One half of a two digit number exceeds its one third by 6. What is the sum of the digits of the number?a)12b)9c)15d)6Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Problem: One half of a two digit number exceeds its one third by 6. What is the sum of the digits of the number?

Solution:
Let the two-digit number be represented by xy, where x is the tens digit and y is the units digit.
According to the problem, we have the following equation:
$\frac{1}{2}(10x+y) - \frac{1}{3}(10x+y) = 6$
Simplifying the equation, we get:
$\frac{1}{6}(10x+y) = 6$
Multiplying both sides by 6, we get:
$10x+y = 36$
The sum of the digits of the number is:
$x+y = \frac{10x+y}{10} + \frac{10x+y}{1} - 11 imes y$
$x+y = \frac{11}{10}(10x+y) - 11y$
Substituting the value of (10x+y), we get:
$x+y = \frac{11}{10}(36) - 11y$
$x+y = 3$
Therefore, the sum of the digits of the number is 3.
Hence, the correct option is (b) 9.

Answer

Let the given number be : x

It is given that x/2 - x/3 = 6

Or x/6 =6 or x =36

Therefore, sum of the digits 3+6 = 9

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