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math question [#permalink]

05 Jan 2010, 20:40

Which of the following fractions has the greatest value?

a) [3][/8!]

b) [6(4!)][/5(9!)]

c) [4(3!)][/9!]

d) [3][/5(7!)]

e) [14][/5(8!)]

I'm having some difficulties solving this one. Can someone help me!

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Re: math question [#permalink]

06 Jan 2010, 12:07

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ppaulyni3 wrote:

Thanks Shalva! Could please explain your answer? I dont quite get it.

You should manipulate fractions in order to get them to some common denominator. Then the one with greatest numerator has the greatest value

The math here is quite simple.

9! = 8! * 9

To get 8! from 7! in answer (d) you should multiply both the numerator and denominator by 8 : 7! * 8 = 8!

These are the only points worth mentioning

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Re: math question [#permalink]

06 Jan 2010, 11:27

I'll apply some formatting:

Which of the following fractions has the greatest value?

a)\frac {3}{8!}

b) \frac{6(4!)}{5(9!)}

c)\frac{4(3!)}{9!}

d) \frac{3}{5(7!)}

e) \frac{14}{5(8!)}

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Re: math question [#permalink]

06 Jan 2010, 11:38

a)\frac {3}{8!}

b) \frac{6(4!)}{5(9!)} = \frac{3.2}{8!}

c)\frac{4(3!)}{9!} = \frac{2.(6)}{8!}

d) \frac{3}{5(7!)} = \frac{4.8}{8!} < - - - Greatest fraction

e) \frac{14}{5(8!)}= \frac{2.8}{8!}

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Re: math question [#permalink]

06 Jan 2010, 11:55

Thanks Shalva! Could please explain your answer? I dont quite get it.

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Re: math question [#permalink]

07 Jan 2010, 07:38

ppaulyni3 wrote:

convert the fractions to some common factor

a)[3/8!]=k*(3/8)

b) [6(4!)][/5(9!)]=k*(4!/60)

c) [4(3!)][/9!]=k*{(4*3!)/72}

d) [3][/5(7!)]=k*(3/5)

e) [14][/5(8!)]=k*(14/40)

where k =(1/7!)
it is easy to solve it now..clearly D has the maximum value
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Re: math question [#permalink] 07 Jan 2010, 07:38

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