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Two positive numbers differ by 12 and their reciprocals differ by 4/5

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

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New post 17 Nov 2014, 11:31
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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.

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

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New post 17 Nov 2014, 20:25
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Answer = C = 15

Let the integers be a & b

a>b

so, \(\frac{1}{a} < \frac{1}{b}\)

a - b = 12

\(\frac{1}{b} - \frac{1}{a} = \frac{4}{5}\)

\(\frac{a-b}{ab} = \frac{4}{5}\)

\(ab = \frac{5*12}{4} = 15\)
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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5  [#permalink]

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New post 17 Nov 2014, 23:10
2
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.



If 2 numbers are x and y --> y-x = 12, 1/x-1/y = 4/5
So (y-x)/xy= 4/5.
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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New post 17 Nov 2014, 23:24
2
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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New post 18 Nov 2014, 02:09
1
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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New post 18 Nov 2014, 06:53
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 08 Mar 2016, 07:24
1
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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New post 26 Jul 2017, 05:13
1
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]

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Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5 &nbs [#permalink] 20 Oct 2018, 01:48
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