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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D

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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post Updated on: 27 Jul 2017, 23:00
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A
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Question Stats:

67% (01:44) correct 33% (02:09) wrong based on 134 sessions

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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D, and 5C = 12A, then the arrangement of the four numbers from greatest to least is

A. CDAB
B. BACD
C. DCAB
D. DCBA
E. BDAC

Originally posted by shinrai15 on 27 Jul 2017, 21:53.
Last edited by Bunuel on 27 Jul 2017, 23:00, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post 27 Jul 2017, 22:32
From equation 1 , A=9/4B,
From equation 3, A=5/12C,
From equation 3, C=11/17D.
On solving all we get
A=9/4B=5/12C=55/204D

WHICH FINALLY YIELDS
204A=459B=85C=55D.

THUS B IS GREATEST AN ANSWER IS BACD. OPTION B



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Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post 27 Jul 2017, 22:45
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1
4A = 9B -> A = 9B/4 If B=4, A=9 - Hence, A>B
5C = 12A -> A = 5C/12 If C=12,A=5 - Hence, C>A
17C = 11D -> C = 11D/17 If D=17, C=11 - Hence, D>C

Hence D > C, C > A and A > B

So, the sequence should be D,C,A and B(Option C)
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Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post 27 Jul 2017, 22:49
pushpitkc wrote:
4A = 9B -> A = 9B/4 If B=4, A=9 - Hence, A>B
5C = 12A -> A = 5C/12 If C=12,A=5 - Hence, C>A
17C = 11D -> C = 11D/17 If D=17, C=11 - Hence, D>C

Hence D > C, C > A and B > A.

To check for the bigger number among B and C
B = 4A/9 and C = 12A/5
If A = 45,
B = 20 and C = 108 - Hence C>A>B

So, the sequence should be D,C,A and B(Option C)

My bad !! I took it totally opposite. I need to be careful in real exam for such mistakes.
Thanks pushpit. Your solutions at amazing.



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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post Updated on: 30 Jul 2017, 14:16
pushpitkc wrote:
4A = 9B -> A = 9B/4 If B=4, A=9 - Hence, A>B
5C = 12A -> A = 5C/12 If C=12,A=5 - Hence, C>A
17C = 11D -> C = 11D/17 If D=17, C=11 - Hence, D>C

Hence D > C, C > A and B > A.

To check for the bigger number among B and C
B = 4A/9 and C = 12A/5
If A = 45,
B = 20 and C = 108 - Hence C>A>B

So, the sequence should be D,C,A and B(Option C)

pushpitkc , I think you have a typo that throws off your subsequent analysis and has you adding what I think is an unnecessary step.

Initially you write:

Quote:
4A = 9B -> A = 9B/4 If B=4, A=9 - Hence, A>B

But then you write:

Quote:
Hence D > C, C > A and B > A[typo?]


Then you revert back to A > B while comparing B and C.

I'm pretty sure there is no need to compare B and C. (See my solution below.)

C > A, A > B, hence C > B.

I might be missing something.

Originally posted by generis on 30 Jul 2017, 14:07.
Last edited by generis on 30 Jul 2017, 14:16, edited 1 time in total.
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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post 30 Jul 2017, 14:08
shinrai15 wrote:
If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D, and 5C = 12A, then the arrangement of the four numbers from greatest to least is

A. CDAB
B. BACD
C. DCAB
D. DCBA
E. BDAC

If 4A = 9B:

A = \(\frac{9}{4}\)B

B = \(\frac{4}{9}\)A

A > B

If 17C = 11D

C = \(\frac{11}{17}\)D

D = \(\frac{17}{11}\)C

D > C

If 5C = 12A

C = \(\frac{12}{5}\)A

A = \(\frac{5}{12}\)C

C > A
_____
THUS
A > B
D > C
C > A


If you're tracking on C (greater than A and B), you'll remember that you discovered D > C, and you are done. D>C>A>B. Or:

From last and first inequalities we know
C > A, and
A > B, so

We have C > A > B.

Anything bigger than C? From second inequality, D > C.

Hence D>C>A>B

Answer C

P.S. In three or four inequalities such as those above, if a variable shows up only once on LHS or RHS, it's either the biggest or the smallest. So you can look for the variables that show up once and work from greatest to least or vice versa. You'd see B and D here. If B, ask: anything smaller? No. It's the smallest. If D, ask: anything larger? No. It's the largest. You can either work in order from there or take biggest and smallest and figure out relationship of middle two.
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Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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New post 03 Sep 2018, 05:32
shinrai15 wrote:
If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D, and 5C = 12A, then the arrangement of the four numbers from greatest to least is

A. CDAB
B. BACD
C. DCAB
D. DCBA
E. BDAC

Approximately
A= 10/4B or A= 2.5 B
C=10/20D OR C=.5D
C=10/5A or C=2A
DCAB IS THE ANSWER
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Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D &nbs [#permalink] 03 Sep 2018, 05:32
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If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D

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