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by Deepthi Subbu » Tue Jan 04, 2011 2:32 am
If 42.42 = k(14 + m/50), where k and m are positive integers and m < 50, then what is the value of
k + m ?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

OA E

Unlike the substitution method , is there a shortcut to solve the problem ?
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by bblast » Tue Jan 04, 2011 4:35 am
substitution for me, just looking at the question u realize m/50 should be equal to 0.14.
therefore m = 7 and k=3.

(took me more than a minute to figure out though :/)

I dnt think an equation approach would be helpful here. Tried but failed.

good question.
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by anshumishra » Tue Jan 04, 2011 5:29 am
Deepthi Subbu wrote:If 42.42 = k(14 + m/50), where k and m are positive integers and m < 50, then what is the value of
k + m ?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

OA E

Unlike the substitution method , is there a shortcut to solve the problem ?
If 14k = 42 and mk/50 = 42/100 = 21/50, then it will satisfy the given equation.

14k = 42 => k = 3
mk = 21 => 3m = 21 => m =7

Hence; m+k = 10.
Thanks
Anshu

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by fskilnik@GMATH » Wed Jan 05, 2011 5:26 am
anshumishra wrote:If 14k = 42 and mk/50 = 42/100 = 21/50, then it will satisfy the given equation.
That´s right, Anshu!

Let us prove that 42+0.42 is the only possible "break" for the conditions given in the question stem!

(a) km/50 is at least 0.2 (why?) therefore
(b) 42.42 = k (14+m/50) = 14k + km/50 >= 14k +0.2 implies that k is necessarily 1,2 or 3 (explain).
(c) k cannot be 1, otherwise m would be much greater than 50 (verify that)
(c) k cannot be 2, otherwise m would be the non-integer 25*14.42 (verify that) then
(d) we are sure k = 3

Conclusion: Anshu´s solution follows without any other alternative!

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by djyogi » Thu Feb 23, 2012 5:29 am
We can also notice that if 42.42 is written as a 14k +mk/50, then mk/50 is a remainder, which equals to 42/100 (it is a remainder, because in general view m=nq (q is constant) + r, when m is divided by n and a remainder is r). then we solve for mk and can factor the mk value of 21 into primes, which yields only sum of 10.
please, tell me if that is legible approach? for me it is much easier than picking values for k.
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by LalaB » Thu Feb 23, 2012 10:54 am
djyogi wrote:We can also notice that if 42.42 is written as a 14k +mk/50, then mk/50 is a remainder, which equals to 42/100 (it is a remainder, because in general view m=nq (q is constant) + r, when m is divided by n and a remainder is r). then we solve for mk and can factor the mk value of 21 into primes, which yields only sum of 10.
please, tell me if that is legible approach? for me it is much easier than picking values for k.
I used the same logic, and think that everything is ok with our solution :)
0.42/k=m/50 km=21
since it was stated, that both k and m are integers, then k+m=7+3=10
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by fskilnik@GMATH » Sun Feb 26, 2012 2:21 pm
djyogi wrote:We can also notice that if 42.42 is written as a 14k +mk/50, then mk/50 is a remainder, which equals to 42/100 (it is a remainder, because in general view m=nq (q is constant) + r, when m is divided by n and a remainder is r)
Hi djyogi and LalaB!

Your idea is nice, but you cannot use the division-algorithm to say that mk/50 is a remainder, because (for start) all terms related to "m = nq + r" must be necessarily integers.

Please note that I did NOT "pick numbers", I simply restricted the initial possibilities for k, and then I could PROVE that there was just one possible choice for it. From that, we can be sure that your "division-algorithm analogous idea" is really legible (because the 42.42 = 42+0.42 is really the only possibility), and the rest of the reasoning follows (as pointed by both of you).

Regards,
Fabio.
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by djyogi » Mon Feb 27, 2012 4:14 am
thank you fskilnik for explanation
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by fskilnik@GMATH » Mon Feb 27, 2012 7:53 am
djyogi wrote:thank you fskilnik for explanation
My pleasure! Good studies and success in your GMAT!
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