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mgodwin16
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Reciprocal of averages

by mgodwin16 Tue Dec 18, 2012 10:40 am

The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x - y = ?

The correct answer is: [(a - b)2]/[2(a + b)]

Although I thought the answer was: 0

Picking numbers: a =2 and b=4 x will equal 3
"The reciprocal of number y is going to equal 1/3 (average of reciprocal a and b). So if 1/3 is the reciprocal of y, that would mean y=3...and x - y = 0. Obviously this is not correct, hoping somebody could shed some light on this. Thanks
vijaykumar.kondepudi
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Re: Reciprocal of averages

by vijaykumar.kondepudi Tue Dec 18, 2012 11:18 am

Hi,
This is how I did

Equation 1 :

X = (a+b)/2

Equation 2:

1/Y = (1/a + 1/b)/2 = (a+b)/2ab

Therefore, Y = 2ab/(a+b)

X -Y= (a+b)/2 - ( 2ab/(a+b))
= (a-b)^2/2(a+b)

Hope this helps.
RonPurewal
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Re: Reciprocal of averages

by RonPurewal Sat Dec 22, 2012 6:52 pm

hi mgodwin216,
per forum rules, please post the question with all of its answer choices.
(this is the rule on this forum anyway, but it's especially important on this problem because your method of "plugging in" doesn't work without the answer choices.)

your error is highlighted in red below:

mgodwin16 Wrote:The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x - y = ?

The correct answer is: [(a - b)2]/[2(a + b)]

Although I thought the answer was: 0

Picking numbers: a =2 and b=4 x will equal 3
"The reciprocal of number y is going to equal 1/3 (average of reciprocal a and b). So if 1/3 is the reciprocal of y, that would mean y=3...and x - y = 0. Obviously this is not correct, hoping somebody could shed some light on this. Thanks


the reciprocal of y is the average of 1/2 and 1/4, which is not 1/3.

the average of 1/2 and 1/4 is 3/8. (you can find this by the traditional method of finding averages, or, more efficiently, by thinking about the fact that 3/8 is halfway between 1/2 and 1/4.)
so, with your numbers, y should be 8/3.

--

also, just as importantly, you should strongly suspect that seemingly "obvious" answers (like 0 in this problem) are incorrect, and are there as traps.
Sage Pearce-Higgins
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Re: Reciprocal of averages

by Sage Pearce-Higgins Fri Feb 01, 2019 1:22 pm

Perhaps it's easiest to think about it like this: when you flip one side of the equation, you flip the whole side. So, sure, if x = y, then 1/x = 1/y. However, if a = b + c, then 1/a doesn't necessarily equal 1/b + 1/c, but 1/a = 1/(b + c). So, in your case, if (1/2a) + (1/2b) = (1/y), then y = 1/(1/2a + 1/2b) and that's a pretty nasty multi-story fraction to deal with. That's why, if we want to flip equations, then we'll typically "unify" each side into a single fraction before flipping.

To make this clearer, think of some example with numbers: it's clear that 1/2 = 1/4 + 1/4, now I'm sure you can see that it's not the case that 2 = 4 + 4. However, if we "unify" the right side into a single fraction - that's 2/4 - before we flip it then we have 2 = 4/2, which is true.