A two digit number of exceeds the sum of the digits of that number by $18$ . If the digit at the unit's place is double the digit in the ten's place , what is the number ?

24

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42

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48

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Solution

The correct option is A $24$
Let the ten's place digit be $x$ and unit place digit be $y .$
Therefore, the two digit number $= 10 x + y$
According to question,
$10 x + y = x + y + 18 9 x = 18 x = 2$

Also, $y = 2 x y = 2 \times 2 y = 4$
So, the two digit number $= 10 x + y = 10 \times 2 + 4 = 24$

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A two digit number of exceeds the sum of the digits of that number by 18. If the digit at the unit's place is double the digit in the ten's place , what is the number ?

The correct option is A 24Let the ten's place digit be x and unit place digit be y.Therefore, the two digit number =10x+yAccording to question,10x+y=x+y+189x=18x=2Also, y=2xy=2×2y=4So, the two digit number =10x+y=10×2+4=24

A two digit number of exceeds the sum of the digits of that number by $18$ . If the digit at the unit's place is double the digit in the ten's place , what is the number ?

The correct option is A $24$
Let the ten's place digit be $x$ and unit place digit be $y .$
Therefore, the two digit number $= 10 x + y$
According to question,
$10 x + y = x + y + 18 9 x = 18 x = 2$

Also, $y = 2 x y = 2 \times 2 y = 4$
So, the two digit number $= 10 x + y = 10 \times 2 + 4 = 24$