Quantitative Aptitude - Arithmetic Ability Questions

125 over 1000 in Simplest Form means 1251000 in its simple fraction form.

Now, to get the simplest form of 125/1000, find the HCF or GCD of both numerator and denominator i.e, 125 and 1000.

HCF of 125, 1000 = 125

Then, divide both numerator and denominator by 125

i.e, 1251251000125 = 18

Hence, 18 is the simplest form of 125 over 1000.

Required average = (67 * 2 + 35 * 2 + 6 * 3) / (2 + 2 + 3)

                           = (134 + 70 + 18) / 7  =  222 / 7 = 31(5/7) years.

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x 

=>  D= 25 days 

Now , the work done in 25 days = 25x 

Total work = 175x

Therefore, workdone before increasing the no of workers = 25x175x×100 % = 1427%

Time taken by Akash = 4 h

Time taken by Prakash = 3.5 h

For your convenience take the product of times taken by both as a distance.

Then  the distance = 14km

Since, Akash covers half of the distance  in 2 hours(i.e at 8 am)

Now, the rest half (i.e 7 km) will be coverd by both prakash and akash

Time taken by them = 7/7.5 = 56 min

Thus , they will cross each other at  8 : 56am.

Present population =  160000 *  (1 + 3/100)(1 + 5/200)(1 + 5/100)= 177366.

A : B = 200 : 169.

A : C = 200 : 182.

C/B = (C/A*A/B) = (182/200*200/169) = 182:169

When C covers 182 m, B covers 169 m.

When C covers 350 m, B covers (169/180*350)m  = 325 m

Therefore, C beats B by (350 - 325) m = 25 m.

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671