Quantitative Aptitude - Arithmetic Ability Questions

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

FV=P(1+r/n)^nt

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

Choose 2 juniors and 2 seniors.

14C2*23C2 = 23023

As the distance travelled is constant, the time taken is inversely proportional to speed.

Let 'u' be the speed of the current and 'v' be the speed of the boat.

Speed of the boat downstream = v + u & upstream = v - u

=> v+u/v-u = 30/20 => v/u = 5 => v = 5u

=> v + u = 6u = 10/20/60 miles/hr

=> 6u = 30 m/h

=> u = 5 m/h

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

Given exp = 18225102+18225104+18225106+18225108

               = 1822510+18225102+18225103+18225104

               = 13510+135100+1351000+13510000

               = 13.5+1.35+0.135+0.0135=14.9985