myhelper: S.chand mathematics solution class 8 chapter 9 simple interest and compound interest

EXERCISE 9 C

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Question 1

What total amount will Ravi get in 3 years if he invest 5000 at 5% per annum , compounded annually ?

Sol :

Principal(P) = 5000 , Rate(R) = 5% and Time period(n) = 3 years

$\text{Amount}=\text{P}\left(1+\dfrac{\text{R}}{100}\right)^n$

$=5000\left(1+\dfrac{5}{100}\right)^3$

$=5000\left(1+\dfrac{1}{20}\right)^3$

$=5000\left(\dfrac{20+1}{20}\right)^3$

$=5000\left(\dfrac{21}{20}\right)^3$

$=5000\times\dfrac{9261}{8000}$

$=\dfrac{5\times 9261}{8}$

= 5×1157.625

= 5788.125

= 5788.13 (approx)

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Question 2
A person borrowed 7500 at 16% per annum compound interest . What is the amount of compound interest he has to pay at the end of 2 years to clear the loan ?

Sol :

Principal(P) = 7500 , Rate(R) = 16% and Time period(n) = 2 years

$\text{Amount}=\text{P}\left(1+\dfrac{\text{R}}{100}\right)^n$

$=7500\left(1+\dfrac{16}{100}\right)^2$

$=7500\left(1+\dfrac{4}{25}\right)^2$

$=7500\times\left(\dfrac{25+4}{25}\right)^2$

$=7500\times\left(\dfrac{29}{25}\right)^2$

$=7500\times\dfrac{841}{625}$

= 12×841

Amount(A) = 10092

Also , C.I = A - P

C.I = 10092 - 7500

=2592

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Question 3

How much will 25000 amount to in 2 years if the rates for successive years be 4% per annum and 5% per annum respectively?

Sol :

Amount(A) = ? , Principal(P) = 25000 , Rate(r₁) = 4% , Rate(r₂) = 5% and Time period(n) = 2 years

$\text{Amount}=\text{P}\left(1+\dfrac{r_{1}}{100}\right)\left(1+\dfrac{r_{1}}{100}\right)$

$\text{Amount}=25000\left(1+\dfrac{4}{100}\right)\left(1+\dfrac{5}{100}\right)$

$\text{Amount}=25000\times\left(\dfrac{104}{100}\right)\times\left(\dfrac{105}{100}\right)$

$\text{Amount}=25\times\left(\dfrac{104}{10}\right)\times{105}$

$\text{Amount}=25\times\left(\dfrac{52}{5}\right)\times{105}$

Amount(A) = 25×52×21

=27300

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Question 4

Find the compound interest on 1000 at the rate of 20% per annum for 18 months when the interest is compound half- yearly .

Sol :

Principal(P) = 1000 , Rate(R) = 20% , Time period(n) = 18 months or $=\dfrac{18}{12}years=\dfrac{3}{2}years$ , and Amount(A) = ?

Here interest is compounded half-yearly that's why Rate is taken as R/2 and for n we take 2n

$\text{A}=\text{P}\left(1+\dfrac{\dfrac{\text{R}}{2}}{100}\right)^{2n}$

$\text{A}=1000\left(1+\dfrac{\dfrac{20}{2}}{100}\right)^{2\times \frac{3}{2}}$

$\text{A}=1000\left(1+\dfrac{10}{100}\right)^{3}$

$\text{A}=1000\times\left(\dfrac{100+10}{100}\right)^{3}$

$\text{A}=1000\times\left(\dfrac{11}{10}\right)^{3}$

$\text{A}=1000\times\left(\dfrac{1331}{1000}\right)$

A = 1331

Also ,C.I = A - P

C.I = 1331 - 1000

=331

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Question 5

What is the compound interest on 16000 for 9 months at 20% per annum , interest being compounded quarterly ?

Sol :

Principal(P) = 16000 , Rate(R) = 20% , Time period(n) = 9 months or $=\dfrac{9}{12}years=\dfrac{3}{4}years$ , and Amount(A) = ?

Here interest is compounded quarterly that's why Rate is taken as R/4 and for n we take 4n

$\text{A}=\text{P}\times\left(1+\dfrac{\dfrac{\text{R}}{4}}{100}\right)^{4n}$

$\text{A}=16000\times\left(1+\dfrac{\dfrac{20}{4}}{100}\right)^{4\times \frac{3}{4}}$

$\text{A}=16000\times\left(1+\dfrac{5}{100}\right)^3$

$\text{A}=16000\times\left(\dfrac{105}{100}\right)^3$

$\text{A}=16000\times\left(\dfrac{1157625}{1000000}\right)$

= 18522

Also ,C.I = A - P

C.I = 18522 - 16000

= 2522

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Question 6

The simple interest on certain sum of money at 6% per annum for 3 years is 432 . What is the compound interest on the same sum for 2 years at 5% per annum , compounded annually ?

Sol :

Case 1 :

Principal (P) = ?

Time period (T) = 3 years

(Rate)R = 6%

S.I. = 432

$\text{S.I}=\dfrac{\text{P}\times\text{R}\times\text{T} }{100}$

$432=\dfrac{\text{P}\times6\times3}{100}$

$P=\dfrac{432\times100}{18}$

P = 2400

Case 2 :

Principal (P) = 2400 (from above)

(Rate)R = 5%

Time period (n) = 2 years

$\text{Amount}=\text{P}\left(1+\dfrac{\text{R}}{100}\right)^n$

$\text{Amount}=2400\left(1+\dfrac{5}{100}\right)^2$

$\text{Amount}=2400\times\left(\dfrac{105}{100}\right)^2$

$\text{Amount}=2400\times\dfrac{11025}{10000}$

A = 2646

Also , C.I = A - P

C.I = 2646 - 2400

= 246

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Question 7

Bela took a study loan of 100000 at 12% per annum compounded half-yearly for a year. How much did she pay at the end of the year ?

Sol :

Principal(P) = 100000 , Rate(R) = 12% , Time period(n) = 1 year and Amount(A) = ?

Here interest is compounded half-yearly that's why Rate is taken as R/2 and for n we take 2n

$\text{A}=\text{P}\left(1+\dfrac{\dfrac{\text{R}}{2}}{100}\right)^{2n}$

$\text{A}=\text{100000}\left(1+\dfrac{\dfrac{12}{2}}{100}\right)^{2}$

$\text{A}=\text{100000}\left(1+\dfrac{6}{100}\right)^{2}$

$\text{A}=\text{100000}\times\left(\dfrac{106}{100}\right)^{2}$

$\text{A}=\text{100000}\times\dfrac{11236}{10000}$

= 112360

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Question 8

The population of a village increases by 5% annually . If its percent population is 4000 , what will be its population after 2 years ?

Sol :

Principal(P) = 4000 , Rate(R) = 5% , Time period(n) = 2 year and Amount(A) = ?

$\text{A}=\text{P}\left(1+\dfrac{\text{R}}{100}\right)^n$

$\text{A}=4000\left(1+\dfrac{5}{100}\right)^2$

$\text{A}=4000\left(\dfrac{105}{100}\right)^2$

$\text{A}=4000\times\dfrac{11025}{10000}$

= 4410

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Question 9

Two years ago , the value of a machine was 62500 . If its value depreciates by 4% every year , what is its present value ?

Sol :

Principal(P) = 62500 , Rate(R) = 4% , Time period(n) = 2 year and Amount(A) = ?

$\text{A}=\text{P}\left(1-\dfrac{\text{R}}{100}\right)^n$

$\text{A}=62500\left(1-\dfrac{4}{100}\right)^2$

$\text{A}=62500\left(1-\dfrac{1}{25}\right)^2$

$\text{A}=62500\left(\dfrac{24}{25}\right)^2$

$\text{A}=62500\times\dfrac{576}{625}$

= 57600

S.chand mathematics solution class 8 chapter 9 simple interest and compound interest

EXERCISE 9 C

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