# CBSE Class 10 Maths Previous Question Paper 2017 Solution

## CBSE Class 10 Maths Previous Question Paper 2017 Solution MATHEMATICS

SECTION – A

1) The ratio of the height of a tower and the length of its shadow on the ground is √3:1. What is the angle of elevation of the sun ?

Ans: tan θ=AB/BC=√3/1

-> θ=60°

2) Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere ?

Ans: 2/3 πr3=3 πr2->r=9/2 units

d = 9 units

3) A number is chosen at random from the numbers -3,-2,-1,0,1,2,3  What will be the probability that square of this number is less then or equal to 1 ?

Ans: Favourable outcomes are –1, 0, 1

d = 9 units

4) If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k ?

Ans: (4-1)2+(k-0)2=5

->k=≠4

SECTION – B

5) Find the roots of the quadratic equation √2x2+7x+5√2=0

Ans: √2x2+7x+5√2=0

-> √2x2+2x+5x+5√2=0

-> (√2x+5)(x+√2)=0

-> x=-5/√2,- √2

or -5√2/2, -√2

6) Find how many integers between 200 and 500 are divisible by 8.

Ans: A.P. formed is 208, 216, 224, …, 496

an=496

-> 208+(n-1)x8=496

->n=37

7) Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other.

Ans: ∠PAQ=∠OBS=90°

But these are alternate interior angle

\ PQ || RS

8) Find the value of k for which the equation x2+k(2x+k-1)+2=0 has real and equal roots.

Ans:

x2+k(2x+k-1)+2=0

-> x2+2kx+(k2-k+2)=0

For, equal roots, b2-4ac=0

-> 4k2-4k2+4k-8=0

-> k=2

9) Draw a line segment of length 8 cm and divide it internally in the ratio 4:5

Ans: Correct construction

10) In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA 12 = cm,

QC = QD = 3 cm, then find PC + PD. PA=PC+CA=PC+CQ

-> 12=PC+3->PC=9CM

PD=9CM

∴ PC+PD=18CM

SECTION – C

11) If mth term of an A.P. is 1/n and nth term is 1/m, then find the sum of its first mn terms.

Ans: am=1/n->a+(m-1)d=1/n  ….(1)

an=1/m->a+(n-1)d+1/m …(2)

solving (1) and (2), a=1/mn and d=1/mn

Smn=mn/2[2×1/mn+(mn-1)x1/mn]

=1/2(mn+1)

12) Find the sum of n terms of the series (4-1/n)+(4-2/n)+(4-3/n)+……

Ans: Sn=(4-1/n)+(4-2/n)+(4-3/n)+… upto n terms

=(4+4+…+4)-1/n(1+2+3+…+n)

=4n-1/nxn(n+1)/2

=7n-1/2

13) If the equation (1+m2)x2+2mcx+c2-a2=0 has equal roots then show that c2=a2(1+m2)

Ans: (1+m2)x2+2mcx+c2-a2=0

Eor equal roots, B2-4AC=0

-> 4m2c2-4(1+m2)(c2-a2)=0

->m2c2-c2-m2c2+a2+m2a2=0

->c2=a2(1+m2)

14) The 3/4th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.

Ans: 3/4x volume of conical vessel=volume of cylindrical vessel

Let the height of cylindrical vessel be h

->3/4×1/3x πx5x5x24= πx10x10xh

->h=3/2cm or 1.5cm

15) In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD 2 = cm, find the area of the shaded region. Ans: =(22/7×3.5×3.5/4-1/2×3.5×2)cm2

=49/8 or 6.125 cm2

16) Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ =2 ∠ OPQ

Ans: Let ∠OPQ= θ

->∠TPQ=90°- θ=∠TQP

∠TPQ+∠TQP+∠PTQ=180°

->90°- θ+90°- θ+ ∠PTQ=180°

->∠PTQ=2θ

=2∠OPQ

17) Show that △ ABC, where A(-2,0), b(2,0) and △ PQR where p(-4,0), R(0,4) are similar triangles.

Ans; A(-2,0),B(2,0),C(0,2)

AB=4 units, BC=2√2units, AC=2√2units

p(-4,0),q(4,0),r(0,4)

pq=8 units, qr=4√2 units, pr=4√2 units

∴ AB/PQ=BC/QR=AC/PR=1/2

∴△ABC ~△PQR

18) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3,-2). If the third vertex is (7/2,y), ind the value of y.

Ans: ar(△ABC)=5 SQ.units

->1/2[2(-2-y)+3(y-1)+7/2(1+2)]=5

->y+7/2=10

->y=13/2

19) Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7

(ii) have a product less than 16

(iii) is a doublet of odd numbers.

Ans: Total number of outcomes = 36

(i) Favourable outcomes are

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

(2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., 15

P(sum less than 7) =15/36 or 5/12

(ii) Favourable outcomes 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) (4, 1)

(4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25

P(product less than 16) = 25/36

(iii) Favourable outcomes are

P(doublet of odd number) =3/36 or 1/12

20) A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

Ans: Let the speed of boat be x m/min

CD = 2x

150/y= tan 60 θ ->y=150/√3=50√3

150/y= tan 45θ ->150=50√3+2x

->x=25(3-√3)

∴ speed=25(3-√3)m/min

=1500(3-√3)m/hr

SECTION – D

21) Construct an isosceles triangle with base 8 cm and altitude 4 cm. Construct another triangle whose sides are 2/3 times the corresponding sides of the isosceles triangle.

Ans: Correct construction of given triangle

Correct construction of similar triangle

22) Prove that the lengths of tangents drawn from an external point to a circle are equal.

Ans: Correct figure, given, to prove and construction Correct proof

23) The ratio of the sums of first m and first n terms of an A. P. is m2:n2. Show that the ratio of its mth and nth terms is (2m-1)(2n-1).

Ans: Sm/Sn=m2/n2->m/2[2a+(m-1)d]/n/2[2a+(n-1)d]=m2/n2

->2a+(m-1)d/2a+(n-1)d=m/n

Solving we get d = 2a

am/an=a+(m-1)d/a+(n-1)d=a+(m-1)x2a/a+(n-1)x2a

=2m-1/2n-1

24) Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.

Ans: Let the speed of stream be x km/hr.

Speed of boat upstream = (15 – x) km/hr.

Speed of boat downstream = (15 + x) km/hr.

30/15-x+30/15+x=4(1/2)=9/2

->30(15+x+15-x)/(15-x)(15+x)=9/2

->200=225-x2

x=5(rejecting-5)

Speed of stream = 5 km/hr

25) If a≠ b≠0, prove that the points (a,a2),(b,b2)(0,0) will not be collinear.

Ans: Area of traingle with vertices (a,a2),(b,b2)and (0,0) is

1/2[a(b2)+b(-a2)+0]

=1/2ab(b-a) ≠o as a≠b≠0

Given points are not collinear

26) The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

Ans: 5/10=r1/r2

->r2=2r1

Ratio of volumes of two parts

=Volume of smaller cone/Volume of frustum

=1/3 πxr21 x5/1/3x πx5[r21 +r22  +r1r2]= r21/ r21+4r21+2r21

=1/7

27) Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25.

Ans: For Peter,

Total number of outcomes = 36

Favourable outcome is (5, 5)

P(Peter getting the number 25) =1/36

For Rina, Total number of outcomes = 6

Favourable outcome is 5.

P (Rina getting the number 25) =1/6

Rina has the better chance

28) A chord PQ of a circle of radius 10 cm subtends an angle of 60°at the centre of circle. Find the area of major and minor segments of the circle.

Ans: Area of minor segment

=22/7x10x10x(60/360-3/4x10x10

=10×10[22/7×1/6-√3/4]

=100/84(44-21√3)cm2 or 25/21(44-21√3)cm2

Area of major segment

=[22/7x10x10-100/84(44-21√3)cm2

=100/84(220+21√3)cm2

or 25/21(220+21√3)cm2

29) The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°.Find the height of the cloud from the surface of water.

Ans: h/x=tan 30° ->x=h√3

60+60+h/x=tan 60°

->120+h/x=√3

->120+h=h√3x√3

->h=60

height of cloud from surface of water = (60 + 60)m = 120 m

30) In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O′ are centres of the circles. Find the area of shaded region.

: =Area of square + Area of 2 major sectors.

=[28×28+2×22/7x14x14x270°/360°]cm2

=28×28(1+33/28)=1708cm2

31) In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park. Write your views on recycling of water.

Ans: Volume of water in cylindrical tank.

= Volume of water in park.

->22/7x1x1x5=25x20xh, where h is the height of standing water.

-> h=11/350m or 22/7cm

Conservation of water or any other relevant value.