Two-Digit Number from Digit-Sum Conditions A two-digit number equals seven times the sum of its digits. If each digit is increased by 2, the resulting number is 4 more than six times the sum of its new digits. Find the original number.

Introduction / Context:
This is a standard two-equation, two-unknowns problem using digit variables for a two-digit number. Such problems appear frequently in aptitude tests, requiring careful translation from words to equations.

Given Data / Assumptions:

• Let tens digit be a and units digit be b; number = 10a + b.
• Condition 1: 10a + b = 7(a + b).
• Condition 2: New number after increasing each digit by 2 is 10(a + 2) + (b + 2), which equals 4 more than 6 times the sum of new digits.

Concept / Approach:
Convert each statement to an equation and solve simultaneously for integers a and b between 0 and 9, with a ≥ 1 for a valid two-digit number.

Step-by-Step Solution:
From Condition 1: 10a + b = 7a + 7b → 3a = 6b → a = 2b.From Condition 2: 10a + b + 22 = 6(a + b + 4) + 4 = 6a + 6b + 28.Rearrange: 4a − 5b = 6.Substitute a = 2b → 8b − 5b = 6 → 3b = 6 → b = 2 → a = 4.Number = 42.

Verification / Alternative check:
Check Condition 1: 42 = 7*(4 + 2) = 42. Check Condition 2: New number 64; sum of new digits = 6 + 4 = 10; 6*10 + 4 = 64. Both satisfied.

Why Other Options Are Wrong:
24, 48, and 36 fail one or both equations; “Data adequate” is not an answer value and does not identify the number.

Common Pitfalls:
Misplacing constants (22 vs 28) in the second equation; assuming a = b instead of using a = 2b from the first equation.

Final Answer:
42