The digit at the ten's place of a $2$ -digit number is double of the digit at the unit's place. If the sum of this number and the number formed by reversing the digits is $33$ . Find the number.

20

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21

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22

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Solution

The correct option is B $21$
Let the unit place digit $= x$
Ten's place digit $= 2 x$
Number formed $=$ $10 \times 2 x + x = 20 x + x = 21 x$
Reversing the digit:
Unit digit $= 2 x$
Ten's digit $= x$
New number formed $=$ $10 \times x + 2 x = 12 x$
The sum of the number is $33.$
$21 x + 12 x = 33$
$33 x = 33$
$x = 1$
So, the number is $21.$

So, option B is correct.

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The digit at the ten's place of a 2-digit number is double of the digit at the unit's place. If the sum of this number and the number formed by reversing the digits is 33. Find the number.

The correct option is B 21Let the unit place digit =xTen's place digit =2xNumber formed = 10×2x+x=20x+x=21xReversing the digit:Unit digit =2xTen's digit =xNew number formed = 10×x+2x=12xThe sum of the number is 33.21x+12x=3333x=33x=1So, the number is 21.So, option B is correct.

The digit at the ten's place of a $2$ -digit number is double of the digit at the unit's place. If the sum of this number and the number formed by reversing the digits is $33$ . Find the number.