1000+ Accenture Aptitude Questions and Answers Pdf - 1

Question: 1

3897 × 999 = ?

(A) 3639403

(B) 3791203

(C) 3883203

(D) 3893103

Ans: D

3897 × 999 = 3897 × (1000 – 1)

= (3897 × 1000) – (3897 × 1)

= 3897000 - 3897 = 3893103.

Question: 2

If a and b be positive integers such that a2 - b2 = 19, then the value of a is

(A) 7

(B) 8

(C) 9

(D) 10

Ans: D

(a2 - b2) = 19 ⇒ (a + b) (a - b) = 19

Clearly, a = 10 and b = 9.

Question: 3

The number of zeros at the end of 60! is

(A) 10

(B) 12

(C) 14

(D) 16

Ans: C

Clearly, highest power of 2 is much higher as compared to that of 5 in 60!, so

Required number of zeros = Height power of

5 = $$[{60}/{5}]$$ + $$[{60}/{5^2}]$$ = 12 + 2 = 14.

Question: 4

If a number is divisible by both 11 and 13, then it must be necessarily

(A) 429

(B) divisible by (11 + 13)

(C) divisible by (13 – 11)

(D) divisible by (11 × 13)

Ans: D

We know that 11 and 13 are co-prime.

So, a number divisible by both 11 and 13 will be divisible by (11 × 13).

Question: 5

The smallest number by which 66 must be multiplied to make the result divisible by 18 is

(A) 3

(B) 6

(C) 9

(D) 18

Ans: A

66 = 11 × 6

In order to get a number divisible by 18, the above product must be multiplied by 3.

Hence 66 must be multiplied by 3.

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