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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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17 Nov 2014, 11:31
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85% (01:20) correct 15% (02:37) wrong based on 170 sessions

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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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17 Nov 2014, 20:25
11

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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17 Nov 2014, 23:10
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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.

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.

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

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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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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.

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Posts: 50615
Re: Two positive numbers differ by 12 and their reciprocals differ by 4/5  [#permalink]

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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:
$$\frac{12}{xy} = \frac{4}{5}$$
$$60 = 4xy$$
$$15 = xy$$

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

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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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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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20 Oct 2018, 01:48
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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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