Question

The digit at ones place of a 2-digit number is four times the digit at tens place. The
the given number.
ined by
number obtained by reversing the digits exceeds the given number by 54. Find
​

Answer 1

Given :-

• The digit at ones place of a 2-digit number is four times the digit at tens place. The the given number obtained by reversing the digits exceeds the given number by 54.

To find :-

• Required numbers

Solution :-

Let the tens digit be x then ones digit be y

• Original number = 10x + y

The digit at ones place of a 2-digit number is four times the digit at tens place.

→ y = 4x ---(i)

The the given number obtained by reversing the digits exceeds the given number by 54.

• Reversed number = 10y + x

→ 10x + y + 54 = 10y + x

→ 10x - x + y - 10y = - 54

→ 9x - 9y = - 54

→ 9(x - y) = - 54

→ x - y = - 6 --(ii)

Put the value of y in equation (ii)

→ x - 4x = - 6

→ - 3x = - 6

→ x = 2

Substitute the value of x in equation (i)

→ y = 4x

→ y = 4 × 2

→ y = 8

Hence,

• Tens digit = x = 2
• Ones digit = y = 8

Therefore,

• Original number = 10x + y = 28
• Reversed number = 10y + x = 82

Answer 2

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The digit at ones place of a $2$-digit number is four times the digit at tens place. The number obtained by reversing the digits exceeds the given number by $54$. Find the numbers.

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• The digit at ones place of a $2$-digit number is four times the digit at tens place.

• The number obtained by reversing the digits exceeds the given number by $54$.

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The number.

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Let the tens digit be $x$.

So, ones digit $= 4x$

According to condition,

∴ ${[10(4x)+x]-(10x+4x) = 54}$

➳ $(40x+x)-(14x) = 54$

➳ $41x-14x = 54$

➳ $27x = 54$

➳ ${x = {\frac{54}{27}}}$

➳ $x = 2$

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$x = 2</p><p>$

Ones digit $= 4x = (4×2) = 8$

Tens digit $= x = 2$

So, the number $= 28$.

The original number

$= (10x+4x)$

By substituting $</em><em>x</em><em>$ with $</em><em>2</em><em>$.

$(10x+4x)$

$= [(10×2)+(4×2)]$

$= (20+8)$

$= 28$

The number with reversed digits

$= [10(4x)+x]$

By substituting $</em><em>x$ with $</em><em>2</em><em>$.

$[10(4x)+x]$

$= [10(4×2)+2]$

$= [(10×8)+2]$

$= (80+2)$

$= 82$

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The original number number is $28$.

The number with reversed digits is $82$.

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The number $= 28$

Number with reversed digits $= 82$

∴ $82-28 = 54$

➳ $54 = 54$

So, L.H.S = R.H.S.

Hence, verified.

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