anupam87
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.
Sounds frustrating! Here is one way. About 30 seconds if just a few set members are listed. If all are listed, maybe 75 seconds.
There's a pattern you should see quickly.
Start by finding the greatest value of \(x\).
Doing so is smart because we know that \(x\) must be even, positive, and
less than 20. Use that high-end limit.
The greatest that \(x\) can be is 18. Work backwards from 18.
First fraction? \(\frac{x}{x+2}=\frac{18}{20}\)
Next? Well, the next \(x\) must be 16 (even, positive, < 20).
Its denominator, \(x+2\), is 18: \(\frac{16}{18}\)
Next? \(x\) must be 14. The denominator is 16: \(\frac{14}{16}\)
List just those three: \(\frac{18}{20}*\frac{16}{18}*\frac{14}{16}\)
18 and 18 cancel
16 and 16 cancel
20 remains. 14 remains.
So the
first denominator in the list will remain
And the
last numerator in the list will remain.
We know the first denominator. What is the last numerator? The smallest \(x\).
Smallest integer that \(x\) could be? Positive, even ...
2?
Yes. So the last member of the set at the end of this list is \(\frac{x}{x+2}=\frac{2}{4}\)
First denominator (20) and last numerator (2) will remain:
\(\frac{2}{20}=\frac{1}{10}\)
Answer B
Another pattern: numerators and denominators decend in order.
Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2
Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4
Same conclusion as above. All but the
first denominator and the
last numerator will cancel.
\(\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=\)
\(\frac{2}{20}=\frac{1}{10}\)
Answer B
I hope that helps.

Let me know if you have questions.