# 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

Answers

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

(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

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

=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.

**)**

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
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