Quantitative Aptitude - Arithmetic Ability Questions

when the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 liters of milk.

 Therefore, profit percentage = (5/20) x 100 =25 %       [ Profit %= (Profit/C.P)*100]

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of  interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

Since S.P. = 5 (M.P. - S.P.) => 5 M.P. = 6 S.P.
M.P. = (6/5) S.P.
Since the percentage discount = Percentage profit,
65S.P - S.P65S.P×100 = S.P -C.PC.P×100
Therefore, 1/6 = (S.P/C.P) – 1 => S.P = (7/6) C.P
M.P = 6/5 S.P = 6/5 x 7/6 C.P = 7/5 C.P
Therefore, Ratio of discount to C.P
= (M.P – S.P)/C.P = [{(7/5)C.P - (7/6)C.P} / C.P.]
= 7/30

Therefore, the ratio is 7 : 30

Time taken to meet for the first time anywhere on the track

= length of the track / relative speed

= 300 / (15 + 25)5/18 = 300x 18 / 40 x 5 = 27 seconds.

Cost of x metres = Rs. d.

Cost of 1 metre = Rs.(d/x)

Cost of y metres = Rs.[(d/x).y] = Rs.(yd/x)

Let the Required cost be X paise.  Less Weight , Less cost (Direct Proportion)

Therefore, 250 : 200 :: 60 : x

=> (250 * x) = (200 * 60)

=> x= 48

area of theplot = 110 * 65 = 7150 sq.m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m
area of the path = 7150 - 6300 =850 sq.m
cost of gravelling the path = 850 x (80/100) = 680 Rs