Solved Examples(Set 6) - Simple Interest

Answer: Option A

Explanation:

Solution 1

Let investments in A,B,C be $x,y,z$

simple interest on $x$ at $10\%$ for $1$ year
+ simple interest on $y$ at $12\%$ for $1$ year
+ simple interest on $z$ at $15\%$ for $1$ year
$=3200$

$\dfrac{x×10×1}{100}+\dfrac{y×12×1}{100}+\dfrac{z×15×1}{100}=3200\\\Rightarrow 10x+12y+15z=320000 \quad \cdots(1)$

investment in Scheme C = $240\%$ of investment in scheme B
$\Rightarrow z=\dfrac{240y}{100}=\dfrac{12y}{5}\cdots\cdots(2)$

investment in Scheme C = $150\%$ of investment in scheme A
$\Rightarrow z=\dfrac{150x}{100}=\dfrac{3x}{2}\\\Rightarrow x=\dfrac{2z}{3}=\dfrac{2}{3}×\dfrac{12y}{5}=\dfrac{8y}{5} \cdots(3)$

Using $(2)$ and $(3)$ in $(1)$

$10x+12y+15z=320000\\\Rightarrow 10\left(\dfrac{8y}{5}\right)+12y+15\left(\dfrac{12y}{5}\right)=320000\\\Rightarrow 16y+12y+36y=320000\\\Rightarrow 64y=320000\\\Rightarrow y=5000$

That is, amount invested in Scheme B $=₹5000$

Solution 2

scheme C investment $=150\%$ of scheme A investment
scheme C investment $=240\%$ of scheme B investment

Therefore, ratio of investments of A,B,C
$=100:150×\dfrac{100}{240}:150\\=8:5:12$
hint

Ratio of rate of interest of A,B,C
$=10:12:15$

simple interest $\propto$ PR (because T is constant here)
Therefore, ratio of simple interests in A,B,C
$=8×10:5×12:12×15=4:3:9$

Interest on B $=3200×\dfrac{3}{4+3+9}=600$

Investment on B $=\dfrac{100×600}{12×1}=5000$

Answer: Option B

Explanation:

Solution 1

Let the partial amount lent at $5\%$ and $8\%$ be $x$ and $(1550-x)$ respectively.

simple interest on $x$ at $5\%$ for $3$ years + simple interest on $(1550-x)$ at $8\%$ for $3$ years $=300$

$\dfrac{x×5×3}{100}+\dfrac{(1550-x)×8×3}{100}=300\\\Rightarrow 5x+8(1550-x)=10000\\\Rightarrow 5x+12400-8x=10000\\\Rightarrow 3x=2400\\\Rightarrow x=800$

Required ratio $=x:(1550-x)=800:(1550-800)=800:750=16:15$

Solution 2

simple interest on $1550$ at $5\%$ for $3$ years $=\dfrac{1550×5×3}{100}=232.5$

Therefore, part amount invested at $8\%$ has earned the additional interest of $(300-232.5)$ at the interest rate of $(8-5)\%$ in $3$ years.

Therefore, part amount invested at $8\%$ $=\dfrac{100×67.5}{3×3}=750$

Part amount invested at $5\%$ $=1550-750=800$

Required ratio $=800:750=16:15$

Answer: Option A

Explanation:

Solution 1

Let the sum be $x.$ Assume that it will treble in $n$ years.

Note that when the money doubles, simple interest is $(2x-x)$ and when the money trebles, simple interest is $(3x-x)$

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

Therefore,
$(2x-x):(3x-x)=12:n\\\Rightarrow x:2x=12:n\\\Rightarrow n=24$

Solution 2

Given that the sum doubles in $12$ years. This means, simple interest is equal to the sum in $12$ years.

Therefore, simple interest will be double the sum in $24$ years. That is, it will treble in $24$ years.

Answer: Option C

Explanation:

Solution 1

Let the first sum and second sum be $4x$ and $5x$ respectively (hint).

Simple interest on first sum at $6\%$ for $2$ years + simple interest on second sum at $7\%$ for $2$ years $=354$
$\Rightarrow \dfrac{4x×6×2}{100}+\dfrac{5x×7×2}{100}=354\\\Rightarrow 4x×6+5x×7=17700\\\Rightarrow 59x=17700\\\Rightarrow x=300$

Total sum invested $=4x+5x=9x=2700$

Solution 2

Let the sum invested at $6\%$ and $7\%$ be $x$ and $y$ respectively.

Simple interest on $x$ at $6\%$ for $2$ years + simple interest on $y$ at $7\%$ for $2$ years $=354$
$\Rightarrow \dfrac{x×6×2}{100}+\dfrac{y×7×2}{100}=354\\\Rightarrow 6x+7y=17700 ~\cdots(1)$

One-forth of the first sum is equal to one-fifth of the second sum
$\Rightarrow \dfrac{x}{4}=\dfrac{y}{5}\\\Rightarrow x=\dfrac{4y}{5}\quad \cdots (2)$

Solving $(1)$ and $(2),$
$6x+7y=17700\\\Rightarrow 6\left(\dfrac{4y}{5}\right)+7y=17700\\\Rightarrow 24y+35y=17700×5\\\Rightarrow 59y=17700×5\\\Rightarrow y=1500$

$x=\dfrac{4y}{5}=\dfrac{4×1500}{5}=1200$

Total sum invested $=x+y=1500+1200=2700$

Answer: Option A

Explanation:

Solution 1

$\text{P}=\dfrac{100×101.20}{5×1}=2024$

Simple interest on $₹2024$ at $6\%$ per annum for $1$ year $=\dfrac{2024×6×1}{100}=121.44$

Additional simple interest $=121.44-101.20=20.24$

Solution 2

$\text{P}=\dfrac{100×101.20}{5×1}=2024$

Additional interest rate $=6\%-5\%=1\%$

Additional simple interest = simple Interest on $₹2024$ at $1\%$ per annum for $1$ year
$=\dfrac{2024×1×1}{100}=20.24$

Solution 3

Additional interest rate $=6\%-5\%=1\%$

$5\%\implies 101.20\\1\% \implies \dfrac{101.20×1}{5}=20.24$

Therefore, additional simple interest is $20.24$