CBSE Class 10 Mathematics Previous Year Question Paper 2017 All India Set-3

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Q 1. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?


Q. 2 If a tower 30 m high, casts a shadow  begin mathsize 12px style 10 square root of 3 end style   m long on the ground, then what is the angle of elevation of the sun?


Q 3. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP.


Q 4. What is the common difference of an A.P. in which a21 – a7 = 84?


Q 5. A circle touches all the four sides of a quadrilateral ABCD. Prove that  AB + CD = BC + DA


Q 6. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.


Q 7. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.


Q 8. If the distances of P(x, y) from A (5, 1) and B (-1, 5) are equal, then prove that 3x = 2y.


Q 9. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.


Q 10. For what value of n, are the nth terms of two A.Ps 63, 65, 67… and 3, 10, 17… equal?


Q 11. On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.


Q 12. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.


Q 13 Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.

 

Q 14. In what ratio does the point begin mathsize 12px style open parentheses 24 over 11 comma straight y close parentheses end style  divides the line segment joining the points P (2, -2) and Q (3, 7)? Also find the value of y. 


Q 15. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?

 

Q 16. In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. begin mathsize 12px style open parentheses Use space straight pi  =  fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close parentheses end style

 

 Q 17. The dimensions of a solid iron cuboid are 4.4 m ⨯ 2.6 m ⨯ 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.

 

Q 18. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

 

Q 19. How many terms of an A.P. 9, 17, 25… must be taken to give a sum of 636?


Q 20.  If the roots of the equation (a2 + b2) x2 – 2(ac + bd) x + (c2 + d2) = 0 are equal, prove that begin mathsize 12px style straight a over straight b equals straight c over straight d end style.

 

Q 21. If the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear, then find the value of k.

 

Q 22.  Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are begin mathsize 12px style 3 over 4 end style times the corresponding sides of the ∆ ABC.

 

Q 23. Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product

 

Q 24. In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangents AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

 

Q 25.  In a rain–water harvesting system, the rain-water from a roof of 22 m ⨯ 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.

 

Q 26. Prove that the lengths of two tangents drawn from an external point to a circle are equal.

 

Q 27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their 9th terms.

 

Q 28. Solve for x:

begin mathsize 12px style fraction numerator straight x minus 1 over denominator 2 straight x plus 1 end fraction plus fraction numerator 2 straight x plus 1 over denominator straight x minus 1 end fraction equals 2 comma space where space straight x space not equal to  -  1 half , 1 end style


Q 29. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?

 

Q 30. From the top of a tower, 100 m high, a man observe two cars on the opposite sides of the tower and in same straight line with its base, with its base, with angles of depression 30° and 45°. Find the distance between the cars. begin mathsize 12px style open square brackets Take space square root of 3 equals 1.732 close square brackets end style

 

Q 31. In the given figure, O is centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

Q 1. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?

Solution:

Let the total number of rotten apples in a heap = n

Total number of apples in a heap = 900

Probability of selecting a rotten apple from a heap = 0.18

Now,

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis selecting space straight a space rotten space apple ) =  fraction numerator Number space of space rotten space apples over denominator Total space number space of space apples end fraction end cell row cell rightwards double arrow 0.18 = straight n over 900 end cell end table end style

⇒ n = 0.18 ⨯ 900

⇒ n = 162

Hence, the number of rotten apples is 162.

 

 

Q. 2 If a tower 30 m high, casts a shadow begin mathsize 12px style 10 square root of 3 end style m long on the ground, then what is the angle of elevation of the sun?

Solution:

Let AB be the tower and BC be its shadow.

Error converting from MathML to accessible text.

In ∆ABC,

Error converting from MathML to accessible text.

 

 

Q 3. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP.

Solution:

In the figure, PA and PB are two tangents from an external point P to a circle with centre O and radius = a

∠APB = 60°   (given)

⇒ ∠APO = 30°  (tangents are equally inclined to the line joining the point and the centre)

Now, OA ⏊ AP

In right – angled triangle OAP,

 Error converting from MathML to accessible text.

⇒ OP = 2a

 

 

 

 

Q 4. What is the common difference of an A.P. in which a21 – a7 = 84?

Solution:

Let a be the first term and d be the common difference of the given A.P.

∴ a21 – a= 84

⇒ (a + 20d) – (a + 6d) = 84

⇒ a + 20d – a - 6d = 84

⇒ 14d = 84

⇒ d = 6

Hence, the common difference is 6.

 

Q 5. A circle touches all the four sides of a quadrilateral ABCD. Prove that  AB + CD = BC + DA

Solution:

Since tangents drawn from an external point to a circle are equal in length, we have

AP = AS …. (i)

BP = BQ …. (ii)

CR = CQ …. (iii)

DR = DS …. (iv)

Adding (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

⇒ AB + CD = BC + DA

 

 

Q 6. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

Solution:

Let AB be a chord of circle with centre O.

Let AP and BP be two tangents at A and B respectively.

Suppose the tangents meet at point P. Join OP.

Suppose OP meets AB at C.

Now, in ∆PCA and ∆PCB,

PA = PB …. (tangents from an external point are equal)

∠APC = ∠BPC …. (PA and PB are equally inclined to OP)

PC = PC …. (common)

Hence, ∆PAC ≅ ∆PBC …. (by SAS congruence criterion)

⇒ ∠PAC = ∠PBC

 

 

Q 7. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.

Solution:

Since a line is intersecting y-axis at P and x-axis at Q, 

Coordinates of P = (0, y) and coordinates of Q = (x, 0)

Let R be the mid-point of PQ.

begin mathsize 12px style table attributes columnalign left end attributes row cell Then ,  co minus ordinates space of space straight R  =  open parentheses fraction numerator 0 plus straight x over denominator 2 end fraction comma fraction numerator straight y plus 0 over denominator 2 end fraction close parentheses equals left parenthesis 2 comma negative 5 right parenthesis end cell row cell rightwards double arrow open parentheses straight x over 2 comma straight y over 2 close parentheses equals left parenthesis 2 comma negative 5 right parenthesis end cell row cell rightwards double arrow straight x over 2 equals 2 space and space straight y over 2 equals negative 5 end cell end table end style

⇒ x = 4 and y = -10

Hence, co-ordinates of P are (0,-10) and co-ordinates of Q are (4, 0).

 

 

 

Q 8. If the distances of P(x, y) from A (5, 1) and B (-1, 5) are equal, then prove that 3x = 2y.

Solution:

Given, P(x, y) is equidistant from A (5, 1) and B (-1, 5)

Now, AP = BP

begin mathsize 12px style rightwards double arrow square root of left parenthesis 5 minus straight x right parenthesis squared plus left parenthesis 1 minus straight y right parenthesis squared end root equals square root of left parenthesis negative 1 minus straight x right parenthesis squared plus left parenthesis 5 minus straight y right parenthesis squared end root end style

⇒ (5 - x) 2 + (1 - y) 2 = (-1 – x) 2 + (5 - y)2

⇒ (25 + x2 – 10x) + (1 + y2 - 2y) = (1 + x2 + 2x ) + (25 + y2 - 10y)

⇒ x2 + y2 - 10x - 2y + 26 = x2 + y2 + 2x - 10y + 26

⇒ -10x – 2x = -10y + 2y

⇒ -12x = -8y

⇒ 3x = 2y …..(Dividing throughout by -4)

 

 

 

Q 9. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.

 Solution:

Given, px2  – 14x + 8 = 0

Here, a = p, b = -14, c = 8

Let α and β be the roots of the given quadratic equation.

Then, β = 6α

Error converting from MathML to accessible text.

 

 

 

Q 10. For what value of n, are the nth terms of two A.Ps 63, 65, 67… and 3, 10, 17… equal?

Solution:

For A.P. 63, 65, 67… We have

First term = 63 and common difference = 65 – 63 =2

Hence, nth term = an = 63 + (n – 1)2

⇒ an = 63 + 2n – 2 = 2n + 61

For A.P. 3, 10, 17, … we have

First term = 3 and common difference = 10 – 3 = 7

Hence, nth term = an’ = 3 + (n – 1)7

⇒ an’ = 3 + 7n – 7 = 7n - 4

The two A. Ps will have identical nth term, if

an = an

⇒ 2n + 61 = 7n – 4

⇒ 5n = 65

⇒ n = 13

 

 

Q 11. On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

Solution:

Let AB be the tower with height h.

Let x be the angle of elevation from C.

So, the angle of elevation from D is (90- x).

....(Since the angles of elevation from C and D are complementary)

In CAB,

begin mathsize 12px style table attributes columnalign left end attributes row cell tan space straight x  =  AB over AC end cell row cell rightwards double arrow space tan space straight x  =  straight h over 4 space            . space. space. space. space. space left parenthesis straight i right parenthesis end cell end table end style

In DAB

Error converting from MathML to accessible text.

From (i) and (ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell tan space straight x space cross times space cot space straight x  =  straight h over 4 cross times straight h over 16 end cell row cell rightwards double arrow   1  =  straight h squared over 64 end cell row cell rightwards double arrow straight h squared equals 64 end cell row cell rightwards double arrow straight h equals square root of 64 end cell end table end style

h = 8 m

Hence, the height of the tower is 8 m.




Q 12. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Solution:

Let the number of black balls in the bag be x.

Number of white balls = 15

Hence, total number of balls in the bag = x + 15

Given, P (black ball) 3 ⨯ P (white ball)

 begin mathsize 12px style table attributes columnalign left end attributes row cell rightwards double arrow fraction numerator straight x over denominator straight x plus 15 end fraction equals 3 cross times fraction numerator 15 over denominator straight x plus 15 end fraction end cell row cell rightwards double arrow fraction numerator straight x over denominator straight x plus 15 end fraction equals fraction numerator 45 over denominator straight x plus 15 end fraction end cell end table end style

⇒ x = 45

Thus, the number of black balls in the bag is 45.

 

 

Q 13. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.

Solution:

Error converting from MathML to accessible text.

Radius of semi-circle E = 4.5 cm

Now, area of the shaded region

= Area of semi-circle (E + B) –Area of semi-circle (A + C) –Area of circle D

Error converting from MathML to accessible text.

= 11.25π – 2.25π – 5.0625π

= 3.9375 π

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= 12.375 cm2

 

 

Q 14. In what ratio does the point begin mathsize 12px style open parentheses 24 over 11 comma straight y close parentheses end style divides the line segment joining the points P (2, -2) and Q (3, 7)? Also find the value of y.

Solution:

Error converting from MathML to accessible text.

33k + 22 = 24k + 24

begin mathsize 12px style rightwards double arrow 9 straight k space equals space 2 rightwards double arrow straight k space equals 2 over 9 end style

Hence, the ratio is 2: 9.

begin mathsize 12px style table attributes columnalign left end attributes row cell Also comma    fraction numerator 7 straight k minus 2 over denominator straight k plus 1 end fraction equals straight y rightwards double arrow fraction numerator 7 cross times 2 over 9 minus 2 over denominator 2 over 9 plus 1 end fraction    equals    straight y end cell row cell rightwards double arrow straight y  =   fraction numerator fraction numerator 14 minus 18 over denominator 9 end fraction over denominator fraction numerator 2 plus 9 over denominator 9 end fraction end fraction    equals    fraction numerator negative 4 over denominator 9 end fraction cross times 9 over 11 equals negative 4 over 11 end cell end table end style

 

 

Q 15. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?

Solution:

We have,

Width of the canal = 5.4 m,

Depth of the canal = 1.8 m

It is given that the water is flowing with a speed of 25 km/hr.

Therefore,

Length of the water columns formed in 40 mins

Error converting from MathML to accessible text.

This volume = volume of cuboid (10 cm of standing water is required for irrigation)

This volume = base area of field ⨯ 0.1 m

 Error converting from MathML to accessible text.

Hence, the canal irrigates 1620000 m2 area in 40 mins

 

 

Q 16. In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. begin mathsize 12px style open parentheses Use space straight pi  =  fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close parentheses end style

Solution:

We have,

Area of the region ABCD

= Area of sector AOB – Area of sector COD

begin mathsize 12px style table attributes columnalign left end attributes row cell straight equals open parentheses 60 over 360 cross times 22 over 7 cross times 42 cross times 42 minus 60 over 360 cross times 22 over 7 cross times 21 cross times 21 close parentheses space cm to the power of straight 2 end cell row cell straight equals open parentheses 1 over 6 cross times 22 cross times 6 cross times 42 minus 1 over 6 cross times 22 cross times 3 cross times 21 close parentheses space cm to the power of straight 2 end cell end table end style

= (22 42 – 11 21) cm2

= (924 – 231) cm2

= 693 cm2

begin mathsize 12px style Area space of space circular space ring  =  open parentheses 22 over 7 cross times 42 cross times 42 minus 22 over 7 cross times 21 cross times 21 close parentheses space cm squared end style

= (22 ⨯ 6 ⨯ 42 – 22 ⨯ 3 ⨯ 21) cm2

= (5544 – 1386) cm2

= 4158 cm2

Hence, Required shaded region = Area of circular ring - Area of region ABCD

= (4158 – 693) cm2

= 3465 cm2

 

 

Q 17. The dimensions of a solid iron cuboid are 4.4 m ⨯ 2.6 m ⨯ 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.

Solution:

Let the length of the pipe be h cm

Then, volume of iron pipe = volume of iron the block.

Volume of the block = (4.4 ⨯ 2.6 ⨯ 1) m3 = (440 ⨯ 260 ⨯ 100) cm3

r = internal radius of the pipe = 30 cm

R = External radius of the pipe = (30 + 5) cm = 35 cm

∴ Volume of the iron pipe = (External volume) – (Internal Volume)

= πR2h – πr2h

= π (R2 – r2) h

= π(R + r) (R - r) h

=π (35 + 30) (35 – 30) h

= π ⨯ 65 ⨯ 5 ⨯ h

Now, Volume of iron in the pipe = Volume of iron in the block

⇒   π ⨯ 65 ⨯ 5 ⨯ h = 440 ⨯ 260 ⨯ 100

Error converting from MathML to accessible text.

Thus, the length of the pipe is 112 m.



 

Q 18. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Radius of common base = 3.5 cm

Total height of toy = 15.5 cm

Height of cone = 15.5 – 3.5 = 12 cm

For cone,

l2 = r2 + h2

⇒ l2 = (3.5)+ (12)2

⇒ l2 = 12.25 + 144

⇒ l2 = 156.25

 Error converting from MathML to accessible text.

∴ Total surface area of the toy

= Curved surface area of cone + Curved surface area of hemisphere

= πrl + 2πr2

Error converting from MathML to accessible text.

= 214.5 cm2




Q 19. How many terms of an A.P. 9, 17, 25… must be taken to give a sum of 636?

Solution:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a2 – a1 = 17 – 9 = 8

begin mathsize 12px style table attributes columnalign left end attributes row cell straight S subscript straight n equals straight n over 2 open square brackets 2 straight a left parenthesis straight n minus 1 right parenthesis straight d close square brackets end cell row cell rightwards double arrow 636   straight n over 2 open square brackets 2 cross times 9 plus left parenthesis straight n minus 1 right parenthesis 8 close square brackets end cell end table end style

636 = n [9 + (n - 1) 4]

636 = n [9 + 4n – 4]

⇒ 636 = n (4n + 5)

⇒ 4n2 + 5n – 636 = 0

⇒ 4n2 + 53n – 48n – 636 = 0

⇒ n (4n + 53) – 12 (4n + 53) = 0

⇒ (4n + 53) (n – 12) = 0

⇒ 4n +53 = 0 or n – 12 = 0

Error converting from MathML to accessible text.

Since number of terms can neither be negative nor fractional, we have n = 12




Q 20.  If the roots of the equation (a2 + b2) x2 – 2(ac + bd) x + (c2 + d2) = 0 are equal, prove that begin mathsize 12px style straight a over straight b equals straight c over straight d end style.

We have

(a+ b2) x2 - 2(ac + bd) x + (c+ d2) = 0

The discriminant of the given equation is given by

D = [-2(ac + bd)]2 – 4 ⨯ (a+ b2) ⨯ (c2 + d2)

⇒ D = 4 (ac + bd)2 – 4 (a2c2 + a2d2 + b2c2 + b2d2)

⇒ D = 4 (a2c+ b2d2 + 2abcd) – 4 (a2c+ a2d2 + b2c2 + b2d2)

⇒ D = 4 (a2c+ b2d2 + 2abcd – a2c2 – a2d- b2c- b2d2)

⇒ D = 4(2abcd – a2d2 – b2c2)

⇒ D = -4 [(ad)2 + (bc)2 – 2(ad) (bc)]

⇒ D = -4 (ad – bc)2

The given equation will have equal roots if D = 0

⇒ -4 (ad – bc)2 = 0

⇒ (ad – bc)2 = 0

⇒ ad – bc = 0

⇒ad = bc

begin mathsize 12px style text ⇒ end text straight a over straight b equals straight c over straight d end style

 

 


Q 21. If the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear, then find the value of k.

Solution:

Given points are A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k)

These points will be collinear, if area of the triangle formed by them is zero.

We have,

i.e,
{[(k + 1) (2k + 3) + (3k) (5k) + (5k – 1) (2k)] - [(3k) (2k) + (5k – 1) (2k+3) +(k +1) (5k)]}]=0

⇒ {(2k2 + 5k + 3 + 15k2 + 10k2 -2k) – (6k2 + 10k2 +13k – 3+ 5k2 + 5k)} = 0

⇒ (27k2 + 3k + 3) – (21k2 + 18k – 3) = 0

⇒ 27k2 + 3k + 3 – 21k2 – 18k + 3 = 0

⇒ 6k2 – 15k + 6 = 0

⇒ 2k2 – 5k + 2 = 0

⇒ 2k2 – 4k – k + 2 = 0

⇒ (k – 2) (2k – 1) = 0

⇒ k – 2 = 0 or 2k – 1 = 0

Error converting from MathML to accessible text.

 

 

Q 22.  Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are begin mathsize 12px style 3 over 4 end style times the corresponding sides of the ∆ ABC.

Solution:

Steps of construction:

  1. Draw BC = 7cm
  2. At B, construct ∠CBX = 45°  and at C construct ∠BCY = 180° - (45° + 105°) = 30°
  3. Let BX and CY intersect at A. ∆ABC so obtained is the given triangle.
  4. Construct an acute angle ∠CBZ at opposite side of vertex A and ∆ABC.
  5. Mark – off four points (greater of 4 and 3 in 3/4) points, B1, B2, B3, B4 on BZ such that BB1 = B1B2 = B2B3 = B3B4
  6. Join B4 to C.
  7. Draw B3C' parallel to B4C which meets BC at C.
  8. From C', draw C'A' parallel to CA meeting BA at A'.

Thus, ∆A'BC' is the required triangle, each of whose sides is times the corresponding sides of ∆ABC .




Q 23. Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product

Solution:

Elementary events associated to the random experiment of throwing two dice are:

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

∴ Total number of elementary events = 6 ⨯ 6 = 36

(i)Let A be the event of getting an even number as the product.

i.e., 2,4,6,8,10,12,

Elementary events favourable to event A are:

(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),

(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)

Total number of favourable events = 18

begin mathsize 12px style Hence ,  required space probablity  =  18 over 36 equals 1 half end style

(ii)Let B be the event of getting an even number as the sum.

i.e., 2,4,6,8,10,12,16,18,20,24,30,36

Elementary events favourable to event B are:

(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),

(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),

(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),

(6,4),(6,5),(6,6)

Total number of favourable events = 27

begin mathsize 12px style Hence ,  required space probablity space =  27 over 36 equals 3 over 4 end style



Q 24. In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangents AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.


Solution:

Since tangents drawn from an external point to a circle are equal.

Therefore, AP = AC.

Thus, in triangles AOP and AOC, we have

AP = AC

AO = AO [Common side]

OP = OC [Radii of the same circle]

So, by SSS- criterion of congruence,

we have

∆AOP ≅ ∆AOC

⇒  ∠PAO = ∠CAO

⇒ ∠PAC = 2∠CAO

Similarly, we can prove that ∠QBO = ∠CBO

⇒ ∠CBQ = 2∠CBO

Now, ∠PAC + ∠CBQ = 180°   [Sum of the interior angle on the same side of transversal is 180°]

⇒ 2∠CAO + 2∠CBO = 180°    [using equations (i) and (ii)]

⇒ ∠CAO + ∠CBO = 90°

⇒ 180° - ∠AOB = 90°   [Since ∠CAO, ∠CBO and ∠AOB are angles of a Triangle, ∠CAO + ∠CBO + ∠AOB = 180°]

⇒∠AOB = 90°

 

 

Q 25.  In a rain–water harvesting system, the rain-water from a roof of 22 m ⨯ 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5m. If the tank is full, find the rainfall in cm. Write your views on water conservation.

Solution:

We have,             

r = Radius of cylindrical vessel = 1m

h = Height of cylindrical vessel = 3.5 m

Let the rainfall be x m.

Then, volume of the water

= Volume of cuboid of base 22 m 20 m and height metres

= (22 20 x) m3

Since the vessel is just full of the water that drains out of the roof into the vessel,

Volume of the water = Volume of the cylindrical vessel

22 20 x = 11

Error converting from MathML to accessible text.

Thus, the rainfall is 2.5 cm.


 

Q 26. Prove that the lengths of two tangents drawn from an external point to a circle are equal.

Solution:

Given: AP and AQ are two tangents from a point A to a circle C(O, r)

To prove: AP = AQ

Construction: Join OP, OQ and OA

Proof:

In ∆OPA and ∆OQA,

∠OPA = ∠OQA = 90°   ….(Tangent at any point of a circle is perpendicular to the radius through the point of contact)

OP = OQ                       ……(Radii of a circle)

OA = OA                        …….(Common)

Hence, by RHS-criterion of congruence, we have

∆OPA ≅ ∆OQA

⇒ AP = AQ                   ………(c.p.c.t)

 

 

Q 27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their 9th terms.

Solution:

Let a1, a2, be the first terms and d1, d2 the common differences of the two given A.P’s

Then, sum of their n terms is given by

Error converting from MathML to accessible text.

In order to find the ratio of the mth terms of the two given A.P’s

We replace n by [2m- 1] in equation (i)

 In order to find the ratio of the 9th terms of the two given A.P’s

We replace n by 17 [2 9 – 1] in equation (i)

begin mathsize 12px style table attributes columnalign left end attributes row cell fraction numerator 2 straight a subscript 1 plus left parenthesis 17 minus 1 right parenthesis straight d subscript 1 over denominator 2 straight a subscript 2 plus left parenthesis 17 minus 1 right parenthesis straight d subscript 2 end fraction equals fraction numerator 7 cross times 17 plus 1 over denominator 4 cross times 17 plus 27 end fraction end cell row cell rightwards double arrow fraction numerator 2 straight a subscript 1 plus 16 straight d subscript 1 over denominator 2 straight a subscript 2 plus 16 straight d subscript 2 end fraction equals 120 over 95 end cell row cell rightwards double arrow fraction numerator straight a subscript 1 plus 8 straight d subscript 1 over denominator straight a subscript 2 plus 8 straight d subscript 2 end fraction equals 24 over 19 end cell end table end style

Thus, the ratio of their 9th terms is 24:19



 

Q 28. Solve for x:

begin mathsize 12px style fraction numerator straight x minus 1 over denominator 2 straight x plus 1 end fraction plus fraction numerator 2 straight x plus 1 over denominator straight x minus 1 end fraction equals 2 comma space where space straight x space not equal to  -  1 half , 1 end style

Solution:

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⇒ 5x2 + 2x + 2 = 4x2 – 2x – 2
⇒ 5x2 + 2x + 2 – 4x2 + 2x + 2 = 0
⇒ x2 + 4x + 4 = 0
⇒ x2 + 2x + 2x + 4 = 0
⇒x x + 2 + 2x + 2 = 0
⇒ x + 2 x + 2 = 0
⇒ x + 22=0
⇒ x + 2 = 0
⇒ x = -2


Q 29. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?

Solution:

Suppose B alone takes x days to finish the work.

Then, A alone can finish it in (x – 6) days

Now,

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8x – 24 = x2 – 6x
⇒ x2 – 6x – 8x + 24 = 0
⇒ x2 – 14x + 24 = 0
⇒ x2 – 12x – 2x + 24 = 0
⇒ x (x – 12) – 2 (x – 12) = 0
⇒ (x – 12) (x – 2) = 0
⇒ x – 12 = 0 or x – 2 = 0
⇒ x = 12 or x = 2

But x cannot be less than 6.

So, x = 12

Hence B alone can finish the work in 12 days




Q 30. From the top of a tower, 100 m high, a man observe two cars on the opposite sides of the tower and in same straight line with its base, with its base, with angles of depression 30° and 45°. Find the distance between the cars. begin mathsize 12px style open square brackets Take space square root of 3 equals 1.732 close square brackets end style

Solution:

The man is at the top of the tower AB.

In right angled triangle ∆ABX and ∆ABY

Error converting from MathML to accessible text.

 

 

Q 31. In the given figure, O is centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

Solution:

AC = 24 cm, AB = 7 cm

Since BC is the diameter of the circle,

So, ∠BAC = 90°

In right ∆BAC,

BC2 = AC2 + AB2

⇒ BC2 = 242 + 72 

⇒ BC2 = 625

⇒ BC = 25 cm

So, the radius of the circle = OC = 12.5 cm

Area of the shaded region

= area of the circle – area of ∆BAC -  Area of sector CD

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= 491.07 – 84 – 122.77

= 284.3 cm2 (approximately)

Hence, the area of the shaded region is 284.3 cm2 approximately

 

 


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