2012 class 10 all india set-1 section-a

8/12/2019 2012 Class 10 All India Set-1 Section-A

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CBSE X Mathematics All India 2012 Solution (SET 1)

Section A

Q1. The roots of the quadratic equation 2x2x6 = 0 are

(A)3

2,2

(B)3

2,2

(C)3

2,2

(D)

3

2, 2

Solution:

The given quadratic equation is 2x2x6 = 0.

Roots of the equation can be found by factorizing it as follows:

2

2

2 6 0

2 4 3 6 0

2 2 3 2 0

2 2 3 0

2 0 or 2 3 0

32 or

2

x x

x x x

x x x

x x

x x

x x

The roots of given quadratic equation are 2 and 3

2 .

Hence, the correct answer is B.

Q2. If the nth

term of an A.P. is (2n+ 1), then the sum of its first three terms is

(A) 6n+ 3

(B) 15

(C) 12

(D) 21


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Solution:

Given:The nth

term of A.P. i.e., an= 2n+ 1To find:Sum of first three terms

On putting n= 1, 2 and 3, we obtain:

1

2

3

1 2 3

2 1 1 2 1 3

2 2 1 4 1 5

2 3 1 6 1 7

Sum of first three terms 3 5 7 15

a

a

a

a a a


Q3. From a point Q, 13 cm away from the centre of a circle, the length of tangent PQ to the

circle is 12 cm. The radius of the circle (in cm) is

(A) 25

(B) 313

(C) 5

(D) 1

Solution:

The given information can be represented diagrammatically as follows:

Let O be the centre of the circle.Given:PQ = 12 cm and OQ = 13 cm.

To find: Radius of the circle

PQ is a tangent drawn from the external point Q to the circle.

OPQ = 90 (Radius is perpendicular to the tangent at the point of contact)


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On applying Pythagoras theorem in OPQ, we obtain:2 2 2

2 2 2

2 2 2

2 2 2

2 2

OQ OP + PQ

OP = OQ PQ

OP =(13cm) (12cm)

OP =169cm 144cm

OP =25cm

OP=5cm

Thus, the radius of circle is 5 cm.

Hence, the correct answer is C.

Q4. In Figure 1, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC= 4 cm, then the length of AP (in cm) is

(A) 7.5

(B) 15

(C) 10

(D) 9

Solution:


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AP, AQ and BC are tangents to the circle. Suppose BC touch the circle at R.It is given that AB = 5 cm, AC = 6 cm and BC = 4 cm.

We know that, the lengths of tangents drawn from an external point to a circle are equal.

AP = AQ (1)PB = BR (2)

CQ = CR ... (3)

2AP = AP + AP

2AP = AP + AQ [Using (1)]

2AP = (AB + PB) + (AC + CQ)

2AP = (AB + BR) + (AC + CR) [Using (2) and (3)]

2AP = AB + (BR + CR) + AC

2AP = AB + BC + AC

2AP = 5 cm + 4 cm + 6 cm

2AP = 15 cmAP = 7.5 cmThus, the length of AP is 7.5 cm.

Hence, the correct answer is A.

Q5. The circumference of a circle is 22 cm. The area of its quadrant (in cm2) is

(A)77

2

(B)77

4

(C)77

8

(D)77

16

Solution:Let the radius of the circle be r cm.Given: Circumference of circle = 22 cm


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2 22cm Circumference of circle = 222

cm2

22 7cm

2 22

7cm

2

r r

r

r

r

Area of quadrant of circle

2

2

2

1Area of circle

4

1

4

1 22 7 7cm4 7 2 2

77cm

8

r


Q6. A solid right circular cone is cut into two parts at the middle of its height by a plane

parallel to its base. The ratio of the volume of the smaller cone to the whole cone is

(A) 1 : 2

(B) 1 : 4

(C) 1 : 6

(D) 1 : 8

Solution:

Here is the link for the solution.

http://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-height

Hence, the correct answer is D.
http://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-heighthttp://www.meritnation.com/discuss/question/1867566/what-is-the-ratio-of-the-small-cone-to-the-whole-cone-after-when-the-cone-is-cut-into-two-parts-at-the-middle-of-its-height


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Q7. A kite is flying at a height of 30 m from the ground. The length of string from the kite to

the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation ofthe kite at the ground is

(A) 45

(B) 30

(C) 60

(D) 90

Solution:

Let A be the position of the kite and AC be the length of the string of the kite.Suppose be the angle of elevation of the kite at the ground.

It is given that, AB = 30 m and AC = 60 m.

In right ABC:

AB Perpendicularsin sin

AC Hypotenuse

30sin

60

1sin

2

sin sin 30

30

Thus, the angle of elevation of the kite at the ground is 30.



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Q8. The Distance of the point (3, 4) from thex-axis is

(A) 3

(B) 3

(C) 4

(D) 5

Solution:

Let B (3, 4) be the given point. Suppose the perpendicular from B on thex-axis

intersects at A.

Point A lies on thex-axis, so the coordinate of point A is (3, 0).

Distance of the given point from thex-axis = AB

Using distance formula, we have

2 2 2 2

AB 3 3 4 0 3 3 4 0 16 4 units

The distance of the point (3, 4) from thex- axis is 4 units.


Q9. In Figure 2, P (5,3) and Q (3,y) are the points of trisection of the line segment joining

A (7,2) and B (1,5). Thenyequals

(A) 2

(B) 4

(C) 4

(D)5

2

Solution:

It is given that P and Q are the points of trisection of line segment AB.

AP = PQ = QB


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Now, AP = QB

2 2 2 2

2 2 2 2

2

2

5 7 3 2 1 3 5 [Using distanceformula]

2 1 2 5

4 1 4 25 10

5 10 29

y

y

y y

y y

Squaring on both sides, we get

2

2

2

5 10 29

10 24 0

6 4 24 0

6 4 6 0

4 6 0

4 0 or 6 0

4 or 6

y y

y y

y y y

y y y

y y

y y

y y

From the obtained values ofy, 4 matches with the option C.


Q10. Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random fromthe bag. The probability of getting a card with a prime number is

(A)1

2

(B)2

5

(C)3

10

(D) 59

Solution:

Number on the cards are 2, 3, 4, , 11.

Sample space associated with the experiment, S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Total number of outcomes = 10

Let A be the event of drawing a card with a prime number.


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The cards with prime number are 2, 3, 5, 7 and 11.

Number of outcomes in favour of event A = 5

Number of outcomes in favour of A 5 1Required probability P(A)

Total number of outcomes 10 2

Hence, the correct answer is A.

2012 class 10 all india set-1 section-a

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