# Maharashtra Board Practice Set 41 Class 7 Maths Solutions Chapter 10 Bank and Simple Interest

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 41 Answers Solutions Chapter 10 Bank and Simple Interest.

## Bank and Simple Interest Class 7 Practice Set 41 Answers Solutions Chapter 10

Question 1.
If the interest on Rs 1700 is Rs 340 for 2 years, the rate of interest must be__.
(A) 12%
(B) 15%
(C) 4%
(D) 10%
Solution:
(D) 10%

Hint:
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$340=\frac{1700 \times R \times 2}{100}$$
∴ R = 10%

Question 2.
If the interest on Rs 3000 is Rs 600 at a certain rate for a certain number of years, what would the interest be on Rs 1500 under the same conditions?
(A) Rs 300
(B) Rs 1000
(C) Rs 700
(D) Rs 500
Solution:
(A) Rs 300

Hint:
The interest on Rs 3000 at certain rate of interest is Rs 600.
Let us suppose the interest on Rs 1500 at the same rate is x.
∴ $$\frac{600}{3000}=\frac{x}{1500}$$
∴ x = Rs 300

Question 3.
Javed deposited Rs 12000 at 9 p.c.p.a in a bank for some years, and withdrew his interest every year. At the end of the period, he had received altogether Rs 17,400. For how many years had he deposited his money?
Solution:
Here, P = Rs 12000, R = 9 p.c.p.a and amount = Rs 17400
Amount = Principal + Interest
∴17400 = 12000 + Interest
∴Interest = 17400 – 12000 = Rs 5400
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
$$5400=\frac{12000 \times 9 \times \mathrm{T}}{100}$$
∴ $$\frac{5400 \times 100}{12000 \times 9}=\mathrm{T}$$
∴ T = 5 years
∴ Javed had deposited the amount for 5 years.

Question 4.
Lataben borrowed some money from a bank at a rate of 10 p.c.p.a interest for $$2\frac { 1 }{ 2 }$$ years to start a cottage industry. If she paid Rs 10250 as total interest, how much money had she borrowed?
Solution:
Here, R = 10 p.c.p.a, T = 2.5 years, I = Rs 10250 ∴ P = Rs 41000
∴ Lataben had borrowed an amount of Rs 41000 from the bank.

Question 5.
Fill in the blanks in the table.

 Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years ii. 6% 4 years Rs 1200 iii. Rs 8000 5% Rs 800 iv. 5% Rs 6000 Rs 18000 v. $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400

Solution:
i. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
= $$\frac{4200 \times 7 \times 3}{100}$$
= Rs 882
Amount = Principal + interest
= 4200 + 882
= Rs 5082

ii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$1200=\frac{\mathrm{P} \times 6 \times 4}{100}$$
∴ $$\frac{1200 \times 100}{6 \times 4}=\mathrm{P}$$
∴ P = Rs 5000
Amount = Principal + interest
= 5000 + 1200
= Rs 6200

iii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$800=\frac{8000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{800 \times 100}{8000 \times 5}=\mathrm{T}$$
∴ T = 2 years
Amount = Principal + interest
= 8000 + 800
= Rs 8800

iv. Amount = Principal + interest
∴ 18000 = Principal + 6000
∴ Principal = Rs 12000
$$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$6000=\frac{12000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{6000 \times 100}{12000 \times 5}=\mathrm{T}$$
∴ T = 10 years

v. R = $$2\frac { 1 }{ 2 }$$ % = 2.5 %
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$2400=\frac{\mathrm{P} \times 2.5 \times 5}{100}$$
∴ $$2400=\frac{P \times 25 \times 5}{100 \times 10}$$
∴ $$\frac{2400 \times 10 \times 100}{25 \times 5}=P$$
∴ P = Rs 19200
Amount = Principal + interest
= 19200 + 2400
= Rs 21600

 Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years Rs 882 Rs 5082 ii. Rs 5000 6% 4 years Rs 1200 Rs 6200 iii. Rs 8000 5% 2 years Rs 800 Rs 8800 iv. Rs 12000 5% 10 years Rs 6000 Rs 18000 v. Rs 19200 $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400 Rs 21600

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 41 Intext Questions and Activities

Question 1.