Percentage Class 8 RS Aggarwal Exe-6B Goyal Brothers ICSE Maths Solutions. Ch-6. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Percentage Class 8 RS Aggarwal Exe-6B Goyal Brothers ICSE Maths Solutions Ch-6
| Board | ICSE |
| Publications | Goyal Brothers |
| Subject | Maths |
| Class | 8th |
| Writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-6 | Percentage |
| Exe-6B | Word Problems on Percentage |
| Edition | 2024-2025 |
Word Problems on Percentage
(Class 8 RS Aggarwal Exe-6B Goyal Brothers ICSE Maths Solutions. Ch-6)
Page- 92, 93
Exercise- 6B
Que-1: 36% of the students in a school are girls. If the number of the boys is 1440, find the total strength of the school.
Sol: Let the total strength of the school be x
We are given that the no. of boys is 1440
Based on a general assumption that the genders of students of the school are boys and girls, the no. of girls is x-1440
Also, we are given that 36% of all students are girls
x – 1440 = 36% of x
x – 1440 = 36/100 × x
x – 1440 = 0.36x
x – 0.36x = 1440
0.64x = 1440
x = 1440/0.64
x = (1440×100)/64
x = 2250.
Que-2: Geeta saves 18% of her monthly salary. If she spends Rs10250 per month, what is her monthly salary ?
Sol: Geeta saves her 18% salary and spends 10250 per month. Let, the monthly salary of Geeta be y.She saves 18% of her monthly salary. So, the rest of the salary she spends.
So, percentage of salary spent, 100 – 18 = 82%
She also spends 10250 per month.
So, 82% of her salary will be 10250.
y × (82 / 100) = 10250
82y= 10250 × 100
82y = 1025000
y = 1025000/82
y = 12500
Que-3: In an examination, a student has to secure 40% marks to pass. Rahul gets 178 marks and fails by 32 marks. What are the maximum marks ?
Sol: If the student have to pass, he should get the mark 178 + 32 = 210 (as he fails by 32,if he get 32 more to 178 he will pass exam)
40% mark is 210.
let the 100% mark is ‘X’.
To find out maximum mark is,
40% of maximum mark = 100% of 210
=> 40% of ‘X’ = 100%*210.
(40/100)X = (100/100)*210
=> 0.4X =210
=> X = 210/0.4
=> X = 525
Que-4: 8% of the students in a class remained absent on a day. If 633 attended the school on that day, how many remained absent ?
Sol: Let the total no. of students be x.
If 8% were absent that day 92% must be present.
therefore, 92x/100 = 1633
x = (1633)(100)/92
x = 1775
No. of absentees = Total no. of students – no. of present
= 1775 – 1633 = 142
Hence, 142 students were absent.
Que-5: On increasing the price of an article by 14%, it becomes Rs1995. What was its original price ?
Sol: Let the increasing price of an article is x
x + 14x % = 1995
x + (14x/100) = 1995
(100x+14x)/100 = 1995
114x/100 = 1995
114x = 199500
x = 199500/114
x = Rs 1750.
Que-6: On decreasing the price of an article by 6%, it becomes Rs1551. What was its original price ?
Sol: let the price of an article = 100
decreasing 6% of it
100 x 6% = 94
according to the given question
94% = 1551
100% = (100×1551)/94
= 155100/94
= 1650 .
hence , original price = Rs 1650 .
Que-7: Reenu reduced her weight by 15%. If now she weighs 52.7 kg, what was her earlier weight ?
Sol: Since Reenu reduced 15% of her earlier weight, her reduced weight is 52.7 kg.
=> x – 15% of x = 52.7
=> x – x(15/100) = 52.7
=> x – 3x/20 = 52.7
=> (20x – 3x)/20 = 52.7
=> 17x = 52.7 × 20
=> 17x = 1054
=> x = 1054/17
=> x = 62 kg.
Que-8: Two candidates contested an election. One of them secured 58% votes and won the election by a margin of 2560 votes. How many votes were polled in all ?
Sol: It is given that one voter secured 58% votes
Then other voters secured (100-58)% = 42% votes
Difference between both voters = (58-42)% = 16%
So, 16% of total votes = 2560
Total votes = x
16% of x = 2560
16/100 × x = 2560
x = (2560×100)/16
x = 16000.
Que-9: In an examination, Preeti scored 60 out of 75 in Science, 84 out of 100 in Mathematics, 36 out of 50 in Hindi and 30 out of 45 in English.
(i) In which subject her performance is worst ?
(ii) In which subject her performance is the best ?
(iii) What is her aggregate percentage of marks ?
Sol: Percentage in Science
= 60/75 x 100 = 80%
Percentage in Mathematics
= 84/100 x 100 = 84%
Percentage in Hindi
= 36/50 x 100 = 72%
Percentage in English
= 30/45 x 100 = 200/3 = 66*(2/3)%
(i) In English her performance is worst.
(ii) In mathematics her performance is best.
(iii) Sum of marks obtained/ Sum of Total number of marks x 100
(60+84+36+30)/(75+100+50+45) x 100
= 210/270 x 100
= 700/9 = 77*(7/9) %.
Que-10: The price of an article is increased by 25%. By how much per cent must this new value be decreased to restore it to its former value ?
Sol: Lets the price of the article = ₹100
The price of the article is increased by 25%
Increase in price of the article is
= 25% of ₹100
= 25/100 x ₹100
= ₹25
Therefore, the new increased price = ₹100 + ₹25 = ₹125
So, if we subtract ₹25 from ₹125, we get the original price of article.
Therefore, decrease in percent is
= (Decrease in price/new price) x 100
= (25/125) x 100
= 20%.
Que-11: The price of an article is reduced by 10%. By how much per cent this value be increased to restore it to its former value ?
Sol: Let the original price of an article is ‘x’ and the new price of an article is ‘y’.
According to the question,
x = y – 10% of y
⇒ x = y – 10y/100
⇒ x = y – y/10
⇒ x = 9y/10
⇒ y = 10x/9
∴ Total increase is x/9.
So, Percentage increase = (x/9) × (100/x)
⇒ 100/9
= 11*(1/9) %.
Que-12: The price of tea is increased by 20%. By how much per cent a housewife should reduced the consumption of tea so as not to increase the expenditure on tea ?
Sol: ⇒ Let the original price of the tea be Rs.100.
⇒ The price of the tea is increased by 20%.
⇒ Increased price of the tea = 100+(20/100) = 100+20 = Rs120.
⇒ Increase in price = 120−100 = Rs.20
⇒ Tea of cost Rs.20 is to decreased out of the increased cost of Rs.120.
⇒ Percentage decrease = 20/120×100 = 50/3%.
= 16*(2/3)%.
Que-13: A man gave 35% of his money to his elder son and 40% of the remainder to the younger son. Now, he is left with Rs11700. How much money had he ?
Sol: Elder son share = 35%
Remainder = ( 100 – 35 ) = 65 %
Younger son share = 40 % of 65
= ( 40 × 65 ) / 100
= 26 %
The percentage left with the man = 100 – 35 – 26
= 100 – 61
= 39%
According to the problem given ,
39% of total amount = Rs 11700
Total amount = ( 11700) / 39%
= ( 11700 × 100 ) / 39
= Rs 30000
Que-14: 5% of the population of a town were killed in an earthquake and 8% of the remainder left the town. If the population of the town now is 43700, what was its population at the beginning ?
Sol: Let original no. of people be x.
After Earthquake :
population = x – (5% of x)
= 95x/100
After people left town = (95x/100) − {8% of (95x/100)}
= 1748x/2000
Given : 43700 = 1748x/2000
⇒ x = (43700×2000)/1748
= 50000
Que-15: A and B are two towers. The height of tower A is 20% less than that of B. How much per cent is B’s height more than that of A ?
Sol: It is given that the height of A is 20% less than the height of B.
Let the height of b is x. So, the height of A is
A = x – (20/100 × x)
= x – 0.2x = 0.8x
We have to find by how much per cent is B’s height more than that of A.
% = (Height of B – Height of A)/(Height of A) x 100
% = (x-0.8x)/(0.8x) x 100
% = (0.2x/0.8x) × 100
% = 1/4 × 100
% = 25%.
Que-16: In an examination, 30% of the candidates failed in English, 35% failed in G.K. and 27% failed in both the subjects. If 310 candidates passed in both, how many candidates appeared in the examination ?
Sol: Given : In a examination 30% students failed in English and 27 % failed in English and G.K. , So
Percentage of students failed only in English = 30% – 27% = 3 %
Also given : In a examination 35% students failed in G.K. and 27 % failed in English and G.K. , So
Percentage of students failed only in G.K. = 35% – 27% = 8 %
Then,
Percentage of students failed in one or both subjects = 3 % + 8 % + 27 % = 38 %
So,
Percentage of students passed only in both subjects = 100% – 38% = 62 %
Also given : Total students passed in both subjects = 310 , We assume total students appeared for examination = x , So
62 % of x = 310
62 × x/100 =310
⇒ x = (310×100)/62
⇒ x = (10×100)/2
⇒ x = 10×50
⇒ x = 500
Que-17: The value of a car depreciates annually by 10%. If the present value of the car be Rs650000, find its value after 2 years.
Sol: Present value = ₹ 650000
Decrease in first year = 10% = 10% of 650000
= (10/100) * 650000 = ₹65000
So, value after first year = ₹650000-65000 = ₹585000
Decrease in second year = 10%
= 10% of 585000 = ₹58500
Value after second year = 585000 – 58500 = ₹526500
Answer : Hence , the car’s value of two years is ₹526500
Que-18: The population of a village increases by 7% every year. If the present population is 80000, then what will be the population after two years ?
Sol: Given that, the population of a village increases at a rate of 7% every year. The present population of the village is 90,000.
To find out: Population of the village after 2 years.
We know that, population after ‘n’ years = P(1+r/100)^n
Here, P = 80,000, rate of increase (r) = 7% and number of years n = 2.
∴ The population after two years = 80000(1+7/100)²
= 80000 (107/100)²
= 80000 × 107/100 × 107/100
= 103041
Que-19: A student was asked to multiply a number by 5/3. He multiplied it by 3/5 instead. Find the percentage error in the calculation.
Sol: Let the number be ‘x’
Then, according to the given data,
(5x/3 – 3x/5)/5x/3×100
⇒ 16/25×100
⇒ 64%
Que-20: In an election between two candidates, 10% of the voters did not cast their votes. 10% of the votes polled were found invalid. The successful candidate got 54%of the valid votes and won by a majority of 1620 votes. Find the number of voters enrolled on the voters list.
Sol: Let the total number of voters be x.
As 10% of the voters did not cast their votes
Then, Votes polled = 90% of x.
10% of the votes polled were found invalid
Valid votes = 90% of (90% of x).
54% of [90% of (90% of x)] − 46% of [90% of (90% of x)] = 1620
=> 8% of [90% of (90% of x)] = 1620
⇒ (8/100)×(90/100)×(90/100)×x = 1620
⇒ x = (1620×100×100×100)/(8×90×90)
= 25000
–: End of Percentage Class 8 RS Aggarwal Exe-6B ICSE Maths Solutions Ch-6. :–
Return to :- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions
Thanks
Please share with yours friends if you find it helpful
