Asif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find 1 the difference in amounts he would be paying after 1 and a half years if the interest is
(1) compounded annually
. (ii) compounded half yearly.
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Asif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find 1 the difference in amounts he would be paying after 1 and a half years if the interest is
(1) compounded annually
. (ii) compounded half yearly.
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Answer:
(i) = Rs.88000+Rs.4400 = 92400Rs.
(ii) = Rs.92610
Thus, the difference between the two amounts = Rs.92610−Rs.92400 =Rs.210
Step-by-step explanation:
Verified answer
Given :-
• Principal = ₹ 80,000 ( P )
• Rate = 10% per annum ( R )
[tex]• \: Time \: = 1 \frac{1}{2} \: years \: ( \: T \: )[/tex]
To find :-
★ The difference in amounts he would be paying after 1 and a half years if the interest is
(1) compounded annually
(2) compounded half yearly.
Solution :-
Here we have ,
P = 80,000
R = 10%
[tex]T \: = 1 \frac{1}{2} [/tex]
★ As we know ,
[tex]A = P \: {(1 + \frac{R}{100} )}^{n} [/tex]
where ,
A = Compound interest
P = principal
R = rate
n = number of years
so , putting the values
[tex]A = 80000 {(1 + \frac{10}{100} ) }^{1} [/tex]
[tex]A = 80000 {(1 + \frac{1}{10} )}^{1} [/tex]
[tex]A = 80000 \: {( \frac{10 + 1}{10} )}^{1} \: \: \: \:by \: LCM[/tex]
[tex]A = 80000( \frac{11}{10} )[/tex]
[tex]A = 8000 \times 11[/tex]
[tex]A = 88000[/tex]
This compound interest was for 1 yrs but we have to take for 1 and a half years
So
Remaining half yrs
For remaining half yrs we will take simple interest
[tex]SI = \frac{P× R× T}{100} [/tex]
where
SI = Simple interest
P = principal
R = rate of interest
T = Time
Note :- This time principal will be 88000
so , putting the values
[tex]S.I = \frac{88000 \times 10 \times 1}{100\times 2 } [/tex]
[tex]S.I = 880 \times 5 \: \: \: \: \: By \: cancellation[/tex]
[tex]S.I = 4400[/tex]
Now total amount =
88000 + 4400
92400
So amount after one and a half years if the interest is compounded annually is ₹ 92400
Now,
Compounded half - yearly
P = 80000
[tex]R = 10\% = \frac{10}{2} = 5[/tex]
[tex]T = 1\frac{1}{2} = \frac{3}{2} = \frac{3}{2} \times 2 = 3[/tex]
Using
[tex]A = P \: {(1 + \frac{R}{100} )}^{n} [/tex]
[tex]A = 80000 {(1 + \frac{5}{100} )}^{3} [/tex]
[tex]A = 80000 {(1 + \frac{1}{20} )}^{3} [/tex]
[tex]A = 80000 {( \frac{20 + 1}{20} })^{3} \: \: \: \: by \: lcm[/tex]
[tex]A = 80000 {( \frac{21}{20} })^{3} [/tex]
[tex]A = 80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} [/tex]
[tex]A = 80 \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} [/tex]
[tex]A = \frac{40 \times 21 \times 21 \times 21}{2 \times 2} [/tex]
[tex]A = \frac{20 \times 21 \times 21 \times 21}{2} [/tex]
[tex]A = 10 \times 21 \times 21 \times 21[/tex]
[tex]A = 92610[/tex]
Now,
Difference
92610 - 92400
210 Ans
Formula used :-
• [tex]A = P \: {(1 + \frac{R}{100} )}^{n} [/tex]
• [tex]SI = \frac{P× R× T}{100} [/tex]