mcqs Math Simple Interest Robert took a loan of $4200 at the rate of 9% simple interest per annum. If he paid an amount of $7980 to clear the loan, then find the time period of the loan. 7989

Question: Robert took a loan of $4200 at the rate of 9% simple interest per annum. If he paid an amount of $7980 to clear the loan, then find the time period of the loan.

(A)  7.5
(B)  12.5
(C)  10
(D)  15

Correct Answer 10

Solution And Explanation

Solution

Given,

Principal (P) = $4200

Rate of Simple Interest (R) = 9% per annum

Amount (A) = $7980

Thus, time (T) = ?

Method (1) Using Formula

We know that, Amount (A) = Principal (P) + Simple Interest (SI)

⇒ Simple Interest (SI) = Amount – Principal

⇒ SI = $7980 – $4200 = $3780

Thus, Simple Interest = $3780

Time (T) = ^{100 × Simple Interest}/_{Principal × Rate of Interest}

⇒ T = ^{100 × SI}/_{P × R}

Thus, Time (T) = ^{100 × 3780}/_{4200 × 9}

= ³⁷⁸⁰⁰⁰/₃₇₈₀₀

= 10 years (using formula)

Thus, Time (T) = 10 years (from time taken before calculation)Answer

Here, we have

Principal (P) = $4200

Rate of Simple Interest (R) = 9% per annum

Simple Interest = $3780 (As calculated above by subtracting Principal from the Amount given)

We know that, interest is calculated on the basis of the Principal.

This means Simple Interest for 1 year = Rate of simple interest × Principal

Thus, Simple Interest for 1 year = 9% of Principal

= 9% of $4200

= ⁹/₁₀₀ × 4200

= ^{9 × 4200}/₁₀₀

= ³⁷⁸⁰⁰/₁₀₀ = 378

Thus, simple Interest for 1 year = $378

Now,

∵ If the simple Interest is $378, then the time = 1 year

∴ If the simple Interest is $1, then the time = ¹/₃₇₈ years

∴ If the simple Interest is $3780, then the time = ¹/₃₇₈ × 3780 years

= ^{1 × 3780}/₃₇₈ years

= ³⁷⁸⁰/₃₇₈ = 10 years

Thus, time (T) = 10 years Answer