Question 1

# Statement-1: S3 = 55 × 29.Statement-2: S1 = 90 × 28 and S2 = 10 × 28. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.  Statement-1 is true, Statement-2 is false. Statement-1 is false, Statement-2 is true.

Solution

C.

Statement-1 is true, Statement-2 is false.

Question 2

## A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after 18 Months 19 Months 20 Months 21 Months

Solution

D.

21 Months

a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms =
n/2[400 + (n -1)40] = 11040 - 400
200n + 20n2 - 20n = 10640
20n2 + 180 n - 10640 = 0
n2 + 9n - 532 = 0
(n + 28) (n - 19) = 0
n = 19
∴ no. of months = 19 + 2 = 21

Question 3

## A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP with common difference -2, then the time taken by him to count all notes is 24 min 34 min 125 min 135 min

Solution

B.

34 min

Let the first term of an AP be a and common difference be d and number of terms be n, then
tn = a + (n-1)d and Sn = n/2 [ 2a + (n-1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference -2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0
⇒ (n-24)n-125) = 0
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.

Question 4

## For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then b , c and a are in G.P b, c and a are in A.P a, b and c are in A.P a, b and c are in G.P

Solution

B.

b, c and a are in A.P

225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
(15a)2 + (3b)2 + (5c)2 – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP

Question 5

Solution

D.

H.P.