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Intern  B
Joined: 08 Sep 2016
Posts: 32
If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D  [#permalink]

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9 00:00

Difficulty:   45% (medium)

Question Stats: 69% (01:49) correct 31% (02:17) wrong based on 166 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, 22:53.
Last edited by Bunuel on 28 Jul 2017, 00:00, 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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3
2
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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Joined: 02 Nov 2015
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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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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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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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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.
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Originally posted by generis on 30 Jul 2017, 15:07.
Last edited by generis on 30 Jul 2017, 15:16, 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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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

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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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
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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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_________________ Re: If A, B, C and D are positive integers such that 4A = 9B, 17C = 11D   [#permalink] 25 Nov 2019, 11:38
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