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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Algebra.



Two positive numbers differ by 12 and their reciprocals differ by \(\frac{4}{5}\). What is their product?

A. \(\frac{2}{15}\)
B. \(\frac{48}{5}\)
C. \(15\)
D. \(42\)
E. \(60\)

Kudos for a correct solution.



let two positive numbers be x and x+12.
Need to find the product x(x+12)
then 1/x - 1/(x+12) = 4/5
solve to get 12/x(x+12) = 4/5 ----> x(x+12) = 15
Hence C
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Let's assume two numbers are x and y -->
As per the question
y-x = 12,
1/x-1/y = 4/5 =>(y-x)/xy= 4/5.
Put the value of y-x
xy = 12*5/4 = 15.

Answer C.
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Official Solution:

Two positive numbers differ by 12 and their reciprocals differ by \(\frac{4}{5}\). What is their product?

A. \(\frac{2}{15}\)
B. \(\frac{48}{5}\)
C. \(15\)
D. \(42\)
E. \(60\)


Don’t be afraid to assign variables even when none are given in the problem. "Two positive numbers differ by 12" can be written as:
\(x - y = 12\)

And "their reciprocals differ by \(\frac{4}{5}\)" can be written as:
\(\frac{1}{y} - \frac{1}{x} = \frac{4}{5}\)

(Note: Here, we’ve assigned \(x\) as the bigger of the two numbers and \(y\) as the smaller, so we’ve intuited that \(\frac{1}{y}\) is the larger reciprocal and \(\frac{1}{x}\) the smaller, and so arranged them in that order to write \(\frac{1}{y} - \frac{1}{x} = \frac{4}{5}\)).

Now we have a system of two variables and two equations. Note that it is NOT necessary to solve for \(x\) and \(y\), since we are being asked for the product, \(xy\).

First, let’s simplify the second equation by finding a common denominator for the terms on the left:
\(\frac{x}{xy} - \frac{y}{xy} = \frac{4}{5}\)
\(\frac{x - y}{xy} = \frac{4}{5}\)

Note that the denominator is \(xy\), which is exactly the quantity we want to find.

Since we know from the first equation that \(x - y = 12\), substitute 12:
\(\frac{12}{xy} = \frac{4}{5}\)
\(60 = 4xy\)
\(15 = xy\)

Answer: C.
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Bunuel wrote:

Tough and Tricky questions: Algebra.



Two positive numbers differ by 12 and their reciprocals differ by \(\frac{4}{5}\). What is their product?

A. \(\frac{2}{15}\)
B. \(\frac{48}{5}\)
C. \(15\)
D. \(42\)
E. \(60\)

Kudos for a correct solution.


Another great question...
the only thing to lookout here is that if x>y => 1/x<1/y (for both x>0 and y>0)
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Considering x>y and they are the two numbers.

Solve 1/y-1/x=4/5 which will give x-y/xy=4/5 ->eqtn 1.
Now plug in x-y=12 from problem statement.
Solve for xy in eqtn 1.

Option C.
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
let the two number be x and y , from question we get x-y=12
now 1/x -1/y =4/5
solving this by putting value of x-y=12 we get xy=15
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
Very good question indeed, looks simple not only to answer, but also get the question wrong. I overlooked the simple fact mentioned below and chose the wrong answer choice. :(

stonecold wrote:
Another great question...
the only thing to lookout here is that if x>y => 1/x<1/y (for both x>0 and y>0)
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 [#permalink]
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