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### Access Answers of Maths Selina Class 9 Chapter 3- Compound Interest [Using Formula]

**Exercise 3(A)**

**1. Find the amount and the compound interest on â‚¹12,000 in 3 years at 5% compounded annually.**

**Solution:**

Given: P = â‚¹12,000: n = 3 years and r = 5%

We know that,

Amount = P(1 + r/100)^{n}Â

= 12000 (1 + 5/100)^{3}

= 12000 (21/20)^{3}

= â‚¹13,891.50

Therefore,

Compound Interest (C.I.) = â‚¹13,891.50 – â‚¹12,000

= â‚¹1,891.50

**2. Calculate the amount of â‚¹15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% and 10% respectively.**

**Solution:**

Given: P = â‚¹15,000; n = 2 years; r_{1}Â = 8% and r_{2}Â = 10%

We know that,

Amount = P(1 + r_{1}/100)(1 + r_{2}/100)

= 15000(1 + 8/100)(1 + 10/100)

= 15000(27/25)(11/10)

= â‚¹17,820

Therefore, the amount after 2 years is â‚¹17,820

**3. Calculate the compound interest accrued on â‚¹6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.**

**Solution:**

Given: P = â‚¹ 6,000; n = 3 years; r_{1 }= 5%; r_{2 }= 8% and r_{3}Â =10%

We know that,

Amount = P(1 + r_{1}/100)(1 + r_{2}/100)

= 6000(1 + 5/100)(1 + 8/100)(1 + 10/100)

= 6000(21/20)(27/25)(11/10)

= â‚¹7,484.40

Therefore,

C.I. = â‚¹7,484.40 – â‚¹6,000 = â‚¹1,484.40

**4. What sum of money will amount to â‚¹5,445 in 2 years at 10% per annum compound interest?**

**Solution: **

Given: Amount = â‚¹5,445; n = 2 years and r = 10%

We know that,

A =Â P(1 + r/100)^{n}

5445 =Â P(1 + 10/100)^{2}

5445 =Â P(11/10)^{2}

P = 5445(10/11)^{2}

= â‚¹4,500

Therefore, the principal amount is â‚¹4,500

**5. On what sum of money will the compound interest for 2 years at 5% per annum amount to â‚¹768.75?**

**Solution:**

Given: C.I. = â‚¹768.75; n = 2 years and r = 5%

We know that,

A = P(1 + r/100)^{n}

= P(1 + 5/100)^{2}

^{ }= P(21/20)^{2} = 441P/400

Thus,

A – P = C.I

441P/400 â€“ P = â‚¹768.75

41P/400 = â‚¹768.75

P = (768.75 x 400)/41

= â‚¹7,500

Therefore, the initial sum is â‚¹7500

**6. Find the sum on which the compound interest for 3 years at 10% per annum amounts to â‚¹1,655.**

**Solution:**

Given: C.I. = â‚¹1,655; n = 3 years and r = 10%

We know that,

A =Â P(1 + r/100)^{n}

= P(1 + 10/100)^{3}

= P(11/10)^{3}

= 1331P/ 1000

So,

C.I. = A – P

1655 = 1331P/ 1000Â â€“ P

1655 = 331P/ 1000

P = (1655 x 1000)/ 331

= 5000

Therefore, the initial sum is â‚¹5,000

**7. What principal will amount to â‚¹9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively?**

**Solution: **

Given: Amount = â‚¹9,856; n = 2 years; r_{1}Â =10% and r_{2}Â =12%

We know that,

Amount = P(1 + r_{1}/100) (1 + r_{2}/100)

9856 = P(1 + 10/100) (1 + 12/100)

9856 = P(11/10) (28/25)

So,

P = (9856 x 10 x 25)/(11 x 28)

= 8000

Therefore, the principal is â‚¹8,000

**8. On a certain sum, the compound interest in 2 years amounts toÂ â‚¹4,240. If the rate of interest for the successive years is 10% and 15% respectively, find the sum.**

**Solution: **

Given: C.I. = â‚¹4,240; n = 2 years; r_{1}Â = 10% and r_{2}Â = 15%

We know that,

A = P(1 + r_{1}/100) (1 + r_{2}/100)

Now,

A = P + C.I.

So,

(P + 4240) = P(1 + 10/100) (1 + 15/100)

(P + 4240) = P(11/10) (23/20)

P + 4240 = P(1.265)

1.265P â€“ P = 4240

P = 4240/0.265

= 16000

Therefore, the initial sum is â‚¹16,000

**Â **

**9. At what per cent per annum willÂ â‚¹6,000 amount toÂ â‚¹6,615 in 2 years when interest is compounded annually?**

**Solution: **

Given: P = â‚¹6000**; **A = â‚¹6,615 and n = 2 years

We know that,

A = P(1 + r/100)^{n}

6615 = 6000(1 + r/100)^{2}

(1 + r/100)^{2} = 6615/6000

(1 + r/100)^{2} = 1.1025

Taking square root on both sides, we get

(1 + r/100) = 1.05

1 + r/100 = 21/20

r/100 = 21/20 â€“ 1

r/100 = 1/20

r = 5

Therefore, the rate of compound interest is 5%

**Â **

**10. At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years?**

**Solution: **

Letâ€™s assume the principal to â‚¹y

Then, the amount will be â‚¹1.44y

And, n = 2 years

We know that,

A = P(1 + r/100)^{n}

1.44y = y(1 + r/100)^{2}

1.44y/y = (1 + r/100)^{2}

1.44 = (1 + r/100)^{2}

Taking square root on both sides, we get

1 + r/100 = 1.2

r/100 = 1.2 â€“ 1 = 0.2

r/100 = 2/10

r = 20

Therefore, the rate of compound interest is 20%

**11. At what rate per cent will a sum of â‚¹4,000 yield â‚¹1,324 as compound interest in 3 years?**

**Solution:**

Given: P = â‚¹4,000; C.I. = â‚¹1,324 and n = 3

Now, A = P + C.I.

So,

A = â‚¹4,000 + â‚¹1,324 = â‚¹5,324

We know that,

A = P(1 + r/100)^{n}

5324 = 4000(1 + r/100)^{3}

5324/4000 = (1 + r/100)^{3}

1331/1000 = (1 + r/100)^{3}

Taking cube root on both sides, we get

11/10 = (1 + r/100)

r/100 = 11/10 â€“ 1

r/100 = 1/10

r = 10

Therefore, the rate of compound interest is 10%

**12. A person invests â‚¹5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to â‚¹6,272. Calculate:**

**(i) the rate of interest per annum.**

**(ii) the amount at the end of the third year.**

**Solution:**

Given: P = â‚¹5,000; A = â‚¹6,272 and n = 2years

(i) We know that,

A = P(1 + r/100)^{n}

6272 = 5000(1 + r/100)^{2}

6272/5000 = (1 + r/100)^{2}

784/625 = (1 + r/100)^{2}

(28/25)^{2} = (1 + r/100)^{2}

Taking square root on both side, we get

28/25 = 1 + r/100

r/100 = 28/25 â€“ 1 = 3/25

r = (3 x 100)/25

Thus, r = 12%

(ii) Amount at the third year

A = 5000(1 + 12/100)^{3}

= 5000(28/25)^{3}

Therefore,

A = â‚¹7,024.64

**13. In how many years will â‚¹7,000 amount to â‚¹9,317 at 10% per annum compound interest?**

**Solution:**

Given: P = â‚¹7,000; A = â‚¹9,317 and r = 10%

We know that,

A = P(1 + r/100)^{n}

9317 = 7000(1 + 10/100)^{n}

9317/7000 = (11/10)^{n}

1331/1000 = (11/10)^{n}

(11/10)^{3} = (11/10)^{n}

On comparing, we have

n = 3

Therefore, the number of years is 3

**14. Find the time, in years, in which Rs4,000 will produce Rs630.50 as compound interest at 5% compounded annually.**

**Solution: **

Given: P = â‚¹4,000; C.I. = â‚¹630.50 and r = 5%

We know that,

C.I. = P[(1 + r/100)^{n} – 1]

630.50 = 4000[(1 + 5/100)^{n} – 1]

630.50/4000 = (1 + 5/100)^{n} â€“ 1

1261/8000 = (21/20)^{n} â€“ 1

1261/8000 + 1 = (21/20)^{n}

9261/8000 = (21/20)^{n}

(21/20)^{3} = (21/20)^{n}

On comparing, we have

n = 3

Therefore, the time in years is 3

**15. Divide â‚¹28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.**

**Solution: **

Letâ€™s assume the share of A as â‚¹y

Share of B = â‚¹(28,730 – y)

Rate of interest = 10%

Then, according to question

Amount of A in 3 years = Amount of B in 5 years

y(1 + 10/100)^{3} = (28730 – y)(1 + 10/100)^{5}

y = (28730 – y)(1 + 10/100)^{2}

y = (28730 – y)(11/10)^{2}

y = (121/100)(28730 – y)

100y = 121 (28730 – y)

100y = 121 x 28730 â€“ 121y

221y = (121 x 28730)

y = (121 x 28730)/ 221

= 15730

Therefore, the share of A = â‚¹15,730** **and share of B = â‚¹28,730 – â‚¹15,730 = â‚¹13,000

**16. A sum of â‚¹44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10% per annum compound interest, they will receive equal amounts on reaching 16 years of age.**

**(i) What is the share of each out of â‚¹44,200?**

**(ii) What will each receive, when 16 years old?**

**Solution: **

(i) Letâ€™s assume the share of John = â‚¹y

So, the share of Smith = â‚¹(44,200 – y)

Rate of interest = 10%

According to question, we have

Amount of John in 4 years = Amount of Smith in 2 years

y(1 + 10/100)^{4} = (44200 – y)(1 + 10/100)^{2}

y(1 + 10/100)^{2} = (44200 – y)

y(11/10)^{2} = (44200 – y)

121y/100 = (44200 â€“ y)

121y = 100(44200 – y)

121y + 100y = 4420000

221y = 4420000

y = 20000

Therefore, share of John = â‚¹20,000 and

Share of Smith= â‚¹44,200 – â‚¹20,000 = â‚¹24,200

(ii) Amount that each will receive

= 20000(1 + 10/100)^{4}

= 20000(11/10)^{4}

= 29282

Therefore, the amount that each will receive is â‚¹29,282

**17. The simple interest on a certain sum of money and at 10% per annum is â‚¹6,000 in 2 years, Find:**

**(i) theÂ sum.**

**(ii) theÂ amount due to the end of 3 years and at the same rate of interest compounded annually.**

**(iii) theÂ compound interest earned in 3 years.Â **

**Solution:**

(i) Given: S.I. = â‚¹6000; n = 2 years and R = 10%

We know that,

I = PTR/100

So,

P = (I x 100)/(R x T)

= (6000 x 100)/ (10 x 2)

= 30000

Thus, the sum of money is â‚¹30,000

(ii) Now, P = â‚¹30,000; n = 3 years and r = 10%

We know that,

A = P(1 + r/100)^{n}

= 30000(1 + 10/100)^{3}

= 30000(11/10)^{3}

= 30 x 11^{3}

= 39930

Thus, the amount is â‚¹39,930

(iii) The C.I. earned in 3 years = A â€“ P = â‚¹39,930 â€“ â‚¹30,000 = â‚¹9,930

**18. Find the difference between compound interest and simple interest on â‚¹8,000 in 2 years and at 5% per annum.**

**Solution:**

Given: P = â‚¹8000, R = 5% and T = 2 years

To calculate simple interest,

S.I. = (P x R x T)/100

= (8000 x 5 x 2)/100

= â‚¹800

To calculate compound interest,

A = P(1 + r/100)^{n}

= 8000(1 + 5/100)^{2}

= 8000(105/100)^{2}

= 8000(21/20)^{2}

= 8820

Thus, the amount is â‚¹8820

So,

C.I. = A â€“ P

= â‚¹(8820 â€“ 8000)

= â‚¹820

Thus, the compound interest is â‚¹820

**Exercise 3(B)**

**1. The difference between simple interest and compound interest on a certain sum isÂ â‚¹54.40 for 2 years at 8 per cent per annum. Find the sum.**

**Solution: **

Letâ€™s assume the principal (P) = x

R = 8%

T = 2 years

Now,

The simple interest is calculated as

S.I. = (x Ã— 8 x 2)/100

= 4x/25

The compound interest is calculated as

C.I. = A â€“ P

= x(1 + 8/100)^{2} â€“ x

= x[(1 + 2/25)^{2} â€“ 1]

= x[(27/25)^{2} – 1]

= 104x/625

Given, C.I. = S.I. = 54.40

104x/625 â€“ 4x/25 = 54.40

x (104/625 â€“ 4/25) = 54.40

x (104/625 â€“ 100/625) = 54.40

x (4/625) = 54.40

x = (54.40 x 625)/4

= 8500

Thus, the principal sum is â‚¹8,500

**2. A sum of money, invested at compound interest, amounts toÂ â‚¹19,360 in 2 years and toÂ â‚¹23,425.60 in 4 years. Find the rate per cent and the original sum of money.**

**Solution: **

Given: Amount after 2 years = â‚¹19360; So, n = 2 years and

Amount after 4 years = â‚¹23,425.60; So, n = 4

Letâ€™s assume the principal as X and the rate of C.I. as R

Now, we have

X(1 + R/100)^{2} = 19360 … (1)

And,

X(1 + R/100)^{4} = 23425.60 â€¦ (2)

Performing (2)Â _{}(1), we have

(1 + R/100)^{2} = 23425.60/19360

= 2342560/1936000

= 14641/12100

= (121/110)^{2}

Now, taking square root on both sides we get

(1 + R/100) = 121/110

R/100 = 121/110 â€“ 1

R = (11/110) x 100

= 10

Therefore, the rate of C.I. is 10%

Now,

Form (1), we have

X(1 + 10/100)^{2} 19360

X(11/10)^{2} = 19360

X = (19360 x 10 x 10)/ (11 x 11)

= 16000

Thus, the principal sum is â‚¹16,000

**3. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money becomes twenty-seven times of itself at the same rate of interest p.a.**

**Solution:**

Letâ€™s assume the principal as x

Amount (A) = 3x, n = 8 years, R =?

We know that,

A = P(1 + R/100)^{n}

Now,

**Case I:**

3x = x(1 + R/100)^{8}

Taking the 8^{th} root on both sides, we have

3^{1/8} = (1 + R/100) â€¦ (1)

**Case II:**

P = x, A = 27x, T =?

27x = x(1 + R/100)^{T}

27^{1/T} = 1 + R/100 â€¦ (2)

From (1) and (2), we have

3^{1/8} = 27^{1/T}

3^{1/8} = (3^{3})^{1/T}

3^{1/8} = 3^{3/T}

On comparing the exponents,

1/8 = 3/T

T = 3 x 8 = 24

Thus, it will take 24 years for the money to become twenty-seven times of itself at the same rate of interest p.a.** **

**4. On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest onÂ **â‚¹**9,430 for 10 years, both at the rate of 5 per cent per annum?**

**Solution:**

Given: P = â‚¹9430, R = 5% and n = 10 years

Now, the simple interest is calculated as

SI = PNR/100

SI =Â (943 x 5 x 10)/100

= â‚¹4,715

Now, letâ€™s assume a sum(principal) x

We have,

CI = â‚¹4,715; T = 2 years and R = 5%

We know that,

CI = A â€“ P

4715 = x(1 + R/100)^{T} â€“ x

= x(1 + 5/100)^{2} â€“ x

= x[(21/20)^{2} – 1]

= x[(441 – 400)/400]

= 41x/400

x = (4715 Ã— 400)/41

= 46000

Thus, the principal sum is â‚¹46,000

**5. Kamal and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand receivedÂ **â‚¹**15 more than Kamal. Find the amount of money lent by each and the interest received.**

**Solution:**

Letâ€™s assume the principal as Rs. 100, R = 5% and for T = 2 years

Then,

For Kamal,

SI = (100 x 5 x 2)/100

= â‚¹10Â

And,

For Anand,

A = P (1 + R/100)^{T}

= 100 (1 + 5/100)^{T}

= 100 (21/20)^{2}

= â‚¹441/4

So,

CI = 441/4 â€“ 100

= (441 – 400)/4

= â‚¹41/4

The difference of CI and SI =Â 41/4 â€“ 10

_{ }= (41 – 40)/4

= â‚¹1/4Â

Now,

When the difference is â‚¹1/4, the principal is â‚¹100

So,

If the difference is 1, the principal = 100Â x 4

= â‚¹400

And if the difference is Rs, 15, the principal = 100Â x 4Â x 15

= â‚¹6000

Hence,

For Kamal, interest =Â (6000 x 5 x 2)/100

= â‚¹600

Â

For Anand, interest = 6000 (1 + 5/100)^{2} â€“ 6000

= 6000 [(21/20)^{2} – 1]

= 6000 (441/400 – 1)

= 6000 (41/400)

= â‚¹615

**6. Simple interest on a sum of money for 2 years at 4% isÂ **â‚¹**450. Find compound interest of the same sum and at the same rate for 2 years.**

**Solution:**

Given: SI = Rs. 450; R = 4%; T = 2 years

Now,

P = (SI x 100)/(R x T)

= (450 x 100)/(4 x 2)

= â‚¹5,625Â

Â

Now, P = â‚¹5,625, R = 4% and T = 2 years

So, the amount is calculated as

A = 5625(1 + 4/100)^{2}

= 5625(26/25)^{2}

= 3802500/625

= â‚¹6084

Hence,

CI = A – P

= 6084 – 5625

= â‚¹459

**7. Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum byÂ **â‚¹**228. Find the sum.**

**Solution:**

Letâ€™s consider the principal as P,

Given: R = 4% and T = 4 years for simple interest and R = 5% and T = 3 years for compound interest

Now,

SI = (P x 4 x 4)/100

= 4P/25

And,

CP = P (1 + 5/100)^{3} â€“ P

= P [(21/20)^{3} – 1]

= P (9261/8000 – 1)

= 1261P/8000

Given: SI – CI = â‚¹228

So,

4P/25 – 1261P/8000 = 228

(4 x 320P â€“ 1261P)/8000 = 228

19P = 228 x 8000

P = (228 x 8000)/19

= 96000

Thus, the principal sum is â‚¹96000

**8. Compound interest on a certain sum of money at 5% per annum for two years isÂ **â‚¹**246. Calculate simple interest on the same sum for 3 years at 6% per annum.**

**Solution:**

Given: CI = â‚¹246, R = 5% and T = 2 years

CI = A â€“ P

246 = P (1 + 5/100)^{2} â€“ P

246 = P [(21/20)^{2} – 1]

246 = P (41/400)

P = (246 x 400)/41

= 2400

Â

Now, P = â‚¹2400, R = 6% and T = 3 years

SI = (2400 x 6 x 3)/100

= â‚¹432

Hence, the simple interest is â‚¹432

**9. A certain sum of money amounts to **â‚¹**23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.**

**Solution:**

Letâ€™s the sum (principle) asÂ *x*

Given: Amount = â‚¹23,400; R = 10% and T = 3 years

Now,

SI = (x Ã— 10 Ã— 3)/100

= 3x/10

We know that,

Amount = Principle + Interest

23400 =Â *xÂ *+Â 3x/10

234000 = (10x + 3x)

x = 234000/13Â

*xÂ *= 18000

Now,

Principle = â‚¹18000, R = 10% and n = 2 years

So,

A = P (1 + R/100)^{n}

= 18000 (1 + 10/100)^{2}

= 18000 (11/10)^{2}

= 18000 (121/100)

= 21780

Hence, the amount is â‚¹21,780

**10. MohitÂ borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying **â‚¹**12,600 at the end of the first year and **â‚¹**17,640 at the end of the second year. Find the sum borrowed.**

**Solution:**

Letâ€™s assume the principal sum to be P

Now,

For the payment of â‚¹12,600 at the end of the first year, we have

A = â‚¹12,600; R = 5% and n = 1 year

So,

A = P(1 + R/100)^{n}

12600 = P(1 + 5/100)^{1}

12600 = P(21/20)

P = (12600 x 20)/21

= â‚¹12,000

Now, for the payment of â‚¹17,640 at the end of second year

A = â‚¹17,640; R = 5% and n = 2 years

We know that,

A = P(1 + R/100)^{n}

17640 =** **P(1 + 5/100)^{2}

17640 = P(21/20)^{2}

P = (20/21)^{2} x 17640

= 16000

Thus, the sum borrowed = â‚¹(12,000 + 16,000) = â‚¹28,000

**Exercise 3(C)**

**1. If the interest is compounded half-yearly, calculate the amount when principal is â‚¹7,400; the rate of interest is 5% per annum and the duration is one year.**

**Solution:**

Given: P = â‚¹7,400; r = 5% p.a. and n = 1 year

As the interest is compounded half-yearly, we have

A = P [1 + r/(2 x 100)] ^{n x 2}

= 7400 [1 + 5/(2 x 100)] ^{1×2}

= 7400 (1 + 1/40)^{2}

= 7400 (41/40)^{2}

= 7774.63

Hence, the amount is â‚¹7,774.63

**2. Find the difference between the compound interest compounded yearly and half-yearly on â‚¹10,000 for 18 months at 10% per annum.**

**Solution:**

Given: P = â‚¹10,000; n = 18 months = 1Â½ year and r = 10%p.a.

Now,

(i) When interest is compounded yearly

For 1 year,

A = P(1 + r/100)^{n}

= 10000 (1 + 10/100)^{1}

= 10000 (11/10)

= 11000

Hence, after the first year the amount is â‚¹11,000

Â

For Â½ year,

P = â‚¹11,000; n = 1/2 year and r = 10%

A = P[1 + r/(2 x100)]^{n x 2}

= 11000 [1 + 10/(2 x100)]^{1/2 x 2}

= 11000 (21/20)^{1}

= 11550

So, after 1Â½ year the amount is â‚¹11,550

Hence, the C.I = â‚¹11,550 â€“ â‚¹10,000 = â‚¹1,550

(ii) When interest is compounded half-yearly

A = P [1 + r/(2 x 100)]^{n x 2}

= 10000[1 + 10/(2 x 100)]^{3/2 x 2}

= 10000(21/20)^{3}

^{ } = 11,576.25

Hence, after 1Â½ years the amount when compounded half-yearly is â‚¹11,576.25

So,

C.I = â‚¹11,576.25 â€“ â‚¹10,000 = â‚¹1,576.25

Therefore, the difference between both C.I = â‚¹1,576.25 â€“ â‚¹1,550

= â‚¹26.25

Â

**3. A man borrowedÂ Rs.16,000 for 3 years under the following terms:**

**(i) 20% simple interest for the first 2 years.**

**(ii) 20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.**

**Find the total amount to be paid at the end of the three years.**

**Solution:**

Given: P = â‚¹16,000; N = 3 years

For the first 2 years, R = 20%

So,

S.I = (P x N x R)/100

= (16000 x 2 X 20)/100

= 6400

Hence, the amount after 2 year will be (P + S.I) = â‚¹(16,000 + 6400) = â‚¹22,400

This is the amount at the end of 2 years.

Now, for the remaining 1 year the interest is compounded half-yearly

So,

A = P[1 + r/(2 x 100)]^{n x 2}

** ^{ } **= 22400(1 + 20/200)

^{2}

= 22400(11/10)^{2}

Â = 27104Â

Therefore,

The total amount to be paid at the end of the three years isÂ â‚¹27,104.

**Â **

**4. What sum of money will amount toÂ **â‚¹**27,783 in one and a half years at 10% per annum compounded half yearly?**

**Solution:**

Given: A = â‚¹27,783; N = 1Â½ years and R = 10% compounded half-yearly

So,

A = P[1 + r/(2 x 100)]^{n x 2}

27783 = P[1 + 10/200]^{3/2 x 2}

27783 = P(1 + 1/20)^{3}

27783 = P(21/20)^{3}

P = 27783 x (20/21)^{3}

= 24000**Â **

Therefore,

The sum ofÂ â‚¹24,000 amounts toÂ â‚¹27,783 in one and a half years if compounded half yearly at 10% per annum.

**5. Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets **â‚¹**33 more than Ashok in 18 months, calculate the money invested.**

**Solution:**

Let P = Rs y; n = 18 months= 1Â½ year and r = 20% p.a.

(i) For Ashok (interest is compounded yearly)

For 1 year,

A = P(1 + r/100)^{n}

= y(1 + 20/100)^{1}

= (6/5)y

For Â½ year,

Now, P = (6/5)y; n = Â½ year and r = 20%

A = P[1 + r/(2 x 100)]^{n x 2}

= (6/5)y[1 + 20/(2 x100)]^{1/2 x 2}

^{ } = (66/50)y

(ii) For Geeta (interest is compounded half-yearly)

A = P[1 + r/(2 x 100)]^{n x 2}

= y[1 + 20/(2 x 100)]^{3/2 x 2}

= y(11/10)^{3}

= (1331/1000)y

Then, according to question

(1331/1000)y â€“ (66/50)y = 33

(11/1000)y = 33

y = (33 x 1000)/11

= 3000

Therefore, the money invested by each person is â‚¹3,000

**6. At what rate of interest per annum will a sum ofÂ **â‚¹**62,500Â earnÂ a compound interest ofÂ **â‚¹**5,100 in one year? The interest is to be compounded half-yearly.**

**Solution:**

Given: P = â‚¹62,500; C.I **= **â‚¹5100 and N = 1 (compounded half-yearly)

So,

C.I = P[1 + r/(2 x 100)^{2xn }– 1]

5100 = 62500[(1 + r/200)^{2} â€“ 1]

(1 + r/200)^{2} = 67600/62500

1 + r/200 = 260/250

r = 8

Therefore, the rate of interest is 8%

**7. In what time will **â‚¹**1,500 yield **â‚¹**496.50 as compound interest at 20% per year compounded half-yearly?**

**Solution:**

Given: P = â‚¹1,500; C.I. = â‚¹496.50 and r = 20% (compounded semi-annually)

Then,

C.I. = P[{1 + r/(2 x 100)^{nx2}} – 1]

496.50 = 1500{[1 + 20/(2 x 100)]^{nx2}) â€“ 1}

496.50/1500 = (11/10)^{2n} â€“ 1

331/1000 + 1 = (11/10)^{2n}

1331/1000 = (11/10)^{2n}

(11/10)^{3} = (11/10)^{2n}

On comparing, we get

2n = 3

n = 1Â½ years

Therefore, the time taken is 1Â½ years.

**8. Calculate the C.I. on **â‚¹**3,500 at 6% per annum for 3 years, the interest being compounded half-yearly.**

**Do not use mathematical tables. Use the necessary information from the following:**

**(1.06) ^{3}Â = 1.191016; (1.03)^{3}Â = 1.092727**

**(1.06) ^{6}Â = 1.418519; (1.03)^{6}Â = 1.194052**

**Solution:**

Given: P = â‚¹3,500; r = 6% and n = 3 years

As the interest is being compounded half-yearly,

Then,

C.I. = P[1 + r/(2 x 100)^{nx2} â€“ 1]

= 3500 {[1 + 6/(2 x 100)]^{3 x 2} – 1}

= 3500 [(103/100)^{6} – 1]

= 3500 [(1.03)^{6} â€“ 1]

= 3500 (1.194052 – 1)

= 3500 x 0.194052

= 679.18

Therefore,

The C.I. is â‚¹679.18

**9. Find the difference between compound interest and simple interest on **â‚¹**12,000 and inÂ 1Â½ years at 10% compounded yearly.**

**Solution:**

Given: P = â‚¹12,000; n =Â 1Â½ years and r = 10%

So,

S.I. = (P x R x T)/100

= (12000 x 10 x 3/2)/100

= 1800

Hence, the S.I. is â‚¹1,800

Now, calculating C.I.

For 1 year

P= â‚¹12,000; n = 1 year and r = 10%

Then,

A = P(1 + r/100)^{n}

= 12000(1 + 10/100)^{1}

= 13200

And, for next 1/2 year

P = â‚¹13,200; n = Â½ year and r = 10%

Then,

A = P[1 + r(2 x 100)]^{n x 2}

= 13200 [1 + 10/(2 x 100)]^{1/2 x 2}

= 13860

Hence, the C.I. =Â â‚¹13,860 – â‚¹12,000 = â‚¹1,860

So, Difference between C.I. and S.I = â‚¹1,860 – â‚¹1,800

= â‚¹60

**10. Find the difference between compound interest and simple interest on **â‚¹**12,000 and in 1Â½ years at 10% compounded half-yearly.**

**Solution:**

Given: P = â‚¹12,000; n =Â 1Â½ years and r = 10%

Then,

S.I. = (P x R x T)/100

= (12000 x 10 x 3/2)/100

= 1800

Next,

Calculating C.I.(compounded half-yearly)

A = P[1 + r/(2 x 100)]^{n x 2}

= 12000[1 + 10/(2 x 100)]^{3/2 x 2}

= 12000(21/20)^{3}

= 13891.50

So, the C.I. = â‚¹13,891.50 – â‚¹12,000 = â‚¹1891.50

Hence, the difference between C.I. and S.I = â‚¹1,891.50 – â‚¹1,800

= â‚¹91.50

**Exercise 3(D)**

**1. The cost of a machine is supposed to depreciate each year at 12% of its value at the beginning of the year. If the machine is valued at **â‚¹**44,000 at the beginning of 2008, find its value:**

**(i) at the end of 2009.**

**(ii) at the beginning of 2007.**

**Solution:**

Given:

Cost of machine in 2008 (P) = â‚¹44,000 and its depreciation rate = 12%

Then,

(i) The cost of machine at the end of 2009 will be

= P(1 â€“ r/100)^{n}

= 44000 (1 â€“ 12/100)^{2}

= 44000 (88/100)^{2}

= 34073.60

Hence, the cost of the machine at the end of 2009 is â‚¹34,073.60

(ii)Â The cost of machine at the beginning of 2007(P)

A = P(1 â€“ 4/100)^{n}

44000 = P(1 â€“ 12/100)^{1}

44000 = P(88/100)

P = (44000 x 100)/88

= 50000

Hence, the cost of the machine at the beginning of 2007 is â‚¹50,000

**2. The value of an article decreases for two years at the rate of 10% per year and then in the third year it increases by 10%. Find the original value of the article, if its value at the end of 3 years isÂ â‚¹40,095.**

**Solution:**

Letâ€™s assume x to be the value of the article

Given, the value of an article decreases for two years at the rate of 10% per year.

So,

The value of the article at the end of the 1^{st}Â year will be

x – 10% of x = x â€“ 0.10x = 0.90x

And,

The value of the article at the end of the 2^{nd}Â year will be

0.90x – 10% of (0.90x) = 0.90x â€“ 0.90x = 0.81x

Now,

The value of the article increases in the 3^{rd}Â year by 10%.

So, the value of the article at the end of 3^{rdÂ }year will be

0.81x + 10% of (0.81x) = 0.81x + 0.081 = 0.891x

Given that, the value of the article at the end of 3 years isÂ â‚¹40,095

Then,

0.891x = 40095

x = 40095/0.891

= 45000

Therefore,

The original value of the article isÂ â‚¹45,000.

**3, According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088?**

**Solution:**

Given: Population in 2009 (P) = 64,000

Letâ€™s assume that after â€˜nâ€™ years its population to be 74,088 (A)

Also, given

Growth rate = 5% per annum

We know that,

A = P(1 + r/100)^{n}

74088 = 64000(1 + 5/100)^{n}

74088/64000 = (21/20)^{n}

9261/8000 = (21/20)^{n}

(21/20)^{3} = (21/20)^{n}

On comparing, we get

n = 3 years

Therefore, it took 3 years after 2009 for the population of the town to reach 74,088

**4. The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.**

**Solution:**

Letâ€™s assume the population in the beginning of 1998 be P

Given, the population at the end of 1999 = 2,85,120 (A)

According to the question,

r_{1}Â = -12% and r_{2}Â = +8%

Then,

A = P(1 â€“ r_{1}/100) (1 + r_{2}/100)

285120 = P(1 â€“ 12/100) (1 + 8/100)

285120 = P(22/25) (27/25)

P = (285120 x 25 x 25)/(22 x 27)

= 300000

Therefore, the population of the town at the beginning of 1998 was 3,00,000

**5. A sum of money, invested at compound interest, amounts to **â‚¹**16,500 in 1 year and to **â‚¹**19,965 in 3 years. Find the rate per cent and the original sum of money invested.**

**Solution:**

Letâ€™s assume the sum of money to be P and the rate of interest = r%

Given, the amount after 1year = â‚¹16,500

And, the amount after 3years = â‚¹19,965

For 1year

A = P(1 + r/100)^{n}

16500 = P(1 + r/100)^{1} â€¦ (1)

For 3years

A = P(1 + r/100)^{n}

19965 = P(1 + r/100)^{3} â€¦ (2)

On dividing equationsÂ (2) by (1), we have

121/100 = (1 + r/100)^{2}

(11/10)^{2} = (1 + r/100)^{2}

On comparing, we have

11/10 = 1 + r/100

11/10 x 100 = 100 + r

110 = 100 + r

r = 10%

Putting the value of r in equationÂ (1), we get

16500 = P(1 + 10/100)

P = (16500 x 10)/11

= 15000

Therefore, the rate of interest and original sum of invested are 10% and â‚¹15,000

**6. The difference between C.I. and S.I. on **â‚¹**7,500 for two years is **â‚¹**12 at the same rate of interest per annum. Find the rate of interest.**

**Solution:**

Given: P = â‚¹7,500 and time (n) = 2 years

Letâ€™s assume the rate of interest as y%

Now,

S.I. = P x R x T/100

= (7500 x y x 2)/100

= 150y

And,

C.I. = P(1 + r/100)^{n} â€“ P

= 7500(1 + y/100)^{2} â€“ 7500

Itâ€™s given that, C.I. – S.I. = â‚¹12

7500(1 + y/100)^{2} â€“ 7500 – 150y = 12

7500(1 + y^{2}/10000 + 2y/100) â€“ 7500 – 150y = 12

7500 + 75y^{2}/100 + 150y â€“ 7500 – 150y = 12

75y^{2}/100 = 12

3y^{2}/4 = 12

y^{2} = (12 x 4)/3 = 16

y = 4%

Hence, the rate of interest is 4%

**7. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.**

**Solution:**

Letâ€™s assume the principal to be â‚¹y and rate = r%

From the question, according to 1^{st}Â condition

Amount in 10 years = â‚¹3y

We know that,

A = P(1 + r/100)^{n}

3y = y(1 + r/100)^{10}

3 = (1 + r/100)^{10} â€¦ (1)

According to 2^{nd}Â condition

Letâ€™s consider that after n years amount will be â‚¹27y

A = P(1 + r/100)^{n}

27y = y(1 + r/100)^{n}

3^{3} = (1 + r/100)^{n}

Using (1), we have

[(1 + r/100)^{10}]^{3} = (1 + r/100)^{n}

(1 + r/100)^{30} = (1 + r/100)^{n}

On comparing,

n = 30

Hence, it will 30 years for the money to become 27 times of itself.

**8. Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by payingÂ **â‚¹**19,360 at the end of the second year andÂ **â‚¹**31,944 at the end of the third year he clears the debt; find the sum borrowed by him.**

**Solution:**

Letâ€™s assume the sum which Sharma borrowed as P

Given that the rate of interest is 10% and compounded annually

Now, according to the question

At the end of the two years the amount will

A_{1} = P(1 + r/100)^{n}

= P(1 + 10/100)^{2}

= P(11/10)^{2}

And,

Mr. Sharma paidÂ â‚¹19,360 at the end of the second year.

So, for the third year the principal will be (A_{1}Â – â‚¹19,360)

Also given, he cleared the debt by payingÂ â‚¹31,944 at the end of the third year.

Now,

A_{2}^{ }= P(1 + r/100)^{n}

31944 = P[(1 + 10/100)^{2} â€“ 19360] (1 + 10/100)^{1}

29040 = P[(11/10)^{2} â€“ 19360]

P(11/10)^{2} = 48400

P = 48400 x (10/11)^{2}

= 40000

Therefore, Mr. Sharma borrowedÂ â‚¹40,000.

**9. The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is **â‚¹**15. Find the sum of money lent out.**

**Solution:**

Letâ€™s assume the sum of money to be â‚¹y

Calculating the S.I.

S.I. = (P x R x T)/100

= (y x 10 x 1)/100

= y/10

And,

Calculating C.I.(compounded half-yearly)

C.I. = P {[1 + r/(2 x 100)]^{nx2} â€“ 1}

= y {[1 + 10/(2 x 100)]^{1×2} â€“ 1}

= y[(21/20)^{2} – 1]

= (41/400)y

Itâ€™s given that C.I. â€“ S.I. = â‚¹15

So,

(41/400)y â€“ y/10 = 15

y/400 = 15

y = 6000

Therefore, the sum of money lent out is â‚¹6,000

**10. The ages ofÂ PramodÂ andÂ RohitÂ are 16 years and 18 years respectively. InÂ what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?**

**Solution:**

Letâ€™s assume that â‚¹x andÂ â‚¹y to be the money invested byÂ PramodÂ andÂ RohitÂ respectively such that they will get the same sum on attaining the age of 25 years.

Now,

PramodÂ will attain the age of 25 yearsÂ after (25 â€“ 16) = 9 years

RohitÂ will attain the age of 25 yearsÂ after (25 -18) = 7 years

So, we have

x (1 + 5/100)^{9} = y(1 + 5/100)^{7}

x/y = 1/(1 + 5/100)^{2}

x/y = 400/441

Â

Therefore,

PramodÂ andÂ RohitÂ should invest in the ratio 400:441 respectively such that they will get the same sum on attaining the age of 25.

**Exercise 3(E)**

**1. Simple interest on a sum of money for 2 years at 4% isÂ **â‚¹**450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half yearly.**

**Solution:**

Given: S.I. = â‚¹450; N = 2 years and rate(R) = 4%

Letâ€™s consider the principal to be P

Now, we have

P = (S.I x 100)/(R x T)

= (450 x 100)/(4 x 2)

= 5625

Thus, the principal is â‚¹5625

Now,

When the interest is compounded half-yearly

Â P =Â â‚¹5,625; n = 1 year and r = 4%

A = P[1 + r/(2 x 100)]^{n x 2}

= 5625(1 + 4/200)^{1 x 2}

= 5625(51/50)^{2}

= 5852.25

Hence,

C.I. = â‚¹5852.25 â€“ â‚¹5,625

= â‚¹227.25

**2. Find the compound interest to the nearest rupee on **â‚¹**10,800 for 2Â½ years at 10% per annum.**

**Solution:**

Given: P = â‚¹10,800; Time (N) = 2Â½ years and rate = 10% p.a

For 2years,

We know that,

A = P(1 + r/100)^{n}

= 10800(1 + 10/100)^{2}

= 13068

And,

For Â½ year

A = P[1 + r/(2 x 100)]^{n x 2}

= 13,068(1 + 10/200)^{1/2 x 2}

= 13068 x 21/20

= 13721.40

~ 13721 (nearest rupee)

Hence,

C.I. = A â€“ P = â‚¹(13,721 â€“ 10,800) = â‚¹2,921**Â **

**Â **

**3. The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value isÂ **â‚¹**97,200, find:**

**i.Â Its value after 2 years.**

**ii.Â Its value when it was purchased.**

**Solution:**

Given,

Present value of machine(P) = â‚¹97,200

Depreciation rate = 10%

(i) Value of machine after 2 years = P(1 â€“ r/100)^{n}

= 97200 (1 â€“ 10/100)^{2}

= 97200 x (9/10)^{2}

= 78732

Thus, the value of the machine after 2 years is â‚¹78,732

(ii) Now,

Â To calculate the cost 2 years ago

We know that,

A = P(1 â€“ r/100)^{n}

97200 = P(1 â€“ 10/100)^{2}

97200 = P(9/10)^{2}

P = 97200 x (10/9)^{2}

= 1,20,000

Thus, the value of the machine when it was purchased was â‚¹1,20,000

**Â **

**4. AnujÂ and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh receivedÂ **â‚¹**64 more thanÂ Anuj. Find the money lent by each and interest received.**

**Solution:**

Letâ€™s assume the sum of money lent by both asÂ â‚¹y

Then,

ForÂ Anuj

P =Â â‚¹y; rate = 8% and time = 2 years

So,

S.I. = (P x R x T)/100

= (y x 8 x 2)/100

= 4y/25

For Rajesh

P =Â â‚¹y; rate = 8% and time = 2 years

C.I. = P [{1 + r/(100)}^{n} – 1]

= y [{1 + 8/(100)}^{2} – 1]

= 104y/625

Now,

Itâ€™s given that, difference in the interests i.e. C.I.Â – S.I. =Â â‚¹64

So,

104y/625 â€“ 4y/25 = 64

(104y â€“ 100y)/625 = 64

4y/625 = 64

y = (64 x 625)/4

= 10,000

Therefore,

The interest received by Anuj = (4 x â‚¹10,000)/25 = â‚¹1600

The interest received by Rajesh = (104 x â‚¹10,000)/625 = â‚¹1664

**Â **

**5. Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest onÂ **â‚¹**4,715 for 5 years, both at the rate of 5% per annum.**

**Solution:**

Given: Principal = â‚¹4,715; time = 5 years and rate = 5% p.a.

Letâ€™s assume the sum of money as P

So,

S.I. = (P x R x T)/100

= (4715 x 5 x 5)/100

= â‚¹1178.75

Then, C.I. = â‚¹1,178.75 x 4 =Â â‚¹4,715 (according to question)

We have, time = 2 years and rate = 5%

C.I. = P [(1 + r/100)^{n} – 1]

4715 = P[(1 + 5/100)^{2} – 1]

4715 = P(41/400)

P = (4715 x 400)/41

= 46000

Therefore, the sum of money is â‚¹46,000

**6. A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted toÂ **â‚¹**4,950, find the sum invested.**

**Solution:**

Given: C.I. for the 2^{nd}Â year =Â â‚¹4,950 and rate = 15%

Then,

C.I. = P [(1 + r/100)^{n} – 1]

4950 = P [(1 + 15/100)^{1} â€“ 1]

4950 = P(3/20)

P = (4950 x 20)/3

= 33000

Then, the amount at the end of 2^{nd} year is â‚¹33,000**Â **

So, for the first 2 years

A =Â Rs.33,000; r_{1}Â =10%

A = P(1 + r_{1}/100)

33000 = P (1 + 10/100)

33000 = P (1 + 11/10)

P = (33000 x 10)/11

= 30,000

Thus, the sum invested isÂ â‚¹30,000.

**7. A sum of money is invested at 10% per annum compounded half yearly. If the difference of amounts at the end of 6 months and 12 months is **â‚¹**189, find the sum of money invested.**

**Solution:**

Letâ€™s assume the sum of money to beÂ â‚¹y

And, given rate = 10% p.a. compounded half yearly

Now, for first 6 months

A = P[1 + r/(2 x 100)]^{n x 2}

= y[1 + 10/(2 x 100]^{1/2 x 2}

= y(1 + 10/200)^{1}

= (21/20)y

And,

For first 12 months

A = P[1 + r/(2 x 100)]^{n x 2}

= y[1 + 10/(2 x 100]^{1 x 2}

= y(1 + 10/200)^{2}

= (441/400)y

Also given, the difference between the above amounts =Â â‚¹189

So,

(441/400)y â€“ (21/20)y = 189

(21/400)y = 189

y = (189 x 400)/21

y = 3600

Thus, the sum of money invested is â‚¹3,600

**8. RohitÂ borrowsÂ **â‚¹**86,000 fromÂ ArunÂ for two years at 5% per annum simple interest. He immediately lends out this money toÂ AkshayÂ at 5% compound interest compounded annually for the same period. CalculateÂ Rohit’sÂ profit in the transaction at the end of two years.**

**Solution:**

Given: P =Â â‚¹86,000; time = 2 years and rate = 5% p.a.

Calculating S.I.

S.I. = (P x R x T)/100

= (86000 x 5 x 2)/100

= 8600

Â

Calculating C.I.

C.I. = P [(1 + r/100)^{n} – 1]

= 86000 [(1 + 5/100)^{2} – 1]

= 86000 (41/40)

= 8815

Thus, the profit = C.I. – S.I. =Â â‚¹(8,815 –Â 8,600) =Â â‚¹215

**9. The simple interest on a certain sum of money for 3 years at 5% per annum isÂ **â‚¹**1,200. Find the amount and the compound interest due on this sum of money at the same rate and after 2 years. Interest is reckoned annually.**

**Solution:**

Letâ€™s assumeÂ â‚¹P to be the sum of money

Rate = 5% p.a., S.I. =Â â‚¹1,200 and n = 3years.

Then,

1200 = (P x 5 x 3)/100

x = 1200/15

= 8000

So,

The amount due and the compound interest on this sum of money at the same rate and after 2 years

Â P =Â â‚¹8,000; rate = 5% p.a. and n = 3 years

We know that,

A = P (1 + r/100)^{n}

= 8000 (1 + 5/100)^{2}

= 8000 (1.1025)

= 8820

Hence, C.I. = A â€“ P = â‚¹(8,820 â€“ 8,000) = â‚¹820**Â **

The amount due after 2 years isÂ â‚¹8,820 and the compound interest isÂ â‚¹820

**10. Nikita investsÂ **â‚¹**6,000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts toÂ **â‚¹**6,720. Calculate:**

**(a) The rate of interest.**

**(b) The amount at the end of the second year.**

**Solution:**

Letâ€™s assume x% to be the rate of interest

P = â‚¹6000, n = 2 year and A = â‚¹6720

For the first year

A = P (1 + r/100)^{n}

6720 = 6000 (1 + x/100)^{1}

6720 = 6000 + 10x

6720 â€“ 6000 = 60x

x = 720/10

= 12

Hence,

The rate of interest is 12%.

So,

The amount at the end of the second year will be

A = 6000 (1 + 12/100)^{2}

= 6000 (112/100)^{2}

= 7526.40

Therefore,

The amount at the end of the second year isÂ â‚¹7,526.40

## Selina Solutions for Class 9 Maths Chapter 3- Compound Interest [Using Formula]

The Chapter 3, Compound Interest [Using Formula], contains 5 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

3.1 Introduction

3.2 Using Formula

3.3 Inverse Problems

3.4 Miscellaneous Problems

3.5 When the Interest Is Compounded Half-Yearly (Two Times in a Year)

3.6 When the Time is Not an Exact Number of Years And the Interest is Compounded Yearly

3.7 Other Applications of the Formula

## Selina Solutions for Class 9 Maths Chapter 3- Compound Interest [Using Formula]

The Chapter 3 of class 9 is a continuation of chapter 2. In chapter 2 the students were taught to calculate the compound interest as a repeated simple interest computation with a growing principal. This calculation becomes tedious as the number of conversion period increases. To make the calculation of compound interest easier, certain formulae are used. Read and learn the Chapter 3 of Selina textbook to get to know more about Compound Interest [Using Formula]. Learn the Selina Solutions for Class 9 effectively to attain excellent result in the examination.