CBSE [ All India ]_X_Mathematics_2007_Set III

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

Now it is out of Syllabus.

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)).

In ADC and BAC
D = A (given)
C = C (Common)
By AA property,

ADCBAC

CA/BC = DC/CA
CA² = BC.CD.

Out of syllabus.

let a be the 1^st term
And d be the common difference
3^rd term a+2d = 3      [ An = a+(n-1)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

Out of Syllabus.