Solved Examples(Set 9) - Simple Interest

Answer: Option B

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 $=\dfrac{20}{4}\%=5\%$

Answer: Option C

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 $(1)$

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

Answer: Option B

Explanation:

Simple interest at $4\%$ is $₹120$

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

Answer: Option D

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 $=12000+x=12000+8000=20000$

Answer: Option C

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 $=\dfrac{100×1100}{4000×2}=13.75$