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# M16-34

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Math Expert
Joined: 02 Sep 2009
Posts: 52231
M16-34  [#permalink]

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16 Sep 2014, 00:00
00:00

Difficulty:

35% (medium)

Question Stats:

74% (01:17) correct 26% (01:31) wrong based on 100 sessions

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If $$a$$, $$b$$, $$c$$, and $$d$$ are non-zero numbers such that $$\frac{a}{b} = \frac{c}{d}$$ and $$\frac{a}{d} = \frac{b}{c}$$, which of the following must be true?

A. $$|a| = |c|$$
B. $$|b| = |d|$$
C. $$|a| = |d|$$
D. $$|b| = |a|$$
E. $$|b| = |c|$$

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Posts: 52231
Re M16-34  [#permalink]

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16 Sep 2014, 00:00
Official Solution:

If $$a$$, $$b$$, $$c$$, and $$d$$ are non-zero numbers such that $$\frac{a}{b} = \frac{c}{d}$$ and $$\frac{a}{d} = \frac{b}{c}$$, which of the following must be true?

A. $$|a| = |c|$$
B. $$|b| = |d|$$
C. $$|a| = |d|$$
D. $$|b| = |a|$$
E. $$|b| = |c|$$

Given: (i) $$\frac{a}{b} = \frac{c}{d}$$ and (ii) $$\frac{a}{d} = \frac{b}{c}$$.

Now, from (i) $$a=\frac{bc}{d}$$, substitute $$a$$ in (ii) $$\frac{bc}{d^2}=\frac{b}{c}$$. Cross-multiply and reduce by $$b$$: $$c^2=d^2$$, which means that $$|c|=|d|$$, but this option is not listed among the answer choices so we should try to get the one which is listed.

Again, from (i) $$c=\frac{ad}{b}$$, substitute $$c$$ in (ii): $$\frac{a}{d}=\frac{b^2}{ad}$$. Cross-multiply and reduce by $$d$$: $$a^2=b^2$$, which means that $$|a|=|b|$$.

Answer: D
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Re: M16-34  [#permalink]

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03 Jan 2015, 11:13
5
Or cross-multiply directly to find:

ad=bc
ac=bd

In order for those equalities to be true you can see that c=b or a=b because those letters appear on either side of the equality.
There is no c=d in the answers so choose a=b.
Manager
Joined: 24 Nov 2013
Posts: 57
M16-34  [#permalink]

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13 Aug 2015, 21:55
we can also solve by testing numbers
a/b = c/d
let a/b = 1/2 and c/d = 2/4, a=1, b=2,c=2,d=4

since a/d = b/c, we should check whether 1/4 = 2/2...this is not true

lets try making a=b=1 and d=c=3..
a/b = c/d = 1.

a/d = b/c = 1/3

so we need to have a=b and c=d
c=d is not present in the options...hence choose |b|=|a|
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GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
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M16-34  [#permalink]

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07 Sep 2015, 12:03
1
1
hi

elimination:
ad=bc
ac=bd

ad-ac=bc-bd

a(d-c)=b(c-d)
a(d-c)=-b(d-c)

a=-b, hence D. |b|=|a|
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Intern
Joined: 07 Dec 2014
Posts: 9
Re M16-34  [#permalink]

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10 Nov 2015, 21:25
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hi,
i have followed following steps.. let me know if it's correct
a/b = c/d
ad=bc (1)
a/d=b/c
ac=bd (2)
adding (1) and (2)
ad+ac=bc+bd
a(d+c)=b(c+d)
cancelling (d+c) on both sides
a=b or b=a
Is this method correct.. as it is mentioned that a,b,c,d are non zero I have divided (c+d) both the sides.. but still I am not getting mod(a) = mod(b) expression.
Could you please, if this approach is correct
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Re M16-34  [#permalink]

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12 Jul 2016, 03:44
I think this is a high-quality question and I agree with explanation.
Senior Manager
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Re: M16-34  [#permalink]

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05 Apr 2017, 06:06
The answer must be option D. Cross multiply both the statements to get :

1) ad=bc
2) ac=bd

Now do 1/2 to get |c|=|d|. Substitute it in 2) to get |a|=|b| , i.e. option D
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Intern
Joined: 02 Jun 2015
Posts: 23
Location: United States
Re: M16-34  [#permalink]

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05 Jun 2018, 14:30
This is the way I solve it...

a/b = c/d

Substitute: a=1 b=1 c=2 d=2

1/1 = 2/2 other words 1=1

a/d = b/c

Substitute by the same number above

1/2 = 1/2

Checking the answer choices, the only match is D b = a

Hope it helps!

Thanks!
Ale
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Joined: 19 Feb 2018
Posts: 25
Re: M16-34  [#permalink]

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31 Oct 2018, 22:04
is cross multiplication allowed? We don't know the sign of any number. Please suggest.
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Joined: 02 Sep 2009
Posts: 52231
Re: M16-34  [#permalink]

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31 Oct 2018, 22:07
harsh8686 wrote:
is cross multiplication allowed? We don't know the sign of any number. Please suggest.

We are concerned about the sign when dealing with inequalities: we should keep the sign if we multiply by a positive value and flip the sign when we multiply by a negative value.

For equations we can multiply/divide by a variable regardless of its sign (providing it's not 0) or cross-multiply for that matter.

Hope it's clear.
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Re: M16-34  [#permalink]

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31 Oct 2018, 22:26
Thanks for the clarification

Posted from my mobile device
Re: M16-34 &nbs [#permalink] 31 Oct 2018, 22:26
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# M16-34

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