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  1. In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is
    1. 47
    2. 59
    3. 23
    4. 35
Correct Option: A

Since two digit number = 10p + q
According to question , → q = 2p – 1.. (i)
When digits are interchanged then new number = 10q + p
then original number – [new number – original number] = 20
→ 10p + q – [10q + p – (10p + q)] = 20
→ 10p + q – 10q – p + 10p + q = 20
19p – 8q = 20
19p – 8 (2p – 1) = 20 (Using eq. (i))
19p – 16p + 8 = 20
3p = 12 → p = 4
From (i) q = 2 × 4 – 1 ⇒ q = 7
∴ original number = 10p + q = 10 × 4 + 7 = 47



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