Quantitative Aptitude - Arithmetic Ability Questions

Let p be the speed of man in kmph

According to the given data in the question,

Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction

=> 20 x 10/60 = 8/60 (20 + p)

=> 200 = 160 + 8p

=> p = 40/8 = 5 kmph.

Let the ratio amount be 'p'

7p - 3p = 2700

4p = 2700

p = 675

R's Share = 675 × 6 = Rs. 4050

Speed of train relative to jogger = 45 - 9 = 36 km/hr.
= 36 x 5/18 = 10 m/sec.

Distance to be covered = 260 + 140 = 400 m.

Time taken = 400/10 = 40 sec.

let the side of the square be x meters
length of two sides = 2x meters
diagonal = 2x= 1.414x m

saving on 2x meters = .59x m

saving % = 0.59x2x*100%

= 30% (approx)

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.

After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days

-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

Ratio of efficiency of A to B is 3 : 2

Let, In 8 days they complete = (3 + 2) × 8 = 40 units

So, total work = 40 × 12/5 units

Time taken by B alone to complete whole work = 40 ×  12/5 ×2 = 48 days 

P.W=Rs.100×2310100+15×52 = Rs.1680