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An amount of money grows upto Rs. 4000 in 2 years and up to Rs. 8000 in 3 years on compound interest. What is the sum?
- a)Rs. 1600
- b)Rs. 1000
- c)Rs. 1200
- d)Rs. 2400
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An amount of money grows upto Rs. 4000 in 2 years and up to Rs. 8000 i...
Solution:
Let the principal be Rs. x.
Given, the principal amount grows up to Rs. 4000 in 2 years and up to Rs. 8000 in 3 years on compound interest.
Calculation:
Using the formula for compound interest,
Amount after 2 years = x(1 + r/100)²
Amount after 3 years = x(1 + r/100)³
Given, amount after 2 years = Rs. 4000 and amount after 3 years = Rs. 8000
So, we have two equations as below:
x(1 + r/100)² = 4000
x(1 + r/100)³ = 8000
Dividing the second equation by the first equation, we get:
(1 + r/100) = 2
r/100 = 1
r = 100
Substituting the value of r in any of the equations above, we get:
x = 1000
Therefore, the sum is Rs. 1000.
Hence, option (b) is the correct answer.
Let the principal be Rs. x.
Given, the principal amount grows up to Rs. 4000 in 2 years and up to Rs. 8000 in 3 years on compound interest.
Calculation:
Using the formula for compound interest,
Amount after 2 years = x(1 + r/100)²
Amount after 3 years = x(1 + r/100)³
Given, amount after 2 years = Rs. 4000 and amount after 3 years = Rs. 8000
So, we have two equations as below:
x(1 + r/100)² = 4000
x(1 + r/100)³ = 8000
Dividing the second equation by the first equation, we get:
(1 + r/100) = 2
r/100 = 1
r = 100
Substituting the value of r in any of the equations above, we get:
x = 1000
Therefore, the sum is Rs. 1000.
Hence, option (b) is the correct answer.
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Community Answer
An amount of money grows upto Rs. 4000 in 2 years and up to Rs. 8000 i...
Simple intrest and compound interest for 1yr is equal
so
3yr-2yr = 1 yr
8000(A)-4000(P) = 4000(I)
I= PTR/100
4000=(4000×1×r )/100
r =100
P×(1+r/100) ²
P×(1+100/100) ²=4000
P×(200/100) ²=4000
P×4=4000
P= 1000
so
3yr-2yr = 1 yr
8000(A)-4000(P) = 4000(I)
I= PTR/100
4000=(4000×1×r )/100
r =100
P×(1+r/100) ²
P×(1+100/100) ²=4000
P×(200/100) ²=4000
P×4=4000
P= 1000
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