 # CBSE [Delhi]_X_Mathematics_2012_Set I

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• Q1

The roots of the quadratic equation 2x² – x – 6 = 0 are
(A)
–2, 3/2
(B)
2, –3/2
(C)
–2, –3/2
(D) 2, 3/2

Marks:1 • Q2

If the nth term of A.P. is (2n + 1), then the sum of its first three terms is
(A)
6n + 3
(B)
15
(C)
12
(D) 21

Marks:1

nth term of A.P. is (2n + 1).

First term = 2(1) + 1 = 3

Second term = 2(2) + 1 = 5

Third term = 2(3) + 1 = 7

Hence, the sum of first three terms is 3 + 5 + 7 = 15.

Hence, the correct option (B)

• Q3

From a point Q, 13 cm away from the centre of a circle, the length of tangent PQ to the circle is 12cm. the radius of the circle (in cm) is
(A) 25
(B) 313
(C)
5
(D) 1

Marks:1  • 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 Marks:1 We know, AB = 5 BP = AP – 5               …(i)

Also, AC = 6 CQ = AQ – 6             …(ii)

Now, BP = BO, CQ = CO and AP = AQ. (Q tangents drawn to a circle from an external point are equal in length)

BO + OC = 4 or BP + CQ = 4        …(iii)

Put the values of BP and CQ from eq. (i) and (ii) respectively in eq. (iii), we get

AP – 5 + AQ – 6 = 4

AP + AP – 11 = 4

2AP = 15

AP = 7.5 cm.

Hence, the correct option 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

Marks:1 • 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

Marks:1 The given cone is of height 'h' with base radius 'r'.

When the cone is cut into two parts the smaller cone is of height By similar triangle property, we have  • Q7

A kite is flying at a height of 30 m from the ground. The length of the string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is
(A)
45°
(B)
30°
(C)
60°
(D) 90°

Marks:1  • Q8

The distance of the point (–3, 4) from the x-axis is
(A)
3
(B)
–3
(C)
4
(D) 5

Marks:1 Hence, the correct option is (C).

• 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). Then y equals Marks:1 We know,
AB = 5 BP = AP – 5               …(i)

Also,    AC = 6 CQ = AQ – 6             …(ii)

Now,    BP = BO, CQ = CO
and      AP = AQ. (
Q tangents drawn to a circle from an
external point are equal in length)

BO + OC = 4
or BP + CQ = 4        …(iii)

Put the values of BP and CQ from eq. (i) and (ii) respectively in eq. (iii), we get

AP – 5 + AQ – 6 = 4

AP + AP – 11 = 4

2AP = 15

AP = 7.5 cm.

Hence, the correct option is (A).

• Q10

Card bearing numbers 2, 3, 4, …, 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is Marks:1 