The DAV Class 8 Maths Book Solutions Pdf and **DAV Class 8 Maths Chapter 6 Worksheet 3** Solutions of Compound Interest offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 6 WS 3 Solutions

Question 1.

Find the amount for ₹ 15000 at 8% per annum compounded annually for 2 years.

Solution:

Here P = ₹ 15000,

R = 8% p.a.,

T = 2 years

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 15000 \(\left[1+\frac{8}{100}\right]^2\)

= 15000 \(\left[\frac{27}{25}\right]^2\)

= \(\frac{15000 \times 729}{625}\)

= ₹ 17496

Hence, the amount = ₹ 17496.

Question 2.

Find the compound interest on ₹ 11200 at 17% p.a. for 2 years.

Solution:

Here P = ₹ 11200,

R = 17 = % p.a.,

T = 2 years

C.P.= P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) – P

= 11200 \(\left(1+\frac{35}{2 \times 100}\right)^2\) – 11200

= 11200 \(\left(\frac{47}{40}\right)^2\) – 11200

= 11200 \(\left[\frac{47}{40} \times \frac{47}{40}-1\right]\) – 11200

= 11200 \(\left[\frac{2209}{1600}-1\right]\)

= 11200 \(\left[\frac{2209-1600}{1600}\right]\)

= 11200 \(\frac{1120 \times 609}{1600}\)

= ₹ 426.30

Hence, the compound interest is ₹ 426.30.

Question 3.

Ram borrowed a sum of ₹ 30000 from Shyam for 3 years. If the rate of interest is 6% p.a. compounded annually, find the interest paid by Ram to Shyam after 3 years.

Solution:

Here P = ₹ 30000,

R = 6% p.a.,

T = 3 years

∴ C.I. = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\) – P

= 30000 \(\left(1+\frac{6}{100}\right)^3\) – 30000

= 30000 \(\left[\left(1+\frac{6}{100}\right)^3-1\right]\)

= 30000 \(\left[\left(\frac{53}{50}\right)^3-1\right]\)

= 30000 [\(\frac{148877}{125000}\) – 1]

= 30000 \(\left[\frac{148877-125000}{125000}\right]\)

= 30000 × \(\frac{23877}{125000}\)

= ₹ 5730.48

Hence, the required C.I. = ₹ 5730.48.

Question 4.

Nidhi deposited ₹ 7500 in a bank which pays her 4% interest per annum compounded annually. Find the amount and the interest received by her after 3 years.

Solution:

Here P = ₹ 7500,

R = 4% p.a.,

T = 3 years

∴ A = P \(\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}\)

= 7500 \(\left[1+\frac{4}{100}\right]^3\)

= 7500 \(\left[\frac{26}{25}\right]^3\)

= 7500 × \(\frac{26 \times 26 \times 26}{25 \times 25 \times 25}\)

= ₹ 8436.48

∴ C.I.= A – P

= ₹ 8436.48 – ₹ 7500

= ₹ 936.48

Hence the required C.I. = ₹ 936.48.

Question 5.

Find the difference between the compound interest and the simple interest on ₹ 30,000 at 7% p.a. for 3 years.

Solution:

Here P = ₹ 30000,

R = 7% p.a.,

T = 3 years

∴ S.I. = \(\)

= \(\)

= ₹ 6300

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= 30000 [\(\left(1+\frac{7}{100}\right)^3\) – 1]

= 30000 [\(\left(\frac{107}{100}\right)^3\) – 1]

= 30000 [\(\left(\frac{107}{100}\right)^3\) – 1]

= 30000 \(\left[\frac{1225043-1000000}{1000000}\right]\)

= 30000 \(\frac{225043}{1000000}\)

= ₹ 6751.29

C.I. = ₹ 6751.29

C.L – S.I. = ₹ 6751.29 – ₹ 6300

= ₹ 451.29

Hence, the required difference is ₹ 451.29.

Question 6.

Aman borrows ₹ 14500 at 11% p.a. for 3 years at simple interest and Tarun borrows the same amount at 10% p.a. for the same time compounded annually. Who pays more interest and by how much?

Solution:

For Aman:

P = ₹ 14500,

R = 11% p.a.,

T = 3 years

For Tarun:

P = ₹ 14500,

R = 10% p.a.,

T = 3 years

C.I. = P [\(\left(1+\frac{R}{100}\right)^{\mathrm{T}}\) – 1]

= 14500 \(\left[\left(1+\frac{10}{100}\right)^3-1\right]\)

= 14500 [\(\left(\frac{11}{10}\right)^3\) – 1]

= 14500 [(1.1)^{3} – 1]

= 14500 [1.331 – 1]

= 14500 × 0.331

= ₹ 4799.50

Hence, Tarun will pay more interest by ₹ 4799.50 – ₹ 4785 = ₹ 14.50.

Question 7.

The simple interest on a certain sum of money for 2 years at 5\(\frac{1}{2}\) % is ₹ 6600. What will be the compound interest on that sum at the same rate and for the same time period?

Solution:

Here S.I. = ₹ 6600,

R = 5 % = % p.a.,

T = 2 years

P = \(\frac{6600 \times 100}{11}\)

= ₹ 60000

C.I. = P [\(\left(1+\frac{R}{100}\right)^{\mathrm{T}}\) – 1]

= 60,000 [\(\left(1+\frac{11}{2 \times 100}\right)^2\) – 1]

= 60,000 [\(\left(\frac{211}{200}\right)^2\) – 1]

= 60,000 [\(\frac{44521}{40000}\) – 1]

= 60,000 \(\left[\frac{44521-40000}{40000}\right]\)

= 60,000 \(\left[\frac{4521}{40000}\right]\)

= 60,000 × \(\frac{4521}{40000}\)

= \(\frac{13563}{2}\)

= ₹ 6781.50

Hence the required C.I. = ₹ 6781.50.

Question 8.

A certain sum amounts to ₹ 2970.25 in 2 years at 9% p.a. compounded annually. Find the sum.

Sol.

Here A = ₹ 2970.25,

T = 2 years,

R = 9% p.a.

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

⇒ 2970.25 = P \(\left[1+\frac{9}{100}\right]^2\)

⇒ 2970.25 = P \(\left[\frac{109}{100}\right]^2\)

⇒ 2970.25 = P × (1.09)^{2}

⇒ 2970.25 = P × 1.1881

⇒ P = \(\frac{2970.25}{11881}\) = ₹ 2500

Hence, the required sum = ₹ 2500.

Question 9.

On what sum will the compound interest at 7\(\frac{1}{2}\) % p.a. for 3 years compounded annually be ₹ 3101.40?

Solution:

Here C.I. = ₹ 3101.40,

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

= \(\frac{15}{2}\) % p.a.,

T = 3 years

C.I. = P [\(\left(1+\frac{R}{100}\right)^{\mathrm{T}}\) – 1]

⇒ 3101.40 = P [\(\left(1+\frac{15}{2 \times 100}\right)^3\) – 1]

⇒ 3101.40 = P [\(\left(\frac{43}{40}\right)^3\) – 1]

⇒ 3101.40 = P [\(\frac{79507}{64000}\) – 1]

⇒ 3101.40 = P \(\left[\frac{79507-64000}{64000}\right]\)

⇒ 3101.40 = \(\frac{P \times 15507}{64000}\)

⇒ P = \(\frac{310140 \times 64000}{15507}\)

⇒ P = ₹ 12800

Hence the required sum = ₹ 12800

Question 10.

At what rate per cent will a sum of ₹ 640 be compounded to ₹ 774.40 in 2 years?

Solution:

Here P = ₹ 64,

A = ₹ 774.40,

T = 2 years

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

⇒ 774.40 = 640 \(\left[1+\frac{\mathrm{R}}{100}\right]^2\)

⇒ \(\frac{774.40}{640}=\left[1+\frac{\mathrm{R}}{100}\right]^2\)

⇒ 1.21 = \(\left[1+\frac{\mathrm{R}}{100}\right]^2\)

Taking square root on both sides,

1.1 = 1 + \(\frac{R}{100}\)

⇒ \(\frac{R}{100}\) = 1.1 – 1

⇒ \(\frac{R}{100}\) = 0.1

⇒ R = 0.1 × 100 = 1%

Hence, the required rate of interest = ₹ 10% p.a.

Question 11.

At what rate per cent will a sum of ₹ 64000 be compounded to ₹ 68921 in 3 years?

Solution:

Here P = ₹ 64000,

A = ₹ 68921,

T = 3 years

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

⇒ 68921 = 64000 \(\left[1+\frac{\mathrm{R}}{100}\right]^3\)

⇒ \(\frac{68921}{64000}=\left[1+\frac{R}{100}\right]^3\)

⇒ \(\left(\frac{41}{40}\right)^3=\left[1+\frac{R}{100}\right]^3\)

Taking cube root on both sides,

⇒ \(\frac{41}{40}=1+\frac{R}{100}\)

⇒ \(\frac{R}{100}=\frac{41}{40}-1\)

⇒ \(\frac{R}{100}=\frac{1}{40}\)

⇒ R = \(\frac{100}{40}\)

= \(\frac{5}{2} \%=2 \frac{1}{2} \%\)

Hence, the required rate per cent = 2\(\frac{1}{2} \)%.

Question 12.

In how many years will ₹ 8000 amount to ₹ 9261 at 5% per annum compounded annually?

Solution:

Here P = ₹ 8000,

A = ₹ 9261,

R = 5% pa.

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

⇒ 9261 = 8000 \(\left[1+\frac{5}{100}\right]^{\mathrm{T}}\)

⇒ \(\frac{9261}{8000}=\left[\frac{21}{20}\right]^{\mathrm{T}}\)

⇒ \(\left[\frac{21}{20}\right]^3=\left[\frac{21}{20}\right]^{\mathrm{T}}\)

Comparing both sides, we get

T = 3 years

Hence, the required time = 3 years.

Question 13.

In what time will a sum of ₹ 3750 at 20% p.a. compounded annually amount to ₹ 6480?

Solution:

Here P = ₹ 3750,

A = ₹ 6480,

R = 20% p.a.

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

⇒ 6480 = 3750 \(\left[1+\frac{20}{100}\right]^{\mathrm{T}}\)

⇒ \(\frac{6480}{3750}=\left[\frac{6}{5}\right]^{\mathrm{T}}\)

⇒ \(\frac{216}{125}=\left[\frac{6}{5}\right]^{\mathrm{T}}\)

⇒ \(\left[\frac{6}{5}\right]^3=\left[\frac{6}{5}\right]^{\mathrm{T}}\)

Comparing the powers on both sides,

we get T = 3 years

Hence the required time = 3 years.

Question 14.

The difference between the compound interest and simple interest on a certain sum of money at 6\(\frac{2}{3}\)% per annum for 3 years is ₹ 46. Find the sum.

Solution:

Here C.I. – S.I. = ₹ 46,

R = 6%

= — % p.a.,

T = 3 years

S.I. = \(\frac{P \times R \times T}{100}\)

= \(\frac{\mathrm{P} \times 20 \times 3}{3 \times 100}\)

= ₹ \(\frac{P}{5}\)

C.I. = P \(\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-1\right]\)

= P \(\left[\left(1+\frac{20}{3 \times 100}\right)^3-1\right]\)

= P \(\left[\left(\frac{16}{15}\right)^3-1\right]\)

= P \(\left[\frac{4096}{3375}-1\right]\)

= \(\mathrm{P}\left[\frac{721}{3375}\right]=\frac{721}{3375} \mathrm{P}\)

C.I. – S.I. = ₹ 46

\(\frac{721}{3375} P-\frac{P}{5}\) = 46

\(\frac{721 \mathrm{P}-675 \mathrm{P}}{3375}\) = 46

\(\frac{46 \mathrm{P}}{3375}\) = 46

P = \(\frac{46 \times 3375}{46}\)

= ₹ 3375

Hence, the required sum = ₹ 3375.

Question 15.

The difference between the compound interest and simple interest on a certain sum of money at 15% p.a. for 3 years is ₹ 283.50. Find the sum.

Solution:

Here C.I. – S.I. = ₹ 283.50,

R = 15% p.a.,

T = 3 years

Hence, the required sum of money is ₹ 4000.

Question 16.

Find the amount and the compound interest on ₹ 12800 for 1 year at 7 % p.a. compounded semi-annually.

Solution:

Here P = ₹ 12800,

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

= \(\frac{15}{2} \times \frac{1}{2}\)

= \(\frac{15}{4}\) semi-annually

T = 1 × 2 = 2 semi-annuals

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

= 12800 \(\left[1+\frac{15}{4 \times 100}\right]^2\)

= 12800 \(\left[\frac{83}{80}\right]^2\)

= 12800 × \(\frac{6889}{6400}\)

= ₹ 13778

C.I. = ₹ 13778 – ₹ 12800 = ₹ 978

Hence, the required amount = ₹ 13778

and compound interest = ₹ 978.

Question 17.

Mr. Arora borrowed ₹ 40960 from a Bank to start a play school. If the bank charges 12\(\frac{1}{2}\)% p.a. compounded half-yearly, what amount will he have to pay after 1\(\frac{1}{2}\) years?

Solution:

Here P = ₹ 40960,

R = 12\(\frac{1}{2}\)%

= \(\frac{25}{2}\)% p.a.

= \(\frac{25}{4}\)% per semi-annual

T = 1\(\frac{1}{2}\) year

= 1\(\frac{1}{2}\) × 2 = 3 semi-annuals

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

= 40960 \(\left[1+\frac{25}{4 \times 100}\right]^3\)

= 40960 \(\left[\frac{17}{16}\right]^3\)

= \(\frac{40960 \times 4913}{4096}\)

= ₹ 49130

Hence, the required amount = ₹ 49130.

Question 18.

Meera lent out ₹ 20,000 for 9 months at 20% p.a. compounded quarterly to Mrs. Sharma. What amount will she get after the expiry of the period?

Solution:

Here, P = ₹ 20,000

R = 20% p.a

= \(\frac{20}{4}\)

= 5% per quarter

T = \(\frac{9}{12}\) × 4

= 3 quarters

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

= 20,000 \(\left[1+\frac{5}{100}\right]^3\)

= 20,000 \(\left[\frac{21}{20}\right]^3\)

= 20,000 × \(\frac{9261}{8000}\)

= \(\frac{5 \times 9261}{2}\)

= \(\frac{46305}{2}\)

= ₹ 23152.50

Hence, the required amount = ₹ 23152.50.

Question 19.

Find the amount and the compound interest on ₹ 24,000 for 6 months if the interest is payable quarterly at the rate of 20 paise a rupee per annum.

Solution:

Here P = ₹ 24000,

R = 20% p.a.

= \(\frac{20}{4}\) %

= 5% per quarter

T = \(\frac{6}{12}\) × 4

= 2 quarter

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

= 24000 \(\left[1+\frac{5}{100}\right]^2\)

= 24000 \(\left[\frac{21}{20}\right]^2\)

= 24000 × \(\frac{441}{400}\)

= 600 × 441

= 26460

C.I. = A – P

= ₹ 26460 – ₹ 24000

= ₹ 2460

Hence the required amount = ₹ 26460

and C.I. = ₹ 2460.