Question

The two numbers in a ratio are 4:3. When 2 is added to the largest number and 6 is subtracted from the smallest number now the ratio between the new numbers is 7:4. Construct a pair of simultaneous equations to represent the above information by taking the largest number as x and the smallest number as y.

Answer 1

Solution :-

Let the two numbers be x and y.

According to the first condition :-

$: \implies \sf \dfrac{x}{y} = \dfrac{4}{3}$

$: \implies \sf x = \dfrac{4}{3}y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ................(1)$

According to the second condition :-

$: \implies \sf \dfrac{x+2}{y-6} = \dfrac{7}{4}$

$: \implies \sf 4(x+2)=7(y-6)$

$: \implies \sf 4x+8 = 7y-42$

$: \implies \sf 4x-7y = -50 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ................(2)$

Substitute the value of x in eq(2) :-

$: \implies \sf 4 \times \dfrac{4}{3}y-7y = -50$

$: \implies \sf \dfrac{16y}{3}-7y = -50$

$: \implies \sf \dfrac{16y-21y}{3}=-50$

$: \implies \sf 16y-21y = -50 \times 3$

$: \implies \sf -5y = -150$

$: \implies \bf y = 30$

Substitute y = 30 in eq(1) :-

$: \implies \sf x = \dfrac{4}{3} \times 30$

$: \implies \sf x = 4 \times 10$

$: \implies \bf x = 40$

The two numbers are 40 and 30.

Answer 2

Step-by-step explanation:

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$\bf{\bigstar}$ $\text{\large\underline{\bf{Question:-}}}$

$\leadsto$ The two numbers in a ratio are 4:3. When 2 is added to the largest number and 6 is subtracted from the smallest number now the ratio between the new numbers is 7:4. Construct a pair of simultaneous equations to represent the above information by taking the largest number as x and the smallest number as y.

$\bf{\bigstar}$ $\text{\large\underline{\bf{Solution:-}}}$

$\leadsto$ Let we take the two numbers be x and y.

$\bf{\bigstar}$ $\text{\large\underline{\bf{Now,\ according\ to\ the\ conditions:-}}}$

$\leadsto$ $\mathrm{=\ \dfrac{x}{y}\ =\ \dfrac{4}{3}}$

$\leadsto$ $\mathrm{=\ x\ =\ \dfrac{4}{3}y}$ .......................[1].

$\text{\large\underline{\bf{The\ above\ is\ the\ first\ required\ eq.}}}$

$\leadsto$ $\mathrm{=\ \dfrac{x+2}{y-6}\ =\ \dfrac{7}{4}}$

$\leadsto$ $\mathrm{=\ 4(x\ +\ 2)\ =\ 7(y\ -\ 6)}$

$\leadsto$ $\mathrm{=\ 4x\ +\ 8\ =\ 7y\ -\ 42}$

$\leadsto$ $\mathrm{=\ 4x\ -\ 7y\ =\ -50}$ ......................[2].

$\text{\large\underline{\bf{The\ above\ is\ the\ second\ required\ eq.}}}$

$\text{\large\underline{\bf{By\ substituting\ the\ value\ of\ x\ in\ eq.(2),\ we\ get:-}}}$

$\leadsto$ $\mathrm{4\ ×\ \dfrac{4}{3}y\ -7y\ =\ -50}$

$\leadsto$ $\mathrm{\dfrac{16y}{3}\ -7y\ =\ -50}$

$\leadsto$ $\mathrm{\dfrac{16y-21y}{3}\ =\ -50}$

$\leadsto$ $\mathrm{16y\ -\ 21y\ =\ -50\ ×\ 3}$

$\leadsto$ $\mathrm{-5y\ =\ -150}$

$\leadsto$ $\mathrm{y\ =\ \dfrac{-150}{-5}}$

$\leadsto$ $\fbox\purple{y\ =\ 30}$ $\red{\bigstar}$

$\bf{\bigstar}$ $\text{\large\underline{\bf{By\ substituting\ y\ =\ 30\ in\ eq.(1),\ we\ get:-}}}$

$\leadsto$ $\mathrm{x\ =\ \dfrac{4}{3}\ ×\ 30}$

$\leadsto$ $\mathrm{x\ =\ 4\ ×\ 10}$

$\leadsto$ $\fbox\purple{x\ =\ 40}$ $\red{\bigstar}$

$\mathrm{●\ Hence,\ the\ two\ numbers\ are\ 40\ and\ 30\ ●}$

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