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# Solved Examples(Set 9) - Simple Interest

 41. If the simple interest on a certain sum of money for $4$ years is one–fifth of the sum, then the rate of interest per annum is A. $7\%$ B. $5\%$ C. $4\%$ D. $6\%$

Explanation:

## Solution 1

Let the sum be be $x$

Then, simple interest $=\dfrac{x}{5}$

T $=4$

$\text{R}=\dfrac{100×\text{SI}}{\text{PT}}=\dfrac{100×\dfrac{x}{5}}{x×4}=5$

## Solution 2

Total simple interest of $4$ years is one–fifth of the sum, that is, $20\%$ of the sum.

rate of interest per annum $=\dfrac{20}{4}\%=5\%$

 42. If a certain sum is invested for T years at simple interest of $10\%$ per annum, it amounts to $₹400.$ If the rate of interest was $4\%$ per annum, it would have amounted to $₹200.$ Find T. A. $45$ years B. $40$ years C. $50$ Years D. $60$ years

Explanation:

## Solution 1

Refer formula

$\text{T}=\dfrac{\text{P}_1-\text{P}_2}{\text{P}_2\text{R}_1-\text{P}_1\text{R}_2}×100\\=\dfrac{400-200}{200×10-400×4}×100=50$

## Solution 2

Let the sum be $x$

Simple interest on $x$ at $10\%$ per annum in $\text{T}$ years $=400-x$

Simple interest on $x$ at $4\%$ per annum in $\text{T}$ years $=200-x$

simple interest $\propto$ R (because P and T are constants here). Therefore,

$400-x:200-x=10:4\\\Rightarrow 400-x:200-x=5:2\\\Rightarrow 800-2x=1000-5x\\\Rightarrow x=\dfrac{200}{3}$

Simple interest on $\dfrac{200}{3}$ at $10\%$ per annum in $\text{T}$ years $=400-\dfrac{200}{3}=\dfrac{1000}{3}$

Therefore, T $=\dfrac{100×\dfrac{1000}{3}}{\dfrac{200}{3}×10}=50$

## Solution 3

Let the sum be $x$

$x$ amounts to $400$ at $10\%$ per annum in $\text{T}$ years.
Simple interest $=400-x$

$400-x=\dfrac{x×10×\text{T}}{100}~\cdots(1)$

$x$ amounts to $200$ at $4\%$ per annum in $\text{T}$ years.
Simple interest $=200-x$

$200-x=\dfrac{x×4×\text{T}}{100}~\cdots(2)$

Dividing $(1)$ by $(2)$

$\dfrac{400-x}{200-x}=\dfrac{5}{2}\\\Rightarrow 800-2x=1000-5x\\\Rightarrow 3x=200\\\Rightarrow x=\dfrac{200}{3}$

Substituting value of $x$ in $(1)$

$400-\dfrac{200}{3}=\dfrac{\left(\dfrac{200}{3}\right)×10×\text{T}}{100}\\\Rightarrow 400-\dfrac{200}{3}=\dfrac{20\text{T}}{3}\\\Rightarrow 1200-200=20\text{T}\\\Rightarrow 20\text{T}=1000\\\Rightarrow\text{T}=50$

 43. Arun lent a sum of money at simple interest for $6$ years. If the same amount was paid at $4\%$ higher, Arun would have got $₹120$ more. Find the sum. A. $₹600$ B. $₹500$ C. $₹200$ D. $₹400$

Explanation:

Simple interest at $4\%$ is $₹120$

$\text{P}=\dfrac{100×\text{SI}}{\text{RT}}=\dfrac{100×120}{4×6}=500$

 44. Mr.Mani invested an amount of $₹12000$ at a simple interest rate of $10\%$ per annum and another amount at a simple interest rate of $20\%$ per annum. The total interest earned at the end of one year on the total amount invested is $14\%$ per annum. Find the total amount invested. A. $₹25000$ B. $₹15000$ C. $₹10000$ D. $₹20000$

Explanation:

## Solution 1

Refer formula

$14=\dfrac{12000×10+\text{P}_2×20}{12000+\text{P}_2}\\\Rightarrow 12000×14+14\text{P}_2=120000+20\text{P}_2\\\Rightarrow 6\text{P}_2=48000\\\Rightarrow \text{P}_2=8000$

Total amount invested $=12000+8000=20000$

## Solution 2

Let the other amount invested be $x$

Total simple interest $=\dfrac{12000×10×1}{100}+\dfrac{x×20×1}{100}=1200+\dfrac{x}{5}$

But $1200+\dfrac{x}{5}$ is the simple interest on $12000+x$ at $14\%$ per annum for $1$ year. Therefore,

$\Rightarrow 1200+\dfrac{x}{5}=\dfrac{(12000+x)×14×1}{100}\\\Rightarrow 120000+20x=14×12000+14x\\\Rightarrow 6x=48000\\\Rightarrow x=8000$

Total amount invested $=12000+x=12000+8000=20000$

 45. A lent $₹4000$ to B for $2$ years and $₹2000$ to C for $4$ years on simple interest at the same rate of interest and received $₹2200$ in all from both of them as interest. The rate of interest per annum is: A. $15\%$ B. $12\%$ C. $13.75\%$ D. $14\%$

Explanation:

## Solution 1

Let the rate of interest per annum be R%

Simple interest on $₹4000$ for $2$ years at R% + simple interest on $₹2000$ for $4$ years at R% $=2200$

$\Rightarrow \dfrac{4000×\text{R}×2}{100}+\dfrac{2000×\text{R}×4}{100}=2200\\\Rightarrow 160\text{R}=2200\\\Rightarrow \text{R}=\dfrac{2200}{160}=\dfrac{55}{4}=13.75$

## Solution 2

$₹4000$ for $2$ years and $₹2000$ for $4$ years give same simple interest.

Therefore, simple interest on $₹4000$ for $2$ years $=\dfrac{2200}{2}=1100$

R $=\dfrac{100×1100}{4000×2}=13.75$