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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