Question

A number is 4 times another number, If
21 is added to both the numbers then first
number becomes 7/4 times the other
number. What are the numbers?​

Answer 1

Answer:

$\large{\underline{\boxed{\mathfrak\blue{Numbers = 7 \: and \: 28 }}}}$

Given:-

• 1st number = 4 × 2nd number
• If 21 added to both numbers,1st number = 7/4 × 2nd number.

To find:-

• What are the numbers?

Let the 2nd number be y.

$\therefore$ 1st number = 4y.

A/Q,

• 21 addedto both numbers,1st number = 7/4 times 2nd number.

$\twoheadrightarrow\sf 4y + 21 = \dfrac{7}{4}\times (y + 21)\\\\ \twoheadrightarrow\sf 4y + 21 = \dfrac{7(y + 21)}{4} \\\\ \twoheadrightarrow\sf 4(4y + 21) = 7y + 147 \\\\ \twoheadrightarrow\sf 16y + 84 = 7y + 147 \\\\\twoheadrightarrow\sf 16y - 7y = 147 - 84 \\\\\twoheadrightarrow\sf 9y = 63 \\\\ \twoheadrightarrow\sf y = 63/9 \\\\ \twoheadrightarrow\large{\underline{\boxed{\sf\blue{y = 7 }}}}$

$\rule{100}2$

$\star \: \: \sf 1st \: number = 4y = 4(7) = \large{\boxed{\sf\red{28}}}$

$\star \: \: \sf 2nd \: number = y = \large{\boxed{\sf\green{7}}}$

$\therefore{\underline{\text{Numbers = 28 \: \& \: 7 \: respectively }}}$

Answer 2

$\red{Let \: the \: numbers \: be \: x \: and \: y.}$

${x=4y …(1)}$

$\blue{acc. \: to \: the \: que.}$

$(21+x)/(y+21) = 7/4 \:$

$=>4(21+x) = 7(y+21) \:$

$\: =>84 + 4x = 7y +147$

$\bf{=16y - 7y=147 - 84}$

$=>9y=63 \:$

$\: =>y=7$

$\purple{now. \: putting \: the \: value \: of \: y}$

$x = 4y = 4 \times 7 = 28$

$</p><p> \pink{y = 7 } \orange{and } \: \green {x = 28.}</p><p></p><p>$