QUANTITY QUEST FOR IBPS CLERK MAINS
1) . 3, 10, 32, 111, ?, 2315, 13902
(1) 512
(2) 460
(3) 424
(4) 412
(5) None of these

2). 7, 13, 24, 40, 61, ?
(1) 87
(2) 92
(3) 89
(4) 93
(5) None of these

3). In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
1.63
2.90
3.126
4.45
5.135

4). There are 6 boxes numbered 1, 2, 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
1. 5
2. 21
3. 33
4. 60
E. 6

5).Find remainder of (7^99)/6
1) 2
2) 1
3) 4
4) 3
5) None of these

6). what is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 2?
1) 9792
2) 9982
3) 9794
4) 9906
5) None of these

7). in a fraction, numerator is increased by 20% and the denominator is diminished by 30%. The new fraction obtained is 2/3. The original fraction is:
1) 7/18
2) 5/9
3) 6/11
4) 3/5
5) None of these

8). Find a single discount equivalent to a discount series of 25%, 20%, 10%-
1) 23%
2) 46%
3) 48%
4) 55%
5) 50%

9). by selling pen for ` 81, a shopkeeper gain 35%. At what price should he sell the pen to gain 20% on the cost price?
1) Rs. 72
2) Rs. 66
3) Rs. 92
4) Rs. 76
5) None of these

1). (2)
(3+7)*1=10,
(10+6)*2=32
(32+5)*3=111
(111+4)*4=460
(460+3)*5=2315
(2315+2)*6=13902

2). (1)
7+6=13
13+11=24
24+16=40
40+21=61
61+26=87

3). (1)Required number of ways = (7C5 x 3C2) = (7C2 x 3C1)
=63

4). (2)

If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.

5). 2

6). 3
LCM of 6, 9, 12, 17 = 612
Greatest Number of 4 digit divisible by 612 is 9792, to get remainder 2 number should be = 9792 + 2 = 9794

7). 1

8). 2

9). 1

QUANTITY QUEST FOR IBPS CLERK MAINS Reviewed by SSC IBPS on 18:22:00 Rating: 5