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Given two positive numbers that the sum of two numbers is 5 times

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Given two positive numbers that the sum of two numbers is 5 times  [#permalink]

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New post Updated on: 12 Dec 2016, 08:57
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A
B
C
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E

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77% (02:46) correct 23% (02:25) wrong based on 67 sessions

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Given two positive numbers that the sum of them is 5 times the difference between them and the product of them is 24 times the difference between them. What is the value of the smaller one?

A. 5
B. 8
C. 10
D. 12
E. 15

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Originally posted by broall on 12 Dec 2016, 08:05.
Last edited by broall on 12 Dec 2016, 08:57, edited 1 time in total.
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Re: Given two positive numbers that the sum of two numbers is 5 times  [#permalink]

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New post 12 Dec 2016, 08:51
If I understood the question correctly.

We have: \(X + Y = 5*(X - Y)\)

From where \(X = \frac{2}{3} Y\)

Then: \(X*Y = 24(Y - X)\)

Putting out Y from previous equation:

\(\frac{2Y^2}{3} = 24 (Y - \frac{2Y}{3})\)

simplifying this:

\(2Y^2 - 24Y = 0\)

\(Y=0\) or \(Y=12\)

Given that Y is positive we have only one option \(Y=12\)

Hence \(X = 8\) our smaller value

Answer B
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Re: Given two positive numbers that the sum of two numbers is 5 times  [#permalink]

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New post 26 Jul 2017, 05:45
X+Y=5∗(X−Y) -> Solve this to express X in terms of Y [X=3/2Y] and plug it into the below equation
XY=24*(X-Y)
Gives 8,12 so the least no is 8.
Option B.
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Re: Given two positive numbers that the sum of two numbers is 5 times  [#permalink]

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New post 26 Jul 2017, 05:50
nguyendinhtuong wrote:
Given two positive numbers that the sum of them is 5 times the difference between them and the product of them is 24 times the difference between them. What is the value of the smaller one?

A. 5
B. 8
C. 10
D. 12
E. 15


Let the positive numbers be x and y
x+y = 5 (x-y)
4x = 6y
x = 3y/2

xy = 24 (x-y)
3y^2 /2 = 24 (3y/2 - y)
y^2/2 = 8 (y/2)
y (y-8 ) = 0
y = 8
x = 12

Answer B
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Re: Given two positive numbers that the sum of two numbers is 5 times   [#permalink] 26 Jul 2017, 05:50
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