Quantitative Aptitude - Arithmetic Ability Questions

99  = 1 x 3 x 3 x 11

101= 1 x 101

176= 1 x 2x 2 x 2 x 2 x 11

182= 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, 99

      divisors of 101 are 1,101

      divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176

      divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182

Hence , 176 hasthe most number of divisors. 

Let the amount collected be 'A'
Then, from the given conditions,
A x 110/100 x 90/100 = 7920
A x 0.99 = 7920
A = 7920/0.99
=> 8000

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

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =122×129=1211

Let the speed of the stream = x kmph

From the given data,

 129+x+129-x=3 hrs

3x2 = 27

=> x = 3 kmph

Therefore, the speed of the stream = 3 kmph

Let present ages of Sumit=w, Krishna=x, Rishabh=y and Rohit=z , then
w+x+y = 43x3 = 129 ---(i) and
w+y+z = 49x3 = 147 ---(ii)
Subtracting (i) from (ii), (w+y+z) - (w+x+y)=147 -129 , z- x =18 ---(iii)
Given Rohit's age => z = 54, so from (iii), x = 54-18 = 36