Quantitative Aptitude - Arithmetic Ability Questions

area of the plot = 120m x 60m = 7200 sq.m
area of plot excluding gravel = 117m x 57m = 6669 sq.m
area of gravel = 7200 sq.m - 6669 sq.m = 531 sq.m
cost of building it = 531 sq.m x 65 ps = 34515 ps
In Rs. = 34515/100 = Rs. 345.15.

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

⇒14+112=13

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is: 

64 − 27 = 37

A loss of Rs.37

Work done by the third pipe in 1 min = 1/50 - (1/60 + 1/75) = - 1/100.
[-ve sign means emptying]

The third pipe alone can empty the cistern in 100 min.

Let the expenditure on the 95 students be 'x'

Then,

95x + 500 = 120(x – 5)
95x + 500 = 120x - 600
120x - 95x = 1100
25x = 1100
=> x = 44

Therefore, total expenditure of the hostel becomes
95 x 44 + 500 = 4680.

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.

Given that their average is 52

Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6

6x + 30 = 312

x = 47

So Product = 47 × 57 = 2679

Let the age of sushma be x and
the age of her son is y
Then five years before x-5=5(y-5) ...(1)
Five years hence x+5 = 3(y+5)-8 .....(2)

By soving (1) & (2), we get
5y - 15 = 3y + 7
y = 11 => x = 35

Therefore, the age of Sushma = 35 and her son = 11.