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CBSE Class 10 Mathematics Solved Question Paper 2012

Multiple Choice Questions

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

1 : 4

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

$V o l u m e o f t h e o r i g i n a l c y l i n d e r = {πr}^{2} h Radius of the new cylinder = \frac{r}{2} Height of the new cylinder = h Volume of the new cylinder = π {(\frac{r}{2})}^{2} h = \frac{{πr}^{2} h}{4} Required ratio = \frac{Volume of the new cylinder}{Volume of the original cylinder} = \frac{\frac{{πr}^{2} h}{4}}{{πr}^{2} h} = \frac{1}{4} = 1 : 4$

Short Answer Type

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 $\Rightarrow$ AB² = AC²

Using distance formula, we have

$\Rightarrow$ ( 0 - 3 )² + (2 - p )² = ( 0 - p )² = ( 2 - 5 )²

$\Rightarrow$ 9 + 4 + p² - 4p = p² + 9

$\Rightarrow$ 4 - 4p = 0

$\Rightarrow$ p = 1

Hence, the value of p = 1.

Multiple Choice Questions

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 )

( 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

$(\frac{(1 x 1) + (2 x 4)}{2 + 1}, \frac{(1 x 3) + (2 x 6)}{2 + 1}) = (\frac{1 + 8}{3}, \frac{3 + 12}{3}) = (3, 5)$

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

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

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,

$A r e a o f t h e l a r g e r c i r c l e = π (r_{1})^{2} + π (r_{2})^{2} = π (5)^{2} + π (12)^{2} = π (25 + 144) = 169 π = π (13)^{2}$

$∴$ Radius of the larger circle = 13 cm

Hence, the diameter of larger circle = 26 cm.

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

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 $\Rightarrow$ RC = 11 cm - 4 cm = 7 cm

$\Rightarrow CQ = 7 cm$

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

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

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{12}$

$\frac{1}{6}$

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) }

$∴ Required probability = \frac{Number of favourable outcomes}{Total number of outcomes} = \frac{6}{36} = \frac{1}{6}$

Short Answer Type

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 }

$∴ Required probability = \frac{Number of favourable outcomes}{total number of outcomes} = \frac{4}{50} = \frac{2}{25}$

Multiple Choice Questions

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

30⁰

Let AB be the tower and BC be its shadow. Let $θ$ be the angle of elevation of the sun.

According to the given information,

$BC = \sqrt{3} AB . . . . . . . . . . (1) In ∆ ABC, tanθ = \frac{AB}{BC} = \frac{AB}{\sqrt{3} AB} = \frac{1}{\sqrt{3}} [Using (1)] We know that \tan 30^{0} = \frac{1}{\sqrt{3}} ∴ θ = 30^{0}$

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

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

18
13.5
12
9

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

$∴$ 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.

10.

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

It is given that 1 is the root of the equations

${ay}^{2} + ay + 3 = 0 and y^{2} + y + b = 0 .$

Therefore, y = 1 will satisfy both the equations.

$∴ a (1)^{2} + a (1) + 3 = 0 \Rightarrow a + a + 3 = 0 \Rightarrow 2 a + 3 = 0 \Rightarrow a = \frac{- 3}{2} Also, (1)^{2} + (1) + b = 0 \Rightarrow 1 + 1 + b = 0 \Rightarrow b = - 2 ∴ ab = - \frac{3}{2} x - 2 = 3$