Quantitative Aptitude - Arithmetic Ability Questions

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 – 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.

P = 68000, R =503%  & T = 9 months (3/4 years)  

 S.I=p×r×t100=(6800×503×34)100=Rs.8500

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.

When profit is calculated on Marked Price (M.P) then,

• C.P = M.P - P%

Let M.P = 100

 => C.P = 100 - 30 = 70

 But S.P = Rs. 80 as he gave 20% discount,

 Now, Actual Profit = 80-7070x100

  = 100/7 %

One hour = 60 min

18/60 = 3/10

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

Let the fixed expenditure of the hotel be Rs.x and the variable expenditure ( which is dependent on the guest ) is Rs.y , then 

          x + 10y = 600        ---------(1)

          x + 20y = 800       ----------(2)

From (1) & (2)

10y = 200 

=>  y = Rs. 20   and    x= 400

Hence the total expenditure when there are 40 guests = 400 + 40 x 20 = 1200

Therefore,  average expenditure = 1200/40 = Rs. 30