CAT Questions

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The given number series 8, 18, 56, 232, 1165 

8

8 x 2 + 2 = 18

18 x 3 + 3 = 57 (not 56)

57 x 4 + 4 = 232

232 x 5 + 5 = 1165

Hence, the odd man in the given series is 56.

The sum of square of two no is 116 i.e no lie 1-11, because 11x11 =121 which is greater than 116
again after cheaking all possible pair of no only 4 and 10 satisfy this condsn , again the progression is of positive integer so 10 cant not be 1st term
so the 1st term is 4,
but from the 2nd we cant able to find the 1st term

so only 1st statement is sufficient to answer the question

$$\begin{gathered} 2 = 12 + 1 \\ 3 = 22 - 1 \\ 10 = 32 + 1 \\ Now 102 - 1 = 99 \\ 9802 = 992 + 1 \end{gathered}$$

If the period between the two months is divisible by 7, then that two months will have the same calender .

 
(a). Oct + Nov  = 31 + 30 = 61 (not divisible by 7)

(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31  = 214 (not divisible by 7)

(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)

(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)

Hence, April and July months will have the same calendar.

The given number series is 3, 11, 80, 291, 803

$$\begin{gathered} 3 \\ 3 + 23 = 3 + 8 = 11 \\ 11 + 43 = 11 + 64 = 75 (not 80) \\ 75 + 63 = 75 + 216 = 291 \\ 291 + 83 = 291 + 512 = 803 \end{gathered}$$

Only one such symbol is there in the given arrangement.

It is 4@5.

The pattern of given series is :

5

14 = (5x3) - 1

45 = (14x3) + 3

130 = (45x3) - 5

(?) = (130x3) + 7

=> (?) = 397

1182 = (397x3) - 9

Thus, the missing number is 397.

The given number series is 5, 8, 40, 90, 664

5

5 - 1 x 2 = 8

8 - 3 x 4 = 20 (NOT 40)

20 - 5 x 6 = 90

90 - 7 x 8 = 664