Percentage Class 8 MCQs RS Aggarwal Exe-6C Goyal Brothers ICSE Maths Solutions. We provide step by step Solutions of council prescribe textbook / publications to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Percentage Class 8 MCQs RS Aggarwal Exe-6C Goyal Brothers ICSE Maths Solutions Ch-6
| Board | ICSE |
| Publications | Goyal Brothers Prakshan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-6 | Percentage |
| Exe-6C | Multiple Choice Questions |
| Edition | 2024-2025 |
MCQs on Percentage With Answer / Solutions
( Class 8 MCQs RS Aggarwal Exe-6C Goyal Brothers ICSE Maths Solutions )
Exercise- 6C Multiple Choice Questions :
Que-1: What per cent of 7.2 kg is 18g ?. (a) 25% (b) 2.5% (c) 0.25% (d) 0.025%
Sol: (c) 0.25%
Reason : So , let’s change 7.2 kg in gram and we find 7200 gm .
Suppose , 18 gm is x percent of 7200 gm . Then equation is ..
7200 × x/100 = 18
72 × x = 18
x = 18 / 72
x = 1/4 = 0.25%.
Que-2: A number decreased by 27*(1/2)% gives 87. The number is
(a) 58 (b) 110 (c) 120 (d) 135
Sol: (c) 120
Reason : Let the number be x.
x decreases by 27 1/2% to give 87
therefore
x – 27 1/2% of x = 87
x – 55/2 * 1/100 * x = 87
x – 55x/200 = 87
200x – 55x / 200 = 87
145x = 17400
x = 17400/145 = 120
Therefore the number is 120.
Que-3: A number increased by 37*(1/2)% gives 33. The number is
(a) 22 (b) 24 (c) 25 (d) 27
Sol: (b) 24
Reason : Let the number be ‘x’
x + 37*(1/2)% of x = 33
137*(1/2)% of x = 33
x = 24
Hence, the number is 24.
Que-4: If 37*(1/2)% of a number is 900, then 62*(1/2)% of the number will be (a) 540 (b) 1200 (c) 1350 (d) 1500
Sol: (d) 1500
Reason : Let the no be X
37*(1/2) % of X = 900
(75/2) /100 *X = 900
75/200 *X = 900
X = 900*200/75
X = 2400
Now 62 1/2% of 2400
= 1500
Que-5: By how much per cent is four-fifths of 70 lesser than five-seventh of 112? (a) 24% (b) 30% (c) 36% (d) 42%
Sol: (b) 30%
Reason : Four fifth of 70
= 4/5 × 70
= 56
five seventh of 112
= 5/7 × 112
= 80
% lesser
= (80 – 56)/80 × 100%
= 24/80 × 100%
= 30%
Que-6: If 11% of a number exceeds 7% of the same by 18, the number is
(a) 72 (b) 360 (c) 450 (d) 720
Sol: (c) 450
Reason : Let the number be x.
Then, 11% of x = 11x/100
And,
7% of x = 7x/100
Now, according to the question:
(11x/100) = (7x/100) + 18
So, transposing 7x/100 to the left hand side, we have:
(11x/100) – (7x/100) = 18
4x/100 = 18
4x = 18×100 = 1800
x = 1800/4 = 450
Thus, the required number is 450.
Que-7: 96% of the population of a town is 23040. The total population of the town is (a) 24000 (b) 24936 (c) 25640 (d) 32256
Sol: (a) 24000
Reason : 96% of the population of a village is 23040.
Let the total population of the village be x.
According to the question,
⇒ 96% of x = 23040
⇒ 96/100 × x = 23040
⇒ x = 23040×100/96
⇒ x = 24000
Que-8: In an examination, 65% of the total examines passed. If the number of failures is 168, the number of those who passed is (a) 218 (b) 312 (c) 480 (d) 648
Sol: (b) 312
Reason : Given that 65% of the total examinees passed, the percentage of examinees who failed is:
100% − 65% = 35%
The number of failures is also given as 168, so we can write:
35% of x = 168
This can be expressed as:
35/100 × x = 168
Solving for x:
x = (168×100)/35 = 480
Thus, the total number of examinees is 480. Now, to find the number of examinees who passed, we calculate 65% of the total:
65/100 × 480 = 312.
Que-9: The price of an item is increased by 20% and then decreased by 20%. The final price as compared to the original is (a) 20% loss (b) 20% more (c) 4% less (d) 4% more
Sol: (c) 4% less
Reason : Let the price of apples be Rs. 100
After increase it is 100 × 120%
⇒ Rs. 120
After decrease it is 120 × 80%
⇒ Rs. 96
So, decrease = 100 – 96
⇒ Rs. 4
% of decrease = (4/100) × 100
⇒ 4%
Que-10: A customer asks for the production of x number of goods. The company produces y number of goods daily out of which z% are unit of sale. The order will be completed in
(a) {100x/y(z-1)} days (b) {x/100y(1-z)} days (c) {100yz/x} days (d) {100x/y(100-z)} days
Sol: (d) {100x/y(100-z)} days
Reason : According to the question daily supply is,
⇒ (100−z)% of y
⇒ {(100−z)y}/100
⇒ Thus, required number of days,
⇒ {100x/y(100−z)} days.
Que-11: A man’s wages were decreased by 50%. Again the reduced wages were increased by 50%. He has a loss of (a) 0% (b) 2.5% (c) 0.25% (d) 25%
Sol: (d) 25%
Reason : Let the wages of man is Rs. 100.
Then wages decreased by 50%
So wages =100−50 = 50
The reduced wages were increased by 50%.
Then wages = (50×150)/100 = 75
So, % of wages loss = 100−75 = 25%.
Que-12: If A’s income is 30% more than B’s, then how much percent is B’s income less than A’s ? (a) 23*(1/13)% (b) 25% (c) 30% (d) 33*(1/3)%
Sol: (a) 23*(1/13)%
Reason : Let B′s salary is Rs100 per month then A′s salary is Rs.130 per month.
Hence B′s salary is Rs 30 less than A′s salary.
If A′s salary is 100 Rs\month then B′s salary
= (30/130×100) = 23*(1/13)% is less than of A.
Que-13: If A’s income is 30% less than B’s, then how much percent is B’s income more than A’s? (a) 30% (b) 32*(1/10)% (c) 42*(6/7)% (d) 51%
Sol: (c) 42*(6/7)%
Reason : Let B′s income be 100
Then, A′s income = 100−30 = 70
⇒ % B′s income more than A′s = (30/70)×100%
= 42.857%
= 42*(6/7)%
Que-14: If the numerator of the fraction is increased by 20% and its denominator be diminished by 10%, the value of the fraction is 16/21. The original fraction is (a) 2/3 (b) 3/5 (c) 4/7 (d) 5/7
Sol: (c) 4/7
Reason : Let the fraction be x/y.
Then from the given condition, (x+20)/(y−10) = 16/21
⇒ (x+x/5)/(y−y/10) = 16/21
⇒ (6x/5)/(9y/10) = 16/21
⇒ x/y = (16/21) × (9/10) × (5/6) = 4/7
Que-15: In an examination, it is required to get 36% of maximum marks to pass. A student got 113 marks and was declared failed by 85 marks. The maximum marks are (a) 500 (b) 550 (c) 640 (d) 720
Sol: (b) 550
Reason : Required Pass percentage = 36% of maximum marks
Marks obtained by the student = 113
The student got failed by 85 marks
Let the maximum marks be x
According to the question
⇒ 36% of x = 113 + 85
⇒ x = 198 × 100/36
⇒ x = 550 marks
Que-16: In an class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girl is
(a) 11:14 (b) 14:11 (c) 25:28 (d) 28:25
Sol: (b) 14:11
Reason : Let the number of boys be x and the number of girls be y.
Then, x = y + 12% (x+y)
⇒ x = y + 0.12x + 0.12y
⇒ x − 0.12x = 1.12y
⇒ 0.88x = 1.12y
⇒ x/y = 112/88 = 14/11
⇒ x:y = 14:11.
— : End of Percentage Class 8 MCQs RS Aggarwal Exe-6C Goyal Brothers ICSE Maths Solutions. :–
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