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sujarajan9
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maths mean question 29 Oct 2014, 01:37
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Amol was asked to calculate the arithmetic mean of 10 positive integers each of which had two digts .By mistake,he intercjanged the two digits ,say a and b,with one of these ten integers.As a result ,his answer for the arithmetic mean was 1.8 more than what it should have been ,then b-a =
A)1
B)2
C)3
D)NONE OF THESE
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BrushMyQuant
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maths mean question 31 Oct 2014, 11:35
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sujarajan9
Amol was asked to calculate the arithmetic mean of 10 positive integers each of which had two digts .By mistake,he intercjanged the two digits ,say a and b,with one of these ten integers.As a result ,his answer for the arithmetic mean was 1.8 more than what it should have been ,then b-a =
A)1
B)2
C)3
D)NONE OF THESE
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What is the answer to this question? Is it B? And is the question framed properly? Or the question was framed as " he interchanged the two digits of one of the two digit number?"
Considering that he interchanged the two digits of one of the numbers, the solution is as follows:--
supposed the number which was interchanged is ba, if we interchange the digits then the number will become ab
Average before interchanging = (ba + x)/10 , where x is sum of other 9 numbers
Average after interchanging = (ab + x)/10

Given that average before interchanging = average after interchanging + 1.8
(ba + x)/10 = (ab+x)/10 + 1.8
or
ba + x = ab + x + 18
ba can be written as 10b + a and ab can be written as 10a + b
10b + a + x = 10a + b + x + 18
or
b-a = 2
So, Answer will be B
hope it helps!