Quantitative Aptitude - Arithmetic Ability Questions

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

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

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 x and y be the respective km's travelled by man via taxi and by his own car.
Given x + y = 90 => x = 90 - y
But according to the question,
7x + 6y = 595
7(90-y) + 6y = 595
=> 630 - 7y + 6y = 595
=> y = 630 - 595 = 35
=> x = 90 - 35 = 55
Therefore, the distance travelled by taxi is 55 kms.