Quantitative Aptitude - Arithmetic Ability Questions

Let the length of the train be 'L' mts

let the speed of the train be 'S' m/s

Given it crosses a pole in 10 sec=> L/S = 10 ......(1)
Given it takes 20 sec (double of pole) to cross a platform of length 200 mts

=> (L + 200)/S = 20

=> L/S + 200/S = 20

But from (1) L/S = 10

=> 200/S = 20 - 10

=> S = 20 m/s

Then, from (1)

=> L = 10 x 20 = 200 mts.

Hence, the length of the train = 200 mts.

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies

Given exp . is like this 854*854*854-276*276*276854*854+854*276+276*276=?

so ans = 854 - 276 = 578

A's cattle = 27 x 19 = 513

B's cattle = 21 x 17 = 357

C's cattle = 24 x 23 = 552

Ratio of cattle = 513 : 357 : 552

                    = 171 : 119 : 184

payment to be made by A = 

                                     = Rs.85.50

payment to be made by B = 

                                     = Rs.59.50

payment to be made by C = 

                                     = Rs.92

Speed on return trip = 150% of 50 = 75 km/hr.

Average speed = (2 x 50 x 75)/(50 + 75) = 60 km/hr.

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

100 P2 = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

Let the total quantity of gold be p kgs.

According to the question,

50% of p + 25% of p + 12.5% of p = 130900

87.5% of p = 130900

p = 130900 x 100/87.5  = 149600 kgs.

Therefore, the initial gold quantity that king has = p = 149600 kgs.