# DAV Class 7 Maths Chapter 5 Worksheet 5 Solutions

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

## DAV Class 7 Maths Ch 5 WS 5 Solutions

Question 1.
Find the unknown quantity in each of the following.

(i) Here P = ?, R = 12%, T = 2$$\frac{1}{2}$$ = $$\frac{5}{2}$$ years, S.I = ₹ 1200, A = ?
S.I = $$\frac{P \times R \times T}{100}$$
⇒ 1200 = $$\frac{P \times 12 \times 5}{100 \times 2}$$
⇒ P = $$\frac{1200 \times 100 \times 2}{12 \times 5}$$
⇒ P = ₹ 4000
A = P + S.I.
= ₹ 4000 + ₹ 1200 = ₹ 5200
Hence P = ₹ 4000, A = ₹ 5200

(ii) Here P = ?, R = 3.5%, T = 2 years, A = ₹ 535, S.I. = ?
S.I = $$\frac{P \times R \times T}{100}$$
⇒ S.I = $$\frac{P \times 3.5 \times 2}{100}$$
⇒ S.I = $$\frac{7 P}{100}$$

A = P + S.I
⇒ 535 = P + $$\frac{7 P}{100}$$
⇒ 535 = $$\frac{107 \mathrm{P}}{100}$$

P = $$\frac{535 \times 100}{107}$$ = 500
S.I. = A – P
⇒ S.I = ₹ 535 – ₹ 500
= ₹ 35
Hence P = ₹ 500, S.I = ₹ 35

(iii) Here P = ?, R = 4%, T = 3 years, S.I = ₹ 120, A = ?
S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
120 = $$\frac{P \times 4 \times 3}{100}$$
P = $$\frac{120 \times 100}{4 \times 3}$$ = ₹ 1000
A = P + S.I = ₹ 1000 + ₹ 120 = ₹ 1120
Hence P = ₹ 1000 and A = ₹ 1120

(iv) Here P = ₹ 450, R = ?, T = 3$$\frac{1}{2}=\frac{7}{2}$$years, S.I = ₹ 189, A = ?
S.I = $$\frac{P \times R \times T}{100}$$
⇒ 189 = $$\frac{450 \times R \times 7}{100 \times 2}$$
⇒ R = $$\frac{189 \times 100 \times 2}{450 \times 7}$$ = 12%
A = P + S.I = ₹ 450 + ₹ 189
= ₹ 639
Hence R = 12% and A = ₹ 639

(v) Here P = 850, R = 6%, T = ?, S.I = ₹ 178.50, A = ?
S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
⇒ 178.50 = $$\frac{850 \times 6 \times T}{100}$$
⇒ T = $$\frac{178.50 \times 100}{580 \times 6}$$
⇒ T = $$\frac{7}{2}$$ = 3 $$\frac{1}{2}$$ years
A = P + S.I
= ₹ 850 + ₹ 178.50
= ₹ 1028.50
Hence T = 3$$\frac{1}{2}$$ years and A = ₹ 1028.50

(vi) Here P = ₹ 5400, R = ?, T = 3 years, S.I = ? A = ₹ 6210
S.I = A – P
S.I = 6210 – 5400
= 810

S.I = $$\frac{P \times R \times T}{100}$$
⇒ 810 = $$\frac{5400 \times \mathrm{R} \times 3}{100}$$
⇒ R = $$\frac{810 \times 100}{5400 \times 3}$$
⇒R = 4%
Hence R = 4% and S.I = ₹ 810

Question 2.
Find the sum of money that amounts to ₹ 5850 in 6 years at 5% P.a.
Here, A = ₹ 5850, T = 6 years, R = 5% P.a., P = ?

Hence the sum = ₹ 4500

Question 3.
What sum will earn an interest oft 480 in 2y years at the rate of 3% p.a. simple interest?
Here, S.I. = ₹ 480, T = 2$$\frac{1}{2}=\frac{5}{2}$$ years, R = 3%, P = ?
S.I = $$\frac{P \times R \times T}{100}$$
480 = $$\frac{P \times 3 \times 5}{100 \times 2}$$
⇒ P = $$\frac{480 \times 100 \times 2}{3 \times 5}$$
⇒ P = ₹ 6400
Hence the sum = ₹ 6400

Question 4.
Mr. Mehta borrowed a sum of money at 8% P.a. If he paid ₹ 640 as interest after 5$$\frac{1}{3}$$ years (5 years 4 months). Find the sum borrowed by him.
Here R = 8%, S.I. = ₹ 640, T = 5$$\frac{1}{3}=\frac{16}{3}$$ years, P = ?
S.I = $$\frac{P \times R \times T}{100}$$
640 = $$\frac{\mathrm{P} \times 8 \times 16}{100 \times 3}$$
⇒ P = $$\frac{640 \times 100 \times 3}{8 \times 16}$$
⇒ P = ₹ 1500
Hence the sum of money borrowed was ₹ 1500

Question 5.
Simple interest on a sum of money is $$\frac{9}{16}$$ of the sum. If the rate is 4$$\frac{1}{2}$$% p.a., find the time.
Let the principal be ₹ x
S.I = ₹ $$\frac{9}{16}$$ x
R = 4$$\frac{1}{2}$$% = $$\frac{9}{2}$$% P.a., T = ?
S.I = $$\frac{P \times R \times T}{100}$$
$$\frac{9}{16} x=\frac{x \times 9 \times \mathrm{T}}{100 \times 2}$$
T = $$\frac{9}{16} x \times \frac{100 \times 2}{x \times 9}=\frac{25}{2}$$ = 12$$\frac{1}{2}$$ years

Question 6.
At what rate percent p.a. will a sum of money double itself in 8 years ?
Let the principal be ₹ x
Amount = ₹ 2x
S.I. = A – P
= ₹ 2x – ₹ x
= ₹ x

T = 8 years, R = ?
S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
x = $$\frac{x \times \mathrm{R} \times 8}{100}$$
R = $$\frac{100 \times x}{x \times 8}=\frac{25}{2}$$% or 12$$\frac{1}{2}$$% P.a

Question 7.
Rahul borrowed ₹ 50,000 from a bank on 1st March 2002 and paid ₹ 53150 on 6th October 2002. Find the rate of interest charged by the bank.
T = March = 30 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Aug. = 31 days
Sept. = 30 days
Oct. = 6 days
Total = 219 days
= $$\frac{219}{365}=\frac{3}{5}$$ years
S.I. = A – P
= ₹ 53150 – ₹ 50,000
= ₹ 3150

S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$

3150 = $$\frac{50,000 \times \mathrm{R} \times 3}{100 \times 5}$$

R = $$\frac{3150 \times 100 \times 5}{50,000 \times 3}=\frac{21}{2}$$ = 10$$\frac{1}{2}$$% P.a.

Question 8.
In what time will a sum of money double itself at 15% p.a. 1
Let the sum be ₹ x
∴ Amount = ₹ 2x
S.I. = A – P = ₹ 2x – ₹ x = ₹ x
S.I = $$\frac{P \times R \times T}{100}$$
⇒ x = $$\frac{x \times 15 \times T}{100}$$
⇒ T = $$\frac{100 \times x}{15 \times x}=\frac{20}{3}$$ = 6$$\frac{2}{3}$$ years

Question 9.
In what time will the simple interest on ₹ 400 at 10% P.a. be the same as the simple interest on ₹ 1000 for 4 years at 4% P.a.?
S.I on ₹ 400 = $$\frac{P \times R \times T}{100}=\frac{400 \times 10 \times T}{100}$$ = ₹ 40 T
S.I on ₹ 1000 = $$\frac{1000 \times 4 \times 4}{100}$$ = ₹ 160
40 T = 160
∴ T = $$\frac{160}{40}$$ = 4 years

Question 10.
Mr. Jane donates ₹ 1 lakh to a school and the interest on it is to be used for awarding five scholarships of equal value. If the value of each scholarship is ₹ 1,500, find the rate of interest.
Here, Principal = ₹ 1,00,000
Value of one scholarship = ₹ 1500
∴ Value of five scholarships of equal value = ₹ 5 × 1500 = ₹ 7500 = S.I.
Time = 1 year
Now, S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
⇒ 7500 = $$\frac{100000 \times \mathrm{R} \times 1}{100}$$
⇒ R = $$\frac{7500 \times 100}{100000}$$
⇒ R = $$\frac{7500}{1000}=\frac{75}{10}$$ = 7.5%
So, rate of interest is 7.5% per annum.

### DAV Class 7 Maths Chapter 5 Value Based Questions

Question 1.
In a survey it was found that out of 125 people in a park, 12% jog, 16% do yoga and rest prefer to walk.
(i) Find the number of people who prefer to walk.
(ii) Discuss the importance of exercises like jogging, yoga, walking, etc.
(i) Total number of people in park = 125
Number of people who prefer to walk = [100 – (12 + 16)]% = (100 -28)%
= 72% of the total number of people
= $$\frac{72}{100}$$ × 125 = $$\frac{72}{4}$$ × 5
= 18 × 5 = 90
So, 90 people prefer to walk.

(ii) Importance of Exercises: All these exercises keep us active, fit, physically and mentally alert.

Question 2.
The library teacher of a school keeps record of students of each class about their reading habits. The results of Class-Vll students for the month of April are as follows :

 Books read Percentage 0 14 1-3 28 4-6 26 More than 6 32

(i) If total number of students ofClass-VII is 500, how many students had read books in the month of April ?
(ii) What is the importance of reading books?
Solution:
(i) Total number of students in class = 500
∴ Number of students who read books = (28 + 26 + 32)% = 86% of the total students
= $$\frac{86}{100}$$ × 500 = 86 × 5 = 430
So, 430 students read books in the month of April.

(ii) Importance of Reading Books: Reading books increases vocabulary, improves thought process etc.