Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In 4 years Rs. 6000 amounts to 8000. In what time will Rs. 525 amount to Rs. 700 at the same rate?

seo-qna
SearchIcon
Answer
VerifiedVerified
376.8k+ views
Hint: In this problem, we are given that in 4 years Rs. 6000 amounts to 8000, we have to find in what time will Rs. 525 amount to Rs. 700 at the same rate. Here we can first find the rate at which Rs.6000 amounts to 8000 in 4 years using the formula \[Rate=\dfrac{S.I\times 100}{P\times T}\]. We can then find the time taken for Rs. 525 amounts to Rs. 700 at the same rate, using the same formula.

Complete step by step answer:
Here we are given that in 4 years Rs. 6000 amounts to 8000 and we have to find in what time Rs. 525 amounts to Rs. 700 at the same rate.
We can now find the rate for the given Principal amount, P = Rs.6000, Time, T = 4 years.
Simple interest = A – P
\[\Rightarrow S.I=8000-6000=2000\]
We know that,
\[Rate=\dfrac{S.I\times 100}{P\times T}\]
We can now substitute the above values in the above formula, we get
\[\Rightarrow Rate=\dfrac{2000\times 100}{6000\times 4}=\dfrac{50}{6}=\dfrac{25}{3}\%\]
The rate is \[\dfrac{25}{3}\%\].
We can now find the time taken for Rs. 525 amounts to Rs. 700 at the same rate.
Where Principle amount, P = Rs.525, Amount, A= Rs.700, Rate, R = \[\dfrac{25}{3}\%\]
 and Simple interest, S.I = 700 – 525 = 175.
We can now substitute these values in \[Time=\dfrac{S.I\times 100}{P\times R}\], we get
\[\Rightarrow Time=\dfrac{175\times 100}{525\times \dfrac{25}{3}}=\dfrac{175\times 100\times 3}{525\times 25}\]
We can now simplify the above step, we get
\[\Rightarrow Time=\dfrac{175\times 12}{525}=\dfrac{12}{3}=4\]
Therefore, the time taken is 4 years.

Note: We should always remember that the initial amount deposited is Principle and the simple interest is equal to the difference of the present amount and the principle amount. We should also remember the formula \[Rate=\dfrac{S.I\times 100}{P\times T}\], to find the required answer for the given problem.