Lucky1994 wrote:
MentorTutoringCan you please help me in solving this question?
Hello,
Lucky1994, and thank you for tagging me. I will admit to taking a brute force approach to this one, using a little number sense and leaning on the answers in a similar way to what
mau5 had done. I will simply walk you through that process, even if, again, it may not be a great method to copy.
1) If
\(\frac{1}{a}+\frac{1}{b}=\frac{2}{15}\),
then we can anticipate that our different
integer combinations will play off of multiples of
\(\frac{2}{15}\). I created a list of several to get started:
\(\frac{2}{15}\),
\(\frac{4}{30}\),
\(\frac{6}{45}\),
\(\frac{8}{60}\),
\(\frac{10}{75}\),
\(\frac{12}{90}\).
2) For
\(\frac{1}{a}+\frac{1}{b}=\frac{2}{15}\),
if
a and
b were the same, it is clear that they will each be 15.
This is our first solution: (15, 15).
3) To test
\(\frac{4}{30}\),
we cannot resort to
\(\frac{2}{30}+\frac{2}{30}\),
since each fraction would reduce to
\(\frac{1}{15}\),
and we have already exhausted that combination. Thus, the only other positive integers available to us are 1 and 3:
\(\frac{1}{30}+\frac{3}{30}=\frac{4}{30}\).
How do we derive the values of
a and
b? We simply divide the
denominator of each fraction by its
numerator and take the integer value.
\(\frac{30}{1}=30\),
and
\(\frac{30}{3}=10\).
Since it makes no difference which unknown takes the value of 30 or 10, we derive
two more valid combinations: (10, 30) and (30, 10).
4) To test
\(\frac{6}{45}\),
we cannot use 3 and 3, since, again, we have already exhausting the matching-value solution, and here is another important realization: neither can we use any even-integer combination, such as 2 and 4, since the denominator of our fraction will
not be evenly divisible by these values. Hence, it only makes sense to try 1 and 5:
\(\frac{1}{45}+\frac{5}{45}=\frac{6}{45}\).
Using our mental math approach of dividing the denominator by the numerator of each fraction, we get another combination: 45 and 9.
Add these two solutions to your tally: (9, 45) and (45, 9).
At this point, we have derived five solutions, and a quick check against the answers reveals that only 5 or 9 will make any sense. Why? Because, once more, we have already exhausted the solution in which our two unknowns equal each other, and we have already seen how to derive our other "mirrored" solutions. Thus, we can
anticipate that we are either sitting on the answer already or need to find just one more winning combination.
5) To test
\(\frac{8}{60}\),
we cannot use 4 and 4. The other combinations are 5/3, 6/2, and 7/1. I start with the larger number because it is easier to assess whether dividing the denominator by such a value will yield an integer. We can see that both 5/3 and 6/2 will work, but not 7/1, since 60 is not divisible evenly by 7 but has all the other numbers as factors.
\(\frac{5}{60}+\frac{3}{60}=\frac{8}{60}\).
From this, we can derive
\(\frac{60}{5}=12\)
and
\(\frac{60}{3}=20\).
(12, 20) and (20, 12) represent two more valid solutions, and that ticks the tally beyond 5.
We can thus conclude that the answer will be 9, without doing any more work. What about the other combination, though, 6 and 2? If you are curious, you can run it through for a quick check:
\(\frac{6}{60}+\frac{2}{60}=\frac{8}{60}\).
From this, we can derive
\(\frac{60}{6}=10\)
and
\(\frac{60}{2}=30\).
Although this is a valid combination, it is an extraneous solution, since we have already worked out these values above. Notice that we do not even need to test either of our other fractions, but as an added bonus, I will show you how I checked them anyway.
---UNNECESSARY EXTRA WORK---
6) To test
\(\frac{10}{75}\),
we cannot use 5 and 5. The other combinations are 6/4, 7/3, 8/2, and 9/1. We can eliminate the even-numbered ones, since 75 is not divisible evenly by an even number. We can also see that neither 7 nor 9 will fit into 75 without getting into decimal territory. This is how I checked:
\(7*10=70\),
so 75 is
not evenly divisible by 7;
\(9*8=72\),
so 75 is
not evenly divisible by 9.
This one was an easy fraction to test. No valid combination of
a and
b values exists.
7) To test
\(\frac{12}{90}\),
we cannot use 6 and 6. The other combinations are 7/5, 8/4, 9/3, 10/2, and 11/1. We can run the larger numbers through in a similar manner to what we had just done with 75:
\(7*10=70\),
and since
\(70+7*3=91\),
90 is
not evenly divisible by 7;
\(8*10=80\),
and since
\(80+8=88\),
it is clear that 90 will
not be evenly divisible by 8;
\(\frac{90}{9}=10\),
and
\(\frac{90}{3}=30\).
Thus, 9 and 3 will work.
\(\frac{90}{10}=9\),
and
\(\frac{90}{2}=45\).
Thus, 10 and 2 will also work.
\(11*8=88\),
so it is clear that 90 will
not be evenly divisible by 11.
Now we can test our first combination, 9 and 3:
\(\frac{9}{90}+\frac{3}{90}=\frac{12}{90}\).
From this, we can derive
\(\frac{90}{9}=10\)
and
\(\frac{90}{3}=30\).
Of course, we know already that these values work. How about our other combination, 10 and 2?
\(\frac{10}{90}+\frac{2}{90}=\frac{12}{90}\).
From this, we can derive
\(\frac{90}{10}=9\)
and
\(\frac{90}{2}=45\).
We came up short again, since we know already that these values work.
Honestly, I look to do as little work as I can think to do on Quant questions, but this one took me a good chunk of time, about 3 minutes, using the above approach. Please note that
my approach above is not the quickest way to solve the problem, but the number sense you can apply may prove helpful in other types of questions, so I would not say that the above is a wasted effort.
I hope that helps. Phew! That has to be my longest post ever!
- Andrew
close
18 Oct 2013, 17:58
78%
(02:30)
wrong 
