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The product of two positive numbers is 616. If the ratio of the differ

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Bunuel
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The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 04:22
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The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

A. 95
B. 85
C. 58
D. 50
E. 40


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D
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 05:15
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Bunuel wrote:
The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

A. 95
B. 85
C. 58
D. 50
E. 40


[textarea]


a*b = 616
616 = 1*616
2*308
4*154
7*88
8*77
11*56
14*44
22*28

\(\frac{a^{3} - b^{3} }{ (a-b)^{3}} = \frac{157}{3}\)

It seems possible only with 22*28 because only these are so close that their cubes will be close to cube of their difference
Also, the difference of these numbers is a multiple of 3 which is necessary to meet the condition of denominator

11304/216 = 157/3

i.e. a and b are 22 and 28

i.e. Sum = 22+28 = 50

Answer: Option D
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 05:27
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(a^3-b^3)/(a-b)^3=157/3

(a-b)(a^2+ab+b^2)/(a-b)^3=157/3

(a^2+ab+b^2)/(a-b)^2=157/3

3(a^2+b^2)+3ab=157(a^2+b^2) -314ab

317ab =154(a^2+b^2)

Now ab =616
Therefore

a^2 +b^2 =1268

Therefore
a=28 b=22

Hence a+b=50
Option D is the answer

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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 05:51
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In my earlier answer we can also say
(a+b)^2=a^2+b^2+ 2ab
(a+b)^2= 1268+ 2×616
=1268 + 1232
=2500
a+b =50
If you have a problem in understanding how value of a and b has come otherwise you can make quadratic equation also

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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 06:19
gurmukh wrote:
In my earlier answer we can also say
(a+b)^2=a^2+b^2+ 2ab
(a+b)^2= 1268+ 2×616
=1268 + 1232
=2500
a+b =50
If you have a problem in understanding how value of a and b has come otherwise you can make quadratic equation also

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How did you get the value of (a^2+b^2)??? am i missing something
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 06:39
That I have taken out in my earlier answer, in this reply I just wanted to explain how after getting the value of a^2+b^2 you can take out the value of a +b
Which I have straight written down in my earlier answer

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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 06:43
jackfr2 wrote:
gurmukh wrote:
In my earlier answer we can also say
(a+b)^2=a^2+b^2+ 2ab
(a+b)^2= 1268+ 2×616
=1268 + 1232
=2500
a+b =50
If you have a problem in understanding how value of a and b has come otherwise you can make quadratic equation also

Posted from my mobile device





How did you get the value of (a^2+b^2)??? am i missing something

I have given answer in my earlier post before this. Please look into it
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The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   06 Apr 2020, 07:54
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Let the number be a and b.
So, ab = 616
Given, (a^3 –b^3)/(a-b)^3 = 157/3
or,(a-b)*(a^2 + b^2 +ab)/(a-b)*(a^2 + b^2 -2ab) = 157/3
or, (a^2 + b^2+ab)/(a^2 + b^2-2ab) = 157/3
or, (a^2 + b^2-2ab+3ab)/(a^2 + b^2-2ab) = 157/3
or,(3ab)/(a^2 + b^2-2ab) = 154/3
or,(a^2 + b^2-2ab)=3*3*616/154 = 36
or,(a-b)^2 =36
or,(a+b)^2-4ab =36
or,(a+b)^2=36+4*(616)
or, (a+b)=50

correct answer D
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   07 Apr 2020, 03:14
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Alternatively

\(xy=616\) -----(1)
\(\frac{(x^3-y^3)}{(x-y)^3}=\frac{157}{3}\)
\(\frac{(x-y)(x^2+xy+y^2)}{(x-y)^3}=\frac{157}{3}\)
\(3(x^2+xy+y^2)=157(x-y)^2\)
\(3x^2+3xy+3y^2=157x^2-314xy+154y^2\)
\(154x^2-317xy+154y^2=0\)
\(x^2+y^2=\frac{317}{154}xy\)
\((x+y)^2=2xy+\frac{317}{154}xy\)
\((x+y)^2=\frac{625}{154}xy\)
But \(xy=616\)
Hence \((x+y)^2=25^2*\frac{616}{154}\)
\((x+y)^2=25^2*2^2\)
So \(x+y=2*25=50\)

The answer is D.
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The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   07 Apr 2020, 03:45
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Bunuel wrote:
The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

A. 95
B. 85
C. 58
D. 50
E. 40


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Solution:


Let, a and b be the positive numbers such that a > b
    o \(a*b = 616\)
    o \(\frac{(a^3-b^3)}{(a-b)^3} =\frac{157}{3}\)
    o \(3(a^3-b^3 )=157(a-b)^3\)
    o \(3(a-b)(a^2+ab+b^2 )=157(a-b)^3 \ \ \ \ [ Identity: \ a^3+b^3 = (a-b)(a^2+ab+b^2)]\)
    o \(3(a^2+ab+b^2 )=157(a-b)^2\)
    o \(3a^2+3ab+3b^2=157a^2+157b^2-157*2*ab \ \ \ \ [Identity: \ (a+b)^2 = a^2+b^2+2ab]\)
    o \(317ab=154a^2+154b^2\)
    o \(a^2+b^2=\frac{314*616}{154}=1268 \ \ \ \ [ab = 616]\)
    o \(a^2+b^2+2ab=1268+2*616\)
    o \((a+b)^2=2500\)
Taking square root on both the sides, we get
    o \((a+b) = 50\)
Hence, the correct answer Option D.
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink] New post   10 Feb 2022, 10:14
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Re: The product of two positive numbers is 616. If the ratio of the differ [#permalink]
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