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Number System

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by harsh.champ » Fri Feb 19, 2010 6:10 am
A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the student interchanges the digits of one number say mn for nm. Due to this, the new calculated average becomes 9/5 less than the previous one. What is the difference between the two digits m and n?

A. 2
B. 4
C. 6
D. 8
E. 10

The OA is A.
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Source: — Problem Solving |

by anish13 » Fri Feb 19, 2010 6:32 am
Let the tens digit be 10t and units digit be u.

so the average is changing because of the reversal of the 2 numbers so we can effectively say,

[(10t+u)-(10u+t)]/10 = 9/5 (substracting the 2 averages will cancel out the rest of the integers since they are the same)

solving further we get,

t-u=2,

Hence A.
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by shashank.ism » Fri Feb 19, 2010 6:38 am
harsh.champ wrote:A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the student interchanges the digits of one number say mn for nm. Due to this, the new calculated average becomes 9/5 less than the previous one. What is the difference between the two digits m and n?

A. 2
B. 4
C. 6
D. 8
E. 10

The OA is A.
let the sum of other numbers = X
so (X + 10m +n)/10 is the initial avg
(X + 10n +m)/10 is the mistaken avg
(X + 10m +n)/10-(X + 10n +m)/10 = 9/5
--> (X + 10m +n)-(X + 10n +m)= 18
--> 9m -9n =18 -->[spoiler] m-n =2 Ans A[/spoiler]
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