 # CBSE [Delhi]_X_Mathematics_2011_Set II

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• Q1

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) units
(C) 4 units
(D) 7 units

Marks:1

So, correct option is A
Let r be the radius of the circle.

then, perimeter of circle=area of circle

2 r= r2

2=r.
Thus, r=2 units.

Thus, the radius of circle is 2 units.

• Q2

In AP, if a=-10, n=6 and an=10, then the value of d is
(A)   0
(B)   4
(C)  -4
(D) 10/3

Marks:1

So, correct option is B
Since,
an = a + (n-1)d

10 =-10 + (6-1)d

20 = 5d d=4.

Thus, d is 4.

• Q3

Which of the following cannot be the probability of an event?
(A) 1.5
(B) 3/5
(C) 25%
(D) 0.3

Marks:1

The probability of an event is always greater than or equal to zero and less than or equal to one i.e., 0
£P(E)£
1.
Here, 3/5=0.6 and 25%=0.25

Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and are less than or equal to 1. But 1.5 is greater than 1. Thus, 1.5 can not be the probability of an event.

• Q4

The mid-point of segment AB is the point P (0,4). If the coordinates of B are (-2,3) then the coordinates of A are
(A)    (2, 5)
(B) (-2, -5)
(C)    (2, 9)
(D) (-2,11)

Marks:1 • Q5

The point P which divides the line segment joining the points A(2,-5) and B(5,2) in the ratio 2:3 lies in the quadrant.
(A)  I
(B) II
(C)III
(D)IV

Marks:1 • Q6

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 45°. The height of the tower (in metres) is
(A) 15
(B) 30
(C) 30 3
(D) 10 3

Marks:1

The correct option is B Let AB be the tower and P be the point on the ground. It is given that BP=30 m, P=45°

Now, AB/BP=tan45° AB/30m=1 AB=30m.

Thus, the height of the tower is 30m.

• Q7

A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then the water level
rises (in cm) by
(A) 3
(B) 4
(C) 5
(D) 6

Marks:1 • Q8

In figure, PA and PB are tangents to the circles with centre O. If APB=60°, then OAB is (A) 30°
(B) 60°
(C) 90°
(D) 15°

Marks:1

Thus, the correct option is A.

PA and PB are tangents to the circle from an external point O.

Therefore, PA=PB.  PAB is an isosceles triangle where PAB= PBA P+ PAB+ PBA=180°   [Angle sum property of triangle]

60°+2ÐPAB=180°

2ÐPAB=180°-60°

=120°

ÐPAB=120°/2

=60°

It is known that the radius is perpendicular to the tangent at the point of contact.  OAB=90° PAB+ OAB=90° OAB=90° - 60°

=30°

• Q9

In figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If AOB=100°, then BAT is equal to (A) 100°
(B) 40°
(C) 50°
(D) 90°

Marks:1

Thus, the correct option is C.
It is given that AOB=100°

DAOB is isosceles because OA=OB= radius  OAB= OBA AOB+ OAB+ OBA=180°

100°+ OAB+ OAB=180°

2 OAB=180° - 100° OAB=80°/2

=40°

Now, OAT=90° [AT is tangent and OA is radius]

Thus, BAT= OAT- OAB

=90°-40°

=50°.

• Q10

The roots of the equation x2+x-p(p+1)=0, where p is a constant, are
(A)    p,p+1
(B)  -p, p+1
(C)   p,-(p+1)
(D) –p,-(p+1)

Marks:1 