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# Solved Examples(Set 2) - Simple Interest

 6. A person borrows $₹5000$ for $2$ years at $4\%$ per annum simple interest. He immediately lends it to another person at $6\dfrac{1}{4}\%$ per annum for $2$ years. Find his gain in the transaction per year. A. $₹112.50$ B. $₹167.50$ C. $₹225$ D. $₹150$

Explanation:

## Solution 1

Difference in interest rate
$=6\dfrac{1}{4}\%-4\%=2\dfrac{1}{4}\%=\dfrac{9}{4}\%$

Gain per year
= simple interest on $5000$ at $\dfrac{9}{4}\%$ for $1$ year
$=\dfrac{5000×\dfrac{9}{4}×1}{100}=112.5$

## Solution 2

Simple interest he pays
$=\dfrac{5000×4×2}{100}=400$

Simple interest he gets
$=\dfrac{5000×\dfrac{25}{4}×2}{100}=625$

Gain in $2$ years $= 625-400=225$

Gain per year $=\dfrac{225}{2}=112.5$

 7. What is the ratio of simple interest earned by a certain amount at the same rate of interest for $5$ years and that for $15$ years? A. $3:1$ B. $1:3$ C. $3:2$ D. $2:3$

Explanation:

## Solution 1

Required ratio
$=\dfrac{\text{PR}×5}{100}:\dfrac{\text{PR}×15}{100}\\=5:15=1:3$

## Solution 2

Simple interest $=\dfrac{\text{PRT}}{100}.$ In this case, P and R are constants and therefore simple interest $\propto \text{T}$

Therefore, required ratio
$=\text{T}_1:\text{T}_2=5:15=1:3$

 8. A sum of money amounts to $₹9800$ after $5$ years and $₹12005$ after $8$ years at the same rate of simple interest. The rate of interest per annum is A. $15\%$ B. $5\%$ C. $12\%$ D. $8\%$

Explanation:

Simple interest for $3$ years
$=12005-9800=2205$

Simple interest for $5$ years
$=\dfrac{2205}{3}×5=3675$

Sum of money $=9800-3675=6125$

$\text{R}=\dfrac{100×2205}{6125×3}=12$

 9. A certain amount earns simple interest of $₹1200$ after $10$ years. Had the interest been $2\%$ more, how much more interest would it have earned? A. None of these B. $₹120$ C. Cannot be determined D. $₹25$

Explanation:

Required details cannot be found out with the given information.
 10. A man took loan from a bank at the rate of $8\%$ per annum simple interest. After $4$ years he had to pay $₹6200$ interest only for the period. What was the principal amount borrowed by him? A. $₹20245$ B. $₹18230$ C. $₹17322$ D. $₹19375$
$\text{P}=\dfrac{100×\text{SI}}{\text{RT}}=\dfrac{100×6200}{8×4}=19375$