# CBSE [ All India ]_X_Mathematics_2007_Set III

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• Q1

A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is (a) White ball (b) neither red nor white.

Marks:2

Number of red balls = 5
Number of green balls = 4
Number of white balls = 7
Total numbers of balls = 16
(a)      P (white) = 7/16
(b)       P (red) = 5/16
P (Neither red nor white) = 1- [P(W) + P(R)]
= 1 - (7/16+5/16)
= 1-12/16
= 4/16
= 1/4

• Q2

Find the mean of the following distribution:

 Class Frequency 0-10 10-20 20-30 30-40 40-50 8 12 10 11 9

Marks:2

 Class fi xi fixi 0-10 10-20 20-30 30-40 40-50 8 12 10 11 9 5 15 25 35 45 40 180 250 385 405 fi = 50 fixi = 1260

• Q3

Find the GCD of

Marks:2

Now it is out of Syllabus.

• Q4

Solve for x and y:
8x - 9y = 6xy
10x + 6y = 19xy

Marks:2

• Q5

Solve for x and y:

Marks:2

• Q6

In figure 1, two circles touch each other externally at C. Prove that the common tangent at C bisects the other two common tangents.

Marks:2

Given: Two circle with center A and B having three common tangents, PQ, LH, EF
To prove: PE = EQ and LF = FH
Proof:
PE = EC ( two tangents to a circle from one point are equal)
EQ = EC (two tangents to a circle from one point are equal)
From above two statements we conclude that,
PE = EQ ...(1)
FL = FC (two tangents to a circle from one point are equal)
FH = FC (two tangents to a circle from one point are equal)
From above two statements we conclude that,
LF = FH ...(2)
Hence, common tangent at C bisects PQ and LH (from (1) and (2)).

• Q7

D is any point on the side BC of a triangle ABC such that ADC = BAC. Prove that CA2 = BCBD.

Marks:2

D = A (given)
C = C (Common)
By AA property,

CA/BC = DC/CA
CA2 = BC.CD.

• Q8

A ceiling fan is marked at Rs. 970 cash or for Rs. 210 as cash down payment followed by three equal installments of Rs. 260. Find the rate of interest charged under the installment plane.

Marks:2

Out of syllabus.

• Q9

The Third term of an A.P is 3 and 11th terms is -21. Find its first term and common difference.

Marks:2

let a be the 1st term
And d be the common difference
3rd term a+2d = 3      [ An = a+(n-1)d]
11th term = a+10d = -21
on subtracting both the equation
-8d = 24
d = -3
putting the value of d in (1) we get
a+2(-3) = 3
a -6 = 3
a = 9 ,d = -3