Percentage Class 7 RS Aggarwal Exe-9B Goyal Brothers ICSE Foundation Maths Solutions. We provide step by step Solutions of lesson/ Chapter-9 Percentage for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-7 Mathematics.

**Percentage Class 7 RS Aggarwal Exe-9B Goyal Brothers ICSE Maths Solutions**

Board | ICSE |

Publications | Goyal brothers Prakashan |

Subject | Maths |

Class | 7th |

Chapter-9 | Percentage |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-9B |

Academic Session | 2024 – 2025 |

**Percentage Class 7 **Word Problems Practice Questions with Solutions

RS Aggarwal Exe-9B Goyal Brothers ICSE Foundation Maths Solutions

**Question-1: If 16% of a number is 60, find the number.**

Answer-Let the number be x

(16/100) * x = 60

x = (60*100) / 16

x = 375.

**Question-2: If 13(1/3)% of a number is 90, find the number.**

Answer-Let the number be x

40/3 % of x = 90

Hence 40 / (3 * 100)] * x = 90

40x / 300 = 90

So 40x = 90 * 300

40x =27000

x = 675.

**Question-3: If 0.6% of a number is 15, find the number.**

Answer-Let the number be x

0.6% of x = 15

[6/(10 * 100)] * x = 15

6x = 15 * 1000

6x = 15000

x = 2500.

**Question-4: If 3/4% of a number is 9, find the number.**

Answer- Let the number be x

0.75% of x = 9

[75 / (100 * 100)] * x = 9

75x = 9 * 10000

75x = 90000

x = 1200.

**Question-5: There are 42 boys and 18 girls in a class. What is the percentage of boys in the class?**

Answer- Total no. of boys = 42

Total no. of girls = 18

total no. of students = 42 + 18 = 60

percentage of boys = (total no. of boys / total no. of students) * 100 = (42 / 60) * 100 = 70%

Hence, percentage of boys in class is 70%.

**Question-6: A team won 6 hockey matches and lost 9 matches. What per cent of the matches did the team win?**

Answer- no. of matches won by hockey team = 6

no. of matches lost by hockey team = 9

total no. of matches played by hockey team = 9 + 6 =15

required percentage = (no. of matches won / no.of matches played) * 100 = (6 / 15) * 100 = 40%

Hence, 40% is the percentage of winning match.

**Question-7: A batsman scored 75 runs which included 3 boundaries and 8 sixes. What per cent of his total score did he make by running between the wickets?**

Answer- Total score of batsman = 75 runs

no. of boundaries = 3

runs from boundaries = 3 * 4 = 12

no. of sixes = 8

runs from sixes = 6 * 8 = 48

Total runs from boundaries and sixes = 48 + 12 = 60 runs

Total runs by the batsman = (total score) – (score from boundaries and sixes) = 75 – 60 = 15

percentage of his score made by runnning between wickets = [(15 / 75) * 100] % = 20%

Hence, 20% is his total score made by running between wickets.

**Question-8: 12% of a sum of money is Rs 42. What is 20% of the same sum?**

Answer- Let the sum of money be x

12% of x is 42

(12/100) * x = 42

12x = 4200

x= 350

The 20% of same sum is

= 20% of 350, = (20/100)*350

= Rs 70.

**Question-9: 6¼% of a weight is 0.25 kg. What is 45% of this weight?**

Answer- Let the weight be x

(25/4) % of x = 0.25kg

(25x / 4) ×100 = 0.25

25x / 4 = 0.25 * 10025 x = 25 * 4

x = 100 / 25 = 4 kg

45% of 4kg = (45 / 100) * 4 =1.8kg

Hence, the weight of 45% is 1.8kg

**Question-10: In a class of 60 pupils, 15% remained absent on a rainy day. How many pupils were present in the class on that day?**

Answer- total no. of pupils in class = 60

Number of pupils absent on a particular day = 15% of 60 = (15/100) * 60 = 9

Therefore, the number of pupils present =60−9 = 51 pupils

Hence, 51 pupils were present in the class on that day.

**Question-11: The monthly income of Mr. Amit Goel is Rs 30400. He saves 12.5% of his income and the rest he spends. How much does he spend each month?**

Answer- Money earned = Rs 30400

Money saved = 12.5% of his income = (12.5/100) * 30400 = 3800

Money Spent = 30400 – 3800 = Rs 26600

Hence, he spent Rs 26600 each month.

**Question-12: An ore contains 15% iron. How much ore will be required to get 18 kg of iron?**

Answer- 100 kg of an ore contains 15% of iron

Let mass required to obtain 18 kg of iron be x

Then, (15/100) * x = 18 kg

x = (18 * 100)/15 = 120 kg.

Hence, mass required to obtain 18 kg of iron is 120 kg.

**Question-13: A property dealer charges a commission of 2% on the first Rs 25000 and 1.5% on the remainder. What commission does he charge for selling a plot of land for Rs 130000?**

Answer- commission on Rs 25000 = 2% of Rs 25000 = (2/100) * 25000 = Rs 500

commission on (Rs 130000 – Rs 25000) = 1.5% of Rs 105000 = (1.5/100) * 105000 = Rs 1575

total commission = Rs 500 + Rs 1575 = Rs 2075

Hence, he charges Rs 2075 comission for selling a plot of land for Rs 130000.

**Question-14: In an examination, the maximum marks are 850. Rohit gets 34% marks and fails by 17 marks. Find (i) the passing marks and (ii) the minimum percentage for passing the examination.**

Answer-14 (i)the passing marks = (34% of 850) + 17 = 289 + 17 = 306.

(ii) the minimum percentage = [(306/850) * 100]% = 36%.

**Question-15: A student secure 90%, 60% and 54 % marks in three test papers with maximum marks 100. 150 and 200 respectively. Find his aggregate percentage.**

Answer-: Let the total scored = (90% of 100) + (60% of 150 ) + (54% of 200)

= [(90/100) * 100] + [(60/100) * 150] + [(54/100) * 200] = 288

Total maximum marks = (100+150+200) = 450

Aggregate percentage = [(288/450) * 100]% = 64%

Hence, his aggregate percentage = 64%.

**Question-16: In an examination, 5% of the applicants were found ineligible 85 % of the eligible candidates belonged to the general category. If 4275 candidates belonged to other categories, then how many candidates applied for the examination?**

Answer- Let the number of candidates applied for the examination be x

Number of ineligible candidates = 95% of (100 – 5%)

Eligible candidates of other categories = 15% of (95% of x) = (15/100) * [(95/100) * x] = (57/400) * x

Given that Candidates belonged to other categories = 4275.

(57/400) * x = 4275

x = (4275 * 400) / 57 = 30000

Hence, 30000 candidates applied for the examination.

**Question-17: Gun-powder contains 75% nitre, 10% sulphur and the rest of it is charcoal. Find the quantity of charcoal in 8 kg of gun-powder.**

Answer- In a gunpowder,

Nitre = 75%

Sulphur = 10%

Then charcoal = 100- (75+10)= 15%

So the amount of charcoal in 8 kg gunpowder = 15% of 8 = (15/100) * 8 = (3/5)* 2 = 6/5 = 1.2 kg.

Hence, the quantity of charcoal in 8 kg of gun-powder is 1.2 kg.

**Question-18: An alloy consists of 13 parts of copper, 7 parts of zinc and 5 parts of nickel. Find the percentage of copper in the alloy.**

Answer- copper = 13 parts , zinc = 7 parts , nickel = 5 parts

total alloy = 25 parts

now, percentage of copper in alloy = (13/25) * 100 = 52%

Hence, percentage of copper is 52%.

**Question-19: Two candidates A and B contested an election. The total votes polled were 9650. If A got 54% of the votes, find the number of votes received by each.**

Answer- no. of votes received by A = (54/100) * 9650 = 5211

So, the remaining votes are gained by B = (total – votes gained by A)

= 9650 – 5211 = 4439

Hence, votes gained by A and B are 5211 and 4439 respectively.

**Question-20: At an election between two candidates, 68 votes were declared invalid. The winning candidate secures 52% of the valid votes and wins by 354 votes. Find the total number of votes polled.**

Answer-20 Percentage of votes of winning candidate = 52%

Percentage of votes of losing candidate = 100-52 = 48%

It is given that winner wins by 354 votes

so, 52% – 48% = 4%

Percentage of votes the winner won = 4%

4% = 354

1% = 354/4

100% = 354 * 100/4 = 8850

Total number of votes polled = 8850 + 68 = 8918

Hence, 8918 votes were pulled.

**Question-21: The salary of Mrs Sarita is ₹ 32000 per month. 10% of it is deducted by the employer as provident fund. Of the remaining money, she spends 20% on house rent, 46% on food, 14% on the education of children and 10% on other expenses. Rest she saves. Find**

(i) how much is credited each month to her Provident Fund Account.

(ii) how much is spent on food.

(iii) how much is paid as house rent.

iv) how much is spent on the education of children.

(v) how much does she save every month.

Answer-21 Given: Sarita salary per month = Rs 32000

1) Provident fund:

provident fund = 32000 * (10/100) = Rs 3200.

Remaining salary:

32000 – 3200 = 28800

2) On food:

28,800 * (46/100) = Rs 13248.

3) On house rent:

28,800 * (20/100) = Rs 5760.

4) On education:

28,800 * (14/100) = Rs 4032.

Given: 10% expenses:

28,800 * (10/100) =Rs 2880

5) Savings:

13248 + 5760 + 4032 + 2880 =Rs 25920

= 28,800 – 25920 = Rs 2880.

**— : End of Percentage Class 7 RS Aggarwal Exe-9B Goyal Brothers ICSE Foundation Maths Solutions :–**

**Return to: – ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions**

Thanks

Please share with yours friends if helpful.