Question

Hii guys .....
Q. The difference between C.I and S.I. on a certain sum at 5% for 2years is rs 1. 50.
Find the sum.
Plz.... Solve this ques. ... With solution.

Answer 1

Given, Time n = 2 years, R = 5%.

(i)

We know that CI = P[(1 + r/100)^n - 1]

⇒ P[(1 + 5/100)^2 - 1]

⇒ P[(21/20)^2 - 1]

⇒ P[441 - 400/400]

⇒ 41P/400

(ii)

We know that SI = PTR/100

⇒ P * 2 * 5/100

⇒ 10P/100

⇒ P/10.

Given that difference between CI and SI is 1.50.

⇒ (41P/400) - (P/10) = 1.50

⇒ 41P - 40P = 600

⇒ P = 600.

Therefore, the sum = 600.

Hope it helps!

Answer 2

HEY DEAR ... ✌️

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=) Given , Time (n) = 2 years , Rate (R) = 5%.

(i) We know that CI = P[(1 + r/100)^n - 1]

⇒ P[(1 + 5/100)^2 - 1]

⇒ P[(21/20)^2 - 1]

⇒ P[441 - 400/400]

⇒ 41P/400

(ii) We know that SI = PTR/100

⇒ P * 2 * 5/100

⇒ 10P/100

⇒ P/10.

Now , Given that difference between CI and SI is 1.50.

⇒ (41P/400) - (P/10) = 1.50

⇒ 41P - 40P = 600

⇒ P = 600.

Therefore, the sum = 600

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HOPE , IT HELPS ... ✌️