# Selina Concise Solutions for Chapter 9 Simple and Compound Interest Class 8 ICSE Mathematics

**Exercise 9A**

**1. Find the interest and the amount on:**

**(i) ₹ 750 in 3 years 4 months at 10% per annum.**

**(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.**

**(iii) ₹ 2,600 in 2 years 3 months at 1% per month.**

**(iv) ₹ 4,000 in 1 years at 2 paise per rupee per month.**

**Solution**

**(i)**Given P = ₹750

∴ Amount (A) = P + I = ₹750 + ₹250 = ₹1000

**(ii)**Principal (P) = ₹5000

Rate (R) = 8% p.a.

Time (T) = 23 December 2011 to 29 July 2012

= 10 × 8 × 3

= ₹240

∴ Amount = P + I = ₹5000 + 240 = ₹5240

Time (T) = 2 years 3 months = 27 months

Rate (R) = 1 % per month

∴ Interest = (P × T × R)/100

∴ Amount = P + I = ₹5000 + 240 = ₹5240

**(iii)**Here P = ₹2,600Time (T) = 2 years 3 months = 27 months

Rate (R) = 1 % per month

∴ Interest = (P × T × R)/100

= (2600 × 27 × 1)/100

= 26 × 27

= Rs. 702

∴ Amount = Rs.(2600 + 702) = Rs. 3302

= 1 year + 12/3 months = 16months

Rate (R) = 2paise per rupee-per month = 2% per month

∴ Interest (I) = (P × R × T)/100

= (4,000 × 2 × 16)/100

= 40 × 32

= Rs.1280

∴ Amount(A) = P + I

= Rs. 4000 + Rs. 1280

= Rs. 5280

Principal (P) = Rs. 24,000

Rate (R) = 7.5% P.A.

Time (T) = 4 years

S.I. = (P × R × T)/100

= Rs. (24,000 × 4 × 7.5)/100

= Rs. 240 × 4 × 7.5

= 240 × 30 = Rs. 7200

Amount needed to clear the debt at the end of 4th year = Rs. 24000 + Rs. 7200 = Rs. 3,1200

Let P = Rs x

Time (T) = 2years

Rate (R) = 0%

∴ Interest = (P × R × T)/100

= (x × 10 × 2)/100 = x/5

x/5 = Rs. 1480

∴ x = 1480 × 5 = Rs. 7400

Hence the money Rs. 7400

Let principal = Rs. P

Time (T) = 6 years 3 months = 6 year + 3/12

year = 75/12 = 25/4 year = 6 ¼ years

Rate (R) = 5%

Simple Interest = Rs. 7,008

We know that

Simple Interest = (P × R × T)/100

⇒ 7,008 = (P × 25/4 × 5)/100

= 26 × 27

= Rs. 702

∴ Amount = Rs.(2600 + 702) = Rs. 3302

**(iv)**Here P = Rs. 4,000, Time (T) = 1 1/3 year= 1 year + 12/3 months = 16months

Rate (R) = 2paise per rupee-per month = 2% per month

∴ Interest (I) = (P × R × T)/100

= (4,000 × 2 × 16)/100

= 40 × 32

= Rs.1280

∴ Amount(A) = P + I

= Rs. 4000 + Rs. 1280

= Rs. 5280

**2. Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt?**

**Solution**Principal (P) = Rs. 24,000

Rate (R) = 7.5% P.A.

Time (T) = 4 years

S.I. = (P × R × T)/100

= Rs. (24,000 × 4 × 7.5)/100

= Rs. 240 × 4 × 7.5

= 240 × 30 = Rs. 7200

Amount needed to clear the debt at the end of 4th year = Rs. 24000 + Rs. 7200 = Rs. 3,1200

**3. The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 per cent per year. Find the sum of money.**

**Solution**Let P = Rs x

Time (T) = 2years

Rate (R) = 0%

∴ Interest = (P × R × T)/100

= (x × 10 × 2)/100 = x/5

x/5 = Rs. 1480

**(Given)**∴ x = 1480 × 5 = Rs. 7400

Hence the money Rs. 7400

**4. On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year?**

**Solution**Let principal = Rs. P

Time (T) = 6 years 3 months = 6 year + 3/12

year = 75/12 = 25/4 year = 6 ¼ years

Rate (R) = 5%

Simple Interest = Rs. 7,008

We know that

Simple Interest = (P × R × T)/100

⇒ 7,008 = (P × 25/4 × 5)/100

⇒ P = (7008 × 100 × 4)/(25 × 5)

= (7008 × 16)/5

= 112128/5

= Rs. 22425.60

Let Principal = Rs. P, Time (T) = 4 years

Rate = 61/4 = 25/4%

Simple Interest = (P × R × T)/100

= (P × 25/4 × 4)/100

= (7008 × 16)/5

= 112128/5

= Rs. 22425.60

**5. Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% per annum.**

**Solution**Let Principal = Rs. P, Time (T) = 4 years

Rate = 61/4 = 25/4%

Simple Interest = (P × R × T)/100

= (P × 25/4 × 4)/100

= P/4

∴ Amount = P + P/4 = 5P/4

⇒ 5P/4 = 4000

⇒ 5P = 4 × 4000

P = (4 × 4000)/5

= 4 × 800

⇒ P = Rs. 3200

Here, principal = Rs. 3200

R = (100 × 1)/(P × T)

∴ Amount = P + P/4 = 5P/4

⇒ 5P/4 = 4000

⇒ 5P = 4 × 4000

P = (4 × 4000)/5

= 4 × 800

⇒ P = Rs. 3200

Here, principal = Rs. 3200

**6.****(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years?****(ii) At what rate per cent per year will a sum double itself in 6 years?**

**Solution****(i)**P = Rs. 630, I = Rs. 126, T = 4 yearsR = (100 × 1)/(P × T)

= (100 × 126)/(630 × 4)

= 100/20

= 5%

∴ Amount = 2 × Rs. 100 = Rs. 200

Interest = A - P

= Rs. 200 - Rs. 100

= Rs. 100

T = 6 ¼ years = 25/4 years

R = (100 × 1)/(P × T)

= (100 × 100)/(100 × 25/4)%

= (100 × 100)/100 × 4/25

= 16%

S.I. = Rs. 399

R = 7%

We know that :

T = (100 × I)/(P × R)

= (100 × 399)/(950 × 7)

= (10 × 21)/(5 ×7)

= 2 × 3

= 6years

P = Rs. 1200

I = A – P

= Rs. 1536 – Rs. 1200

We know that :

T = (100 × I)/(P × R)

= (100 × 336)/(1200 × 3.5)

= (100 × 336 × 10)/(1200 × 35)

= (28 × 10)/35

= 8years

Let P = Rs. 8

S.I. = Rs. 3/8 × 8

= Rs. 3

T = 61/4 years = 25/4 years

We know that :

R = (100 × I)/(P × T)

= (100 × 3)/(8 × 25/4)

= (100 × 3)/8 × 4/25

= 2 × 3

= 6%

A = Rs. 10210.20

R = 5% P.A.

T = May + June + July + August + Sept.+ Oct.

= 7 + 30 + 31 + 31 + 30 + 17

= 146/365days = 2/5 year

We know that:

P + I = A

⇒ P = Rs. 10010

∴ Money to be borrowed = Rs. 10010

Let P = Rs. 8

Interest = Rs. 8 × 5/8 = Rs. 5

R = 6%

T = (100 × I)/(P × R)

= (100 × 5)/(8 × 6)

= 500/48

= 125/12 years

= 10 5/ 12 years

= 10 years 5 months

[∵ 5/12 year = 5/12 × 12 months = 5 months]

∴ Time = 10 years 5 months

R = 6%

T = 3 years

S.I. = (P × R × T)/100

= Rs. (7000 × 6 × 3)/100

= Rs. 1260

R = 5%

T = 3 years

S.I. = (P × R × T)/100

= Rs. (9500 × 5 × 3)/100

= Rs. 1425

Total income from the interest = Rs. 1260 + Rs. 1425

= Rs. 2685

Total sum borrowed by Raj = Rs. 8000

In the First case:

P = Rs. 4500

R = 5%

T = 4 years

S.I. = (P × R × T)/100

= Rs. (4500 × 5 × 4)/100

= Rs. 900

In the Second case:

P = Rs. 8000 - Rs. 4500

= 3500

R = 6%

T = 4 years

S.I. = (P × R × T)/100

= 35 × 6 × 4 = Rs. 840

= Rs. (3500 × 6 × 4)/100

= 35 × 6 × 4 = Rs. 840

Total interest paid by Raj = Rs. 900 + Rs. 840

= Rs. 1740

In the first case:

P = Rs. 4800

R = x%

T = 4 ½ years = 9/2 years

Interest = P × R × T)/100

= Rs. (4800 × x × 9)/(100 × 2)

= Rs. 24 × x × 9

= Rs. 216x

In the second case:

P = Rs. 2500

R = x %

T = 6 years

Interest = (P × R × T)/100

= Rs. (2500 × x × 6)/100

= Rs. 25 × x × 6

= Rs. 150x

According to statement,

Interest in first case + Interest in second case = Rs. 2196

∴ Rs. 216x + Rs. 150x = Rs. 2196

⇒ Rs. 366x = Rs. 2196

⇒ x = 2196/366

⇒ x = 6

∴ Rate = 6%

P = Rs.2550

R = 7.5%

T = 8 months = 8/12 years

= 2/3 years

S.I. = (P × R × T)/100

= Rs. 2550 × 7.5 × 2/3 × 1/100

= Rs. (2550 × 7.5 × 2)/(3 × 100)

= Rs. (2550 × 5)/100

= Rs. 12750/100

= Rs. 127.50

Amount = P + I

= Rs. 2550 + Rs. 127.50

= Rs. 2677.50

Mohan paid in cash = Rs. 1422.50

Price of the television = Amount – Paid in cash

= Rs. 2677.50 – Rs. 1422.50

= Rs. 1255

= 5%

**(ii)**Let P = Rs. 100∴ Amount = 2 × Rs. 100 = Rs. 200

Interest = A - P

= Rs. 200 - Rs. 100

= Rs. 100

T = 6 ¼ years = 25/4 years

R = (100 × 1)/(P × T)

= (100 × 100)/(100 × 25/4)%

= (100 × 100)/100 × 4/25

= 16%

**7.****(i) In how many years will Rs.950 produce Rs.399 as simple interest at 7% ?****(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.**

**Solution****(i)**P = Rs. 950S.I. = Rs. 399

R = 7%

We know that :

T = (100 × I)/(P × R)

= (100 × 399)/(950 × 7)

= (10 × 21)/(5 ×7)

= 2 × 3

= 6years

**(ii)**A = Rs. 1536P = Rs. 1200

I = A – P

= Rs. 1536 – Rs. 1200

We know that :

T = (100 × I)/(P × R)

= (100 × 336)/(1200 × 3.5)

= (100 × 336 × 10)/(1200 × 35)

**[∵ 1/3.5 = 10/35]**= (28 × 10)/35

= 8years

**8. The simple interest on a certain sum of money is 3/8 of the sum in 6 1/4 years. Find the rate percent charged.**

**Solution**Let P = Rs. 8

S.I. = Rs. 3/8 × 8

= Rs. 3

T = 61/4 years = 25/4 years

We know that :

R = (100 × I)/(P × T)

= (100 × 3)/(8 × 25/4)

= (100 × 3)/8 × 4/25

= 2 × 3

= 6%

**9. What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.**

**Solution**A = Rs. 10210.20

R = 5% P.A.

T = May + June + July + August + Sept.+ Oct.

= 7 + 30 + 31 + 31 + 30 + 17

= 146/365days = 2/5 year

We know that:

P + I = A

⇒ P = Rs. 10010

∴ Money to be borrowed = Rs. 10010

**10. In what time will the interest on a certain sum of money at 6% be of itself?**

**Solution**Let P = Rs. 8

Interest = Rs. 8 × 5/8 = Rs. 5

R = 6%

T = (100 × I)/(P × R)

= (100 × 5)/(8 × 6)

= 500/48

= 125/12 years

= 10 5/ 12 years

= 10 years 5 months

[∵ 5/12 year = 5/12 × 12 months = 5 months]

∴ Time = 10 years 5 months

**11. Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.**

**Solution****In Ist case:**

P = Rs. 7000R = 6%

T = 3 years

S.I. = (P × R × T)/100

= Rs. (7000 × 6 × 3)/100

= Rs. 1260

**In IInd case:**

P = Rs. 9500R = 5%

T = 3 years

S.I. = (P × R × T)/100

= Rs. (9500 × 5 × 3)/100

= Rs. 1425

Total income from the interest = Rs. 1260 + Rs. 1425

= Rs. 2685

**12. Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and remainder at 6%. Find the total interest paid by him in 4 years.**

**Solution**Total sum borrowed by Raj = Rs. 8000

In the First case:

P = Rs. 4500

R = 5%

T = 4 years

S.I. = (P × R × T)/100

= Rs. (4500 × 5 × 4)/100

= Rs. 900

In the Second case:

P = Rs. 8000 - Rs. 4500

= 3500

R = 6%

T = 4 years

S.I. = (P × R × T)/100

= 35 × 6 × 4 = Rs. 840

= Rs. (3500 × 6 × 4)/100

= 35 × 6 × 4 = Rs. 840

Total interest paid by Raj = Rs. 900 + Rs. 840

= Rs. 1740

**13. Mohan lends Rs.4800 to John for 4 1/2 years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.**

**Solution**In the first case:

P = Rs. 4800

R = x%

**(Suppose)**T = 4 ½ years = 9/2 years

Interest = P × R × T)/100

= Rs. (4800 × x × 9)/(100 × 2)

= Rs. 24 × x × 9

= Rs. 216x

In the second case:

P = Rs. 2500

R = x %

T = 6 years

Interest = (P × R × T)/100

= Rs. (2500 × x × 6)/100

= Rs. 25 × x × 6

= Rs. 150x

According to statement,

Interest in first case + Interest in second case = Rs. 2196

∴ Rs. 216x + Rs. 150x = Rs. 2196

⇒ Rs. 366x = Rs. 2196

⇒ x = 2196/366

⇒ x = 6

∴ Rate = 6%

**14. John lent Rs. 2550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.**

**Solution**P = Rs.2550

R = 7.5%

T = 8 months = 8/12 years

= 2/3 years

S.I. = (P × R × T)/100

= Rs. 2550 × 7.5 × 2/3 × 1/100

= Rs. (2550 × 7.5 × 2)/(3 × 100)

= Rs. (2550 × 5)/100

= Rs. 12750/100

= Rs. 127.50

Amount = P + I

= Rs. 2550 + Rs. 127.50

= Rs. 2677.50

Mohan paid in cash = Rs. 1422.50

Price of the television = Amount – Paid in cash

= Rs. 2677.50 – Rs. 1422.50

= Rs. 1255

### Exercise 9 B

**1. The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.**

**Solution**

Let the sum borrowed = Rs. 100

Time = 3 years

Let rate of interest = r%

∴ Interest = (100 × 3 × r)/100

**[∵ S.I. = (P × R × T)/100]**

= 3r

= (0.24)(100)

= 24

(Given)

⇒ r = 24/3 = 8

Hence reqd. rate of interest = 8%

A = Rs. 4620

P = Rs. 3750

I = A – P = Rs. 4620 – Rs. 3750 = Rs. 870

T = 3 years

R = (100 × I)/(P × T)

= (100 × 870)/(3750 × 3)

= (100 × 290)/3750

= (4 × 29)/15

= 116/15

= 7 11/15%

In second case :

P = Rs. 7500

R = 116/15%

T = 5 ½ years = 11/2 years

= Rs. (7500 × 11 × 116)/(2 × 15 × 100)

= (250 × 116 × 11)/100

= 10 × 29 × 11

= 290 × 11

= Rs. 3190

Amount = Rs. 7500 + 3190

= Rs. 10, 690

Let P = Rs. 100

A = Rs. 200

I = Rs. 200 – Rs. 100 = Rs. 100,

T = 8 years

R = (100 × I)/(P × T)

= (100 × 100)/(100 × 8)

= 100/8

= 25/2%

Now, again P = Rs. 100

A = Rs. 300

I = Rs. 300 – Rs. 100

= Rs. 200

R = 25/2%

T = (100 × I)/(P × R)

= (100 × 200)/(100 × 25/2)

= (100 × 200 × 2)/(100 × 25)

= 16 years

So, the given sum of money will become triple in 16 years.

In first case:

A = Rs. 5000

P = Rs. 4000

I = A – P

= Rs. 5000 – Rs. 4000

= Rs. 1000

T = 8 years

R = (100 × I)/(P × R)

= (100 × 1000)/(4000 × 8)

= 25/8%

In the second case:

A = Rs. 2800

P = Rs. 2100

I = Rs. 2800 - Rs. 2100

= Rs. 700

R = 25/8%

T = (100 × I)/(P × R)

= (100 × 700)/(2100 × 25/8)

= (100 × 700 × 8)/(2100 × 25)

= 32/3 years

= 10 2/3 years

10 2/3 × 12 months

= 10 24/3 months

= 10 years 8 months

In first case:

P = Rs. 7500

R = 5%

T = 6 years

Interest = (P × R × T)/100

= Rs. (7500 × 5 × 6)/100

= Rs. 75 × 5 × 6

= Rs. 2250

In second case :

According to the statement, interest = Rs. 2250

R = 6.5% P.A.

T = 4 years

P = (100 × I)/(R × T)

= Rs. (100 × 2250)/(6.5 × 4)

= Rs. 225000/26

= Rs. 112500/13

= Rs. 8653.85

Required principal = Rs. 8653.85

In 6 years sum amounts to = Rs. 4050

In 4 years sum amounts to = Rs. 3825

∴ Interest of 2 years = Rs. 4050 – Rs. 3825

= Rs. 225

Interest of 4 years = Rs. 225/2 × 4

= Rs. 450

(∵ Rs. 225 is interest for 2 years)

Now P = A – I

= Rs. 3825 – Rs. 450

= Rs. 3375

I = Rs. 450

T = 4 years

R = (100 × I)/(P × T)

= (100 × I)/(P × T)

= (100 × 450)/(3375 × 4)

= 45000/13500%

= 450/135%

= 10/3%

= 3 1/3%

∴ R = 3 1/3 %

P = Rs. 3375

P = Rs. 3750

I = Rs. 3750 × 1/5

= Rs. 750

T = 4 years

R = (100 × I)/(P × T)

= (100 × 750)/(3750 × 4)

= (100 × 750)/(3750 × 4)

= 5%

Again, P = Rs. 3750

Interest of 4 years = Rs. 750

Interest of 1 year = Rs. 750/4

Interest of 15 years = Rs.750/4 × 15

= Rs. (750 × 15)/4

= Rs. 5625/2

= Rs. 2812.50

Amount in 15 years will be = Rs. 3750 + Rs. 2812.50

= Rs. 6562.50

∴ Rate = 5%

Amount in 15 years will be = Rs. 6562.50

P = Rs. 1950

A = Rs. 2125.50

R = 5% p.a

I = A – P

= Rs. 2125.50 – Rs. 1950

= Rs. 175.50

T = (100 × I)/(P × R)

= (100 × 175.50)/(1950 × 5)

= 17550/9750

= 1755/975

= 117/65

= 9/5 years

= 4 1/5 years

= 1 years 292 days

∵ 4/5 years = 4/5 × 365 days = 292 days

Jan. + Feb. + March + April + May + June + July + Aug. + Sept. + Oct.

(31 – 5) + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 23

= 292 days

∴ Required date = 23rd October 2012

In first case:

P = Rs. 2400

R = x% (Assume)

T = 3 years

Interest = (P × R × T)/100

= Rs. (2400 × x × 3)/100

= Rs. 72x

In second case:

P = Rs. 2000

R = x%

T = 3 years

Interest = (P × R × T)/100

= Rs. (2000 × x × 3)/100

= Rs. 60x

According to the statement,

72x = 60x + 60

⇒ 72x – 60x = 60

⇒ 12x = 60

⇒ x = 60/12

⇒ x = 5

∴ Rate = 5%

Let one part = Rs. x

∴ Second part = Rs. (15,600 – x)

By the given condition = (x × 5 × 5)/100

= {(15,600 – x) × 9/2 × 6}/100

⇒ 25x = 27 × 15,600 – 27x

⇒ 25x + 27x = 27 × 15,600

⇒ 52x = 27 × 15, 6000

⇒ 52x = 27 × 15,600

⇒ x = (27 × 15,600)/52 = 27 × 300 = 8100

Hence one part = Rs. 8100 and second part = Rs. (15,600 – 8,100)

= Rs. 7,500

(Given)

⇒ r = 24/3 = 8

Hence reqd. rate of interest = 8%

**2. If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:****(i) the rate of interest****(ii) the amount of Rs. 7,500 in 5 years at the same rate of interest**

**Solution****(i)**In first case:A = Rs. 4620

P = Rs. 3750

I = A – P = Rs. 4620 – Rs. 3750 = Rs. 870

T = 3 years

R = (100 × I)/(P × T)

= (100 × 870)/(3750 × 3)

= (100 × 290)/3750

= (4 × 29)/15

= 116/15

= 7 11/15%

In second case :

P = Rs. 7500

R = 116/15%

T = 5 ½ years = 11/2 years

**(ii)**Interest = (P × R × T)/100= Rs. (7500 × 11 × 116)/(2 × 15 × 100)

= (250 × 116 × 11)/100

= 10 × 29 × 11

= 290 × 11

= Rs. 3190

Amount = Rs. 7500 + 3190

= Rs. 10, 690

**3. A sum of money, lent out at simple interest, doubles itself in 8 years. Find :****(i) the rate of interest****(ii) in how many years will the sum become triple (three times) of itself at the same rate per cent?**

**Solution**Let P = Rs. 100

A = Rs. 200

I = Rs. 200 – Rs. 100 = Rs. 100,

T = 8 years

R = (100 × I)/(P × T)

= (100 × 100)/(100 × 8)

= 100/8

= 25/2%

Now, again P = Rs. 100

A = Rs. 300

I = Rs. 300 – Rs. 100

= Rs. 200

R = 25/2%

T = (100 × I)/(P × R)

= (100 × 200)/(100 × 25/2)

= (100 × 200 × 2)/(100 × 25)

= 16 years

So, the given sum of money will become triple in 16 years.

**4. Rupees 4000 amount to Rs.5000 in 8 years; in what time will Rs.2100 amount to Rs.2800 at the same rate?**

**Solution**In first case:

A = Rs. 5000

P = Rs. 4000

I = A – P

= Rs. 5000 – Rs. 4000

= Rs. 1000

T = 8 years

R = (100 × I)/(P × R)

= (100 × 1000)/(4000 × 8)

= 25/8%

In the second case:

A = Rs. 2800

P = Rs. 2100

I = Rs. 2800 - Rs. 2100

= Rs. 700

R = 25/8%

T = (100 × I)/(P × R)

= (100 × 700)/(2100 × 25/8)

= (100 × 700 × 8)/(2100 × 25)

= 32/3 years

= 10 2/3 years

10 2/3 × 12 months

= 10 24/3 months

= 10 years 8 months

**5. What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum?**

**Solution**In first case:

P = Rs. 7500

R = 5%

T = 6 years

Interest = (P × R × T)/100

= Rs. (7500 × 5 × 6)/100

= Rs. 75 × 5 × 6

= Rs. 2250

In second case :

According to the statement, interest = Rs. 2250

R = 6.5% P.A.

T = 4 years

P = (100 × I)/(R × T)

= Rs. (100 × 2250)/(6.5 × 4)

= Rs. 225000/26

= Rs. 112500/13

= Rs. 8653.85

Required principal = Rs. 8653.85

**6. A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.**

**Solution**In 6 years sum amounts to = Rs. 4050

In 4 years sum amounts to = Rs. 3825

∴ Interest of 2 years = Rs. 4050 – Rs. 3825

= Rs. 225

Interest of 4 years = Rs. 225/2 × 4

= Rs. 450

(∵ Rs. 225 is interest for 2 years)

Now P = A – I

= Rs. 3825 – Rs. 450

= Rs. 3375

I = Rs. 450

T = 4 years

R = (100 × I)/(P × T)

= (100 × I)/(P × T)

= (100 × 450)/(3375 × 4)

= 45000/13500%

= 450/135%

= 10/3%

= 3 1/3%

∴ R = 3 1/3 %

P = Rs. 3375

**7. At what rate percent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years? To what will it amount in 15 years?**

**Solution**P = Rs. 3750

I = Rs. 3750 × 1/5

= Rs. 750

T = 4 years

R = (100 × I)/(P × T)

= (100 × 750)/(3750 × 4)

= (100 × 750)/(3750 × 4)

= 5%

Again, P = Rs. 3750

Interest of 4 years = Rs. 750

Interest of 1 year = Rs. 750/4

Interest of 15 years = Rs.750/4 × 15

= Rs. (750 × 15)/4

= Rs. 5625/2

= Rs. 2812.50

Amount in 15 years will be = Rs. 3750 + Rs. 2812.50

= Rs. 6562.50

∴ Rate = 5%

Amount in 15 years will be = Rs. 6562.50

**8. On what date will ₹ 1950 lent on 5th January, 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?**

**Solution**P = Rs. 1950

A = Rs. 2125.50

R = 5% p.a

I = A – P

= Rs. 2125.50 – Rs. 1950

= Rs. 175.50

T = (100 × I)/(P × R)

= (100 × 175.50)/(1950 × 5)

= 17550/9750

= 1755/975

= 117/65

= 9/5 years

= 4 1/5 years

= 1 years 292 days

∵ 4/5 years = 4/5 × 365 days = 292 days

Jan. + Feb. + March + April + May + June + July + Aug. + Sept. + Oct.

(31 – 5) + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 23

= 292 days

∴ Required date = 23rd October 2012

**9. If the interest on Rs.2400 be more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent; find the rate.**

**Solution**In first case:

P = Rs. 2400

R = x% (Assume)

T = 3 years

Interest = (P × R × T)/100

= Rs. (2400 × x × 3)/100

= Rs. 72x

In second case:

P = Rs. 2000

R = x%

**(Rate same as in first case)**T = 3 years

Interest = (P × R × T)/100

= Rs. (2000 × x × 3)/100

= Rs. 60x

According to the statement,

72x = 60x + 60

⇒ 72x – 60x = 60

⇒ 12x = 60

⇒ x = 60/12

⇒ x = 5

∴ Rate = 5%

**10. Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4 per cent for 6 years.**

**Solution**Let one part = Rs. x

∴ Second part = Rs. (15,600 – x)

By the given condition = (x × 5 × 5)/100

= {(15,600 – x) × 9/2 × 6}/100

⇒ 25x = 27 × 15,600 – 27x

⇒ 25x + 27x = 27 × 15,600

⇒ 52x = 27 × 15, 6000

⇒ 52x = 27 × 15,600

⇒ x = (27 × 15,600)/52 = 27 × 300 = 8100

Hence one part = Rs. 8100 and second part = Rs. (15,600 – 8,100)

= Rs. 7,500

### Exercise 9 C

**1. A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:**

**(i) interest for the first year.**

**(ii) principal for the second year.**

**(iii) interest for the second year.**

**(iv) final amount at the end of second year**

**(v) compound interest earned in 2 years.**

**Solution**

**(i)**Here Principal (P) = Rs. 8,000

Rate of interest = 10%

Interest for the first year = (8,000 × 10 × 1)/100

= Rs. 800

**(ii)**∴ Amount = Rs.8,000 + Rs. 800

= Rs. 8,800

Thus principal for the second year = Rs, 8,800

**(iii)**Interest for the second year = (8,800 × 10 × 1)/100

= Rs. 880

**(iv)**Amount at the end of second year = Rs. 8,800 + Rs. 880

= Rs 9,860

**(v)**Hence compound interest earned in 2 years = Rs. 9,680 – Rs. 8,000

= Rs. 1680

**2. A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :**

**(i) the interest of the first year.**

**(ii) the interest of the second year.**

**(iii) the final amount at the end of second year.**

**(iv) the compound interest of two years.**

**Solution**

Here Principal (P) = Rs. 20,000, Time = 1 year

Rate = 8%

**(i)**∴ Interest of the first year = (20,000 × 8 × 1)/100

= Rs.1600

**(ii)**∴ Amount after one year

i.e., Principal for second year = Rs. 20,000 + Rs. 1,600

= Rs. 21,600

∴ Interest for second year = (21,600 × 8 × 1)/100

= 216 × 8

= Rs. 1728

**(iii)**Final amount at the end of second year = Rs.(21,600 + 1728)

= Rs. 23,328

**(iv)**Interest of two years = Rs. 23,328 – Rs. 20,000

= Rs. 3,328

**3. Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.**

**Solution**

**For 1st year**

Principal (P) = Rs. 12,000

Rate (R) = 10%

Time (T) = 1 year

I = Interest = (12,000 × 10 × 1)/100

= 120 × 10

= Rs. 1200

Amount = P + I

= Rs. 12,000 + Rs. 1200

= Rs. 13,200

**For IInd year**

P = Rs. 13,200, R = 10%, Time(T) = 1 year

∴ Interest = (13,200 × 10 × 1)/100

= 132 × 10

= Rs. 1320

∴ Amount in 2 years = Rs.(13,200) + (1320)

= Rs. 14520

Compound interest in 2 years = Rs. 1200 + Rs. 1320

= Rs. 2520

[or directly = Rs. 14520 – Rs. 12000 = Rs. 2520]

**4. Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.**

**Solution**

**For 1st year**

Principal (P) = Rs. 10,000, Rate (R) = 8%

Time (T) = 1 year

∴ Interest = (10,000 × 8 × 1)/100

= 100 × 8

= Rs. 800

**For 2nd year**

P = Rs. 10,000 + Rs. 800 = Rs. 10,800

Rate (R) = 8% Time (T) = 1 year

∴ Interest = (10,800 × 8 × 1)/100

= 108 × 8

= Rs. 864

For 3rd year

∴ P = Rs. 10,800 + Rs. 864 = Rs. 11664

R = 8%

T = 1 year

∴ Interest = (11664 × 8 × 1)/100

= (11664 × 2)/25

= Rs. 933.12

∴ Amount = Rs.11664 + 933.12 = Rs. 12597.12

Hence, required amount = Rs. 12597.12

∴ Compound Interest = Rs. 12597.12 – 10000

= Rs. 2597.12

**5. Calculate the compound interest on Rs. 5,000 in 2 years; if the rates of interest for successive years be 10% and 12% respectively.**

**Solution**

For 1st year

Principal (P) = Rs. 5,000, Rate (R) = 10%

Time (T) = 1 year

∴ Interest = (5,000 × 10 × 1)/100

= 50 × 10

= Rs. 500

∴ Amount at the end of 1st year = Rs. (5000 + 500) = Rs. 5500

For 2nd year

P = Rs. 5500, rate = 12%, T = 1 year

∴ Interest = (5500 × 12 × 1)/100

= 55 × 12

= Rs. 660

∴ Amount at the end of 2nd year = Rs. 5500 + Rs. 660 = Rs. 6160

Hence compound Interest = Rs. 6160 – Rs. 5000

= Rs. 1160

**6. Calculate the compound interest on Rs. 15,000 in 3 years; if the rates of interest for successive years be 6%, 8% and 10% respectively.**

**Solution**

**For 1st year**

Principal (P) = Rs. 15,000 , Rate (R) = 6%

Time (T) = 1 year

∴ Interest = (15,000 × 6 × 1)/100

= 150 × 6

= Rs. 900

∴ Amount at the end of 1st year = Rs. 15,000 + Rs. 900

= Rs.15900

**For 2nd year**

P = Rs. 15900, R = 8%, T = 1 year

∴ Interest = (15,900 × 8 × 1)/100

= 159 × 8

= Rs. 1272

∴ Amount at the end of 2nd year = Rs. (15,900 + 1272)

= Rs. 17172

**For 3rd year**

P = Rs. 17172, R = 10% , T = 1 year

∴ Interest = (17172 × 10 × 1)/100

= Rs. 1717.20

∴ Amount at the end of 3rd year = Rs. (17172 + 1717.20)

= Rs. 18889.20

∴ Compound interest = 18889.20 – 15,000

= Rs. 3889.20

**7. Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.**

**Solution**

**For 1st year**

Principal (P) = Rs. 16,000.

Rate (R) = 5%

Time (T) = 1 year

∴ Interest = (16,000 × 5 × 1)/100

= 160 × 5

= Rs. 800

∴ Amount at the end of 1st year = Rs. (16,000 + 800)

= Rs. 16,800

**For 2nd year**

P = Rs. 16,800

R = 5%

T = 1 year

∴ Interest = (16,800 × 5 × 1)/100

= 168 × 5

= Rs. 840

∴ Amount at the end of 2nd year = Rs.(16,800 + 840)

= Rs. 17640

**For 3rd year**

P = 17640,

R = 5%,

T = 1 year

∴ Interest = (17640 × 5 × 1)/100

= 1764/2

= Rs. 882

∴ Amount at the end of 3rd year = Rs. (17640 + 882)

= Rs. 18522

Hence reqd. amount = Rs. 18522

**8. Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.**

**Solution**

**For 1st year**

Principal = Rs. 40,000,

Rate = 10%,

Time = 1 year

∴ Interest = (40,000 × 10 × 1)/100

= 400 × 10

= Rs. 4000

∴ Amount at the end of 1st year = Rs. (40,000 + 4000)

= Rs.44,000

**For 2nd year**

P = Rs. 44,000,

R = 10%,

T = 1 year

∴ Interest = Rs.(44,000 × 10 × 1)/100

= 440 × 10

= Rs. 4400

Thus interest earned in the second year = Rs. 4400

**9. Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.**

**Solution**

Principal (P) = Rs. 15000

Rate (R) = 6% p.a.

Period (n) = 5 years

Interest for the first year = PRT/100

= (15000 × 6 × 1)/100

= Rs. 900

∴ Amount for the first year = Rs. 15000 + 900

= Rs. 15900

Principal for the second year = Rs. 15900

Interest for the second year = (15900 × 6 × 1)/100

= 159 × 6 = Rs. 954

**10. A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :**

**(i) the interest for the first year.**

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

**(iii) the interest for the second year.**

**(iv) the interest for the third year.**

**Solution**

Principal (P) = Rs. 9600

Rate (R) = 10% p.a.

Period (P) = 3 years

**(i)**∴ Interest for the first year = PRT/100

= (9600 × 10 × 1)/100

= Rs. 960

**(ii)**Amount at the end of first year = P + S.I. = Rs. 9600 + 960

= Rs. 10560

**(iii)**Principal for the second year = Rs. 10560

Interest for the second year = (10560 × 10 × 1)/100

= Rs. 1056

∴ Amount after second year = Rs. 10560 + 1056

= Rs. 11616

**(iv)**Principal for the third year = Rs. 11616

Interest for the third year = (11616 × 10 × 1)/100

= 116.16 × 10

= Rs. 1161.60

**11. A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :**

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

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

**Solution**

Principal (P) = Rs. 5000

Period (T) = 2 years

Amount at the end of one year = Rs. 5600

∴ Interest for the first year = A – P

= Rs. 5600 – 5000

= Rs. 600

**(i)**∴ Rate of interest = (S.I. × 100)/(P × T)

= (600 × 100)/(5000 × 1)

= 12% p.a.

**(ii)**Principal for the second year = Rs. 5600

Interest for the second year = (5600 × 12 × 1)/100

= ₹672

∴ Amount at the end of second year = P + S.I.

= 5600 + 672

= ₹6272

**12. Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.**

**Solution**

Principal (P) = ₹7500

Rate (R) = 8% p.a.

Period (T) = 2 years

∴ Simple Interest = PRT/100

= (7500 × 8 × 2)/100

= ₹ 1200

Interest for the second year = (7500 × 8 × 1)/100

= ₹ 600

∴ Amount at the end of first year = P + S.I.

= ₹ 7500 + ₹ 600

= ₹ 8100

∴ Interest for the second year = (8100 × 8 × 1)/100

= ₹ 648

∴ Total C.I. for 2 years = ₹ 600 + ₹ 648

= ₹1248

∴ Difference between C.I. and S.I. for 2 years = ₹ 1248 - ₹1200

= ₹ 48

**13. Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.**

**Solution**

Principal (P) = ₹8000

Rate (R) = 10% p.a.

Period (T) = 3 years

∴ S.I. for 3 years = PRT/100

= (8000 × 10 × 3)/100

= ₹ 2400

Now, S.I. for 1st year = ₹ (8000 × 10 × 1)/100

= 80 × 10 × 1

= ₹ 800

Amount for the first year = P + S.I.

= ₹ 8000 + ₹ 800

= ₹ 8800

Principal for the second year = ₹8800

Interest for the second year = (8800 × 10 × 1)/100

₹880

∴ Amount after second year = ₹ 8800 + ₹ 880

= ₹ 9680

Principal for the third year = ₹ 9680

Interest for the third year = ₹ (9680 × 10 × 1)/100

= ₹ 968

∴ C.I. for 3 years = ₹800 + ₹880 + ₹968

= ₹ 2648

∴ Difference between C.I. and S.I. for 3 years = ₹2648 - ₹2400

= ₹248

**14. Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?**

**Solution**

**In first case,**

Sum borrowed (P) = ₹40000

Rate (R) = 10 p.a. compounded annually

Time (T) = 2 years

∴ Interest for first year = PRT/100

= ₹ (4000 × 10 × 1)/100

= ₹ 4000

Amount after one year = ₹ 40000 + 4000

= ₹ 44000

Principal for the second year = ₹ 44000

∴ Interest for the second year = (44000 × 10 × 1)/100

= ₹4400

∴ Compound Interest for 2 years = ₹4000 + 4400

= ₹ 8400

Rate (R) = 10.5% p.a.

Time (T) = 2 years

∴ Simple interest = PRT/100

= (40000 × 10.5 × 2)/100

= ₹ (40000 × 105 × 2)/(100 × 10)

= ₹ 8400

In both the cases, interest is same.

Sum borrowed (P) = ₹24000

Rate (R) = 13% p.a.

Time (T) = 2 years

In case of simple interest,

Simple interest for 2 years = PRT/100

= ₹(24000 × 13 × 2)/100

= ₹6240

In case of compound interest,

Interest for the first year = (24000 × 12 × 1)/100

= ₹ 2880

Amount after first year = ₹24000 + 2880

= ₹26880

Interest for the first year = (26880 × 12 × 1)/100

= ₹322560/100

= ₹ 3225. 60

∴ C.I for 2 years = ₹ 2880 + ₹ 3225.60

= 6105.60

Total interest = ₹ 6240 + ₹6105.60

= ₹ 12345.60

Sum borrowed = ₹12000

Rate (R) = 10% p.a. compound annually

Time (T) = 2 years

Interest for the first year = PRT/100

= (12000 × 100 × 1)/100

= ₹1200

Amount paid = ₹ 8000

= ₹5200

= ₹ 520

∴ Amount = ₹ 5200 + ₹ 520

= ₹ 5720

Loan taken (P) = ₹ 16000

Rate (R) = 15% p.a.

Time (T) = 2 years

∴ Interest for the first year = PRT/100

= (16000 × 15 × 1)/100

= ₹ 2400

Amount after one year = ₹ 16000 + ₹ 2400

= ₹ 18400

At the end of one year aount paid back = ₹ 9000

Balance amount = ₹18400 – 9000

= ₹9400

Interest for the second year = (9400 × 15 × 1)/100

= ₹ 1410

Amount after second year = ₹ 9400 + ₹ 1410

= ₹ 10810

Rate (R) = 8% p.a.

Period (T) = 5 years

Interest (I) = ₹ 12000

= ₹(12000 × 100)/(8 × 5)

= ₹30000

Time (T) = 2 years

Principal (P) = ₹ 30000

Interest for the first year = PRT/100

= ₹(30000 × 10 × 1)/100

= ₹3000

∴ Amount after one year = ₹30000 + 3000

= 33000

Principal for the second year = ₹ 33000

Interest for the second year = (33000 × 10 × 1)/100

= ₹ 3300

∴ Compound Interest for two years = ₹3000 + 3300

= ₹6300

Principal (P) = ₹12,000

Rate (R) = 10%

Time (T) = 1 years

Amount = P × {1 + r/(2 × 100)}

= ₹12,000 × (1 + 10/200)

= ₹12,000 × (210/200)

= ₹12,000 × 21/20 × 21/20

= ₹13,230

C.I. = Amount – Principal

= ₹13230 - ₹ 12000

= ₹1230

Principal (P) = ₹8000

Rate (R) = 20%

Time = 1 ½ years = 3/2 years

Amount = Principal × {1 + r/(2 × 100)}

= ₹8000 × (1 + 20/200)

= ₹8000 × (220/200)

= ₹ 8000 × 11/10 × 11/10 × 11/10

= ₹ 10648

C.I. = Amount – Principal

= ₹10648 - ₹8000

= ₹2648

Principal (P) = ₹24,000

Time (T) = 2 years

Rate (R) = 10%

Amount = Principal – {1 + r/(2 × 100)}

= ₹24,000 × (1 + 10/200)

= ₹24,000 × (210/200)

= ₹24,000 × 21/20 × 21/20 × 21/10 × 21/20

= ₹29, 172

C.I. = Amount – Principal

= ₹29,172 - ₹24,000

= ₹5,172

Principal (P) = ₹16,000

Time (T) = 3 years

Rate (R) = 5%

Amount = Principal × {1 + r/(2 × 100)}

= ₹16,000 × (1 + 5/200)

= ₹16,000 × (205/200)

= ₹ 16,000 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40

= ₹18,555

C.I. = Amount – Principal

= ₹18,555 - ₹16,000

= ₹2555

Principal (P) = ₹20,000

Time (T) = 1 ½ years = 3/2 years

Rate (R) = 10%

Amount = P ×{1 + r/(2 × 100)}

= ₹20,000 × (1 + 10/200)

= ₹20,000 × (210/200)

= ₹20,000 × 21/20 × 21/20 × 21/20

= ₹23,152.50

C.I. = Amount – Principal

= ₹23,152.50 - ₹20,000

= ₹3, 152.50

Principal (P) = ₹32,000

Time (T) = 1 year

Rate (R) = 20%

Amount = Principal × {1 + r/(2 × 100)}

= ₹32,000 × (1 + 20/200)

= ₹32,000 × (11/10)

= ₹32,000 × 11/10 × 11/10

= ₹38,720

C.I. = Amount – Principal

= ₹38,720 - ₹32,000

= ₹6,720

C.I = Amount – Principal

= ₹38,720 - ₹ 32,000

= ₹6,720

Principal (P) = ₹4,000

Time (T) = 2 years

Rate (R

Amount = P(1 + R

= ₹4,000 (1 + 10/100)(1 + 15/100)

= ₹4,000 × 11/10 × 23/20

= ₹5060

C.I. = Amount - Principal

= ₹5060 - ₹4000

= ₹1060

Principal (P) = ₹10,000

Time (t) = 3 years

Rate (R

Rate (R

Rate (R

Amount = P(1 + R

= ₹10,000 × (1 + 10/100)(1 + 15/100)(1 + 20/100)

= ₹10,000 × 11/10 × 23/20 × 6/5

= ₹15,180

C.I. = Amount – Principal

= ₹15,180 - ₹10,000

= ₹5180

Rate (R) = 10 p.a. compounded annually

Time (T) = 2 years

∴ Interest for first year = PRT/100

= ₹ (4000 × 10 × 1)/100

= ₹ 4000

Amount after one year = ₹ 40000 + 4000

= ₹ 44000

Principal for the second year = ₹ 44000

∴ Interest for the second year = (44000 × 10 × 1)/100

= ₹4400

∴ Compound Interest for 2 years = ₹4000 + 4400

= ₹ 8400

**In second case,**

Principal (P) = ₹40000Rate (R) = 10.5% p.a.

Time (T) = 2 years

∴ Simple interest = PRT/100

= (40000 × 10.5 × 2)/100

= ₹ (40000 × 105 × 2)/(100 × 10)

= ₹ 8400

In both the cases, interest is same.

**15. Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.**

**Solution**Sum borrowed (P) = ₹24000

Rate (R) = 13% p.a.

Time (T) = 2 years

In case of simple interest,

Simple interest for 2 years = PRT/100

= ₹(24000 × 13 × 2)/100

= ₹6240

In case of compound interest,

Interest for the first year = (24000 × 12 × 1)/100

= ₹ 2880

Amount after first year = ₹24000 + 2880

= ₹26880

Interest for the first year = (26880 × 12 × 1)/100

= ₹322560/100

= ₹ 3225. 60

∴ C.I for 2 years = ₹ 2880 + ₹ 3225.60

= 6105.60

Total interest = ₹ 6240 + ₹6105.60

= ₹ 12345.60

**16. Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:****(i) the amount at the end of first year, before making the repayment.****(ii) the amount at the end of first year, after making the repayment.****(iii) the principal for the second year.****(iv) the amount to be paid at the end of second year, to clear the account.**

**Solution**Sum borrowed = ₹12000

Rate (R) = 10% p.a. compound annually

Time (T) = 2 years

Interest for the first year = PRT/100

= (12000 × 100 × 1)/100

= ₹1200

**(i)**Amount = ₹12000 + 1200 = ₹13200Amount paid = ₹ 8000

**(ii)**Balance amount = ₹13200 – 8000= ₹5200

**(iii)**∴ Principal for the second year = ₹ 5200**(iv)**Interest for the second year = (5200 × 10 × 1)/100= ₹ 520

∴ Amount = ₹ 5200 + ₹ 520

= ₹ 5720

**17. Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How mucH must he pay at the end of second year to clear the debt?**

**Solution**Loan taken (P) = ₹ 16000

Rate (R) = 15% p.a.

Time (T) = 2 years

∴ Interest for the first year = PRT/100

= (16000 × 15 × 1)/100

= ₹ 2400

Amount after one year = ₹ 16000 + ₹ 2400

= ₹ 18400

At the end of one year aount paid back = ₹ 9000

Balance amount = ₹18400 – 9000

= ₹9400

Interest for the second year = (9400 × 15 × 1)/100

= ₹ 1410

Amount after second year = ₹ 9400 + ₹ 1410

= ₹ 10810

**18. A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:****(i) the sum of money.****(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.**

**Solution**Rate (R) = 8% p.a.

Period (T) = 5 years

Interest (I) = ₹ 12000

**(i)**∴ Sum = (I × 100)/(R × T)= ₹(12000 × 100)/(8 × 5)

= ₹30000

**(ii)**Rate (R) = 10% p.a.Time (T) = 2 years

Principal (P) = ₹ 30000

Interest for the first year = PRT/100

= ₹(30000 × 10 × 1)/100

= ₹3000

∴ Amount after one year = ₹30000 + 3000

= 33000

Principal for the second year = ₹ 33000

Interest for the second year = (33000 × 10 × 1)/100

= ₹ 3300

∴ Compound Interest for two years = ₹3000 + 3300

= ₹6300

**19. Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.**

**Solution**Principal (P) = ₹12,000

Rate (R) = 10%

Time (T) = 1 years

Amount = P × {1 + r/(2 × 100)}

^{nx2}= ₹12,000 × (1 + 10/200)

^{2}= ₹12,000 × (210/200)

^{2}= ₹12,000 × 21/20 × 21/20

= ₹13,230

C.I. = Amount – Principal

= ₹13230 - ₹ 12000

= ₹1230

**20. Find the amount and the C.I. on ₹ 8,000 in 1 1/2 year at 20% per year compounded half-yearly.**

**Solution**Principal (P) = ₹8000

Rate (R) = 20%

Time = 1 ½ years = 3/2 years

Amount = Principal × {1 + r/(2 × 100)}

^{nx2}= ₹8000 × (1 + 20/200)

^{3/2x2}= ₹8000 × (220/200)

^{3}= ₹ 8000 × 11/10 × 11/10 × 11/10

= ₹ 10648

C.I. = Amount – Principal

= ₹10648 - ₹8000

= ₹2648

**21. Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.**

**Solution**Principal (P) = ₹24,000

Time (T) = 2 years

Rate (R) = 10%

Amount = Principal – {1 + r/(2 × 100)}

^{nx2}= ₹24,000 × (1 + 10/200)

^{2x2}= ₹24,000 × (210/200)

^{4}= ₹24,000 × 21/20 × 21/20 × 21/10 × 21/20

= ₹29, 172

C.I. = Amount – Principal

= ₹29,172 - ₹24,000

= ₹5,172

**22. Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.**

**Solution**Principal (P) = ₹16,000

Time (T) = 3 years

Rate (R) = 5%

Amount = Principal × {1 + r/(2 × 100)}

^{nx2}= ₹16,000 × (1 + 5/200)

^{3x2}= ₹16,000 × (205/200)

^{6}= ₹ 16,000 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40

= ₹18,555

C.I. = Amount – Principal

= ₹18,555 - ₹16,000

= ₹2555

**23. Find the amount and the compound interest on ₹ 20,000 for 1 ½ years at 10% per annum compounded half-yearly.**

**Solution**Principal (P) = ₹20,000

Time (T) = 1 ½ years = 3/2 years

Rate (R) = 10%

Amount = P ×{1 + r/(2 × 100)}

^{nx2}= ₹20,000 × (1 + 10/200)

^{3/2x2}= ₹20,000 × (210/200)

^{3}= ₹20,000 × 21/20 × 21/20 × 21/20

= ₹23,152.50

C.I. = Amount – Principal

= ₹23,152.50 - ₹20,000

= ₹3, 152.50

**24. Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.**

**Solution**Principal (P) = ₹32,000

Time (T) = 1 year

Rate (R) = 20%

Amount = Principal × {1 + r/(2 × 100)}

^{nx2}= ₹32,000 × (1 + 20/200)

^{1x2}= ₹32,000 × (11/10)

^{2}= ₹32,000 × 11/10 × 11/10

= ₹38,720

C.I. = Amount – Principal

= ₹38,720 - ₹32,000

= ₹6,720

C.I = Amount – Principal

= ₹38,720 - ₹ 32,000

= ₹6,720

**25. Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.**

**Solution**Principal (P) = ₹4,000

Time (T) = 2 years

Rate (R

_{1}) = 10% and rate (R_{2}) = 15%Amount = P(1 + R

_{1}/100) (1 + R_{2}/100)= ₹4,000 (1 + 10/100)(1 + 15/100)

= ₹4,000 × 11/10 × 23/20

= ₹5060

C.I. = Amount - Principal

= ₹5060 - ₹4000

= ₹1060

**26. Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.**

**Solution**Principal (P) = ₹10,000

Time (t) = 3 years

Rate (R

_{1}) = 10%Rate (R

_{2}) = 15%Rate (R

_{3}) = 20%Amount = P(1 + R

_{1}/100) (1 + R_{2}/100)(1 + R_{3}/100)= ₹10,000 × (1 + 10/100)(1 + 15/100)(1 + 20/100)

= ₹10,000 × 11/10 × 23/20 × 6/5

= ₹15,180

C.I. = Amount – Principal

= ₹15,180 - ₹10,000

= ₹5180