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. :–

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