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Gmat750aspirant
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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 21 Mar 2025, 09:37
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C
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65% (hard)

Question Stats:

63% (02:59) correct 37% (02:55) wrong based on 46 sessions

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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64, what is the value of mn?

A) 20
B) 24
C) 30
D) 32
E) 36

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C
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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 21 Mar 2025, 12:04
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Gmat750aspirant
If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64, what is the value of mn?

A) 20
B) 24
C) 30
D) 32
E) 36
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(1/m) + (1/n) = 8/15 => \(\frac{m+n }{ mn} = \frac{8}{15}\), from here we can infer that mn is a positive multiple of 15. Only option C. fits that criteria.

Answer C.
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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 11 Apr 2025, 02:36
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I'm a bit confused about this question, it's not mentioned that m and n are integers then how did you arrive at the conclusion that mn is a integral multiple of 15.
Krunaal
Gmat750aspirant
If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64, what is the value of mn?

A) 20
B) 24
C) 30
D) 32
E) 36
(1/m) + (1/n) = 8/15 => \(\frac{m+n }{ mn} = \frac{8}{15}\), from here we can infer that mn is a positive multiple of 15. Only option C. fits that criteria.

Answer C.
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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 11 Apr 2025, 02:59
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We are given the equations:
\(\frac{1}{m} + \frac{1}{n} = \frac{8}{15}\)
\(m^2 - n^2 = 64\)

Step 1: Express in Terms of a Common Denominator
Rewriting the first equation:
\(\frac{m + n}{mn} = \frac{8}{15}\)

Multiplying both sides by mn:
\(m + n = \frac{8}{15} mn\)

Step 2: Use Difference of Squares
Using the given second equation:
\((m−n)(m+n)=64\)

Substituting \(m + n = \frac{8}{15}\) :
\((m - n) \times \frac{8}{15} mn = 64\)

Multiplying both sides by 15/8:
\((m - n) \times mn = \frac{64 \times 15}{8}\)

\((m - n) \times mn = 120\)

Step 3: Solve for mn
Let \(x=mn\). Then:
\((m−n)x=120\)

From our earlier equation:
\(m + n = \frac{8}{15} x\)

Using the identity:
\((m+n)^2 - (m-n)^2 = 4mn\)

\(\left(\frac{8}{15}x\right)^2 - \left(\frac{120}{x}\right)^2 = 4x\)

Is this solution correct, I'm kinda stuck after this. I try substituting values of mn as 30 but it doesn't work
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If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 11 Apr 2025, 03:02
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napolean92728
I'm a bit confused about this question, it's not mentioned that m and n are integers then how did you arrive at the conclusion that mn is a integral multiple of 15.
Krunaal
Gmat750aspirant
If (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64, what is the value of mn?

A) 20
B) 24
C) 30
D) 32
E) 36
(1/m) + (1/n) = 8/15 => \(\frac{m+n }{ mn} = \frac{8}{15}\), from here we can infer that mn is a positive multiple of 15. Only option C. fits that criteria.

Answer C.
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This is not a good question.

(1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 has no solution over integers.

The solution of (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 is irrational numbers.

And finally, (1/m) + (1/n) = 8/15 and (m^2 - n^2) = 64 and mn =30 have no solution at all.

This Question is Locked Due to Poor Quality
Hi there,
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