Deloitte Quantitative Aptitude Questions
A.16.8 % gain
B. 16.8 % loss
C. 13.6 gain
D. 13.6 loss
So, instead of paying Rs. 130, the tradesman pays only Rs (130 × 64)/100 = Rs 83.2.
Loss suffered by the dealer selling sugar to the tradesman = Rs 100 — Rs 83.2 = Rs 16.8 loss
loss = ( 16.8/100) × 100 = 16.8%.
- 53.2
- 54.68
- 55.55
- 56.72
Total students =50+60+45=160
Marks total =(55×50)+(60×55)+(45×60)
=2750+6000
=8750
Average marks = 8750/16 =54.6875.
A. 10
B. 15
C. 20
D. 25Number of lemons =12
3 parts of lemon =12
1 part of lemon = 12/3 = 4
5 parts of water ≡4×5=20
number of cups of water to be used=20
A. 40 days
B. 20 days
C. 60 days
D. 100 days
K + S + J = 20 days.
Chocolates = LCM (24, 20) = 120C
Efficiency - S + J = 120C/24d = 5 c/d
Efficiency – K +S + J = 120C/20d = 6 c/d
So, K efficiency = 1c/d
K is half efficient as S,
Then Efficiency of S = 2 c/d
So, efficiency of S + K = 3 c/d.
Days by S + k = 120c/(3c/d) = 40 days
A. 7
B. 22
C. 154
D. None
A.12.50%
B.10%
C.8.50%
D.14%
Let Principle = P.
Given, Amount = 2P
n = 8 years.
I = ?
SI = PNR/100.
SI = A-P
SI = 2P-P =P
So, P = P8R/100,
R = 100/8 = 12.5%
A.36 m
B.58 m
C.72 m
D.100 m
A stone is dropped from a height of 5 km.
The distance it falls through varies directly with the square of the time
Let us suppose distance as d and time as t
d
d
where, k is constant of proportionality
If a stone travels 64 m in 4 second.
d=64 and t=4
Substitute into the formula and solve for k
64=16k
k=4 m/s
We get an relation between d and t
d-4t2
We need to find distance covered by stone in the 5th second.
Distance covered in 5th second = d(5)-d(4)
At t=5, d=4*5^2 = 100 m
At t=4, d=64m
Distance covered in 5th second = 100-64 = 36 m
A.100
B.110
C.120
D.130
A.2 log 6 + 1
B.6 log 2 + 1
C.2 log 6 + 2
D.6 log 2 +2
= log 6^2 + log10^2
= 2log6 + 2 log10
= 2log6 + 2.1
= 2log6 + 2
A.1/4
B.1/3
C.1/2
D.2/3
Totally 36 sums that are possible, out of which 18 are even and 18 are odd.
So, probability of even sums = 18/36 = 1/2
A.5%
B.10%
C.15%
D.20%
Tax expense = 10%30000 = 3000.
Remaining amount = 27000.
Rent expense = 1/3(27000) = 9000.
Remaining amount = 18000.
Petrol expense = 1/2(18000) = 9000
Remaining = 9000.
Electricity = 1/3(9000) = 3000.
Remaining is the Rest = Saving = 6000.
X%30000 =6000
X = 20%
A.8
B.10
C.12
D.15
400 = 4*100 = 22* 102
= 22 * (2*5)2 = 24 * 52
So, 400 = 24 * 52
No. of factors = (P+1)*(P+1) = (4+1)*(2+1) = 15.
A.6
B.24
C.120
D.720
A.8
B.16
C.15
D.10
No. of 1-rupee coins = 4x
No. of 2-rupee coins = 8x
No. of 5-rupee coins = 5x.
Given total amount in the bag = Rs. 90
4x coins each of 1 rupee, the total value of 1-rupee coins = 4x*1
8x coins each of 2 rupee, the total value of 2 rupee coins = 8x*2
5x coins each of 5 rupee, the total value of 5 rupee coins = 5x*5
So, 4x + 16x + 25x = 90
45x = 90.
x = 2.
Total number of 5-rupee coins = 5x = 5*2 = 10 coins
A.18
B.28
C.27
D.9
E.30
Total 3 balls to be selected and at least 2 of them are green.
There are 5 red, 4 white and 3 green from which the required balls to be picked.
Different ways = 2 green and 1 red OR 2 green and 1 white OR 3 green
3C2*5C1 or 3C2*4C1 or 3C3 = (3*5) + (3*4) + 1 = 28