Quantitative Aptitude - Arithmetic Ability Questions

n (A) = 34, n (B) = 42, n (A  ∩ B) = 20.

So, n (A  U B) = n (A) + n(B) - n(A ∩B) = 34 + 42 - 20 = 56.

Percentage failed in either or both the subjects = 56.

Hence, percentage passed = (100 - 66)% = 44%.

First,we count the number of odd days for the left over days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result.

Part filled in 2 hours = 2/6=1/3

Remaining part =1-13=23

 (A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

A+B's 1 hour work=112+115=960=320

A+C's 1 hour work=112+120=860=215  

part filled in 2 hrs=320+215=1760  

part filled in 6 hrs=3*1760=1720  

 remaining part =1-1720=320  

Now, it is  the turn of A and B (3/20) part is filled by A and B in 1 hour.

Therefore, Total  time taken to fill the tank =(6+1)hrs= 7 hrs

Double Difference pattern is followed in the above series. +2, +5, + 10, +17, +26

sum = x. amount =7x/6.

S.I. =(7x/6 – x) = x/6 ; time = 3 years

Rate =[ (100 * x) / (x * 6 * 3)]% = 5.5%