The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum
Answers
Given :
The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120.
To find :
Sum of money
Solution :
We have :
CI - SI = 120
Time (n) = 2 years
Rate of interest (r) = 8% per annum
We know,
SI = Prn/100
CI = P(1 + r/100)ⁿ - P
Now atq,
⇒ [P(1 + r/100)ⁿ - P] - Prn/100 = 120
⇒ [P(1 + 8/100)² - P] - (P * 8 * 2)/100 = 120
⇒ [P(1 + 0.08)² - P] - (16p/100) = 120
⇒ [P(1.08)² - P] - 0.16P = 120
⇒ [P * 1.1664 - P] - 0.16P = 120
⇒ [1.1664P - P] - 0.16P = 120
⇒ 0.1664P - 0.16P = 120
⇒ 1664P/10000 - 16P/100 = 120
⇒ (1664P - 1600P)/10000 = 120
⇒ 64P = 1200000
⇒ P = 1200000/64
⇒ P = 18750
Therefore,
Sum of money = Rs. 18750
Answer :-
Here the concept of Simple Interest and Compound interest has been used. According to this, the Simple Interest is the Principal × Rate × Time divided by 100 . Also, here we just have to apply the values, and find the sum. Let's do it.
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★ Formula Used :-
where P is the principal, R is the rate, and T is the time.
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★ Solution :-
Given,
» Difference between SI and CI = 120
» Time Period = T = 2 years
» Rate = R = 8 %
Now according to the question :-
❐ CI - SI = 120
❐ P{1 + (R/100)}² - P - {(P × R × T)/100} = 120
❐ P{1 + (8/100)}² - P - {(P × 8 × 2) /100} = 120
❐ P(1 + 0.08)² - P - (16p/100) = 120
❐ P(1.08)² - P - 0.16P = 120
❐ P(1.1664) - P - 0.16P = 120
❐ 1.1664P - P - 0.16P = 120
❐ 0.1664P - 0.16P = 120
Multiplying all the terms by 10000. Then,
❐ 1664P - 1600P = 120
❐ 64P = 1200000
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• Simple Interest is the interest compound yearly just using the normal method for a simple term.
• Compound Interest is the interest compounded yearly and monthly too but using compound technique and methods for a term.
• Time is the period for which interest is given.
• Rate is the percentage of interest given on the sum.