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Mr. Smith calculated the average of 10 " three digit numbers". But due

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Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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Mr. Smith calculated the average of 10 " three digit numbers". But due to a mistake he reversed the digits of a number and thus his average increased by 29.7. The difference between the unit digit and hundreds digit of that number is :

a) 4
b) 3
c) 2
d) 1
e) 0
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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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Ashishmathew01081987 wrote:
Mr. Smith calculated the average of 10 " three digit numbers". But due to a mistake he reversed the digits of a number and thus his average increased by 29.7. The difference between the unit digit and hundreds digit of that number is :

a) 4
b) 3
c) 2
d) 1
e) 0


Since the average increased by 29.7 and there were a total of 10 numbers, it means the incorrect number was 297 greater than the correct number.

Say, the correct number was abc (where a, b and c are the digits of the 3 digit number)
Then the incorrect number was cba.

100c + 10b + a - (100a + 10b + c) = 297
99c - 99a = 99(c - a) = 297
297 = 99*3 = 99(c - a)
So c - a = 3

Answer (B)
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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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Ashishmathew01081987 wrote:
Mr. Smith calculated the average of 10 " three digit numbers". But due to a mistake he reversed the digits of a number and thus his average increased by 29.7. The difference between the unit digit and hundreds digit of that number is :

a) 4
b) 3
c) 2
d) 1
e) 0


Let the total of first 9 numbers = x

Let the 10th number = abc = 100a + 10b + c (3 digit number expansion)

Let the total of 10 numbers = t

Average \(= \frac{x + abc}{10} = t\)

x + 100a + 10b + c = 10t ................. (1)

10th number reversed = cba = 100c + 10b + a

New Average\(= \frac{x + cba}{10} = t + 29.7\)

x + 100c + 10b + a = 10t + 297 ............... (2)

(2) - (1)

99(c-a) = 297

c-a = 3

Answer = B
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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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New post 10 Sep 2014, 06:49
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average difference is 29.7 for 10 integer. with all 9 number remaining the same, difference between the 10th number will be 297.

a b c
c b a
------
2 9 7
------

in the subtraction, b - b should be 0. but it is 9. hence it has borrowed 1 from a. after borrowing 1, (a-c) = 2.
So adding 1, difference between unit and hundred digits will be, (a-c) = 3

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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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New post 19 Sep 2014, 00:45
WOW!! great approach paresh ..!! :)
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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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New post 09 May 2015, 04:52
One question that came up while solving this problem was which digits were reversed i.e.whether it is between 1000s & 100s, 100s & unit or 1000s & unit. It couldn't be between 100s & unit because then we will get only a 2-digit or 1-digit difference but here it is coming as 297. It couldn't be between 1000s & 100s because then the last digit of difference quoted would have been 0 (eg: 812 - 182 = 630 ). So it must be 1000s & unit. Any other easy ways to identify which positions got swapped OR any other thought process to get rid of this step ?

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Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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New post 10 May 2015, 20:44
anilbhatt1 wrote:
One question that came up while solving this problem was which digits were reversed i.e.whether it is between 1000s & 100s, 100s & unit or 1000s & unit. It couldn't be between 100s & unit because then we will get only a 2-digit or 1-digit difference but here it is coming as 297. It couldn't be between 1000s & 100s because then the last digit of difference quoted would have been 0 (eg: 812 - 182 = 630 ). So it must be 1000s & unit. Any other easy ways to identify which positions got swapped OR any other thought process to get rid of this step ?


When you reverse the digits of a 3 digit number, abc, you get cba.
If you reverse the digits of a 4 digit number, abcd, you get dcba.
and so on...
Until and unless you are given some specific information such as "the last two digits of a 4 digit number were reversed" - In that case, abcd becomes abdc.
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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due [#permalink]

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New post 09 Aug 2017, 00:33
Hello from the GMAT Club BumpBot!

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Re: Mr. Smith calculated the average of 10 " three digit numbers". But due   [#permalink] 09 Aug 2017, 00:33
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Mr. Smith calculated the average of 10 " three digit numbers". But due

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