Compound Interest Class 8th Mathematics (new) MHB Solution

Compound Interest

Class 8th Mathematics (new) MHB Solution

Class 8th Mathematics (new) MHB Solution

Practice Set 14.1
Question 1.

Find the amount and the compound interest.



Answer:

(a) Principal = 2000/-, Rate = 5% (p.c.p.a), Duration (n) = 2 years



2


A = 2000 (1+0.05)2


A = 2000 (1.05)2


A = 2000 (1.1025)


∴A= 2205/-


∴ C.I = A – P


∴ C.I = 2205 – 2000


C.I. = 205/-


Amount is 2205/- and Compound interest is 205/- .


b. Principal = 5000/-, Rate = 8% (p.c.p.a), Duration (n) = 3 years



3


A = 5000 (1 + 0.08)3


A = 5000 (1.08)3


A = 5000 (1.259712)


∴A= 6298.56/-


∵ C.I. = A - P


∴ C.I. = 6298.56 - 5000


C.I. = 1298.56/-


Amount is 6298.56/- and Compound interest is 1298.56/- .


c. Principal = 4000/-, Rate = 7.5% (p.c.p.a), Duration (n) = 2 years



A = 4000 (1 + 0.075)2


A = 4000 (1.075)2


A = 4000 (1.155625)


∴A= 4622.5/-


∵ C.I. = A - P


∴ C.I. = 4622.5 - 4000


C.I. = 622.5/-


Amount is 4622.5/- and Compound interest is 622.5/- .



Question 2.

Sameerrao has taken a loan of ₹ 12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?


Answer:

Principal = 12500/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years



3


A = 12500 (1+ 0.12)3


A = 12500 (1.12)3


A = 12500 (1.404928)


A = 17561.60/-


Sameerao has to pay an amount of 17561.60/- .



Question 3.

To start a business Shalaka has taken a loan of ₹ 8000 at a rate of p.c.p.a. After two years how much compound interest will she have to pay?


Answer:

Principal = 8000/-, Rate = 10.5% (p.c.p.a), Duration (n) = 2 years



2


A = 8000 (1+0.105)2


A = 8000 (1.105)2


A = 8000 (1.221025)


∴ A = 9768.2/-


∵ C.I. = A - P


∴ C.I. = 9768.2 - 8000


C.I. = 1768.2/-


∴ Shalaka has to pay a compound interest of 1768.2/- .




Practice Set 14.2
Question 1.

On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.


Answer:

Present number of workers = 320 workers, Rate (increase) = 25% (p.c.p.a), Duration (n) = 2 years


2


A = 320 (1+0.25)2


A = 320 (1.25)2


A = 320 (1.5625)


∴A= 500/-


∴ The number of workers after 2 years will be 500.



Question 2.

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.


Answer:

Present number of sheeps (P) = 200 sheeps, Rate = 8% (p.c.p.a), Duration (n) = 3 years



3


A = 200 (1+0.08)3


A = 200 (1.08)3


A = 200 (1.259712)


∴A= 251.9424


A = 252 sheeps (Rounded off)


∴ The number of sheeps after 3 years is 252.



Question 3.

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.


Answer:

Present Trees (P) = 40000 trees, Rate = 5% (p.c.p.a), Duration (n) = 3 years



3


A = 40000 (1+0.05)3


A = 40000 (1.05)3


A = 40000 (1.157625)


∴A= 46305/-


∴ The expected number of trees after 3 years will be 46305.



Question 4.

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.


Answer:

Principal = 250000/-, Rate (decrement) = 10% (p.c.p.a), Duration (n) = 2 years



2


2


A = 250000 (1-0.1)2


A = 250000 (0.9)2


A = 250000 (0.81)


∴A= 202500/-


∵ C.I. = A - P


∴ Depreciation in Price (C.I.) = 202500 – 250000


Depreciation in Price (C.I.) = -47500/-


(-) sign denotes the depreciation in amount.


∴ Depreciation in Price of the machine after 2 years will be 47500/- .



Question 5.

Find the compound interest if the amount of a certain principal after two years is
₹ 4036.80 at the rate of 16 p.c.p.a.


Answer:

Amount= 4036.80/-, Rate = 16% (p.c.p.a), Duration (n) = 2 years



2


4036.80 = P (1+0.16)2


4036.80 = P (1.16)2


4036.80 = P (1.3456)


∴ P = 3000/-


∵ C.I. = A - P


∴ C.I. = 4036.80 - 3000


C.I. = 1036.80/-


Compound interest is 1036.80/- .



Question 6.

A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.


Answer:

Principal = 15000/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years



3


A = 15000 (1+0.12)3


A = 15000 (1.12)3


A = 15000 (1.404928)


∴A= 21073.92/-


Amount to settle the loan after 3 years is 21073.92/- .



Question 7.

A principal amounts to ₹ 13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.


Answer:

Amount= 13924/-, Rate = 18% (p.c.p.a), Duration (n) = 2 years



2


13924 = P (1+0.18)2


13924 = P (1.18)2


13924 = P (1.3924)


∴ A = 10000/-


∴ The principal is 10000/- .



Question 8.

The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.


Answer:

Present Population (P) = 16000/-, Rate = R% (p.c.p.a), Duration (n) = 2 years


Population after 2 years (A) =17640/-



2





∴R= 5%


∴ The population of that suburb will increase at the rate of 5% .



Question 9.

In how many years ₹ 700 will amount to ₹ 847 at a compound interest rate of 10 p.c.p.a.


Answer:

Principal = 700/-, Rate = 10% (p.c.p.a), Duration (n) = n years Amount = 847/-





1.21 =


1.21 =


∴ n = 2 years


∴ The number of years required to gain an amount of 847/- from a principal of 700/- is 2 .



Question 10.

Find the difference between simple interest and compound interest on ₹ 20000 at 8 p.c.p.a.


Answer:

Principal = 20000/-, Rate = 8% (p.c.p.a), Duration (n) = n years


For the first year, compound interest and simple interest will be same, so it will vary from second year, therefore assuming the duration as 2 years in the same case.




A = 20000 (1+0.08)2


A = 20000 (1.08)2


A = 20000 (1.1664)


∴ A = 23328/-


∵ C.I. = A - P


C.I. = 23328 – 20000


C.I. = 3328/-




S.I. = 3200/-


∴ Difference = C.I. – S.I.


Difference = 3328 – 3200


Difference = 128 /-


∴ The difference between simple interest and compound interest is 128/- .



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