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Set K consists of all fractions of the form x/(x+2) where x is a posit

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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 26 Sep 2018, 17:28
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Question Stats:

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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.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 26 Sep 2018, 19:15
1
anupam87 wrote:
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.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 02 Oct 2018, 19:15
2
anupam87 wrote:
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


Let’s begin by writing out this product:

2/4 x 4/6 x 6/8 x 8/10 x 10/12 x 12/14 x 14/16 x 16/18 x 18/20

So we see that everything cancels except the numerator of the first fraction and the denominator of the last fraction.

Thus, the product of all the fractions in set K is 2/20 = 1/10.

Answer: B
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 02 Oct 2018, 19:26
generis wrote:
anupam87 wrote:
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 (don't write out all the fractions below)

There's a pattern you should see quickly.

Start by finding the greatest value of \(x\).
Finding the greatest value of \(x\) is smart because we have a lot of information:
\(x\) must be even, positive, and less than 20.

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.

\(\frac{18}{20}*\frac{16}{18}*\frac{14}{16}\)

All but the first denominator and the last numerator "cancel."**

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

As mentioned, all but one denominator and 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.

**I wrote only this list of 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 its list will remain
And the last numerator in its list will remain: \(\frac{2}{20}=\frac{1}{10}\)

Why x cannot be 19?, could you help me please?
Thank you.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 02 Oct 2018, 19:44
1
jorgetomas9 wrote:
generis wrote:
anupam87 wrote:
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

Sounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)

There's a pattern you should see quickly.

Start by finding the greatest value of \(x\).
Finding the greatest value of \(x\) is smart because we have a lot of information:
\(x\) must be even, positive, and less than 20.

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.

\(\frac{18}{20}*\frac{16}{18}*\frac{14}{16}\)

All but the first denominator and the last numerator "cancel."**

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

As mentioned, all but one denominator and 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

Why x cannot be 19?, could you help me please?
Thank you.

jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake. :)
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 03 Oct 2018, 08:39
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generis wrote:
jorgetomas9 wrote:
generis wrote:
[
jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake. :)

Thank you very much I haven't seen that one.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 08 Oct 2018, 11:26
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Here is another, simpler approach to solving the problem in under 1 minute.

You can take several values of x and see the pattern.
For x=2, fraction is 1/2
For x=4, fraction is 2/3
For x=6, fraction is 3/4 etc.

So, the product of the 9 fractions would be 9!/10!*1 = 1/10
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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New post 11 Nov 2018, 14:24
1. Solving for X =2,4,6,8 gives us a pattern i.e. 1/2, 2/3 , 3/4 , 4/5 respectively.

2. The last is for 18 which is 9/10

3. See the pattern it's 9! / 10!

Therefore the answer is 1/10
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit   [#permalink] 11 Nov 2018, 14:24
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