Quantitative Aptitude - Arithmetic Ability Questions

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.

Let the number of papers be x. Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11.

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

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

The best approach is to pick a number that satisfies the rules that x and y are greater than 0. So for example you could choose 25 for y.

The value of x before doubling is then 25 - 50/25 = 23

The value after doubling y will be 50 - 50/50 = 49, which is more than double.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

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%