CBSE [ All India ]_X_Mathematics_2007_Set III
To Access the full content, Please Purchase

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:2Answer:
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) = 112/16 = 4/16
= 1/4 
Q2
Find the mean of the following distribution : Class Frequency 010 1020 2030 3040 4050 8 12 10 11 9 Marks:2Answer:
Class f_{i} x_{i} f_{i}x_{i} 010 1020 2030 3040 4050 8 12 10 11 9 5 15 25 35 45 40 180 250 385 405 f_{i} = 50 f_{i}x_{i} = 1260 
Q3
Find the GCD of Marks:2Answer:
Now it is out of Syllabus. 
Q4
Solve for x and y:
8x  9y = 6xy
10x + 6y = 19xyMarks:2Answer:

Q5
Solve for x and y:
Marks:2Answer:

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:2Answer:
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 CA^{2} = BCBD. Marks:2Answer:
In ADC and BAC
D = A (given)
C = C (Common)
By AA property, ADC BAC CA/BC = DC/CA
CA^{2} = 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:2Answer:
Out of syllabus. 
Q9
The Third term of an A.P is 3 and 11^{th} terms is 21. Find its first term and common difference. Marks:2Answer:
let a be the 1^{st} term And d be the common difference 3^{rd} term a+2d = 3 [ An = a+(n1)d] 11^{th} 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 
Q10
Simplify Marks:3Answer:
Out of Syllabus.