## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.1

Question 1.

Express the following percentages as fractions :

(i) 356%

(ii) \(2 \frac{1}{2}\)%

(iii) \(16 \frac{2}{2}\)%

Solution:

Question 2.

Express the following fractions as percentages :

Solution:

Question 3.

Express the following fractions as decimals. Then express the decimals as percentages :

Solution:

Question 4.

Express the following fractions as decimals correct to four decimal places. Then express the decimals as percentages:

Solution:

Question 5.

Express the following ratios as percentages:

(i) 17 : 20

(ii) 13 : 18

(iii) 93 : 80

Solution:

Question 6.

Express the following percentages as decimals:

(i) 20%

(ii) 2%

(iii) \(3 \frac{1}{4}\)%

Solution:

Question 7.

Find the value of:

(i) 27% of ₹50

(ii) \(6 \frac{1}{4}\)% of 25 kg

Solution:

Question 8.

What per cent is :

(i) 300 g of 2 kg

(ii) ₹7·50 of ₹6

Solution:

(i) Required percentage

(ii) Required percentage

Question 9.

What percent of:

(i) 50 kg is 65 kg

(ii) ₹9 is ₹4

Solution:

(i) Let x% of 50 kg is 65 kg

∴ x% of 50 kg = 65 kg

Hence 130% of 50 kg is 65 kg

(ii) Let x% of ₹9 is ₹4

∴ x% of ₹9 = ₹4

Hence \(44 \frac{4}{9}\) % of ₹9 is ₹4.

Question 10.

(i) If \(16 \frac{2}{3}\)% of a number is 25, find the number.

(ii) If 13·25% of a number is 159, find the number.

Solution:

(i) Let the number is x

Hence the number is 150

(ii) Let the number is x

∴ 13·25% of x = 159

Hence the number is 1200.

Question 11.

(i) Increase the number 60 by 30%

(ii) Decrease the number 750 by 10%

Solution:

Question 12.

(i) What number when increased by 15% becomes 299?

(ii) On decreasing the number by 18%, it becomes 697. Find the number.

Solution:

(i) Let the original number = x

Then, New number number = \(\left(1+\frac{15}{100}\right)\) of original number

Hence the original number = 260

(ii) Let the original number = x

Then, new number = \(\left(1-\frac{18}{100}\right)\) of original number

Hence the original number = 850

Question 13.

Mr Khanna spent 83% of his salary and saved ₹1870. Calculate his monthly salary.

Solution:

Mr. Khanna spent of his salary = 83%

∴ Saving = 100 – 83 = 17%

Now, 17% of his salary = ₹ 1870

∴ His salary = ₹ \(\frac{1870 \times 100}{17}\) = ₹11000

Question 14.

In school, 38% of the students are girls. If the number of boys is 1023, find the total strength of the school.

Solution:

Number of girls in school = 38%

∴ Number of boys in school = (100 – 38)% = 62%

Let the total strength of the school = x

∴ 62% of x = 1023

⇒ x = 1650

Hence the totai strength of school = 1650

Question 15.

The price of an article increases from ₹960 to ₹1080. Find the percentage increase in the price.

Solution:

Increase price of an article = 1080 – 960 = 120

% increase price of an article

Question 16.

In a straight contest, the loser polled 42% votes and lost by 14400 votes. Find the total number of votes polled. If the total number of eligible voters was 1 lakh, find what percentage of voters did not vote.

Solution:

Since the losing candidate secured 42% of the votes polled

∴ Winning candidate secures votes

= (100 – 42)% of the votes polled

= 58% of the votes polled

Difference of votes = 58% – 42%

= 16% of the votes polled

we are given:

16% of the votes polled = 14400

⇒ \(\frac{16 \%}{100}\) of the votes polled = 14400

⇒ Votes polled = 14400 × \(\frac{100}{16}\)

= 900 × 100 = 90000

But total numbei of eligible voters

= 100000 (Given)

Number of voter did not vote

= 100000 – 90000 = 10000

% of number of voter did not vote

Question 17.

Out of 8000 candidates, 60% were boys. If 80% of the boys and 90% of the girls passed the exam, find the number of candidates who failed.

Solution:

Total number of candidates = 8000

Number of boys = 60% of 8000

= \(\frac{60}{100}\) × 8000 = 60 × 80 = 4800

Number of girls = 8000 – 4800 = 3200

Number of passed boys = 80% of (Number of boys)

= \(\frac{80}{100}\) × 4800 = 80 × 48 = 3840

Number of passed girls = 90% of (Number of girls)

= \(\frac{90}{100}\) × 3200 =90 × 32 = 2880

∴ Number of passed candidates

= 3840 + 2880 = 6720

Hence, Number of failed candidates

= 8000 – 6720 = 1280

Question 18.

In an exam, \(\frac{1}{4}\) of the students failed both in English and Maths, 35% of students failed in Maths and 30% failed in English.

(i) Find the percentage of students who failed in any of the subjects.

(ii) Find the percentage of students who passed in both subjects.

(iii) If the number of students who failed only in English was 25, find the total number of students.

Solution:

Let the total number of students = x

Number of students who are failed both in English and Maths = \(\frac{1}{4}\) of x = \(\frac{x}{4}\)

Number of those students who are failed in Maths = 35% of x

= \(\frac{35}{100} \text { of } x=\frac{7}{20} \times x=\frac{7 x}{20}\)

Number of those students who are failed in English

= 30% of x = \(\frac{30}{100} \times x=\frac{3}{10} \times x=\frac{3 x}{10}\)

(i) Number of those students who are failed in any of the subject = \(\left(\frac{7 x}{20}+\frac{3 x}{10}\right)-\frac{x}{4}\)

= \(\frac{7 x+6 x}{20}-\frac{x}{4}=\frac{13 x-5 x}{20}=\frac{8 x}{20}\)

% Number of those students who are failed in any of the subject

= \(\frac{\frac{8 x}{20}}{x} \times 100 \%=\frac{8 x}{20} \times \frac{1}{x} \times 100 \%\)

= 8 × 1 × 5% = 40%

(ii) % of Number of those students who passed in both the subjects = (100 – 40)% = 60%

(iii) Given the number of students who failed only in English = 25

∴ \(\frac{3 x}{10}-\frac{x}{4}=25\)

{∴ Number of students who failed only in English =\(\frac{3 x}{10}-\frac{x}{4}\)}

⇒ \(\frac{6 x-5 x}{20}\) = 25 ⇒ x = 25 × 20 ⇒ x = 500

Hence total number of students = 500

Question 19.

On increasing the price of an article by 16%, it becomes ₹1479. What was its original price?

Solution:

Let the original price of an article = ₹ x

Hence the original price of an article = ₹ 1275

Question 20.

Pratibha reduced her weight by 15%. If now she weighs 59.5 kg, what was her earlier weight?

Solution:

Pratibha reduced her weight = 15%

and her present weight = 59.5 kg

Let her original weight = 100

∴ Reduced weight = 100 – 15 = 85%

∴ 85% of her original weight = 59.5 kg

∴ Her original weight = \(\frac{59.5 \times 100}{85}\)kg = 0.7 × 100 = 70 kg

Question 21.

In a sale, a shop reduces all its prices by 15%. Calculate:

(i) the cost of an article which was originally priced at ₹40.

(ii) the original price of an article which was sold for ₹20.40.

Solution:

Rate of reduction = 15%

(i) Original price of an article = ₹40

Rate of reduction = 15%

∴ Reduction = \(\frac{40 \times 15}{100}\) = ₹6

Sale price = ₹40 – 6 = ₹34

(ii) Sale price = ₹20.40

Rate of reduction = 15%

Question 22.

Increase the price of ₹200 by 10% and then decrease the new price by 10%. Is the final price same as the original one ?

Solution:

Rate of increase = 10%

Rate of decrease = 10%

Price of an article = ₹200

Increased price = ₹200 × \(\frac{100+10}{100}\)

= ₹200 × \(\frac{110}{100}\) = ₹220

Now decreased price

No. final price is not the same as the original price.

Question 23.

Chandani purchased some parrots. 20% flew away and 5% died. Of the remaining, 45% were sold. Now 33 parrots remain. How many parrots had Chandani purchased?

Solution:

Let Chandani purchased x parrots.

Then the number of parrots flew away = 20% of x

Hence Chandani purchased 80 parrots.

Question 24.

A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks, gets 18 more marks than the minimum pass marks. Find the maximum marks and the percentage of pass marks.

Solution:

Let the maximum marks = x

Then, the first candidate secured marks = 36% of x

Hence maximum marks = 600

First candidate secured marks = \(\frac{36}{100}\) × 600 = 36 × 6 = 216

Then qualifying marks = 216 + 24 = 240

Percentage of qualifying marks