Question

A positive number is 6times another number. If 24 is added to both the numbers, then one of the new numbers become twice the other new number . what are the number

Answer 1

Given that:

• A positive number is 6 times another number.

To Find:

• What are the number?

Let us assume:

1. First number be x.
2. Second number = 6x

Given: 24 is added to both the numbers, then one of the new number become twice the other new number.

Adding 24 to both the numbers:

1. First number = x + 24
2. Second number = 6x + 24

According to the question.

↣ 2(x + 24) = 6x + 24

↣ 2x + 48 = 6x + 24

↣ 48 - 24 = 6x - 2x

↣ 24 = 4x

↣ 24/4 = x

↣ 6 = x

↣ x = 6

We get:

1. First number = x = 6
2. Second number = 6x = 6(6) = 36

Hence,

• The numbers are 6 and 36.

Answer 2

Answer:

Given :-

• A positive number is 6 times another number.
• 24 is added to both the numbers, then one of the new numbers becomes twice the other new numbers.

To Find :-

• What are the numbers.

Solution :-

Let,

$\mapsto \bf{First\: number =\: y}$

$\mapsto \bf{Second\: number =\: 6y}$

Now,

$\leadsto$ 24 is added to both the numbers :

$\mapsto \sf First\: number =\: y + 24$

$\mapsto \sf Second\: number =\: 6y + 24$

According to the question :

$\implies \sf 6y + 24 =\: y + 24$

$\implies \sf 6y + 24 =\: 2 \times (y + 24)$

$\implies \sf 6y + 24 =\: 2(y + 24)$

$\implies \sf 6y + 24 =\: 2y + 48$

$\implies \sf 6y - 2y =\: 48 - 24$

$\implies \sf 4y =\: 24$

$\implies \sf y =\: \dfrac{\cancel{24}}{\cancel{4}}$

$\implies \sf y =\: \dfrac{\cancel{12}}{\cancel{2}}$

$\implies \sf y =\: \dfrac{6}{1}$

$\implies \sf\bold{\purple{y =\: 6}}$

Hence, the required numbers are :

➲ First Number :

$\longrightarrow \sf First\: number =\: y$

$\longrightarrow \sf\bold{\red{First\: number =\: 6}}$

➲ Second Number :

$\longrightarrow \sf Second\: number =\: 6y$

$\longrightarrow \sf Second\: number =\: 6 \times 6$

$\longrightarrow \sf\bold{\red{Second\: number =\: 36}}$

${\small{\bold{\underline{\therefore\: The\: numbers\: are\: 6\: and\: 36\: respectively\: .}}}}$