Solved Examples(Set 4) - Simple Interest

Answer: Option C

Explanation:

Solution 1

Let the sum lent to C be $x$

Simple interest on $₹1500$ at $8\%$ for $4$ years + simple interest on $x$ at $8\%$ for $4$ years $=₹1400$

$\Rightarrow \dfrac{1500×8×4}{100}+\dfrac{x×8×4}{100}=1400\\\Rightarrow \dfrac{8×4(1500+x)}{100}=1400\\\Rightarrow 1500+x=4375\\\Rightarrow x=2875$

Solution 2

Simple interest received from B
$=\dfrac{1500×8×4}{100}=480$

Simple interest received from C
$=1400-480=920$

Sum lent to C $=\dfrac{100×920}{8×4}=2875$

Solution 3

Simple interest on the total money lent at $8\%$ for $4$ years $=₹1400$

Total money lent
$=\dfrac{100×1400}{8×4}=4375$

Thereofre, money lent to C
$=4375-1500=2875$

Answer: Option C

Explanation:

Solution 1

Refer formula

Let rate of interest be R%

Amount due in $6$ months
$=10$ + simple interest on $₹10$ for six months
$=10+\dfrac{10×\text{R}×\dfrac{1}{2}}{100}=10+\dfrac{\text{R}}{20}$

With the formula mentioned,
$3=\dfrac{100\left(10+\dfrac{\text{R}}{20}\right)}{100×6+\dfrac{\text{R}×6(6-1)}{2×12}}\\\Rightarrow 3=\dfrac{1000+5\text{R}}{600+\dfrac{5\text{R}}{4}}\\\Rightarrow 1800+\dfrac{15\text{R}}{4}=1000+5\text{R}\\\Rightarrow \dfrac{5\text{R}}{4}=800\\\Rightarrow \text{R}=640$

Solution 2

Loan amount $=₹10$

Let rate of interest be R%

Simple Interest on $₹10$ for six months
$=\dfrac{10×\text{R}×\dfrac{1}{2}}{100}=\dfrac{\text{R}}{20}$

Therefore, amount due in $6$ months $=10+\dfrac{\text{R}}{20}$

Payment after first month $=3$
Interest on $₹3$ for remaining five months $=\dfrac{3×\text{R}×5}{12×100}$

Payment after second month $=3$
Interest on $3$ for remaining four months $=\dfrac{3×\text{R}×4}{12×100}$

$\cdots$

Payment after fifth month $=3$
Interest on $3$ for the remaining one month $=\dfrac{3×\text{R}×1}{12×100}$

Payment after sixth month $=3$ and this will close the loan.

Therefore,
$3×6+\dfrac{3×\text{R}(5+4+3+2+1)}{12×100}=10+\dfrac{\text{R}}{20}\\\Rightarrow 8+\dfrac{3\text{R}}{80}=\dfrac{\text{R}}{20}\\\Rightarrow \dfrac{\text{R}}{80}=8\\\Rightarrow \text{R}=640$

Answer: Option D

Explanation:

Solution 1

Let the sum be $x$
Then, simple interest $=\dfrac{x}{4}$
T $=3\dfrac{1}{8}=\dfrac{25}{8}$ years

$\text{R}=\dfrac{100×\dfrac{x}{4}}{x×\dfrac{25}{8}}=8$

Solution 2

SI on $100$ for $\dfrac{25}{8}$ years $=25$ (because, SI is $1/4$ of principal)
$\Rightarrow$ SI on $100$ for $1$ year $=25×\dfrac{8}{25}=8$

Therefore, required interest rate $=8\%$

Solution 3

Since $\text{SI}=\dfrac{\text{PRT}}{100}$ and given that simple interest is $\dfrac{1}{4}$ of the principal,

$\dfrac{\text{RT}}{100}=\dfrac{1}{4}\\\Rightarrow \text{R}×\dfrac{25}{8}×\dfrac{1}{100}=\dfrac{1}{4}\\\Rightarrow \text{R}=8$

Answer: Option B

Explanation:

Solution 1

Refer formula

Let rate of interest be R%

Amount due in $10$ months
$=9$ + simple interest on $₹9$ for ten months
$=9+\dfrac{9×\text{R}×\dfrac{10}{12}}{100}=9+\dfrac{\text{3R}}{40}$

With the formula mentioned,
$1=\dfrac{100\left(9+\dfrac{3\text{R}}{40}\right)}{100×10+\dfrac{\text{R}×10(10-1)}{2×12}}\\\Rightarrow 900+\dfrac{\text{15R}}{2}=1000+\dfrac{15\text{R}}{4}\\\Rightarrow \dfrac{15\text{R}}{4}=100\\\Rightarrow \text{R}=26.67$

Solution 2

Loan amount $=₹9$

Let rate of interest be R%

Simple Interest on $₹9$ for $10$ months
$=\dfrac{9×\text{R}×\dfrac{10}{12}}{100}=\dfrac{\text{3R}}{40}$

Therefore, amount due in $10$ months $=9+\dfrac{3\text{R}}{40}$

Payment after first month $=1$
Interest on $1$ for remaining nine months $=\dfrac{1×\text{R}×9}{12×100}$

Payment after second month $=1$
Interest on $1$ for remaining eight months $=\dfrac{1×\text{R}×8}{12×100}$

$\cdots$

Payment after ninth month $=1$
Interest on $1$ for the remaining one month $=\dfrac{1×\text{R}×1}{12×100}$

Payment after tenth month $=1$ and this will close the loan.

Therefore,
$1×10+\dfrac{\text{R}(9+8+\cdots+1)}{12×100}=9+\dfrac{\text{3R}}{40}\\\Rightarrow 1+\dfrac{3\text{R}}{80}=\dfrac{\text{3R}}{40}\\\Rightarrow \dfrac{3\text{R}}{80}=1\\\Rightarrow \text{R}=26.67$

Answer: Option B

Explanation:

Solution 1

Let the parts be $x,y,z$
R $=5\%$

$x$ + interest on $x$ for $2$ years = $y$ + interest on $y$ for $3$ years = $z$ + interest on $z$ for $4$ years

$\Rightarrow x+\dfrac{x×5×2}{100}=y+\dfrac{y×5×3}{100}=z+\dfrac{z×5×4}{100}\\\Rightarrow x+\dfrac{x}{10}=y+\dfrac{3y}{20}=z+\dfrac{z}{5}\\\Rightarrow \dfrac{11x}{10}=\dfrac{23y}{20}=\dfrac{6z}{5}$

$\text{ Let }\dfrac{11x}{10}=\dfrac{23y}{20}=\dfrac{6z}{5}=k$

Then,
$x=\dfrac{10k}{11}, y=\dfrac{20k}{23}, z=\dfrac{5k}{6}$

$\text{Given, }x+y+z=2379\\\Rightarrow \dfrac{10k}{11}+\dfrac{20k}{23}+\dfrac{5k}{6}=2379\\\Rightarrow 10k×23×6+20k×11×6+5k×11×23=2379×11×23×6\\\Rightarrow 1380k+1320k+1265k=2379×11×23×6\\\Rightarrow 3965k=2379×11×23×6\\\Rightarrow k=\dfrac{2379×11×23×6}{3965}$

$x=\dfrac{10k}{11}=\dfrac{10}{11}×\dfrac{2379×11×23×6}{3965}=828$

Solution 2

Let the parts be $x,y,z$
R $=5\%$

$x\left(1+\dfrac{5×2}{100}\right)=y\left(1+\dfrac{5×3}{100}\right)\\\Rightarrow \dfrac{x}{y}=\dfrac{23}{22}\cdots(1)$

$x\left(1+\dfrac{5×2}{100}\right)=z\left(1+\dfrac{5×4}{100}\right)\\\Rightarrow \dfrac{x}{z}=\dfrac{12}{11}\cdots(2)$

$(1)\Rightarrow \dfrac{x}{y}=\dfrac{23×12}{22×12}\cdots(3)$
$(2)\Rightarrow \dfrac{x}{z}=\dfrac{12×23}{11×23}\cdots(4)$

From $(3)$ and $(4),$
$x:y:z=23×12:22×12:11×23$

Given, $x+y+z=2379$

Therefore,
$x=\dfrac{2379×23×12}{23×12+22×12+11×23}=828$