RD Sharma Solutions Class 8 Compound Interest Exercise 14.3

RD Sharma Class 8 Solutions Chapter 14 Ex 14.3 PDF Free Download

RD Sharma Solutions Class 8 Chapter 14 Exercise 14.3

Exercise 14.3

Q1) On what sum will the compound interest at 5% per annum for 2 years compounded annually be Rs 164?

Solution: Let the sum be Rs y.

As we know that:

C.I = Amount – Principle

= \(P(1+\frac{R}{100})^{n}\) – P

= \(P[(1+\frac{R}{100})^{n}-1]\)

164 = \(y[(1+\frac{5}{100})^{2}-1]\)

164 = \(y[(1.05)^{2}-1]\)

y = \(\frac{164}{0.1025}\)

y = 1600

Therefore, the required sum is Rs 1600.

Q2) Find the principal if the interest compounded annually at the rate of 10% for two years is Rs 210.

Solution:

Let the sum be Rs y.

Now we know that:

C.I = Amount – Principle

210 = \(P(1+\frac{R}{100})^{n}\) – P

210 = \(P[(1+\frac{R}{100})^{n}-1]\)

210 = \(y[(1+\frac{10}{100})^{2}-1]\)

210 = \(y[(1.10)^{2}-1]\)

y = \(\frac{210}{0.21}\)

y = 1000

Therefore, the required amount is Rs 1000.

Q3) A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Solution:

Let the sum be Rs y.

Next,

Amount = \(P(1+\frac{R}{100})^{n}\)

A = \(P[(1+\frac{R}{100})^{n}]\)

756.25 = \(y[(1+\frac{10}{100})^{2}]\)

756.25 = \(y[(1.10)^{2}]\)

y = \(\frac{756.25}{1.21}\)

y = 625

Therefore, the required sum is Rs 625.

Q4) What sum will amount to Rs 4913 in 18 months, if the rate of interest is \(12\frac{1}{2}\)% per annum, compounded half-yearly?

Solution:

Let the sum be Rs y.

Given:

Amount = Rs 4913

Rate of interest = 12.5 %

n = 18 months = 1.5 years (time)

Now we know that:

Amount = \(P(1+\frac{R}{200})^{2n}\)

4913 = \(P(1+\frac{R}{200})^{2n}\)

4913 = \(x(1+\frac{12.5}{200})^{3}\)

4913 = \(P[(1.0625)^{3}]\)

y = \(\frac{4913}{1.1995}\)

y = 4096

Therefore, the required sum is Rs 4096.

 

Q5) The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.

Solution: It is given that

C.I – S.I = Rs 283.50

Rate of Interest = 15 %

n = 3 years(time)

suppose the sum is Rs y.

now we know that:

Amount = \(P(1+\frac{R}{100})^{n}\)

= \(y(1+\frac{5}{100})^{3}\)

= \(y(1.15)^{3}\)   …..(1)

and also,

S.I = \(\frac{PRT}{100}\) = \(\frac{y(15)(3)}{100}\) = 0.45y

Amount = S.I + P = 1.45y   …..(2)

Thus, we have:

\(y(1.15)^{3}\) – 1.45y = 283.50               [From (1) and (2)]

1.523y – 1.45y = 283.50

0.070875y = 283.50

y = \(\frac{283.50}{0.070875}\)

= 4000

Therefore, the sum is Rs 4000.

Q6) Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1290 as interest compounded annually, find the sum she borrowed.

Solution:

Let us suppose that the money borrowed by Rachana is Rs y.

So, we have:

C.I = \(P(1+\frac{R}{100})^{n}\) – P

1290 = \(y[(1+\frac{15}{100})^{2}-1]\)

1290 = y[0.3225]

y = \(\frac{1290}{0.3225}\)

= 4000

Thus, Rachana borrowed Rs 4000.

Q7) The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.

Solution:

Let us assume the time period to be t years.

And, we have:

C.I = \(P(1+\frac{R}{100})^{t}-P\)

163.20 = \(2000(1+\frac{4}{100})^{t}-2000\)

2163.20 = \(2000(1.04)^{t}\)

\((1.04)^{t}=\frac{2163.20}{2000}\)

\((1.04)^{t}=1.0816\)

\((1.04)^{t}=(1.04)^{2}\)

when we compare both the sides, we get:

t = 2 years

Thus, the required time is two years.

Q8) In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest?

Solution:

Let us assume the time period to be t years.

So, then we have:

C.I = \(P(1+\frac{R}{100})^{t}-P\)

6655 = \(5000(1+\frac{10}{100})^{t}-5000\)

11655 = \(5000(1.10)^{t}\)

\((1.1)^{t}=\frac{11655}{5000}\)

\((1.1)^{t}=2.331\)

\((1.1)^{t}=(1.1)^{3}\)

On comparing both the sides, we get:

t = 3 years

Therefore, the required time is three years.

 

Q9) In what time will Rs 4400 become Rs 4576 at 8% per annum interest compounded half-yearly?

Solution:

Let us assume the time period be t years.

R = 8 % = 4 % (Half-yearly – 6 months)

So, we have:

Amount = \(P(1+\frac{R}{100})^{t}\)

4576 = \(4400(1+\frac{4}{100})^{t}\)

4576 = \(4400(1.04)^{t}\)

\((1.04)^{t}=\frac{4576}{4400}\)

\((1.04)^{t}=1.04\)

\((1.04)^{t}=(1.04)^{1}\)

On comparing both the sides, we get:

t = 1 year

Therefore, the required time is 6 months or half year.

Q10) The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.

Solution: It is given that

C.I – S.I = Rs 20

\([P(1+\frac{4}{100})^{2}-P]-\frac{P\times 4\times 2}{100}=20\)

\(P[(1.04)^{2}-P]-0.08P=20\)

0.0816P – 0.08P = 20

0.0016P = 20

P = \(\frac{20}{0.0016}\)

Principal = Rs.12500

Therefore, the required sum is Rs 12500.

Q11) In what time will Rs 1000 amount to Rs 1331 at 10% per annum, compound interest?

Solution:

Let us assume the time to be t years.

So then,

Amount = \(P(1+\frac{10}{100})^{t}\)

1331 = \(1000(1+\frac{10}{100})^{t}\)

\((1.1)^{t}=\frac{1331}{1000}\)

\((1.1)^{t}=1.331\)

\((1.1)^{t}=(1.1)^{3}\)

Let us compare both the sides, we get:

n = 3 years

Therefore, the required time is three years.

Q12) At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years?

Solution:

Amount = \(P(1+\frac{R}{100})^{t}\)

774.40 = \(640(1+\frac{R}{100})^{2}\)

\((1+\frac{R}{100})^{2}=\frac{774.40}{640}\)

\((1+\frac{R}{100})^{2}=1.21 \)

\((1+\frac{R}{100})^{2}=(1.1)^{2}\)

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

\(\frac{R}{100}=0.1\)

R = 10 percent

Therefore, the required R is 10 % per annum.

Q13) Find the rate percent per annum if Rs 2000 amount to Rs 2662 in \(1\frac{1}{2}\) years, interest being compounded half-yearly?

Solution:

Let us consider the rate of interest be R %.

So, then

Amount = \(P(1+\frac{R}{100})^{n}\)

2662 = \(2000(1+\frac{R}{100})^{3}\)

\((1+\frac{R}{100})^{3}=\frac{2662}{2000}\)

\((1+\frac{R}{100})^{3}=1.331\)

\((1+\frac{R}{100})^{3}=(1.1)^{3}\)

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

\(\frac{R}{100}=0.1\)

Rate of Interest = 10 %

As we compounded the R half yearly so, it would be 20 % per annum.

Q14) Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest but paid Rs 200 only as simple interest. Find the sum and the rate of interest.

Solution:

Let the sum be Rs S and the rate of interest be N%.

We know that Kamla paid Rs 200 as simple interest.

\(∴ 200=\frac{SN(2)}{100}\)

SN = 10000 …..(1)

Also, Kamla received Rs 210 as compound interest.

\(∴ 210=S(1+\frac{N}{100})^{2}-1\)

\(210(10000)=S(N^{2}+200N)\)

\(210N=N^{2}+200N\)                              [From (1)]

N = 10% p.a.

Put the equation in (1), we get:

S = 1000

Therefore, the required sum is Rs 1000 and the R is 10%.

Q15) Find the rate percent per annum, if Rs 2000 amount to Rs 2315.25 in a year and a half, interest being compounded six monthly.

Solution:

Consider the rate percent per annum be R.

As interest is compounded every half year (six months), t will be 3 for 1.5 yrs.

So,

Amount = \(P(1+\frac{R}{200})^{t}\)

2315.25 = \(2000(1+\frac{R}{200})^{3}\)

\((1+\frac{R}{200})^{3}=\frac{2315.25}{2000}\)

\((1+\frac{R}{200})^{3}=1.157625\)

\((1+\frac{R}{200})^{3}=(1.05)^{3}\)

\((1+\frac{R}{200})=1.05\)

\(\frac{R}{200}=0.05\)

R = 10%

Therefore, the required R is 10 % per annum.

Q16) Find the rate at which a sum of money will double Itself in 3 years if the interest is compounded annually.

Solution:

Consider the rate percent per annum be R.

So,

Amount = \(P(1+\frac{R}{100})^{t}\)

2P = \(P(1+\frac{R}{100})^{3}\)

\((1+\frac{R}{100})^{3}\) = 2

\((1+\frac{R}{100})\) = 1.2599

\(\frac{R}{100}\) = 0.2599

Rate of Interest = 25.99

Therefore, the required rate is 25.99 % per annum.

Q17) Find the rate at which a sum of money will become four times the original amount in 2 years if the interest is compounded half-yearly.

Solution: Let the rate percent per annum be R.

So then,

Amount = \(P(1+\frac{R}{200})^{2n}\)

4P = \(P(1+\frac{R}{200})^{4}\)

\((1+\frac{R}{200})^{4}\) = 4

\((1+\frac{R}{200})\) = 1.4142

\(\frac{R}{200}\) = 0.4142

Rate of Interest = 82.84

Therefore, the required rate is 82.84 %.

Q18) A certain sum amounts to Rs 6832 in 2 years at 8% compounded interest. Find the sum.

Solution:

Consider the sum to be P.

Therefore, we have:

Amount = \(P(1+\frac{R}{100})^{n}\)

5832 = \(P(1+\frac{8}{100})^{2}\)

5832 = 1.1664P

Principle = \(\frac{5832}{1.1664}\)

= 5000

Therefore, the required sum is Rs 5000.

Q19) The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.

Solution:

Consider the sum to be P.

Therefore, we have:

C.I – S.I = 360

\([P(1+\frac{R}{100})^{n}-P]-\frac{P\times 7.5\times 2}{100}\) = 360

\(P[(1+\frac{7.5}{100})^{2}-1]-\frac{P\times 7.5\times 2}{100}\) = 360

P [1.155625 – 1] – 0.15P = 360

0.155625P – 0.15P = 360

Principle = \(\frac{360}{0.005625}\)

Principle = 64000

Therefore, the required sum is Rs 64000.

Q20) The difference in simple interest and compound interest on a certain sum of money at \(6\frac{2}{3}\)% per annum for 3 years is Rs 46. Determine the sum.

Solution:

It is given that

C.I – S.I = 46

\(P[(1+\frac{R}{100})^{n}-1]-\frac{PRT}{100}=46\)

\(P[(1+\frac{20}{100})^{3}-1]-\frac{P\times 20\times 3}{100}=46\)

\(\frac{4096}{3375}P-\frac{P}{5}-P=46\)

\(\frac{(4096-3375-675)P}{3375}=46\)

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

= 3375

Therefore, the required sum is Rs 3375.

Q21)  Ishita invested a sum of Rs 12000 at 5% per annum compound interest. She received an amount of Rs 13230 after years. Find the value of n.

Solution:

Ampount = \(P(1+\frac{R}{100})^{n}\)

13230 = \(12000(1+\frac{5}{100})^{n}\)

\((1.05)^{n}=\frac{13230}{12000}\)

\((1.05)^{n}=1.1025 \)

\( (1.05)^{n}=(1.05)^{2}\)

after comparing both sides, we get:

n = 2 yrs

Therefore, the value of n is two years.

Q22) At what rate percent per annum will a sum of Rs 4000 yield compound interest of Rs 410 in 2 years?

Solution:

Let us consider the rate percent be R.

Now we know that:

C.I = \(P(1+\frac{R}{100})^{n}\) – P

410 = \(4000(1+\frac{R}{100})^{2}\) – 4000

4410 = \(4000(1+\frac{R}{100})^{2}\)

\((1+\frac{R}{100})^{2}\) = \(\frac{4410}{4000}\)

\((1+\frac{R}{100})^{2}\) = 1.1025

\((1+\frac{R}{100})^{2}\) = \((1.05)^{2}\)

\(1+\frac{R}{100}\) = 1.05

\(\frac{R}{100}\) = 0.05

Rate of Interest = 5 yrs

Therefore, the required rate percent is 5%.

Q23) A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.

Solution:

Amount = \(P(1+\frac{R}{100})^{n}\)

10404 = \(P(1+\frac{2}{100})^{2}\)

10404 = \(P(1.02)^{2}\)

\(P=\frac{10404}{1.0404}\)

Principle = 10000

Therefore, the required sum is Rs 10000.

Q24) In how much time will a sum of Rs 1600 amount to Rs 1852.20 at 5% per annum compound interest?

Solution:

Amount = \(P(1+\frac{R}{100})^{n}\)

1852.20 = \(1600(1+\frac{5}{100})^{n}\)

\((1.05)^{n}=\frac{1852.20}{1600}\)

\((1.05)^{n}=1.157625\)

\( (1.05)^{n}=(1.05)^{3}\)

If we compare both the sides, we get:

n = 3yrs

Therefore, the required time is 3 years.

Q25) At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest?

Solution:

Amount = \(P(1+\frac{R}{100})^{n}\)

1102.50 = \(1000(1+\frac{R}{100})^{2}\)

\((1+0.01R)^{2}=\frac{1102.50}{1000}\)

\((1+0.01R)^{2}=1.1025\)

\( (1+0.01R)^{2}=(1.05)^{2}\)

If we compare both the sides, we get:

1 + 0.01R = 1.05

0.01R = 0.05

R = 5 %

Therefore, the required R is 5%.

 

Q26) The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.

Solution:

C.I = \(P(1+\frac{R}{100})^{n}\) – P

378 = \(1800(1+\frac{10}{100})^{n}\) – 1800

2178 = \(1800(1+\frac{10}{100})^{n}\)

\((1+\frac{10}{100})^{n}\) = \(\frac{2178}{1800}\)

\((1.1)^{n}\) = 1.21

\((1.1)^{n}\) = \((1.1)^{2}\)

If we compare both the sides, we get:

n = 2 yrs

Therefore, the required time is 2 years.

 

Q27) What sum of money will amount to Rs 45582.25 at \(6\frac{3}{4}\)% per annum in two years, interest being compounded annually?

Solution:

Amount = \(P(1+\frac{R}{100})^{n}\)

45582.25 = \(P(1+\frac{27}{400})^{2}\)

\(P(1.0675)^{2}\) = 45582.25

P = \(\frac{45582.25}{1.13955625}\)

Principle = Rs. 40000

Therefore, the required sum is Rs 40000.

Q28) Sum of money amounts to. Rs 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.

Solution:

Amount = \(P(1+\frac{R}{100})^{n}\)

453690 = \(P(1+\frac{6.5}{100})^{2}\)

\(P(1.065)^{2}\) = 453690

Principle = \(\frac{453690}{1.134225}\)

Principle = Rs 400000

Therefore, the required sum is Rs 400000.

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