A number consists of two digits.The digit in ten's place exceeds the digit in unit's place by $2$ .The difference of the digits is $\frac{1}{32}$ of the number.Find the number.

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Solution

Let the digit at unit's place $= x$

Then the digit at ten's place $= x + 2$

Number $= 10 (d i g i t a t t e n^{'} s p l a c e) +$ Digit at unit's place

$= 10 (x + 2) + x = 10 x + 20 + x = 11 x + 20$

Difference of the digits $= \frac{1}{32} (N u m b e r)$

$\Rightarrow 2 = \frac{1}{32} (11 x + 20)$

$\Rightarrow 64 = 11 x + 20$

$\Rightarrow 11 x = 64 - 20 = 44$

$\Rightarrow x = \frac{44}{11} = 4$

Digit at unit's place $= 4$

Digit at ten's place $= 4 + 2 = 6$

Number $= 10 (d i g i t a t t e n^{'} s p l a c e) +$ Digit at unit's place

$= 10 \times 6 + 4 = 60 + 4 = 64$

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A number consists of two digits.The digit in ten's place exceeds the digit in unit's place by 2.The difference of the digits is 132 of the number.Find the number.

A number consists of two digits.The digit in ten's place exceeds the digit in unit's place by $2$ .The difference of the digits is $\frac{1}{32}$ of the number.Find the number.