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.

Answer:

(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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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 ?

Answer:

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.

Answer:

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

Answer:

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.?

Answer:

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.

Answer:

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.

Answer:

(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.