# Solved Example #9

Monday, February 11th, 2013

Question : A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.

Let us say that the 8 numbers are a1 b1, a2 b2, a3 b3 … a8 b8 …{eg: 38}

They actually represent a1*10 + b1, a2*10 + b2 … a8*10 + b8 …{eg: 3*10 + 8}

Sum of the numbers = (a1 + a2 + a3 .. a8)*10 + (b1 + b2 + b3 .. b8)

Reversed numbers will be b1 a1, b2 a2, … b8 a8

Sum of the reversed numbers will be (b1 + b2 + b3 .. b8)*10 + (a1 + a2 + a3 .. a8)

Reverse sum – actual sum = 36 … {Given in the question}

(b1 + b2 + b3 .. b8)*9 – (a1 + a2 + a3… a8)*9 = 36

=> (b1 + b2 + b3 … b8) – (a1 + a2 + a3 … a8) = 4 {9 can be taken common and cancelled from both sides}

=> Sum of the units digits – Sum of the tens digits = 4

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