Quantitative Aptitude - Arithmetic Ability Questions

B covers 28 meters in 7sec. So, B's time over the course = (7/28)*100 =250 sec.

Whereas A's time over the course = 250 -7 = 243 sec.

i.e A's time over the course is 4min , 3 sec.

By replacing the “?” with “x”

(12.5×14÷8.75) x (100/100) + 42 = 50 + X

(125*140)/875 + 42 = 50 + X

140/7 + 42 = 50 + X

20 + 42 = 50 + X

62 = 50 + X

X = 62 - 50

X = 12

Given Number series is :

3   4   14   48   576   27648

It follows a pattern that,

3

4

4 x 3 = 12   but not 14

12 x 4 = 48

48 x 12 = 576

576 x 48 = 27648

Hence, the odd one in the given Number Series is 14.

Here the given series 1, 6,  ?, 15, 45, 66, 91 follows a

Pattern in the series is, +5, +9, +13,...,.. +21, +25

So +4 is getting increased at every term addition.

Missing Number in the series will be 15+(9+4) = 15 + 13 = 28.

S.I. for 1 year =  Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

Aman : Rakhi : Sagar = (70000 x 36) : (105000 x 30) : (140000 x 24) = 12 : 15 : 16.

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.