The DAV Class 7 Maths Solutions and **DAV Class 7 Maths Chapter 5 Worksheet 1** Solutions of Application of Percentage offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 5 WS 1 Solutions

Question 1.

A property dealer charges 2% commission from a buyer or a seller of property. If Mr. Ram Lai paid ₹ 20,000 as a commission for purchasing aflat, what was the value of the flat?

Answer:

Let the value of the flat be ₹ x

Commission = \(\frac{2}{100}\)x

\(\frac{2}{100}\)x = 20000

x = 20000 × \(\frac{100}{2}\)

⇒ x = 10000 × 100 = 1000000

Hence the value of flat = ₹ 1000000

Question 2.

A 3-Star hotel in Delhi charges 10% sales tax on the price of the food taken. Mr. Saxena and his family had food for ₹ 685. Find the total money he had to pay.

Answer:

Price of food = ₹ 685

Sales Tax = \(\frac{10}{100}\) × 685

= ₹ 68.50

∴ Money paid by Mr. Saxena = ₹ (685 + 68.50)

= ₹ 753.50

Question 3.

A painter paints 240 feet of a wall. If this is 48% of the total wall surface, what is the area of the wall in sq. feet?

Answer:

Let the total surface area of the wall be x sq. feet

∴ 48% of x = 240

⇒ \(\frac{48}{100}\) × x

= 240

⇒ x = \(\frac{240 \times 100}{48}\)

= 500 sq.feet

Hence the surface area of the wall = 500 sq. feet.

Question 4.

The salary of a person was increased by 15%. If his salary now is ₹ 9936, what was his initial salary?

Answer:

Let the initial salary be ₹ x

Increase in salary = \(\frac{15}{100}\) × x

= ₹ \(\frac{3}{20}\) x

Now increased salary = x + \(\frac{3}{20}\)x

= ₹ \(\frac{23}{20}\) x

∴ \(\frac{23}{20}\) x = 9936

⇒ x = 9936 × \(\frac{20}{23}\)

= ₹ 8640

Hence the initial salary was ₹ 8640

Question 5.

In a co-education school, 45% of the total students are girls, If there are 440 boys in the school, find the number of girls in the school.

Answer:

Let total number of students be x

Number of girls = \(\frac{45}{100}\) × x

= \(\frac{9}{20}\)x

Number of boys = x – \(\frac{9x}{20}\)

= \(\frac{11 x}{20}\)

\(\frac{11 x}{20}\) = 440

⇒ x = \(\frac{440 \times 20}{11}\)

⇒ x = 800

Number of girls = \(\frac{9}{20}\) × 800 = 360

Hence, the number of girls = 360

Question 6.

A potato seller sells 70% of the total potatoes and still has 150 kg potatoes left with him, Find the weight of potatoes he had originally.

Answer:

Let the weight of potatoes originally be x kg

Weight of potatoes sold = \(\frac{70}{100}\)kg

= \(\frac{7}{10}\)x kg

Left quantities of potatoes = x – \(\frac{7}{10}\)x

= \(\frac{3 x}{10}\)kg

\(\frac{3 x}{10}\) = 150

⇒ x = \(\frac{150 \times 10}{3}\) = 500 kg

Hence the original weight of potatoes = 500 kg

Question 7.

In Rohan’s birthday party, 90% of his friends came and 3 friends did not come. Find the total number of friends Rohan has.

Answer:

Let the total number of friends be x

Number of friends who attended the party = \(\frac{90}{100}\)x = \(\frac{9}{10}\)x

and 3 friends did not attend

\(\frac{9}{10}\)x + 3 = x

x – \(\frac{9x}{10}\) = 3

\(\frac{x}{10}\) = 3

x = 30

Hence the total number of friends = 30

Question 8.

In a quiz competition, Ravi gave answer of 70% of the questions. He failed to give answers of 6 questions. Find the total number of questions asked in the quiz competition.

Answer:

Let the total number of questions be x

∴ Number of answers given = x × \(\frac{70}{100}=\frac{7 x}{10}\)

Number of answers not given = x – \(\frac{7 x}{10}=\frac{3 x}{10}\)

\(\frac{3 x}{10}\) = 6

⇒ x = \(\frac{6 \times 10}{3}\)

x = 20

Hence total number of questions = 20

Question 9.

Out of a class of 45 students, 5 were absent, 30% of the remaining has failed to do homework. Find the number of students who did the homework.

Answer:

Number of students in the class = 45

Number of absent students = 5

Number of present students = 45 – 5 = 40

Number of students who did not do homework = \(\frac{30}{100}\) × 40 = 12

Number of students who did the homework = 40 – 12 = 28

Hence the number of students who did the homework = 28.

### DAV Class 8 Maths Chapter 5 Worksheet 1 Notes

- Percent means per hundred. It is represented by %.

e.g. 19% = \(\frac{19}{100}\) - 4% = \(\frac{4}{100}\)
- 0.6% = \(\frac{0.6}{100}\) etc.
- Percent can be converted into fraction, ratio and decimal.
- If S.P. > C.P., then there is a profit.
- If C.P. > S.P., then there is a loss.
- Profit = S.P. – C.P.
- Loss = C.P. – SP.
- Profit percent or loss percent are always calculated on C.P.
- Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100
- Loss% = \(\frac{\text { Loss }}{\text { C.P. }}\) × 100
- S.P. = C.P. [1 + \(\frac{\mathrm{P} \%}{100}\)]
- S.P. = C.P. [1 – \(\frac{\mathrm{L} \%}{100}\)]

Simple Interest (S.I.) = \(\frac{\text { Principal } \times \text { Rate } \times \text { Time }}{100}\)

Amount = Principal + S.I.

Example 1:

Express the following percents in lowest terms.

(i) 7\(\frac{1}{2}\) %

Answer:

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

= \(\frac{15}{2}\) %

= \(\frac{15}{2} \times \frac{1}{100}\)

= \(\frac{3}{40}\)

(ii) 4%

Answer:

4% = \(\frac{4}{100}\)

= \(\frac{2}{50}\)

(iii) 0.25%

Answer:

0.25% = \(\frac{0.25}{100}\)

= \(\frac{25}{100 \times 100}\)

= \(\frac{1}{400}\)

(iv) 12.5%

Answer:

12.5% = \(\frac{12.5}{100}\)

= \(\frac{125}{100 \times 10}\)

= \(\frac{1}{8}\)

Example 2:

In a class of 40 students, 25% like football, 40% like cricket and rest of the students like tennis to play. Find the number of students who like tennis to play.

Answer:

Number of students who like football = 40 × \(\frac{25}{100}\) = 10

Number of students who like cricket = 40 × \(\frac{40}{100}\) = 16

Number of students who like tennis = 40 – (10 + 16)

= 40 – 26 = 14.

Example 3:

Ramesh purchases sorne goods for ₹ 25000 and spends ₹ 250 as overhead. He sold them at a profit of 12%. Find the selling price of the goods.

Answer:

C.P. = ₹ 25000

Overhead = ₹ 250

Total C.P. = ₹ 25000 + ₹ 250 = ₹ 25250

Profit = \(\frac{12}{100}\) × 25250

= ₹ 3030

Example 4:

Find the simple interest on ₹ 1500 for 3 years at 5% per annum. Also, find the amount.

Solution:

Here P = ₹ 1500,

T = 3 years,

R = 5% p.a.

S.I. = \(\frac{P \times R \times T}{100}\)

S.I. = \(\frac{1500 \times 5 \times 3}{100}\)

= ₹ 225

A = P + S.I.

= ₹ 1500 + ₹ 225 = ₹ 1725.