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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

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D01-43

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D01-43  [#permalink]

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New post 15 Sep 2014, 23:13
4
19
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

38% (01:39) correct 62% (01:34) wrong based on 276 sessions

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Re D01-43  [#permalink]

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New post 15 Sep 2014, 23:13
1
7
Official Solution:


Statement 1: \(xy = 6\) is not sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is -10 when \(x\) is \(-\frac{3}{5}\).

\(5x = y + 7\)

\(x = \frac{y + 7}{5}\) ................i

\(xy = 6\) ........................ ii

Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\)

\(y^2 + 7y = 30\)

\(y^2 + 10y - 3y - 30= 0\)

\(y (y + 10) - 3 (y + 10) = 0\)

\((y + 10) (y - 3) = 0\)

\(y = - 10\) or \(3\)

If \(y = - 10\), \(x = -\frac{3}{5}\). In this case, \(x - y = -\frac{3}{5} - (-10) = \frac{47}{5}\). Yes.

If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.

In each case \((x-y) \gt 0\) and \(\lt 0\). Hence statement 1 is not sufficient.

Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\).

When \(x\) and \(y\) are consecutive integers, either \(x = y+1\) or \(y = x+1\) is possible.

(i) If \(x = y+1\)

\(5x = y + 7\)

\(5(y+1) = y + 7\)

\(4y = 2\)

\(y = \frac{1}{2}\). Then \(x = \frac{3}{2}\). However this is not possible because \(x\) and \(y\) are even not integers. So this option is not valid.

(ii) If \(y = x+1\)

\(5x = y + 7\)

\(5x = x + 1 + 7\)

\(4x = 8\)

\(x = 2\)

Then, \(y = 3\). This is valid because \(x\) and \(y\) are consecutive integers.

So \((x-y) = 2-3 = -1\). Hence statement 2 is sufficient.


Answer: B
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Re: D01-43  [#permalink]

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New post 14 Nov 2014, 08:56
How did we get to y(y+10)-3(y+10) from the step just before this.

Thanks!
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New post 14 Nov 2014, 09:01
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New post 14 Nov 2014, 10:20
Got it! thanks!
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Re: D01-43  [#permalink]

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New post 19 Jan 2015, 05:23
1
if x and y are consecutive then x=y+1 and y=x+1 both can be true either ways both the numbers are consecutive
my solution was like this
given
5x=7+y
x-y=7-4x
therefore, x-y>0 when 7-4x>0
so when x is negative x-y is postive
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D01-43  [#permalink]

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New post 21 Mar 2015, 04:24
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How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.

Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5

Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?

1) Not sufficient: many ways y < or > 7/4

2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient

Is that right?

I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.
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Re: D01-43  [#permalink]

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New post 19 Feb 2016, 11:26
1
The way I thought of it:

I rearranged the question to say 0= y+7-5x

Looking at the statements I skipped straight to B. Since they have to be consecutive integers Y=3 and X = 2 works, . Sufficient.
With statement A) x could be 3 and y could be 2 or vice versa so not sufficient.
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Re: D01-43  [#permalink]

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New post 17 Mar 2016, 12:00
If 5x=y+7, is (x−y)>0?

(1) xy=6
x(5x-7)=6 => 5x^2-7x-6=0 => x=(7+/- Sqrt (49-4*5*(-6)))/10 => (7+/-13)/10 => 2 or -3/5
x=2,y=3, x-y=-1<0
x=-3/5, y=-10, x-y= 47/5=9.4 >0
Insufficient

(2) x and y are consecutive integers with the same sign

Let x=y+1
5x=y+7 => 4y=2 => y=0.5 not integer

Let y=x+1
5x=y+7 => 4x=8 =>x=2 , y=3
x-y= -1<0
SUFFICIENT

Hence B

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Re D01-43  [#permalink]

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New post 14 Jul 2016, 12:52
I think this is a high-quality question and I agree with explanation.
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D01-43  [#permalink]

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New post 05 Sep 2016, 09:51
rephrase as
x- y =7- 4x ( eq.1)
thus .. we can say...
is 7 - 4x > 0..
i.e. x < 7/4..
x < 1.75?

1. x = 2 NO... X = 1.5 YES
2 * 3 = 6
1.4 * 4 =6
INSUFFICIENT

2. x = y + 1
substitu in eq. 1
x - x + 1 = 7 - 4x => x= 3/2= 1.5 so y = 0.5 ... NOT AN INTEGER... reject this case

y = x+1
substiute in eq.1
x - x -1 = 7 - 4x => x = 2 ... so y = 3.. BOTH ARE CONSECUTIVE INTGERS....
Definite NO .. for x < 1.75?

(2) is sufficient
so B
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Re: D01-43  [#permalink]

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New post 09 Sep 2016, 04:28
I like this question. This is, for sure, a great question. Thanks team at GC!
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Re: D01-43  [#permalink]

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New post 01 Nov 2016, 20:57
from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?
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New post 02 Nov 2016, 01:19
phanikrishna wrote:
from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?


Please read the solution carefully. There are TWO sets of (x,y) given which satisfy the equations: x=2 and y=3 AND x=-5/3 and y=-10.

Also, the red part above is wrong. xy=6 has infinitely many solutions for x and y. You considered only positive integers there but x and y can be negative and fractions too.
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D01-43  [#permalink]

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New post 15 Jan 2017, 13:05
madmax1000 wrote:
How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.

Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5

Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?

1) Not sufficient: many ways y < or > 7/4

2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient

Is that right?

I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.



Bunuel, could you please suggest me if the highlight is correct? That "many ways" is without any proof, and as a matter of fact there are only two solutions to the system of equations XY=6 and 5X=Y-7, which in the specific case are one with Y>7/4 and one with Y<7/4. But with the above approach how could I say that the first stm is not suff?

Thank you in advance
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Re: D01-43  [#permalink]

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New post 24 Jan 2017, 01:27
Is there a shortcut for this question ?
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New post 24 Jan 2017, 01:35
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New post 03 Jun 2017, 19:53
Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?
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New post 04 Jun 2017, 05:35
chidananda3891 wrote:
Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?


Which values did you get for (2) and how? There is only set possible as shown in the solution.
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: D01-43  [#permalink]

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New post 28 Sep 2018, 20:34
why does it half to be an integer? isnt this an assumption
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Re: D01-43 &nbs [#permalink] 28 Sep 2018, 20:34

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