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.