CAT Questions

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Let first no. be "k" and multiple be "l"...start value of y with 3
then,
[(k x y) + {l x (l-1)}]
Thus, we get

15 x 3 + 3 x 2 = 51
51 x 4 + 4 x 3 = 216
216 x 5 + 5 x 4 = 1100
1100 x 6 + 6 x 5 = 6630
6630 x 7 + 7 x 6 = 46452

Therefore, the missing number is 6630.

Here in the given series, it follows a pattern that

77 + 2 = 79

79  - 3 = 76

76 + 5 = 81

81 -  7 = 74

74 + 11 = 85

Consecutive adding and subtracting of prime numbers.

So the wrong number in the given series is 75.

Given the number series is 4, 6, 12, 30, 81, 315.

Here it follows a pattern that

4

4 x 1.5 = 6

6 x 2 = 12

12 x 2.5 = 30

30 x 3 = 90 not equals to 81

90 x 3.5 = 315

Hence the odd one in the series is 81.

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 x 2000) - (2000 x 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> 14012 - 6002 = 8010.

The given number series is 

3, 5, 8, 15, 28, 51, ?

Here the pattern it follows is

3

3 + 2 = 5 ( add prime number less than 3 i.e, 2)

5 + 3 = 8 (3 is prime number less than 5)

8 + 7 = 15 (7 is prime number less than 8)

15 + 13 = 28 (13 is prime number less than 15)

28 + 23 = 51 (23 is prime number less than 28)

51 + 47 = 98 (47 is prime number less than 51)

Hence, the next number in the given number series is 98.

1397 x 1397

The terms of the series are products of two consecutive prime numbers i.e.

(2 x 3), (3 x 5), (5 x 7), (7 x 11),.....

So, 165 is wrong and must be replaced by (11 x 13) i.e. 143.

Here the series follows the pattern that, product of succesive prime numbers i.e,

 2 × 3 = 6

 3 × 5 = 15

 5 × 7 = 35

 7 × 11 = 77

 11 × 13 = 143

 13 × 17 = 221

 17 × 19 = 323

So here in place of 77 the number 63 is written.

So the wrong number is 63.