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?

plz help me out☝️
​

Answer 1

EXPLANATION.

• GIVEN

A number is 4 times another number.

21 is added to both the number then first number

become 7/4 times the other number.

Find the number,

According to the question,

Let the number be = x and y

number is 4 times another number

x = 4y .....(1)

21 is added to the number then first number

become 7/4 times the other number.

=> ( x + 21 ) / ( y + 21 ) = 7/4

=> 4( x + 21 ) = 7 ( y + 21 )

=> 4x + 84 = 7y + 147

=> 4x - 7y = 63 ......(2)

From equation (1) and (2) we get,

put the value of x in equation (2)

=> 4( 4y ) - 7y = 63

=> 16y - 7y = 63

=> 9y = 63

=> y = 7

Put value of y in equation (1)

we get,

=> x = 4 X 7

=> x = 28

Therefore,

One number is 7

other number is 28

Two number are = 7 and 28

Answer 2

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• A number is 4 times the another number.

• If 21 is added to both the numbers then first number becomes 7/4 times the other number.

$\large{\blue{\underline{\underline{\bold{Solution:-}}}}}$

• The two numbers

$\large{\green{\underline{\underline{\bold{Solution:-}}}}}$

According to the first condition:-

One number is 4 times the other

• Let the first number be 4x

• The second number = x

According to the second condition:-

By adding 21 to both the numbers.

• The first number = 4x + 21

• The second number = x + 21

Given, first number becomes 7/4 the other number

So:-

$\sf \implies(4x + 21) = \dfrac{7}{4} (x + 21)$

$\sf \implies4(4x + 21) = {7} (x + 21)$

$\sf \implies16x + 84= 7x + 147$

$\sf \implies16x - 7x= 147 - 84$

$\sf \implies9x = 63$

$\sf \implies x = \dfrac{63}{9}$

$\sf \implies x = 7$

Therefore:-

• The second number = x = 7

• The first number = 4x = 4× 7 = 28

Hence;

$\large \boxed{\sf \purple{The \: numbers \: are \: 7 \: and \: 28}}$