If a ten's place is halved and unit's place is doubles of a two - digit…

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If a ten's place is halved and unit's place is doubles of a two - digit number, the numbers obtained is 37 less than the original number. If digit in unit place is 5 less than tens place, then the number is :

  1. A.

    72

  2. B.

    61

  3. C.

    83

  4. D.

    94

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Correct answer: C

Answer: 83

Let the tens digit be x and the units digit be y. The original two-digit number is 10x + y.

After transformation: the tens digit is halved, so its contribution becomes 10*(x/2) = 5x. The units digit is doubled, giving 2y. Thus the new number is 5x + 2y.

Given that the new number is 37 less than the original, we have (10x + y) - (5x + 2y) = 37, which simplifies to 5x - y = 37.

Also given: the units digit is 5 less than the tens digit, so y = x - 5.

  1. Substitute y = x - 5 into 5x - y = 37:

  2. 5x - (x - 5) = 37 ⇒ 4x + 5 = 37 ⇒ 4x = 32 ⇒ x = 8.

  3. Then y = x - 5 = 8 - 5 = 3. The required number is 10*8 + 3 = 83.

Verification: 5x - y = 5*8 - 3 = 40 - 3 = 37, and the units digit is 5 less than the tens digit. Hence 83 satisfies both conditions.

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