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Babita was asked

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Babita was asked [#permalink]

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New post 26 Jan 2017, 00:19
Babita was to calculate the arithmetic mean of ten positive two digit integers. By miatake, she interchanged the two digits, say t & u, in one of these ten integers. As a result, her answer for arithmetic mean was 1.8 more than what it should have been. Then u-t equals

(A) 1
(B) 2
(C) 3
(D) 4

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Re: Babita was asked [#permalink]

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New post 26 Jan 2017, 13:15
The answer is B.
Lets say the 9 integers be Y and the changed integer be X.
So X= 10t+u and X' = 10u+t is the number that was mistakingly calculated.
Now, mean = (X+Y)/10
mean' = (X'+Y)/10
Subtracting,
mean' - mean = (X' - X)/10
1.8= (X'-X)/10
18=(X'-X)
18= (10u+t)-(10t+u)
18= 9u-9t
18= 9(u-t)
2=(u-t)

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Re: Babita was asked [#permalink]

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New post 26 Jan 2017, 20:26
Harshit401 wrote:
Babita was to calculate the arithmetic mean of ten positive two digit integers. By miatake, she interchanged the two digits, say t & u, in one of these ten integers. As a result, her answer for arithmetic mean was 1.8 more than what it should have been. Then u-t equals

(A) 1
(B) 2
(C) 3
(D) 4


Hi Harshit401,

Let the interchanged integer 10t + u.
It is changed to 10u + t.

( 10u + t ) - ( 10t + u ) = 10 ( u - t ) + ( t - u ) = 9 ( u - t ) = 18
Then u - t = 2
The answer is B.

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Re: Babita was asked   [#permalink] 26 Jan 2017, 20:26
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