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# A boy finds an average of 10 positive integers. Each

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Senior Manager
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A boy finds an average of 10 positive integers. Each [#permalink]  21 Dec 2003, 11:27
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A boy finds an average of 10 positive integers. Each integeres contain two digits .By mistake the boy interchanges the digits of one number say ba for ab. Due to this the average become 1.8 less than the previous one.What was the difference of the two digit a and b?
a)4
b)2
c)6
d)8

Any easy & understandable solution?
_________________

shubhangi

Director
Joined: 13 Nov 2003
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2 ??

here's how I did it..

suppose that the sum of 9 integers is A9
S is the average.

{A9+(10X+Y)}/10 = S

after intechanging digits;
{A9+(10Y+X)}/10 = S-1.8

subtract 2nd from 1st, u will get X-Y = 2
Intern
Joined: 04 Oct 2003
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dj - exellent approach. Thanks for the solution.
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Re: PS [#permalink]  22 Dec 2003, 11:44
shubhangi wrote:
A boy finds an average of 10 positive integers. Each integeres contain two digits .By mistake the boy interchanges the digits of one number say ba for ab. Due to this the average become 1.8 less than the previous one.What was the difference of the two digit a and b?
a)4
b)2
c)6
d)8

Any easy & understandable solution?

let the number be represented by 10a + b
a = tens digits
b = unit digit
we know that.
Orginal Average - Mistaken Average = 1.8
[(Sum of 9 nos.) + (10 a+b) / 10] - [Sum of 9 nos +(10b +a) / 10]= 1.8

that is just...

[(10a + b ) - (10b +a)] /10 =1.8 ( 10 is the common denominator..so
we can simplify)
10a + b -10b -a = 18
9a - 9b =18

thus,
a- b =2

recall a = tenths digits, b =units digits

Senior Manager
Joined: 30 Aug 2003
Posts: 325
Location: dallas , tx
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Yuppy!!
2 is right.
thanks both of you .. praet and dj.
_________________

shubhangi

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