Math, asked by ag5117756, 4 months ago

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 –



a) Compound Annually

b) Compounded half yearly. to​

Answers

Answered by pratikshaborkar144
4

Answer:

compounded half yearly to

Answered by Anonymous
36

Given :-

• Arif took a loan of 80,000 from bank

• Rate of interest is 10% per annum

• Time period = 1½ years

Solution :-

a) Here,

Principal = Rs 80,000

Rate = 10% p.a

Time = 1½

Time = 1 + 1/2 years

Firstly, We will find the amount Compounded annually for 1 year

As we know that,

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

Put the required values in the given formula,

A = 80000 ( 1 + 10/100 )^1

A = 80000( 100 + 10 / 100 )^1

A = 80000 * 110/100

A = 88000

Now,

Amount = Principal + SI

Put the required values,

88000 = 80000 + SI

SI = 88000 - 80000

SI = Rs 8000

Thus,

Interest for 1 year = Rs8000

Amount for 1 year = Rs88000

Now,

SI for next 1/2 year

Principal = Amount in previous year

Rate = 10% p. a

Time = 1/2 year

By using formula to calculate SI

SI = P * R * T / 100

SI = 88000 * 10 * ½ / 100

SI = 88000 * 10 * 1 / 100 * 2

SI = 880 * 5

SI = Rs 4400

Therefore,

SI for next year = Rs 4400

Now,

Total SI for 1½ years = 8000 + 4400 = 12400

Thus,

Amount = Principal + SI

Amount = 88000 + 12400

Amount = 92400

Hence, Amount = Rs 92400

Solution 2 :-

Here ,

We have to calculate Compound halfyearly

Therefore,

Principal = 80000

Rate = 10/2 = 5%

[ Compounded half yearly ]

Time = 1½ = 3/2 * 2 = 3 years

[ Compounded half yearly ]

Now,

As we know that,

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

Put the required values,

A = 80000( 1 + 5/100 )^3

A = 80000( 1 + 1/20)^3

A = 80000 ( 21/20)^3

A = 80000 * 21/20 *21/20 *21/20

A = 80000 * 441/400 * 21 / 20

A = 200 * 441 * 21 /20

A = 10 * 441 * 21

A = 210 * 441

A = 92610

Now,

We have to calculate difference between amounts .

Therefore,

= 92610 - 92400

= 210

Hence, The difference between amounts is Rs210 .

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