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# Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what

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Math Expert
Joined: 02 Sep 2009
Posts: 51055
Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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10 Apr 2018, 21:18
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25% (medium)

Question Stats:

77% (01:30) correct 23% (01:48) wrong based on 92 sessions

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Given that $$m^2 - 2m -15 = 0$$ and $$n^2 - 3n - 10 = 0$$, where m ≠ n, what is the product of m and n?

(A) -15
(B) -10
(C) -6
(D) 6
(E) 15

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Re: Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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10 Apr 2018, 22:41
Bunuel wrote:
Given that $$m^2 - 2m -15 = 0$$ and $$n^2 - 3n - 10 = 0$$, where m ≠ n, what is the product of m and n?

(A) -15
(B) -10
(C) -6
(D) 6
(E) 15

Factorizing the two equations
$$m^2 - 2m -15 = 0$$ -> $$m^2 - 5m + 3n -15 = 0$$ -> $$(m-5)(m+3) = 0$$ -> $$m = 5 or -3$$
$$n^2 - 3n - 10 = 0$$ -> $$n^2 - 5n + 2n - 10 = 0$$ -> $$(n-5)(n+2) = 0$$ -> $$n = 5 or -2$$

Therefore, the product of m and n such that m≠n is (-3)(-2) = 6(Option D)
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Joined: 07 Jul 2017
Posts: 1
Re: Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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11 Apr 2018, 01:13
Can answer be -10 and -15 also? the question stated that m ≠ n , doesn’t that means m = 5, n = -2 or m = -3, n = 5 also possible?
Director
Joined: 02 Oct 2017
Posts: 724
Re: Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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21 Apr 2018, 05:17
After solving equation we get
m=5 or -3
n=5 or -2

It is written that m is not equal to n
Then options 6,-10 and -15 also fulfill the criteria..
But otherwise options in this question are not correct.

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Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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21 Apr 2018, 16:26
Bunuel wrote:
Given that $$m^2 - 2m -15 = 0$$ and $$n^2 - 3n - 10 = 0$$, where m ≠ n, what is the product of m and n?

(A) -15
(B) -10
(C) -6
(D) 6
(E) 15

It's given that m is not equal to n.

$$m^2-2m-15=0$$

$$m^2 - 5m + 3m - 15 =0$$

m(m-5) + 3(m-5) = 0

(m-5)(m+3)=0

So, the value of m = 5 or -3.

$$n^2 - 3n - 10 = 0$$

$$n^2 - 5n + 2n -10=0$$

(n-5) (n+2) = 0.

So the value of n = 5 or -2

Note: we know m and n can't be equal as per the direction of the question.

Thus the value of mn will be (-3) (-2) = 6

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Joined: 27 May 2012
Posts: 627
Re: Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what  [#permalink]

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23 Aug 2018, 04:27
linggling wrote:
Can answer be -10 and -15 also? the question stated that m ≠ n , doesn’t that means m = 5, n = -2 or m = -3, n = 5 also possible?

Absolutely you are correct , on solving for m we get m =5 or m=-3
on solving for n we get n =5 or n =-2
now we know that m $$\neq$$ n , hence m and n cannot both be 5, the following pairs of (m n) are possible
(-3 5) (-3,-2) (5, -2) hence product of m and n can be -15, 6, -10 all of these are present in the options .

Please let me know if I am missing anything. Thanks.
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Re: Given that m^2 - 2m -15 = 0 and n^2 - 3n - 10 = 0, where m ≠ n, what &nbs [#permalink] 23 Aug 2018, 04:27
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