The DAV Books Solutions Class 5 Maths and **DAV Class 5 Maths Chapter 13 Brain Teasers** Solutions of Simple Interest offer comprehensive answers to textbook questions.

## DAV Class 5 Maths Ch 13 Brain Teasers Solutions

Question 1.

Tick (✓) the correct answer.

(а) The extra money paid by a bank for every ₹ 100 after one year is called-

(i) Interest

(ii) Rate of interest

(iii) Amount

(iv) Principal

Solution:

(ii) Rate of Interest

(b) Mr Gupta borrowed ₹ 1,75,000 from a bank to buy a car. This amount is called-

(i) Interest

(ii) Rate of interest

(iii) Loan

(iv) Time

Solution:

(iii) Loan

(c) Simple interest for ₹ 1,000 at 10% for 9 months is-

(i) ₹ 75

(ii) ₹ 900

(iii) ₹ 90

(iv) ₹ 1,075

Solution:

(i) ₹ 75

SI = \(\frac{1000 \times 10 \times 9}{100 \times 12}\)

= \(\frac{900}{12}\)

= ₹ 75

(d) If amount = ₹ 1,645.50, simple interest = ₹ 95, principal is-

(i) ₹ 1,740.50

(ii) ₹ 1,550

(iii) ₹ 1,740

(iv) ₹ 1,550.50

Solution:

(iv) ₹ 1,550.50

A = SI + P

1,645.50 = 95 + P

1,645.50 – 95 = P

P = ₹ 1,550.50

(e) If loan = ₹ 1,75,000, Rate of interest = 4%, Time = 2 years. Then amount to be repaid is equal to-

(i) ₹ 1,89,000

(ii) ₹ 1,76,400

(iii) ₹ 1,98,000

(iv) ₹ 1,80,000

Solution:

(i) ₹ 1,89,000

SI = \(\frac{1,75,000 \times 4 \times 2}{100}\) = ₹ 14,000

A = P + I

= 1,75,000 + 14,000

= ₹ 1,89,000

Question 2.

Name the three factors that determine simple interest.

Solution:

- Principal
- Rate of Interest
- Time

Question 3.

Find the amount if the principal is ₹ 625 and the interest is ₹ 55.

Solution:

A = P + I

= ₹ 625 + ₹ 55

= ₹ 680

Question 4.

Calculate the simple interest on ₹ 1,200 at 3\(\frac{1}{2}\)% interest per annum for six months.

Solution:

P = ₹ 1,200

R = 3\(\frac{1}{2}\)% = \(\frac{7}{2}\)%

T = six months = \(\frac{6}{12}\) = \(\frac{1}{2}\) year

SI = PRT

= \(\frac{1,200 \times 7 \times 1}{100 \times 2 \times 2}\)

= ₹ 21

Question 5.

Sohan deposited ₹ 6,000 in a bank at 7\(\frac{1}{2}\)% per annum for three years. Mohan deposited ₹ 6,000 in another bank at 11% per annum for 2\(\frac{1}{2}\) years. Who will get more simple interest?

Solution:

P = ₹ 6,000

R = 7\(\frac{1}{2}\)% = \(\frac{15}{2}\)% = \(\frac{15}{200}\) per annum

T = 3 years

SI got by Sohan = 6,000 × \(\frac{15}{200}\) × 3 = ₹ 1350

P = ₹ 6,000

R = 11% = \(\frac{11}{100}\) per annum

T = \(\frac{5}{2}\) years

SI got by Mohan = \(\frac{6,000 \times 11 \times 5}{100 \times 2}\) = ₹ 1650

Mohan will get more simple interest.

Question 6.

A farmer borrowed ₹ 2,400 at 12% interest per annum. At the end of 3 years, he repaid ₹ 1,200 and a cow for the balance amount. Find the cost of the cow.

Solution:

P = ₹ 2,400

R = 12%

T = 3 years

SI = 2,400 × \(\frac{12}{100}\) × 3

= ₹ 864

A = P + SI

= ₹ 2,400 + ₹ 864

= ₹ 3,264

Cost of cow = ₹ 3,264 – ₹ 1200 = ₹ 2,064

Cost of cow is ₹ 2,064.

Question 7.

Mr Kannan had deposited ₹ 2,500 in a bank at 12% per annum. After 3\(\frac{1}{2}\) years, he withdraws this amount. Out of this money, he want to buy a dressing table costing ₹ 4000. How much more money does he need?

Solution:

P = ₹ 2,500

R = 12% = latex]\frac{12}{100}[/latex] per annum

T = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years

SI = 2,500 × \(\frac{12}{100} \times \frac{7}{2}\) = ₹ 1,050

A = P + I

= ₹ 2,500 + ₹ 1,050

= ₹ 3,550

Cost of table = ₹ 4,000

Amount he needs = ₹ 4,000 – ₹ 3,550 = ₹ 450

He needs ₹ 450 more.

Additional Questions

Question 1.

Calculate simple interest for the following:

(a) P = ₹ 4,000; R = 3% per annum; T = 2 years

Solution:

P = ₹ 4,000

R = 3% per annum = \(\frac{3}{100}\) per annum

T = 2 years

Simple Interest = PRT

= 4,000 × \(\frac{3}{100}\) × 2

= ₹ 240

(b) P = ₹ 1,200; R = 2\(\frac{1}{2}\)% per annum; T = 1\(\frac{1}{2}\) years

Solution:

P = ₹ 1,200

R = 2\(\frac{1}{2}\)% = \(\frac{5}{2}\)% per annum

T = 1\(\frac{1}{2}\) years = \(\frac{3}{2}\) years

Simple Interest = PRT

= 1200 × \(\frac{5}{200} \times \frac{3}{2}\)

= ₹ 45

(c) P = ₹ 1,040; R = 12% per annum; T = 9 months

Solution:

P = ₹ 1040

R = 12% per annum

T = \(\frac{9}{12}\) year

Simple Interest = PRT

= 1,040 × \(\frac{12}{100} \times \frac{9}{12}\)

= \(\frac{468}{5}\)

= ₹ 93.6

Question 2.

A man deposited ₹ 12,000 in a bank at an interest of 9% per annum. How much simple interest will he get at the end of 24 months?

Solution:

Principal = ₹ 12,000

R = 9% per annum = \(\frac{9}{100}\) per annum

T = 24 months = 2 years

SI = PRT

= 12000 × \(\frac{9}{100}\) × 2

= ₹ 2160

He will get a simple interest of ₹ 2160 at the end of 24 months.

Question 3.

Calculate the following:

(a) P = ₹ 1,200; SI = ?; Amount = ₹ 1,450

Solution:

A = P + SI

₹ 1,450 = ₹ 1,200 + SI

SI = ₹ 1,450 – ₹ 1,200 = ₹ 250

(b) P = ?; SI = ₹ 134; Amount = ₹ 3100

Solution:

A = P + SI

P = A – SI

= ₹ 3,100 – ₹ 134

= ₹ 2,966

(c) P = ₹ 1,50,000; SI = ?; Amount = ₹ 1,75,000

Solution:

A = P + SI

SI = A – P

= ₹ 1,75,000 – ₹ 1,50,000

= ₹ 25,000

Question 4.

Find the amount.

(a) P = ₹ 21,000; Interest = ₹ 350

Solution:

A = P + I

= ₹ 21,000 + ₹ 350

= ₹ 21,350

(b) P = ₹ 1,750; Interest = ₹ 250

Solution:

A = P + I

= ₹ 1,750 + ₹ 250

= ₹ 2,000

(c) P = ₹ 7,500; Interest = ₹ 200.50

Solution:

A = P + I

= ₹ 7,500 + ₹ 200.50

= ₹ 7,700.50

Question 5.

Maya deposited ₹ 3,500 in a bank at 14% interest per annum. She withdraws her money after 9 months. Out of this amount, she purchases a centre table worth ₹ 3,800. How much money is left with her?

Solution:

Principal = ₹ 3,500

R = 14% per annum

T = 9 months = \(\frac{9}{12}\) years

SI = PRT

= 3,500 × \(\frac{14}{100} \times \frac{9}{12}\)

= ₹ 367.5

A = P + I

= ₹ 3,500 + ₹ 367.5

= ₹ 3,867.5

Cost of centre table = ₹ 3,800

Money left with her = ₹ 3,867.5 – ₹ 3,800 = ₹ 67.5

Question 6.

Tick (✓) the correct answer.

(a) Mr Sharma borrowed ₹ 20,00,000 from the bank to buy a small truck. This amount is called

(i) Principal

(ii) Rate of Interest

(iii) Loan

(iv) Time

Solution:

(iii) Loan

(b) The extra money paid by the bank is called

(i) Principal

(ii) Rate of Interest

(iii) Loan

(iv) Interest

Solution:

(iv) Interest

(c) Extra money charged by the bank for every hundred rupees deposit is called

(i) Principal

(ii) Amount

(iii) Interest

(iv) Loan

Solution:

(iii) Interest

(d) The period of time for which money is kept in a bank is called

(i) Amount

(ii) Rate of Interest

(iii) Principal

(iv) Time period

Solution:

(iv) Time period