Question

The digit in the tens place of a two digit number is twice the digit in the ones place. If the digits are interchanged then the difference between the two numbers is 18.

Answer 1

Answer:

42

Step-by-step explanation:

  Let the ones digit be a, so the digit at tens place should be 2a, as it's twice of ones digit.

 So number is 2a a, which can also be written as 10( 2a ) + a = 20a + a = 21a

When digits are interchanged, new number is 10a + 2a = 12a

Here,

 Difference in both the numbers is 18

⇒ 21a - 12a = 18

⇒ 9a = 18

⇒ a = 18/9 = 2

        Hence the required number is 21a = 21( 2 ) = 42

Answer 2

⠀⠀⠀ d✮$\purple{\mathbb{ANSWER}}$✮

Let one of the digits of a 2 digit number be $\red x$ and the other digit be $\red y$.

Let $x$ be in the tens place and $y$ be in ones place.

Let the Original number = $\red{10x+y}$

Interchanged number = $\red{10y+x}$

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It is given that:-

• $\red{x = 2y}$
• $\red{10x + y - (10y + x) = 18}$

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$⇢10x + y - 10y - x = 18 \\⇢ 9x - 9y = 18 \\⇢ 9(x - y) = 18 \\ ⇢x - y = \frac{18}{9} = 2$

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Substituting $x = 2y$ in above $x - y = 2$, we get:-

$⇢2y - y = 2 \\⇢ y = 2$

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$⇢x = 2y \\⇢ x = 2 \times 2 \\⇢ x = 4$

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∴ $\red{x=4}$ & $\red{y=2}$

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Original number = $\red{40+2}$= $\red{42}$