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Babette was asked

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by sanju09 » Sun Dec 13, 2009 11:22 pm
Babette was asked to calculate the arithmetic mean of ten positive integers each of which had two digits. By mistake, she interchanged the two digits, say p and q, in one of these ten integers. As a result, her answer for the arithmetic mean was 1.8 more than what it should have been. Then q - p equals

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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Source: — Problem Solving |

by Stuart@KaplanGMAT » Mon Dec 14, 2009 12:21 am
sanju09 wrote:Babette was asked to calculate the arithmetic mean of ten positive integers each of which had two digits. By mistake, she interchanged the two digits, say p and q, in one of these ten integers. As a result, her answer for the arithmetic mean was 1.8 more than what it should have been. Then q - p equals

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Fun question!

It's always good to begin by jotting down any common formulae that apply; in this question he have the average formula:

Average = sum of terms / # of terms

In this question, we can apply the formula to the difference between the average we got and the average we were supposed to get:

1.8 = extra sum of terms/10

18 = extra sum of terms

So, the difference between pq and and qp is 18.

At this point, I'd just experiment to find two numbers that fit; based on the choices, we know that the numbers can be at most 5 digits apart.

35 and 53 are 18 apart.. there we go! 5-3=2, so choose B.

There are actually a lot of different numbers we could have chosen (e.g. 24/42, 79/97...), but of course they always have the same difference.

If we wanted to do the last step with logic, that works too; since the units digit of the difference is 8, the units digit of the "reverse difference" will be 10-8 = 2.

In other words, to get a difference that ends in 8, our units digits have to be x and x-8 (going in cycles of units digits). Our units digits could have been:

9/1
8/0
7/9
6/8
5/7
4/6
3/5
2/4
1/3

When we reverse the order, we always get a difference that ends in 2.
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by viju9162 » Mon Dec 14, 2009 3:29 am
is the OA B ?

Assume Average of 10 +ve integers be "x". and the number with p and q be " 10p+q" ( because it is a +ve 2 digit number).

and S9 be sum of other numbers...

Therefore,

x= S9 + (10p+q) / 10
=> 10x = s9 + 10p + q -- eqn (1)

Now, it states that there is a increase of 1.8 in the average..

Therefore, the equation becomes..

1.8 + x = S9 + ( 10q+p) / 10

by solving, 18 + 10x = S9 + 10q + p ---- eqn (2)

Subtract eqn ( 1) and eqn (2) ..

-18 = 9p - 9q
-18 = 9 ( p-q)
-2 = p-q

therefore..q-p =2.

Regards,
Viju
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by getso » Tue Dec 15, 2009 5:32 am
OA is B
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by rohit.reddy » Tue Dec 15, 2009 9:04 am
1.8+{(const+10p+q)/10}=(const+10q+p)/10


=> q-p=2
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