Question

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?​

Answer 1

Answer:

1st no. = 7

2nd no. = 35

Step-by-step explanation:

let, the 1st no. be x .

So, the other no. be 5x.

Again,

1st no. = x + 21

2nd no. = 5x + 21

According to question,

$5x + 21 = 2(x + 21) \\ \\ 5x + 21 = 2x + 42 \\ \\ 5x - 2x = 42 - 21 \\ \\ 3x = 21 \\ \\ x = 7$

So, The required no. are x = 7 and 5x = 35 .

Answer 2

AnswEr :

$\:\bullet\:\sf\ Let \: the \: first \: number \: be \: \bf\ x$

$\:\bullet\:\sf\ So, \: the \: second \: number \: is \: \bf\ 5x$

$\rule{100}1$

$\normalsize\quad\star\sf\ When \: 21 \: is \: added \: to \: both \: no.s$

$\normalsize\twoheadrightarrow\sf\ First \: number = x + 21$

$\normalsize\twoheadrightarrow\sf\ Second \: number = 5x + 21$

$\underline{\bigstar\:\textsf{According \: to \: the \: question \: now:}}$

$\normalsize\dashrightarrow\sf\ 2(First \: number) = Second \: number$

$\normalsize\dashrightarrow\sf\ 2(x + 21 ) = 5x + 21$

$\normalsize\dashrightarrow\sf\ 2x + 42 = 5x + 21$

$\normalsize\dashrightarrow\sf\ 3x = 21$

$\normalsize\dashrightarrow\sf\ x = \frac{\cancel{21}}{\cancel{3}}$

$\normalsize\dashrightarrow\sf\ x = 7$

$\underline{\bigstar\:\textsf{Calculation \: of \: numbers:}}$

$\normalsize\twoheadrightarrow\sf\ First \: no. = x \\ \\ \normalsize\twoheadrightarrow\sf\ First \: no.= 7$

$\rule{100}1$

$\normalsize\twoheadrightarrow\sf\ Second \: no. = 5x \\ \\ \normalsize\twoheadrightarrow\sf\ Second \: no.= 5 \times\ 7 \\ \\ \normalsize\twoheadrightarrow\sf\ Second \: no. = 35$

$\therefore\:\underline{\textsf{Hence, \: the \: required \: numbers \: are}{\textbf{ 7 \: and \: 35}}}$