# 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     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   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.

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 .

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  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:

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.

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,

⇒ 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  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.

In ∆ABC,

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,

⇒ 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.

⇒ 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

⇒ (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

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,

In DAB

From (i) and (ii)

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)

⇒ 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:

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

= 11.25π – 2.25π – 5.0625π

= 3.9375 π

= 12.375 cm2

Q 14. In what ratio does the point  divides the line segment joining the points P (2, -2) and Q (3, 7)? Also find the value of y.

Solution:

33k + 22 = 24k + 24

Hence, the ratio is 2: 9.

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

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

This volume = base area of field ⨯ 0.1 m

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.

Solution:

We have,

Area of the region ABCD

= Area of sector AOB – Area of sector COD

= (22 42 – 11 21) cm2

= (924 – 231) cm2

= 693 cm2

= (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

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

∴ Total surface area of the toy

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

= πrl + 2πr2

= 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

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

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 .

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 – bc)2

The given equation will have equal roots if D = 0

⇒ -4 (ad – bc)2 = 0

⇒ (ad – bc)2 = 0

⇒ ad – bc = 0

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

Q 22.  Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are  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

(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

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

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

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)

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

Q 28. Solve for x:

Solution:

⇒ 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,

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.

Solution:

The man is at the top of the tower AB.

In right angled triangle ∆ABX and ∆ABY

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 circle – area of ∆BAC -  Area of sector CD

= 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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• Increase paper-solving speed and confidence with weekly tests

### STUDY RESOURCES

REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India.