Numerical Aptitude
Q-1 Find the value of A in 72A4 + 2398 = 9642
(a) 3
(b) 4
(c) 9
(d) 5
Q-2 Find the unit digit in 357 × 421 × 418 × 315
(a) 1
(b) 0
(c) 2
(d) 4
Q-3 If we multiply 15 by a certain number then we obtain 845 . If middle one is wrong then what would be correct answer ?
(a) 815
(b) 825
(c) 865
(d) 895
Q-4 (XX4) . (XX6) = ?
(a) 444464144
(b) 334424
(c) 3324
(d) 44433404
Q-5 600 × 9999
(a) 5999400
(b) 5954400
(c) 5964400
(d) 5964600
Q-6 If n = 2 & n + (n+2) + (n+4) will be divisible by
(a) 2
(b) 8
(c) 5
(d) 9
Answers & Explanations :
1. (B) As 72A4 + 2398 = 9642 , so when we add 4 + 8 then we got 2 in unit place of answer & 9 of second value & 4 of tenth place of value of answer & 1 as carry . Then if we add them we got 9 + 4 +1 = 14 , so clearly answer is 4
2 (B) Just multiply the unit digit each of the numbers
i.e 7 × 1× 8 × 5 = 280 , therefore unit digit is 0.
3(B) First of all we know that 3 × 6 = 18 or 6 = 18 /3
As 15 = 3 × 5 , i.e the number should be divisible by 3 or 5
Here 825 is divisible by both 3 & 5 .
4 (B) Its reason is simple because when we multiply two three digit numbers then its product will be
In 5 or 6 digit .
5(A) 600 × ( 10000 – 1 )
6000000 – 600 = 5999400
6 (A) 3n + 6 & when n = 2 value will be 12
Numerical Aptitude
Q-1 Which number is nearest to 1500 which is exactly divisible by 13 ?
(a) 1508
(b) 1495
(c) 1509
(d) 1511
Q-2 What number should be added to 121 so that it is exactly divisible by 13 ?
(a) 7
(b) 9
(c) 8
(d) 5
Q-3 What number should be subtracted from 121 to make it exactly divisible by 13 ?
(a) 5
(b) 4
(c) 15
(d) 9
Q-4 Find the number at xx place in 365xx2 , which is exactly divisible by 3 ?
(a) 3 , 2
(b) 3, 4
(c) 3 , 7
(d) 3 , 9
Q-5 1 + 2 +3 + 4 + ------------------------------------------------------------------ + 9
(a) 46
(b) 40
(c) 45
(d) 41
Q-6 2 + 4 + 6 + -------------------------------------------------------------------------- 100
(a) 2550
(b) 2650
(c) 2750
(d) 2850
Answers & Explanations :
1(B ) When we divide 1500 by 13 , then remainder is 5 , so the number which are exactly divisible by 13 are 1508 , 1495, so answer is 1495 . But the main question that would arise in your mind would be that from where 1508 , 1495 comes , answer is just simple , 1500 is not completely divisible by 13 , as it leaves 5 as remainder . So in order to find the number which are completely divisible by 13 , either we have to add something or subtract something . Add difference between 13 & 5 i.e 8 so it becomes 1508 . & subtract just remainder i.e 5 so number is 1495 . So nearest is 1495 .
2( B) When we divide 121 by 13 , then we got 4 as remainder therefore just subtract 13 & 4
3(B) Reason is simple, just subtract 4 from 121 because remainder is 4 .
4(a) If we take 3 , 2 , then sum of digit will be 21 which is exactly divisible by 3 .
5(C) 1 + 2 + 3 + ------------------------------------------ + n = n(n+1) / 2
Here n = 9 , so value is 9(9 +1) / 2 = 90 / 2 = 45
6(A) It is an A.P having a = 2 , d = 2
We know nth term = 100
a + ( n – 1 )d = 100
2 + ( n – 1 )2 = 100
n = 50
sum of 100 term is = n / 2 ( First term + Last term )
= 50 / 2 ( 2 + 100)
= 25 ( 102) = 2550
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