Bunuel wrote:
A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?
A. 8
B. 6
C. 4
D. 2
E. 1
I think my method is correct. Please correct me if I'm wrong.
\(\frac{Sum of 10 Positive Integers}{10} = X\) <------ Average #1
Sum of 10 Positive Integers = \(X(10)\) <------ Sum #1\(\frac{Sum of 10 Positive Integers}{10} = (X - 1.8)\) <------ Average #2 with reversed digits.
Sum of 10 Positive Integers = \(10(X - 1.8)\) <------ Sum #2
Sum of 10 Positive Integers = \(10X - 18\)Sum differences:
=
[Sum #2] -
[Sum #1]= (10X - 18) - 10X
= 10X - 18- 10X
= -18Now start taking the lowest double digit numbers and subtracting from their reversed counterparts.
10: 10 - 01 = 9 <----
Nope11: 11 - 11 = 0 <----
Nope12: 12 - 21 = -9 <----
Nope13: 13 - 31 = -18 <----
YASS!!! The difference between digits is (3 - 1) = 2
24: 24 - 42 = -18 <----
YASS!!!The difference between digits is (4 - 2) = 2
35: 35 - 53 = -18 <----
YASS!!!The difference between digits is (5 - 3) = 2
46: 46 - 64 = -18 <----
YASS!!!The difference between digits is (6 - 4) = 2
Answer: D
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