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.