Half of a two-digit number is greater than 10 times its unit digit by 1 more than the ten's digit of that number. If the sum of the digits present in the number is equal to 7, then what is the number?

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162

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179

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Solution

The correct option is B
52

Let the ten's digit be x and unit's be y. Then,

$\frac{10 x + y}{2} - 10 y = (x + 1)$

$\Rightarrow 10 x + y = 20 y + 2 x + 2$

$\Rightarrow 8 x - 19 y = 2 . . . . (i)$

$a n d x + y = 7.... (i i)$

Solving (i) and (ii), we get: x = 5 and y = 2. Hence, required number = 52.

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Half of a two-digit number is greater than 10 times its unit digit by 1 more than the ten's digit of that number. If the sum of the digits present in the number is equal to 7, then what is the number?

The correct option is B 52 Let the ten's digit be x and unit's be y. Then, 10x+y2−10y=(x+1) ⇒10x+y=20y+2x+2 ⇒8x−19y=2....(i) and x+y=7....(ii) Solving (i) and (ii), we get: x = 5 and y = 2. Hence, required number = 52.