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M16-34

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M16-34  [#permalink]

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New post 16 Sep 2014, 00:00
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

74% (01:16) correct 26% (01:31) wrong based on 99 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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Re M16-34  [#permalink]

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New post 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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New post 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.
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M16-34  [#permalink]

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New post 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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M16-34  [#permalink]

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New post 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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Re M16-34  [#permalink]

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

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New post 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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Re: M16-34  [#permalink]

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New post 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!
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Re: M16-34  [#permalink]

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

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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M16-34  [#permalink]

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New post 31 Oct 2018, 22:26
Thanks for the clarification

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