# DAV Class 8 Maths Chapter 6 Worksheet 3 Solutions

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.