The sum of the digits of a two digit number is 10. When the digits are reversed, the number decreases by 54. What is the new (reversed) number?

Introduction / Context:
This is a classic two digit number puzzle that tests algebraic modelling of digits. You are given the sum of the digits and the change in the number when its digits are reversed. The goal is to form simple equations involving the tens and units digits, solve for those digits and then identify the reversed number. Problems like this appear frequently in quantitative aptitude sections.

Given Data / Assumptions:

• Let the original two digit number be 10a + b, where a is the tens digit and b is the units digit.
• The sum of the digits is 10, so a + b = 10.
• When the digits are reversed, the number becomes 10b + a.
• The new number is 54 less than the original number: 10b + a = (10a + b) − 54.
• Digits a and b are integers from 0 to 9, with a not equal to 0 because the number has two digits.

Concept / Approach:
Translate the verbal statements into equations in a and b. The first equation comes from the sum of digits, and the second from the relation between original and reversed numbers. With two equations in two unknowns, we can solve for a and b. Once we know the digits, we can construct both the original and the reversed numbers. Careful handling of the minus 54 condition is crucial to avoid sign errors.

Step-by-Step Solution:
Let the original number be N = 10a + b. Given: a + b = 10. (1) Reversed number is N' = 10b + a. Given that when digits are reversed, the number decreases by 54: N − N' = 54. So (10a + b) − (10b + a) = 54. Simplify: 10a + b − 10b − a = 54 ⇒ 9a − 9b = 54. Divide both sides by 9: a − b = 6. (2) Now solve equations (1) and (2): a + b = 10 and a − b = 6. Add them: 2a = 16 ⇒ a = 8. Substitute a = 8 into a + b = 10: 8 + b = 10 ⇒ b = 2. Original number N = 10a + b = 10 * 8 + 2 = 82. Reversed number N' = 10b + a = 10 * 2 + 8 = 28.

Verification / Alternative check:
Check the conditions with the numbers found. The digits of 82 are 8 and 2, and their sum is 10, which matches the given condition. When digits are reversed, we get 28. The decrease is 82 − 28 = 54, which matches the second condition. Thus both conditions are satisfied exactly, confirming that our solution is correct and unique for two digit numbers with digit sum 10.

Why Other Options Are Wrong:

• 19: The reversed number 19 would correspond to an original number 91, whose digits sum to 10, but 91 − 19 = 72, not 54.
• 37: The reverse of 37 is 73 and 73 − 37 = 36, not 54.
• 46: The reverse 64 gives 64 − 46 = 18, and the digits of 46 sum to 10, but the decrease is not 54.
• 55: The digits sum to 10, but reversing gives the same number and the difference is 0.

Common Pitfalls:
Common mistakes include reversing the decrease relationship and writing 10b + a − (10a + b) = 54, which flips the sign, or using a + b = 54 instead of difference of the numbers. Another pitfall is forgetting that a must be non zero in a two digit number. Always define digits clearly, set up equations carefully and verify that the final numbers satisfy both the digit sum and the difference conditions.

Final Answer:
The new number obtained after reversing the digits is 28.