**1. Calculate the amount and compound interest on**

**(a) ₹ 10,800 for 3 years at 12½ % per annum compounded annually.**

**Solution:**

Given,

Principal (P) = ₹ 10,800

Rate (R) = 12½ % = 25/2 % (annual)

Number of years (n) = 3

Now,

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

= 10800(1 + 25/200)^{3}

= 10800(225/200)^{3}

= 15377.34375

= ₹ 15377.34 (approximately)

Compound Interest = A – P = ₹ (15377.34 – 10800) = ₹ 4,577.34

**(b) ₹ 18000 for 2½ years at 10% per annum compounded annually.**

**Solution:**

Given,

Principal (P) = ₹ 18,000

Rate (R) = 10% annual

Number of years (n) = 2½

[We can calculate the amount for 2 years and 6 months by calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years]

Now,

First, the amount for 2 years has to be calculated

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

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

= 18000(11/10)^{2}

= ₹ 21780

By taking ₹ 21780 as principal, the Simple Interest for the next ½ year will be calculated

Simple interest = (21780 x ½ x 10)/100

= ₹ 1089

Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780

And, interest for the next ½ year = ₹ 1089

Total Compound Interest = ₹ 3780 + ₹ 1089

= ₹ 4,869

Therefore,

Amount, A = P + C.I.

= ₹ 18000 + ₹ 4869

= ₹ 22,869

**(c) ₹ 62500 for 1½ years at 8% per annum compounded half yearly.**

**Solution:**

Given,

Principal (P) = ₹ 62,500

Rate = 8% per annum or 4% per half year

Number of years = 1½

There will be 3 half years in 1½ years

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

= 62500(1 + 4/100)^{3}

= 62500(104/100)^{3}

= 62500(26/25)^{3}

= ₹ 70304

Compound Interest = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804

**(d) ₹ 8000 for 1 year at 9% per annum compound half yearly.**

(You could use the year by year calculation using SI formula to verify)

**Solution:**

Given,

Principal (P) = ₹ 8000

Rate of interest = 9% per annum or 9/2% per half year

Number of years = 1 year

There will be 2 half years in 1 year

Now,

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

= 8000(1 + 9/200)^{2}

= 8000(209/200)^{2}

= 8736.20

Compound Interest = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20

**(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.**

**Solution:**

Given,

Principal (P) = ₹ 10,000

Rate = 8% per annum or 4% per half year

Number of years = 1 year

There are 2 half years in 1 year

Now,

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

= 10000(1 + 4/100)^{2}

= 10000(1 + 1/25)^{2}

= 10000(26/25)^{2}

= ₹ 10816

Compound Interest = A – P = ₹ 10816 – ₹ 10000 = ₹ 816

**2. Kamala borrowed ₹ 26400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?**

**(Hint: Find A for 2 years with interest is compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)**

**Solution:**

Given,

Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

[We can calculate the amount for 2 years and 4 by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years]

First, the amount for 2 years has to be calculated

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

= 26400(1 + 15/100)^{2}

= 26400(1 + 3/20)^{2}

= 26400(23/20)^{2}

= ₹ 34914

Now,

Taking ₹ 34,914 as principal, the S.I. for the next 1/3 years will be-

S.I. = (34914 × 1/3 x 15)/100 = ₹ 1745.70

Interest for the first two years = ₹ (34914 – 26400) = ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

**3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?**

**Solution:**

Given,

For Fabina-

P = ₹ 12,500

R = 12%

T = 3 years

Interest paid by Fabina = (P x R x T)/100

= (12500 x 12 x 3)/100

= 4500

Amount paid by Radha at the end of 3 years

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

A = 12500(1 + 10/100)^{3}

= 12500(110/100)^{3}

= ₹ 16637.50

C.I. = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Thus, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Hence, Fabina will have to pay ₹ 362.50 more.

**4. I borrowed ₹ 12000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?**

**Solution:**

P = ₹ 12000

R = 6% per annum

T = 2 years

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

= (12000 x 6 x 2)/100

= ₹ 1440

To find the compound interest, first we calculate the amount (A).

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

= 12000(1 + 6/100)^{2}

= 12000(106/100)^{2}

= 12000(53/50)^{2}

= ₹ 13483.20

C.I. = A − P

= ₹ 13483.20 − ₹ 12000

= ₹ 1,483.20

C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440

= ₹ 43.20

Hence, the extra amount to be paid is ₹ 43.20.

**5. Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get**

(i) after 6 months?

**Solution:**

P = ₹ 60,000

Rate = 12% per annum = 6% per half year

n = 6 months = 1 half year

Now,

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

= 60000(1 + 6/100)^{1}

= 60000(106/100)

= 60000(53/50)

= ₹ 63600

(ii) after 1

**Solution:**

There are 2 half years in 1 year

So, n = 2

Now,

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

= 60000(1 + 6/100)^{2}

= 60000(106/100)^{2}

= 60000(53/50)^{2}

= ₹ 67416

**6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1½ years if the interest is**

**(i) Compounded annually**

**Solution:**

P = ₹ 80,000

R = 10% per annum

n = 1½ years

We can calculate the amount for 1 year and 6 months by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

Now, the amount for 1 year has to be calculated

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

= 80000(1 + 10/100)^{1}

= 80000 x 11/100

= ₹ 88000

Now, taking ₹ 88,000 as principal, the S.I. for the next ½ year will be calculated as

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

= (88000 x 10 x ½)/100

= ₹ 4400

Interest for the first year = ₹ 88000 – ₹ 80000 = ₹ 8000

And interest for the next ½ year = ₹ 4,400

Total C.I. = ₹ 8,000 + ₹ 4,400 = ₹ 12,400

A = P + C.I.= ₹ (80000 + 12400)

= ₹ 92,400

**(ii) Compounded half yearly**

**Solution:**

The interest is compounded half yearly

Rate = 10% per annum = 5% per half year

There will be three half years in 1½ years

Now,

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

= 80000(1 + 5/100)^{3}

= 80000(105/100)^{3}

= ₹ 92610

Thus, the difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

**7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find**

**(i) The amount credited against her name at the end of the second year**

**Solution:**

Given,

P = ₹ 8,000

R = 5% per annum

n = 2 years

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

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

= 8000(105/100)^{2}

= ₹ 8820

**(ii) The interest for the 3rd year**

**Solution:**

The interest for the next one year, i.e. the third year, has to be calculated by taking ₹8,820 as principal, the S.I. for the next year will be calculated.

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

= (8820 x 5 x 1)/100

= ₹ 441

**8. Find the amount and the compound interest on ₹ 10,000 for 1½ years at 10% per annum, compounded half yearly. Would this interest be more that the interest he would get if it was compounded annually?**

**Solution:**

Given,

P = ₹ 10,000

Rate = 10% per annum = 5% per half year

n = 1½ years

There will 3 half years in 1½ years

Now,

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

= 10000(1 + 5/100)^{3}

= 10000(105/100)^{3}

= ₹ 11576.25

C.I. = A − P

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

We can calculate the amount for 1 year and 6 months by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

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

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

= 10000(110/100)

= ₹ 11000

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

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

= (11000 x 10 x ½)/100

= ₹ 550

So, the interest for the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Hence, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So, the difference between two interests = 1576.25 – 1550 = 26.25

Hence, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.

**9. Find the amount which Ram will get on ₹ 4,096, he gave it for 18 months at 12½ per annum, interest being compounded half yearly.**

**Solution:**

Given,

P = ₹ 4,096

R = 12½ per annum = 25/2 per annum = 25/4 per half year

n = 18 months

There will be 3 half years in 18 months

So,

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

= 4096(1 + 25/(4 x 100))^{ 3}

= 4096 x (1 + 1/16)^{ 3}

= 4096 x (17/16)^{ 3}

= ₹ 4913

Hence, the required amount is ₹ 4,913.

**10. The population of a place increased to 54000 in 2003 at a rate of 5% per annum**

**(i)find the population in 2001**

**Solution:**

Given, population in the year 2003 = 54,000

Now,

54,000 = (Population in 2001) (1 + 5/100)^{2}

54,000 = (Population in 2001) (105/100)^{2}

Population in 2001 = 54000 x (100/105)^{2}

= 48979.59

Hence, the population in the year 2001 was approximately 48,980

**(ii) what would be its population in 2005?**

**Solution:**

Population in 2005 = 54000(1 + 5/100)^{2}

= 54000(105/100)^{2}

= 54000(21/20)^{2}

= 59535

Hence, the population in the year 2005 would be 59,535.

**11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.**

**Solution:**

Given the initial count of bacteria = 5,06,000

Bacteria at the end of 2 hours = 506000(1 + 2.5/100)^{2}

= 506000(1 + 1/40)^{2}

= 506000(41/40)^{2}

= 531616.25

Hence, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).

**12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.**

**Solution:**

Principal = ₹ 42,000

Depreciation = 8% of ₹ 42,000 per year

= (P x R x T)/100

= (42000 x 8 x 1)/100

= ₹ 3360

Hence, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

**👍👍👍**