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..
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Answers
✠ ʟᴇᴛ ᴜs ғɪʀsᴛ ᴜɴᴅᴇʀsᴛᴀɴᴅ:–
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.
✠ ғᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :–
SI = (P × R × T)/100
CI = P × (1 + R/100)ᵗ – P
Where:—
P = principal
R = rate
T = time
✠ ɢɪᴠᴇɴ:–
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
→ P = 1200000/64
→ P = 18750
- 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.
- formula used :–
SI = (P × R × T)/100
CI = P × (1 + R/100)ᵗ – P
- Where:—
P = principal
R = rate
T = time
- 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
→ P = 1200000/64
→ P = 18750
Sum=Rs.18750
- Hence,the sum is
Rs.18750