A certain sum of money become double of itself in 6 years on simple in

1.	A certain sum of money become double of itself in 6 years on simple interest. In how much time it will be 10 times of itself.1. 56 years2. 55 years3. 54 years4. 53 years
Answer» Correct Answer - Option 3 : 54 years Given: Time = 6 years Formula used: SI = (P x R x T)/100, Amount = SI + P where, P → Principal, R → Rate of Interest, T → Time Period of the Loan/deposit in years, S.I. → Simple Interest Calculation: Let, the P = x, A = 2x ⇒ SI = A - P ⇒ SI = 2x - x = x ⇒ x = (x x R x 6)/100 ⇒ R = 50/3% In 10 times Let, P = x, A = 10x ⇒ SI = 10x - x = 9x ⇒ SI = 9x ⇒ 9x = (x x 50 x T)/100 × 3 ⇒ T = 54 years ∴ It will take 54 years to become 10 times of itself.

A certain sum of money become double of itself in 6 years on simple interest. In how much time it will be 10 times of itself.1. 56 years2. 55 years3. 54 years4. 53 years

Answer» Correct Answer - Option 3 : 54 years

Given:

Time = 6 years

Formula used:

SI = (P x R x T)/100,

Amount = SI + P

where,

P → Principal,

R → Rate of Interest,

T → Time Period of the Loan/deposit in years,

S.I. → Simple Interest

Calculation:

Let, the P = x, A = 2x

⇒ SI = A - P

⇒ SI = 2x - x = x

⇒ x = (x x R x 6)/100

⇒ R = 50/3%

In 10 times Let, P = x, A = 10x

⇒ SI = 10x - x = 9x

⇒ SI = 9x

⇒ 9x = (x x 50 x T)/100 × 3

⇒ T = 54 years

∴ It will take 54 years to become 10 times of itself.

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A certain sum of money become double of itself in 6 years on simple interest. In how much time it will be 10 times of itself.1. 56 years2. 55 years3. 54 years4. 53 years

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