Express each of the following as a fraction:
(i) 48% (ii) 220% (iii) 2.5%
(i) 48% means, 48 divided by 100.
So, 48% = 48 /100
= 12 /25
(ii) 220% means, 220 divided by 100.
So, 220% = 220 /100
= 11 /5
(ii) 2.5% means, 2.5 divided by 100.
So, 2.5% = 2.5 /100
= 1 /40
Express each of the following as a decimal:
(i) 6% (ii) 72% (iii) 125%
(i) 6% means, 6 divided by 100.
So, 6% = 6 /100
= 3 /50 = 0.06
(ii) 72% means, 72 divided by 100.
So, 72% = 72 /100
= 18 /25 = 0.72
(iii) 125% means, 125 divided by 100.
So, 125% = 125 /100
= 5 /4 = 1.25
Express each of the following as a percentage:
(i) (ii) (iii)
(i) = ( x 100) %
= (9 x 4) %
= 36%
(ii) = ( x 100) %
= 2.4%
(iii) = ( x 100) %
= (12 x 20) %
= 240%
Convert the ratio 4 : 5 to percentage.
4 : 5 =
= ( x 100) % [Because 100% = 1]
= 80%
Express 125% as a ratio.
125% = 125/100
= 5 /4 [Divided by 25]
= 5 : 4
Which is largest inand
= (20 /3) %
= (20 /3 x 1 /100)
= 1 /15
= 0.06 ____ (i)
= 0.15 ____ (ii)
0.14 ____ (iii)
From equation (i), (ii) and (iii),
0.15 > 0.14 > 0.06
What per cent of 150 is 96?
Percentage = (96 /150 x 100) %
= (96 /3 x 2) % [Divided by 50]
= (32 x 2) %
= 64%
What per cent of 5 kg is 200 g?
5 kg = 5 x 1000
= 5000 g
Now,
Percentage = (200 /5000 x 100) %
= (200 /50) % [Divided by 100]
= 4 %
What per cent of 2 litres is 250 mL?
2 liters = 2 x 1000
= 2000 mL
Now,
Percentage = (250 /2000 x 100) %
= (250 /20) % [Divided by 100]
= 12.5 %
Findof 3600.
= (9 /2) x 100
= 9 /200
Now,
9 /200 of 3600 = 9 /200 x 3600
= 9 x 18 [Divided by 200]
= 162
If 16% of number is 72, find the number.
Let the number = Z
∴ 16% of Z is 72.
⇒ 16 /100 × Z = 72
⇒ 16 Z = 7200
⇒ Z = 7200 /16
⇒ Z = 450
A man saves 18% of his monthly income. If he saves Rs. 3780 per month, what is his monthly income?
Let Rs. Z his monthly income.
∴ Saving = 18% of Rs. Z
⇒ 3780 = 18 /100 × Z
⇒ 3780 = 9 /50 × Z
⇒ Z = 3780 × 50/9
⇒ Z = 420 × 50
[Because 420 × 9 = 3780]
⇒ Z = 21000
Therefore, his monthly income is Rs 21000/-
A football team wins 7 games, which is 35% of total games played. How many games were played in all?
Let, total games played = Z
∴ percentage of games won = 35% of Z
⇒ 7 = 35/100 × Z
⇒ 7 = 7/20 × Z [Divided by 5]
⇒ Z = 7 × 20/7
⇒ Z = 20
Amit was given an increment of 20% on his salary. If his new salary is Rs. 30600, what was his salary before the increment?
Let Amit’s old salary = Z
∴ Salary after increment = (Z + 20Z/100)
Now,
⇒ (Z + 20 Z/100) = 30600
⇒ (100 Z + 20 Z)/100 = 30600
⇒ 120 Z = 30600 × 100
⇒ Z = 25500
Sonal attended her school on 204 days in a full year. If her attendance is 85%, find the number of days on which the school was opened.
Let the number of days the school was opened = Z
∴ Percentage of attendance = 85% of Z
Now,
85% of Z = 204
⇒ 85/100 × Z = 204
⇒ Z = 204 × 100/85
⇒ Z = 204 × 20/17 [Divided by 5]
⇒ Z = 12 × 20
⇒ Z = 240
A’s income is 20% less than that of B. By what per cent is B’s income more than A’s?
Let B’s income = 100
Then, A’s income = (100 – 20) = 80
∴ B’s income more than A’s income = (100 – 80)/80 × 100
= 20/80 × 100
= 1/4 × 100
= 25
The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that the expenditure on it remains unchanged?
Let the consumption of petrol = 1 unit and its cost = Rs.100
∴ New cost of 1 unit of petrol = Rs.110
Now,
Rs.110 will yield 1 unit of petrol.
∴Rs.100 will yield (1/110 × 100)
= 10/11 unit of petrol
Now,
Reduction of consumption = 1 – (10/11)
= 1/11
Percentage of reduction = (1/11 × 100) %
=
The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?
Let population of the town a year ago = Z
∴ Present population = 108% of Z
⇒ 54000 = Z × 108/100
⇒ 54000 = Z × 27/25
⇒ Z = 54000 × 25/27
⇒ Z = 2000 × 25
⇒ Z = 50000
The value of a machine depreciates every year by 20%. If the present value of the machine be Rs. 160000, what was its value last year?
Let the value of machine last year = Z
∴ Present value = (100 – 20) % of Z
⇒ 160000 = 80% of Z
⇒ 160000 = Z × 80/100
⇒ Z = 160000 × 100/80
⇒ Z = 2000 × 100
⇒ Z = 200000
An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.
Given,
Percentage of copper = 40%
Percentage of nickel = 32%
∴ Percentage of zinc = {100 – (40 + 32)} %
= 28 %
Now,
Mass of zinc in 1 kg of the alloy = (28 × 1/100) kg
= 0.28 kg
= 0.28 × 1000 g
= 280 g
Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his daily food intake.
Amount of proteins = 12% of 2600
= 26 × 12
= 312 calories
Amount of fats = 25% of 2600
= 26 × 25
= 650 calories
Amount of carbohydrates = 63% of 2600
= 26 × 63
= 1638 calories
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?
Let the amount of gunpowder which carries 9 kg nitre = Z
∴ 75% of Z = 9 kg
⇒ Z × 75/100 = 9
⇒ Z = 9 × 100/75
⇒ Z = 9 × 4/3
⇒ Z = 12 kg
Now,
Let the amount of gunpowder which carries 2.5 kg sulphur = K
∴ 10% of K = 2.5 kg
⇒ K × 10/100 = 2.5
⇒ K = 2.5 × 100/10
⇒ K = 2.5 × 10
⇒ K = 25 kg
Divide Rs. 7000 among A, B and C such that A gets 50% of what B gets and B gets 50% of what C gets.
Let the amount of money gets by C = Rs. Z
∴ Amount of money B gets = (50% of Rs.Z)
∴ Amount of money A gets = (50% of B)
= (25% of Rs.Z)
Now,
Z + (50% of Rs.Z) + (25% of Rs.Z) = RS.7000
⇒ Z + (Z × 50/100) + (Z × 25/100) = 7000
⇒ Z + 50 Z/100 + 25 Z/100 = 7000
⇒ 175 Z/100 = 7000
⇒ Z = 7000 × 100/175
⇒ Z = 7000 × 4/7
⇒ Z = 4000
∴ C gets = Rs.4000
∴ Amount of money B gets = (50% of Rs.Z)
= (50% of Rs.4000)
= (Rs.4000 × 50/100)
= Rs.2000
∴ Amount of money A gets = (25% of Rs.Z)
= (25% of Rs.4000)
= (Rs.4000 × 25/100)
= Rs.1000
Find the percentage of pure gold in 22-carat gold, if 24-carat gold is 100% pure.
22-carat gold contains 22 parts out of 24 parts.
∴ Percentage of pure gold in 22-carat gold
Hence, 22-carat gold contains of pure gold.
The salary of an officer is increased by 25%. By what per cent should the new salary be decreased to restore the original salary?
Let the original salary = Rs.100
Then,
After increment of 25% = 100 (1 + 25/100)
= 100 (125/100)
= Rs.125
Now,
To restore the original salary,
Let the new salary decreased by Z%
∴ 125(1 – Z/100) = 100
⇒ (1 – Z/100) = 100/125
⇒ (1 – Z/100) = 4/5
⇒ Z/100 = 1/5 [1 – 4/5 = 1/5]
⇒ Z = 100/5
⇒ Z = 20%
Choose the correct answer:
A.
B.
C.
D.
3/5 = (3/5 × 100) %
= (3 × 20) %
= 60%
when expressed as a percentage, is
A.
B.
C.
D.
0.8% = 0.8/100
= 0.008
when expressed as a percentage, is
A.
B.
C.
D.
6 : 5 = 6/5
= (6/5 × 100) % [100% = 1]
= (6 × 20) %
= 120 %
of a number is 9. The number is
A.
B.
C.
D.
Let number = Z
Then,
5% of Z = 9
⇒ 5/100 × Z = 9
⇒ 5 Z = 900
⇒ Z = 180
What per cent of 90 is 120?
A.
B.
C.
D. none of these
Let Z% of 90 is 120
∴ Z/100 × 90 = 120
⇒ 90 Z = 120 × 100
⇒ Z = 12000/90
⇒ Z = 400/3
⇒ Z =
What per cent of 10 kg 250 g?
A.
B.
C.
D.
10 kg = 10 × 1000
= 10000 g
Let Z% of 1000 is 250
∴ Z/100 × 10000 = 250
⇒ 100 Z = 250
⇒ Z = 250/100
⇒ Z = 2.5%
of
A.
B.
C.
D.
Let, 40% of Z = 240
⇒ 40/100 × Z = 240
⇒ Z = 240 × 100/40
⇒ Z = 6 × 100 [40 × 6 = 240]
⇒ Z = 600
?% of 400 = 60
A.
B.
C.
D.
Let, Z% of 400 = 600
⇒ Z/100 × 400 = 60
⇒ 4 Z = 60
⇒ Z = 60/4
⇒ Z = 15
(180% of ?)
A.
B.
C.
D.
Let (180% of Z)
∴ (180/100 × Z)
⇒ (18/10 × Z) = 504 × 2
⇒ Z = 504 × 2 × 10/18
⇒ Z = 504 × 10/9
⇒ Z = 560
20% or Rs. 800 = ?
A.
B.
C.
D. none of these
20% of Rs.800 = 20/100 × 800
= 20 × 8
= 160
In an examination, Nitin gets 98 marks. This amounts to 56% of the maximum marks.
What are the maximum marks?
A. 75
B. 150
C. 175
D. 225
Let the maximum marks = Z
∴ 56% of Z = 98
⇒ Z × 56/100 = 98
⇒ Z = 98 × 100/56
⇒ Z = 7 × 100/4
⇒ Z = 175
A number is first increased by 10% and then reduced by 10%. The number
A. does not change
B. decrease by 1%
C. increased by 1%
D. none of these
Let the number = Z
10% increased by number = Z (1 + 10/100)
= 11Z/10
Now,
10% decreased by number = 11Z/10 (1 - 10/100)
= (11Z/10) (90/100)
= 99Z/100
∴ difference = Z – 99Z/100
= Z/100
Percentage of decreases = Z/100 × 1/Z × 100
= 1%
A period of 4 hours 30 min is what per cent of a day?
A.
B.
C.
D.
4 hours 30 min = (4 × 60) + 30
= 240 + 30
= 270 min
24 hours = 24 × 60
= 1440 min
Now,
Percentage = (270/1440 × 100) %
= (3/16 × 100) %
= (3/4 × 25) %
= (75/4) %
=
In an examination, 65% of the total examines passed. If the number of failures is 420, the total number of examines is
A.
B.
C.
D.
Let the total number of examines = Z
Percentage of examines failed = (100 – 65) % = 35%
∴ 35% of Z = 420
⇒ Z × 35/100 = 420
⇒ Z = 420 × 100/35
⇒ Z = 12 × 100
⇒ Z = 1200
A number exceeds 20% of itself by 40. The number is
A. 50
B. 60
C. 80
D. 320
Let the number = Z
∴ 20% of Z + 40 = Z
⇒ (Z × 20/100) + 40 = Z
⇒ Z/5 + 40 = Z
⇒ Z – Z/5 = 40
⇒ 4Z/5 = 40
⇒ Z = 40 × 5/4
⇒ Z = 50
A number decreased byof itself by 87. The number is
A. 58
B. 110
C. 120
D. 135
Let the number = Z
∴ Z - (of Z) = 87
⇒ Z - (Z × 55/2 × 1/100) = 87
⇒ Z - (Z × 11/2 × 1/20) = 87
⇒ Z- (11Z/40) = 87
⇒ 29Z/40 = 87
⇒ 29Z/40 = 87
⇒ Z = 87 × 40/29
⇒ Z = 120
is what per cent of
A.
B.
C.
D.
Percentage = (0.05/20 × 100) %
= (0.05 × 5) %
= 0.25%
One-third of 1206 is what per cent of
A.
B.
C.
D.
Percentage = {(1/3 × 1206) × (1/134) × 100} %
= {402 × 1/134 × 100} %
= {3 × 100} %
= 300%
x% of y is y% of
A.
B.
C.
D.
Let x% of y is y% of Z
∴ x/100 × y = y/100 × Z
⇒ x y/100 = y/100 × Z
⇒ Z = x y/100 × 100/y
⇒ Z = x
What per cent ofis
A.
B.
C.
D.
Percentage = {(1/35)/(2/7) × 100} %
= {1/35 × 7/2 × 100} %
= {1/5 × 1/2 × 100} %
= {1/5 × 50} %
= 10%
Express:
24% as a fraction;
24% means, 24 divided by 100.
So, 24% = 24/100
= 6/25
Express:
105% as a decimal;
105% means, 105 divided by 100.
So, 105% = 105/100
= 1.05
Express:
4 : 5 as a percentage;
4 : 5 = 4 /5
= (4 /5 x 100) % [Because 100% = 1]
= 80%
Express:
56% as a ratio.
56% means, 56 divided by 100.
So, 56% = 56/100
= 14/25
= 14:25
If 34% of a number is 85, find the number.
Let the number = Z
∴ 34% of Z = 85
⇒ 34/100 x Z = 85
⇒ Z = 85 x 100/34
⇒ Z = 5 x 100/2
⇒ Z = 250
The value of a machine depreciates every year by 10%. If the present value of the machine is Rs.54000, what was its value last year?
Let the value of the machine last year = Z
∴ Present value of the machine = (100 – 10) % of Rs.Z
⇒ 54000 = 90% of Z
⇒ 54000 = Z x 90/100
⇒ Z = 54000 x 100/90
⇒ Z = 600 x 100
⇒ Z = 60000
An alloy contains 30% copper, 42% nickel and rest zinc. Find the mass of zinc in 1 kg of alloy.
Given,
Percentage of copper = 30%
Percentage of nickel = 42%
∴ Percentage of zinc = {100 – (30 + 42)} %
= 28 %
Now,
Mass of zinc in 1 kg of the alloy = (28 × 1/100) kg
= 0.28 kg
= 0.28 × 1000 g
= 280 g
In a class, 60% of the total number of students are boys and there are 14 girls. How many students are there in the class?
Let the total number of students = Z
Percentage of girls = (100 – 60) % = 40%
Now,
Number of girls = 40% of Z
⇒ 14 = Z × 40/100
⇒ Z = 14 × 100/40
⇒ Z = 14 × 5/2
⇒ Z = 35
Which is largest inand
= (25 /3) %
= (25 /3 x 1 /100)
= 8.33 /100
= 0.08 _____ (i)
= 0.16 ______ (ii)
0.15 _____ (iii)
From equation (i), (ii) and (iii),
0.16 > 0.15 > 0.08
What per cent ofis
A. 2.5%
B. 5%
C. 7.5%
D. 10%
Percentage = {(1/45)/(2/9) × 100} %
= {1/45 × 9/2 × 100} %
= {1/5 × 1/2 × 100} %
= {1/5 × 50} %
= 10%
A number decreased by 30% gives 84. The number is
A. 90
B. 110
C. 120
D. 135
Let the number = Z
∴ Z – (30% of Z) = 84
⇒ Z – (Z × 30/100) = 84
⇒ Z – 30 Z/100 = 84
⇒ 70 Z/100 = 84
⇒ Z = 84 × 100/70
⇒ Z = 12 × 10
⇒ Z = 120
(?)% of 320 is 48?
A. 25%
B. 15%
C. 14%
D. 9%
Percentage = (48/320 × 100) %
= (48/32 × 10) %
= (3/2 × 10) %
= 15%
What per cent of 45 is 54?
A.%
B. 104%
C. 108%
D. 120%
Percentage = (54/45 × 100) %
= (54/9 × 20) %
= (6 × 20) %
= 120%
A number exceeds 25% of itself by 60. The number is
A. 75
B. 45
C. 80
D. 65
Let the number = Z
∴ 25% of Z + 60 = Z
⇒ (Z × 25/100) + 60 = Z
⇒ Z/4 + 60 = Z
⇒ Z – Z/4 = 60
⇒ 3Z/4 = 60
⇒ Z = 60 × 4/3
⇒ Z = 80
5% of which number is 12?
A. 120
B. 180
C. 240
D. 320
Let the number = Z
∴ 5% of Z = 12
⇒ Z × 5/100 = 12
⇒ Z = 12 × 100/5
⇒ Z = 12 × 20
⇒ Z = 240
Fill in the blanks.
(i)% of Rs.1200 = ……..
(ii) 240 mL is……% of 3 L.
(iii) If% of 35 is 42, then=……..
(iv)
(v) 120 = (……)% of 80.
(i) 90
% of Rs.1200 = (15/2) % of Rs.1200
= 15/2 × 1/100 × 1200
= 15/2 × 12
= 90
∴ Rs.90
(ii) 8
240 mL = (240/1000) L
Now,
Percentage = (240/1000 × 1/3 × 100) %
= (240/10 × 1/3) %
= (80/10) %
= 8%
(iii) 120
X% of 35 = 42
⇒ 35 × X/100 = 42
⇒ 35X/100 = 42
⇒ X = 42 × 100/35
⇒ X = 6 × 100/5
⇒ X = 120
(iv) 240
12/5 = (12/5 × 100) %
= (12 × 20) %
= 240%
(v) 150
Let the number = Z
∴ 120 = Z% of 80
⇒ 120 = 80 × Z/100
⇒ Z = 120 × 100/80
⇒ Z = 120 × 5/4
⇒ Z = 150
Write ‘T’ for true and ‘F’ for false for each of the following:
(i) 6% of 8 is 48.
(ii) 6 : 5 = 30%.
(iii)= 60%.
(iv) 6 hours = 25% of a day.
(i) False
6% of 8 = 8 × 6/100
= 48/100
= 0.48
(ii) False
6:5 = 6/5
= (6/5 × 100) %
= (6 × 20) %
= 120%
(iii) True
3/5 = 3/5
= (3/5 × 100) %
= (3 × 20) %
= 60%
(iv) True
1 day = 24 hours
6 hours = (6/24 × 100) %
= (1/4 × 100) %
= 25%