﻿ CBSE Class 10 Mathematics Solved Question Paper 2012 | Previous Year Papers | Zigya

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# CBSE Class 10 Mathematics Solved Question Paper 2012

#### Multiple Choice Questions

1.

If 1 is a root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0, then ab equals

• 3

• $-\frac{7}{2}$

• 6

• -3

A.

3

It is given that 1 is the root of the equations

Therefore, y = 1 will satisfy both the equations.

2.

If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, then the diameter of the larger circle (in cm) is

• 34

• 26

• 17

• 14

B.

26

Diameter of two circles are given as 10 cm and 24 cm.

Radius of one circle = r1 = 5 cm,  Radius of other circle = r2 = 12 cm.

According to the given information,

$\therefore$ Radius of the larger circle = 13 cm

Hence, the diameter of larger circle = 26 cm.

3.

Two dice are thrown together. The probability of getting the same number on both dice is:

• $\frac{1}{2}$

• $\frac{1}{3}$

• $\frac{1}{6}$

• $\frac{1}{12}$

C.

$\frac{1}{6}$

When two dice are thrown together, the total number of outcomes is 36.

favourable outcomes = { (1,1),  (2,2),  (3,3),  (4,4),  (5,5),  (6,6) }

#### Short Answer Type

4.

If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5) then find the value of p.

It is given that the point  A(0, 2) is equidistant from the points B(3, p) and C(p, 5).

So, AB = AC $⇒$ AB2 = AC2

Using distance formula, we have

$⇒$ ( 0 - 3 )2 + (2 - p )2 = ( 0 - p )2 = ( 2 - 5 )2

$⇒$ 9 + 4 + p2 - 4p = p2 + 9

$⇒$4 - 4p = 0

$⇒$ p = 1

Hence, the value of p = 1.

#### Multiple Choice Questions

5.

In Fig., the sides AB, BC and CA of a triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then the length of BC (in cm) is • 11

• 10

• 14

• 15

B.

10

it is know that the lengths of tangents drwan from a point outside a circle

are equal in length.

Therefore, we have;

AP = AR     ........(1)  (Tangents drawn from point A)

BP = BQ    .........(2)  (Tangents drawn from point B)

CQ = CR   ..........(3)   (Tangents drawn from point C)

Using the above equations,

AR = 4 cm          ( AP = 4 cm,  given)

BQ = 3 cm         ( BP = 3 cm,  given)

AC = 11 cm $⇒$ RC = 11 cm - 4 cm = 7 cm

$⇒ CQ = 7 cm$

Hence, BC = BQ + CQ = 3 CM + 7 CM = 10 cm.

6.

The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 are

• ( 2, 4 )

• ( 3, 5 )

• ( 4, 2 )

• ( 5, 3 )

B.

( 3, 5 )

It is given that the point P divides AB in the ratio  2 : 1.

Using the section formula, the coordinates of the point P are

Hence, the coordinates of the point P are (3,5).

7.

The length of shadow of a tower on the plane ground is 3 times the height of the tower. The angle of elevation of sun is

• 450

• 300

• 600

• 900

B.

300 Let AB be the tower and BC be its shadow. Let $\mathrm{\theta }$ be the angle of elevation of the sun.

According to the given information,

Hence, the angle of elevation of the sun is 300.

#### Short Answer Type

8.

A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.

Total mumber of outcomes is 50.

Favourable outcomes = { 12, 24, 36, 48 }

#### Multiple Choice Questions

9.

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

• 1 : 2

• 2 : 1

• 1 : 4

• 4 : 1

C.

1 : 4

Let the original radius and the height of the cylinder be r and h respectively.

10.

In Fig., a circle touches the side DF of $\angle$EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of $\angle$EDF (in cm) is: • 18

• 13.5

• 12

• 9

A.

18

It is known that the tangents from an external point to the circle are equal.

$\therefore$ EK = EM,  DK = DH  and FM = FH     .....(1)

Perimeter  of $∆$EDF = ED + DF + FE

= (EK - DK) + (DH + HF) + EM - FM)

= (EK - DH) + (DH + HF) + (EM - FH)         [Using (1)]

= EK + EM

= 2EK = 2(9 CM) = 18 CM

Hence, the perimeter of $∆$EDF is 18 cm.