In a two – digit number, the digit in the unit's place exceeds twice the digit in the ten's place by 1. If the digit in the unit’s place and the ten’s place are interchanged, then the difference between the newly formed number and the original number is less than the original number by 1. What is the original number?

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Solution

The correct option is B
37

Let the ten's digit be x. Then, unit's digit = 2x+1.

$[10 x + (2 x + 1)] - [{10 (2 x + 1) + x} - {10 x + (2 x + 1)}] = 1$

$\Rightarrow (12 x + 1) - (9 x + 9) = 1$

$\Rightarrow x = 3$

So, ten's digit = 3 and unit's digit = 7.

Hence, original number = 37.

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The correct option is B 37 Let the ten's digit be x. Then, unit's digit = 2x+1. [10x+(2x+1)]−[{10(2x+1)+x}−{10x+(2x+1)}]=1 ⇒(12x+1)−(9x+9)=1 ⇒x=3 So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

The correct option is B
37

Let the ten's digit be x. Then, unit's digit = 2x+1.

$[10 x + (2 x + 1)] - [{10 (2 x + 1) + x} - {10 x + (2 x + 1)}] = 1$

$\Rightarrow (12 x + 1) - (9 x + 9) = 1$

$\Rightarrow x = 3$

So, ten's digit = 3 and unit's digit = 7.

Hence, original number = 37.