In a two digit number if unit's place is halved and tens place is doubled ,…
2023
In a two digit number if unit's place is halved and tens place is doubled , difference between the number is 37 . Digit in unit's place is 2 more than tens place , then sum of the digits of a number is
- A.
8
- B.
12
- C.
10
- D.
14
Attempted by 15 students.
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Correct answer: C
Solution: step-by-step.
Let the tens digit be t and the units digit be t + 2 (units is 2 more than tens).
Original number = 10t + (t + 2). After changes: tens doubled → 2t (in tens place), units halved → (t + 2)/2. New number = 20t + (t + 2)/2.
Given that the difference between the numbers is 37, take (new) − (original) = 37:
Compute the difference: (20t + (t+2)/2) − (10t + t + 2) = 10t − (t+2)/2 = 37. Multiply both sides by 2: 20t − (t + 2) = 74.
Simplify: 19t − 2 = 74 ⇒ 19t = 76 ⇒ t = 4. Then units = t + 2 = 6.
Therefore the sum of the digits = 4 + 6 = 10.