prove that .                                                              [4M]

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

Q 5.
If x = a sinθ and y = b tanθ, show that .                                                                                         [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 =

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 = = 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 =  = Rs. 1620
SGST =  = 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

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

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 =

⇒ 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 =

⇒ 47250 = 42000 + 437.5r

⇒ 437.5r = 5250

⇒ r = 12% p.a.

ii. Interest =

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]

Solution:

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:

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:                                                                                                                            [3M]

Solution:

⟹ 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 =

Q 3.  Give your answer correct to two significant decimal places.                                            [3M]

Solution:

⟹ 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

Q 4. Solve for x:                                                                                                [4M]

Solution:

⟹ 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

⟹ 2(x2 + 1) = -5x

⟹ 2x2 + 5x + 2 = 0

⟹ 2x2 + 4x + x + 2 = 0

⟹ 2x(x + 2) + (x + 2) = 0

⟹ (2x + 1)(x + 2) = 0

⟹ 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

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 =  hour
Time taken to cover 60 km when the speed is increased by 15 km/hr =  hour
It is given that the time to cover 60 km reduced by 20 minutes =   hour
According to the question,

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

Length when decreased by 10 m = (x – 10) m
Breadth when increased by 10 m = (y + 10) m =
According to the question,

(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 =
Time taken to return =
According to the question,

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.

When the number of children were 10 more than the original number of children, each child gets = Rs.

According to the question,

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 , find the value of  .                                                                                                    [3M]

Solution:

Q 2. If a, b, c and d are in proportion, prove that   .                                                             [4M]

Solution:

Q 3. If , then find x:y.                                                                                                                     [4M]

Solution:

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

And 112x = y2

Put x =  in 112x = y2, we get

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

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]

Solution:

Q 2. If  and ; find the matrix X such that AX = B.                                                              [4M]

Solution:

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.

Q 3. Given  and , 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

Q 4. Given  and  ,  and A2 = 9A + mI. Find m.                                                                          [4M]

Solution:

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

Q 5.
Solve for x and y :                                                                                          [3M]

Solution:

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:                                                                                                              [3M]

Solution:

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

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,

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

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,

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.

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.

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 .
Solution:
i. Given equation of the line is 4x – 5y = 9
Comparing it with y = mx + c, we get , 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  is given by
∴ y – y1 = m(x – x1)
∴ y – 0 = (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

ii. Inclination is θ = 60˚ and passing through (−4, 2)
slope (m) = tan θ = tan 60˚ =
The equation of the line is
∴ y – y1 = m(x – x1)
∴ y – 2 = (x + 4)
∴
iii. To find the equation of the line passing through the points (3, −4) and (7, 1) is
The equation of the line passing through (3, −4) with slope =  is given by

Q 9. Write the equation of the line whose gradient is  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.

The equation of the line whose gradient is  and passing through (0, 2) is

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.

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

∴ Mid-point of P (6, -11)  and Q (34, 3) is

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
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

∴ 72 – 3x = x

∴ 4x = 72
∴ x = 18 cm
∴ EC = 18 cm

ADE = ABC                     ∵ corresponding angles
AED = ACB                     ∵ corresponding angles
∴ ΔADE ~ ΔABC                        ∵ by AA similarity

iii.

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:

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 =

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

⇒ 2 = × 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

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

⇒ 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, BAC = 40˚ and CBD = 60˚. Find                                                              [4M]

1. BCD
2. BCA
3. ABC
Solution:

AC and BD are the diagonals of a cyclic quadrilateral.
∴ BAC = 40˚ and CBD = 60˚
CAD = CBD = 60˚                 ∵ angle in the same segment
Also, BAC = BDC = 40˚
= 40˚ + 60˚
= 100˚

i. BCD + BAD = 180˚     ∵ opposite angles of a cyclic quadrilateral

BCD + 100˚ = 180˚
BCD  = 80˚

ABCD is an isosceles trapezium and AB || DC
BAC = DCA                 ∵ alternate angles
DCA = 40˚
ABD = ACD = 40˚      ∵ angles in the same segment
BCA = BCD – ACD
= 80˚ – 40˚
= 40˚

iii. ABC = ABD + CBD
= 40˚ + 60˚
= 100˚

iv. ADB = BCA = 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:

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. BDC = 60˚. Find BAO.          [4M]

Solution:

In ΔBDC,
DBC = 90˚                      ∵ tangent radius
BDC = 60˚       …(i)
In ΔBDC,

DCB + DBC + BDC = 180˚         ∵ angle sum property in the triangle
DCB + 90˚ + 60˚ = 180˚
DCB = 30˚
In ΔOEC,
OEC = OCE              ∵ isosceles triangle property
OEC = BCD = 30˚
In ΔAEB,
ADE + BDE = 180˚            ∵ straight line property

ADE + 60˚  = 180˚            ∵ BDE = BDC from (i)
DEA = OEC = 30˚…vertically opposite angle
DAE + ADE + DEA = 180˚    ∵ angle sum property in the triangle
DAE + 120˚ + 30˚ = 180˚
DAE = 30˚
BAO = DAE = 30˚

Q 5.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ACO = 25˚, find                   [3M]

1. BCO
2. AOB
3. APB

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
BCO = ACO = 25˚

ii. In ΔOAC,
AOC = 180˚  –  OAC  – ACO
= 180˚ – 90˚ – 25˚
= 65˚
ΔOAC ≅ ΔOBC
BOC = AOC = 65˚
AOB = AOC + BOC
= 65˚ + 65˚
= 130˚
ii. APB = ½ AOB    ∵ angle subtended at the centre = 2 times angle on the circle
APB = 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                        [4M]

Solution:

Since M is the mid-point of AB,
∴ AM = MB =
Since P and S are the mid-points of AM and MB,
∴ AP = PM = MS = SB =
Radius of a circle = r
∴ CP = r + PQ = r +
∴ CM = DM – DC =  – r

In Δ CMP,
CMP = 90˚
∴ CM2 + MP2 = CP2

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

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 =  = 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

Length of wire required for 40 rounds = 10∏ × 40 = 400∏ cm = 400 × 3.14 = 1256 cm

Now, radius of the wire =

Volume of the wire = ∏r2h =

Weight of the wire = Volume of the wire × 8.88 gm =

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:

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

i. Hence, height of the cone = 17.5 cm

Chapters 13: Trigonometrical Identities

Q 1. Prove that:                                                                                                                 [3M]

Solution:

Q 2.
Evaluate:                                                                                 [3M]

Solution:

Q 3. If sin A + tan A = p and tan A - sin A = q, prove that .                                                              [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)

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 .                                                                                         [3M]

Solution:

Given: x = a sinθ and y = b tanθ

Consider

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.

ii.

Let B be the new position of the boat and θ be 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]

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, ABD = 30° and EBD = 60°.

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

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

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

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.

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

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

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:

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.

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 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

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

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

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

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

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

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.

ii.

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

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

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

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

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

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