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## S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Ex 2(C)

Question 1.

What amount of money should Mohan invest in a bank in order to get ₹ 1323 in 2 years at 5% compounded annually?

Solution:

Amount after 2 years = ₹ 1323

Rate (R) = 5% p.a.

Period = 2 years

Now A = P \(\left(1+\frac{\mathrm{R}}{100}\right)^n\)

⇒ ₹ 1323 = P \(\left(1+\frac{5}{100}\right)^2\)

⇒ ₹ 1323 = P \(\left(\frac{21}{20}\right)^2\)

⇒ P = ₹ 1323 x \(\left(\frac{20}{21}\right)^2\)

= 1323 x \(\frac { 20 }{ 21 }\) x \(\frac { 20 }{ 21 }\) = ₹ 1200

∴ Principal = ₹ 1200

Question 2.

Find the sum which amounts to ₹ 1352 in 2 years at 4% compound interest.

Solution:

Amount after 2 years = ₹ 1352

Rate (R) = 4% p.a.

Now A = P \(\left(1+\frac{\mathrm{R}}{100}\right)^n\)

⇒ ₹ 1352 = P \(\left(1+\frac{4}{100}\right)^2\)

⇒ ₹ 1352 = P x \(\left(\frac{26}{25}\right)^2\)

⇒ P = ₹ 1352 x (\(\frac { 25 }{ 26 }\))²

= ₹ 1352 x \(\frac { 25 }{ 26 }\) x \(\frac { 25 }{ 26 }\) = ₹ 1250

∴ Principal = ₹ 1250

Question 3.

What principal will amount to ₹ 9768 in two years. If the rates of interest for the successive years are 10% p.a. and 11% p.a. respectively.

Solution:

Amount after 2 years = ₹ 9768

Rate of interest for the first year (r_{1}) = 10% p.a.

and for the second year (r_{2}) = 11% p.a.

Now A = P \(\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\)

⇒ 9768 = P \(\left(1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right)\)

⇒ 9768 = P \(\left(\frac{11}{10}\right)\left(\frac{111}{100}\right)\)

∴ P = ₹ 9768 x \(\frac { 10 }{ 11 }\) x \(\frac { 100 }{ 11 }\)= ₹ 8000

∴ Principal = ₹ 8000

Question 4.

On what sum of money does the difference between the simple interest and compound interest in 2 years at 5% per annum is Rs. 15?

Solution:

Let principal (P) = ₹ 100

Rate (R) = 5% p.a.

Period in) = 2 years

Difference between S.I. and C.I. = ₹ 15

Now S.I = \(\frac{\mathrm{PRT}}{100}=\frac{100 \times 5 \times 2}{100}\) = ₹ 10

and by compound interest

Now difference between S.I. and C.I. = \(\frac { 41 }{ 4 }\) – 10

= ₹ \(\frac{41-40}{4}=₹ \frac{1}{4}\)

If difference is ₹ \(\frac { 1 }{ 4 }\), then principal = ₹ 100

If difference is ₹ 1, then principal = ₹\(\frac { 100×4 }{ 1 }\)

and if difference is ₹ 15, then principal

= ₹ \(\frac { 100×4×15 }{ 1 }\) = ₹ 6000

Hence principal = ₹ 6000

Question 5.

The difference between simple and compound interest on the same sum of money at 6\(\frac { 2 }{ 3 }\)% for 3 years is ₹ 184. Determine the sum.

Solution:

Difference between C.I. and S.I. = ₹ 184

Let principal (Sum) = Rs. 100

Rate (R) = 6\(\frac { 2 }{ 3 }\)% = \(\frac { 20 }{ 3 }\) %

Period (n) = 3 years

Simple interest = \(\frac{\mathrm{PRT}}{100}=\frac{100 \times 20 \times 3}{100 \times 3}\)

= ₹ 20

and by compound interest

If difference is ₹\(\frac { 184 }{ 135 }\), then principal = ₹ 100

If difference is ₹ 1 then principal

= \(\frac{100 \times 135}{184}\)

and if difference is ₹ 184, then principal

= \(\frac{100 \times 135 \times 184}{184}\) = ₹ 13500

Question 6.

On what sum of money will the difference between the simple interest and the compound interest for 2 years at 5% per annum be equal to ₹ 50.

Solution:

Let sum (P) = ₹ 100

Rate (R) = 5% p.a.

Period (n) = 2 years

∴ Simple Interest = \(\frac { PRT }{ 100 }\)

= \(\frac{100 \times 5 \times 2}{100}\) = ₹ 10

By compound interest,

If difference is ₹\(\frac { 1 }{ 4 }\), then principal = ₹ 100

and if difference is ₹ 1, then principal

= \(\frac { 100×4 }{ 1 }\)

and if difference is ₹ 50, then principal

= \(\frac{100 \times 4 \times 50}{1}\) = ₹ 20,000

Question 7.

Find the rate per cent per annum, if compounded yearly

(i) Principal = ₹ 196, Amount = ₹ 225, time = 2 years

(ii) Principal = ₹ 3136, Compound interest = ₹ 345, Time = 2 years

Solution:

(i) Principal (P) = ₹ 196

Amount (A) = Rs. 225

Time = 2 years

We know that,

(ii) Principal (P) = 136

C.I. = Rs. 345

∴ Amount (A) = P + C.I. = ₹ 3136 + 345 = ₹ 3481

Time (n) = 2 years

We know that,

Comparing, we get

Question 8.

Hari purchased Relief Bonds for ^1000, a sum which will fetch him ₹ 2000 after 5 years. Find the rate of interest if the interest is compounded half-yearly.

(Given that \(\sqrt[10]{2}\) = 1.072)

Solution:

Principal (P) = ₹ 1000

Amount (A) = ₹ 2000

Period (n) = 5 years = 10 half years

We know that,

\(\frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{2000}{1000}=\left(1+\frac{\mathrm{R}}{100}\right)^{10}\)

⇒ \(\frac{2}{1}=\left(1+\frac{R}{100}\right)^{10}\)

1 + \(\frac { R }{ 100 }\) = \(\sqrt[10]{2}\) = 1.072

\(\frac { R }{ 100 }\) = 1.072 – 1.000 = 0.072

R = 0.072 x 100 = 7.2

∴ Rate half-yearly = 7.2%

and rate annually = 7.2 x 2 = 14.4% p.a.

Question 9.

₹ 8000 became ₹ 9261 in a certain interval of time at the rate of 5% per annum C.l. Find the time.

Solution:

Principal (P) = ₹ 8000

Amount (A) = ₹ 9261

Rate (R) = 5% p.a.

We know that,

\(\frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{9261}{8000}=\left(1+\frac{5}{100}\right)^n\)

⇒ \(\frac{9261}{8000}=\left(\frac{21}{20}\right)^n \Rightarrow\left(\frac{21}{20}\right)^{30}=\left(\frac{21}{20}\right)^n\)

Comparing, we get n = 3

∴ Period = 3 years

Question 10.

In how many years will a sum of ₹ 3000 at 20% per annum compounded semi-annually become ₹ 3993.

Solution:

Principal (P) = ₹ 3000

Amount (A) = ₹ 3993

Rate (R) = 20% p.a. or 10% half-yearly

We know that,

\(\frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{3993}{3000}=\left(1+\frac{10}{100}\right)^n\)

⇒ \(\frac{1331}{1000}=\left(\frac{11}{10}\right)^n \Rightarrow\left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^n\)

Comparing, we get n = 3

∴ Period = 3 half years or 1 \(\frac { 1 }{ 2 }\) years

Question 11.

A sum of money put out at compound interest amounts in 2 years to ₹ 578.40 and in 3 years to ₹ 614.55. Find the rate of interest.

Solution:

Amount for 3 years = ₹ 614.55

Amount for 2 years = ₹ 578.40

Subtracting,

Interest for 1 year = ₹ 36.15

∴ ₹ 36.15 is interest on ₹ 578.40 for 1 year

∴ Rate = \(\frac{\text { Simple interest } \times 100}{\mathrm{P} \times \text { Time }}\)

= \(\frac{36.15 \times 100}{578.40 \times 1}=\frac{25}{4} \%=6 \frac{1}{4} \% \text { p.a. }\)

Question 12.

A sum compounded annually becomes \(\frac { 25 }{ 16 }\) times of itself in 2 years. Determine the rate of interest per annum?

Solution:

Let Principal (P) = ₹ 1

Then Amount (A) = ₹ \(\frac { 25 }{ 16 }\)

Period (n) = 2 years

Let R be the rate of interest per annum, then

\(\frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n\)

\(\frac{25}{16}=\left(1+\frac{\mathrm{R}}{100}\right)^2 \Rightarrow\left(\frac{5}{4}\right)^2=\left(1+\frac{\mathrm{R}}{100}\right)^2\)

Comparing, we get

1 + \(\frac{\mathrm{R}}{100}=\frac{5}{4} \Rightarrow \frac{\mathrm{R}}{100}=\frac{5}{4}-1=\frac{5-4}{4}=\frac{1}{4}\)

∴ R = \(\frac { 1 }{ 4 }\) x 100 = 25

∴ Rate = 25 % p.a.