Question

Divide a $5inch$ line into two parts so that one part is:

(a) $2\frac{1}{4}inches$shorter than the other;

(b) $3$ times the other.

Answer 1

Explanation for the correct answer:

Step 1: Assumtions:

Let $\text{L}$ be a line of length $5inches$.

Let ${\text{L}}_1$ be the longest parts of the line

Let ${\text{L}}_2$ be the shortest parts of the line

So,

$\Rightarrow {\text{L}}_1+{\text{L}}_2=5inches...\left(i\right)$

Step 2: Divide the line into two parts when one part is $2\frac{1}{4}$ inches shorter than the other:

Given,

$$\begin{gathered} {\text{L}}_1-{\text{L}}_2=2\frac{1}{4} \\ \Rightarrow {\text{L}}_1-{\text{L}}_2=\frac{9}{4}inches...\left(ii\right)\left[\because a\frac{c}{b}=\frac{\left(a\times b\right)+c}{b}\right] \end{gathered}$$

Add equation $\left(i\right)\text{and}\left(ii\right)$

we get,

$$\begin{gathered} \Rightarrow 2{\text{L}}_1=5+\frac{9}{4} \\ \Rightarrow 2{\text{L}}_1=\frac{20+9}{4}\left[\text{taking L.C.M in denominator and cross multiply}\right] \\ \Rightarrow {\text{L}}_1=\frac{29}{4\times 2} \\ \Rightarrow {\text{L}}_1=\frac{29}{8}inches \end{gathered}$$

Substitute ${\text{L}}_1=\frac{29}{8}$ in equation $\left(i\right)$.

$$\begin{gathered} \Rightarrow \frac{29}{8}+{\text{L}}_2=5 \\ \Rightarrow {\text{L}}_2=5-\frac{29}{8} \\ \Rightarrow {\text{L}}_2=\frac{40-29}{8} \\ \Rightarrow {\text{L}}_2=\frac{11}{8}inches \end{gathered}$$

Step 3: Divide the line into two parts when one part is $3$ times the other.

Given,

$\Rightarrow {\text{L}}_1=3{\text{L}}_2$

Substitute this in the equation $\left(i\right)$

$$\begin{gathered} \Rightarrow 3{\text{L}}_2+{\text{L}}_2=5 \\ \Rightarrow 4{\text{L}}_2=5 \\ \Rightarrow {\text{L}}_2=\frac{5}{4}inches \end{gathered}$$

Substitute ${\text{L}}_2=\frac{5}{4}$ in equation $\left(i\right)$.

$$\begin{gathered} \Rightarrow {\text{L}}_1+\frac{5}{4}=5 \\ \Rightarrow {\text{L}}_1=5-\frac{5}{4} \\ \Rightarrow {\text{L}}_1=\frac{15}{4}inches \end{gathered}$$

Hence, For (a) the length of ${\text{L}}_1$ and ${\text{L}}_2$ are $\frac{29}{8}inches$ and $\frac{11}{8}inches$,

For (b) the length of ${\text{L}}_1$ and ${\text{L}}_2$ are $\frac{15}{4}inches$ and $\frac{5}{4}inches$.