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

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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?
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shubhangi

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21 Dec 2003, 12:27
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
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21 Dec 2003, 14:02
dj - exellent approach. Thanks for the solution.
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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

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22 Dec 2003, 14:42
Yuppy!!
2 is right.
thanks both of you .. praet and dj.
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shubhangi

22 Dec 2003, 14:42
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