Practice Test: Percentages- 3 - JAMB MCQ

Let original wage = 100.
After reduction, salary = 50.
Now 50% increase of reduced salary = 50 + [(50/100) x 50] = 75
Thus loss = 100 - 75 = 25.
25% will be the loss.

Let the original price of petrol be Rs. 100
original consumption be p litres
original expenditure be Rs. 100p
Updated price = Rs. 125
updated consumption = p - (r/100)p
updated expenditure = 125[p - (pr/100)]
original and updated expenditure must be same
Thus 125[p - (pr/100)] = 100p
solving this, we get r = 20
i.e. reduction in consumption should be 20%

Let's assume the price of 5 pair package be p% of individual 5 pairs
3.40 = p% of 5
p = 68%
i.e (100 - 68) = 32
Thus, 5 pair package is 32% cheaper than the individual 5 pairs

Total amount of metal in the alloy = 25 kg
5 = z% of 25
(5 x 100)/25 = z
= 20%

3 hrs 40 min = 220 min
3 hrs 45.5 min = 225.5 min
225.5 - 220 = 5.5
Now, let's assume, 225.5 is z% of 220
z = 102.5
Percentage difference = 102.5 - 100
= 2.5

Let the number be a
a - 125 = (37.5/100) a
100a - 37.5a = 12500
a = 200
Thus 25% of 200 = 50

30% students either failed or didn't appear in the exam.
Thus the No. of students who passed the exam
= 450 - 135 = 315

Maximum mark be m
passing marks = (30/100)m + 10
marks obtained by paul = (40/100)m = [(30/100)m + 10] + 15
40m/100 = 30m/100 + 25
10m/100 = 25
m = 2500/10
m = 250
Thus passing marks = 85

Let the original number of working hours a day be p hrs.
and original wages per hour be Rs. y
Then, original daily earning = Rs. (py).
Increased working hours = p + (20p/100)
= (120p/100)
New wage per hr = y + 15y/100
= (115y/100)
New daily earning = (120p/100) x (115y/100)
= 1.38py
Assume 1.38py is z% of py
1.38py = z%(py)
z = 138
i.e. the increase in percent = 138 - 100
= 38%

Let the original salary be Rs. 100. New salary = Rs. 90.
Let the increase be p% of 90
[90 + (p/100)90] = 100
90 + 90p/100 = 100
9000 + 90p = 10000
90p = 10000 - 9000
p = 11.11

Total rolls = 40 dozen
20 dozen sold by noon
Remaining rolls = 20 dozen
Rolls sold between noon and closing time (60/100)20 = 1200/100
= 12
Unsold = 20 - 12 = 8 dozen

Number of rotten apples = (30/100) x 450
=135
Number of apples in good condition = 450 - 135 = 315

Let the number be a
(75/100)a + 75 = a
75a + 7500 = 100a
7500 = 25a
a = 300

Sonal's monthly salary = (1.08/12) lacs
= 9000
Thus, the monthly salary of Amit = Rs. 18000
Let's assume, the monthly salary of Mahesh = z
(12/100) z = (16/100) x 18000
z = 24000

(5/7) x 112 = 80
(4/5) x 70 = 56
lets assume 56 is z% lesser than 80
80 - 56 =24
24 = z% of 80
(24 x 100)/80 = z
z = 30