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Lois has x dollars more than Jim has, and together they have

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Manager
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Lois has x dollars more than Jim has, and together they have [#permalink]

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20 Dec 2012, 08:04
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Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x
[Reveal] Spoiler: OA
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Posts: 44657
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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20 Dec 2012, 08:09
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Walkabout wrote:
Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x

Let J be the number of dollars that Jim has. Then Lois has J+x dollars.

Since together they have a total of y dollars, then J+(J+x)=y --> J=(y-x)/2.

Answer: A.
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Posts: 190
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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10 Jan 2014, 03:01
Walkabout wrote:
Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x

1) Set up the equations: [L = J + x], [L + J = y]

2) Substitute L for J + x and insert this in the second equation, solve so you only have J separately on one side ----> J = (y - x)/2
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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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30 May 2014, 06:07
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Walkabout wrote:
Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x

In the exam the answer would be presented as above or as below, the way it is after formatting with math symbols?

(A) $$\frac{y-x}{2}$$
(B) $$y -\frac{x}{2}$$
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Joined: 02 Sep 2009
Posts: 44657
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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30 May 2014, 07:16
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b2bt wrote:
Walkabout wrote:
Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x

In the exam the answer would be presented as above or as below, the way it is after formatting with math symbols?

(A) $$\frac{y-x}{2}$$
(B) $$y -\frac{x}{2}$$

_________
The way you put it.
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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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28 Jun 2016, 05:50
Walkabout wrote:
Lois has x dollars more than Jim has, and together they have a total of y dollars. Which of the following represents the number of dollars that Jim has?

(A) (y - x)/2
(B) y - x/2
(C) y/2 - x
(D) 2y - x
(E) y - 2x

To solve, we will set up two equations. Let's start by defining two variables.

J = number of dollars Jim has

L = number of dollars Lois has

We are given that Lois has x dollars more than Jim. We set up an equation:

L = x + J

We are next given that together they have a total of y dollars. We can set up our second equation:

J + L = y

Since we know that L = x + J, we can substitute x + J for L into the second equation J + L = y.

Notice that, after the substitution, we will only have variables of J, x, and y. Thus, we have:

J + x + J = y

2J + x = y

2J = y – x

J = (y – x)/2

The answer is A.
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Manager
Joined: 09 Aug 2016
Posts: 68
Lois has x dollars more than Jim has, and together they have [#permalink]

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09 Sep 2016, 14:31
I found something interesting with this.

Lets say that someone decides to use numbers to solve this and goes for the following choices:

x = 10, J = 10

Then this will give L = 10 + 10 = 20 and y = 10 + 20 = 30

Answer A) will give (30-10) / 2 = 20 / 2 = 10

Answer E) will give 30 - 2(x) = 30 - 2*(10) = 10

LOL :S what if someone goes for E rather than A ????

Insufficient answer choice E
Math Expert
Joined: 02 Sep 2009
Posts: 44657
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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11 Sep 2016, 03:44
Ndkms wrote:
I found something interesting with this.

Lets say that someone decides to use numbers to solve this and goes for the following choices:

x = 10, J = 10

Then this will give L = 10 + 10 = 20 and y = 10 + 20 = 30

Answer A) will give (30-10) / 2 = 20 / 2 = 10

Answer E) will give 30 - 2(x) = 30 - 2*(10) = 10

LOL :S what if someone goes for E rather than A ????

Insufficient answer choice E

When plugging numbers, it might happen that two or more choices give "correct" answer. If this happens, just pick some other number and check again these "correct" options only.
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Posts: 11
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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11 Sep 2016, 05:31
This is a fairly simple Linear Equation problem.

Let no. of dollars with Jim = Z
Then, no. of dollars with Lois = Z + x
The total no. of dollars with Jim and Lois combined would be Z + Z + x
However, their total no. of dollars is mentioned as y.
Thus, Z + Z + x = y

To find Z from the above equation, just transpose:
2Z = y - x; Thus, Z = (y-x)/2

So, Jim has (y-x)/2 dollars with him
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Manager
Joined: 09 Aug 2016
Posts: 68
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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11 Sep 2016, 06:52
Bunuel wrote:
When plugging numbers, it might happen that two or more choices give "correct" answer. If this happens, just pick some other number and check again these "correct" options only.

This is what concerns me a lot.... The "Plug number" method is far more easier and fast than actually solving the problem. Is quite frustrating that the test actually allows this.

Given that this is the case "plug number" method always requires scan of all answer choices rather than stopping to the first correct answer.
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Joined: 06 Dec 2016
Posts: 253
Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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13 Apr 2017, 11:29
Let's say
Lois has $10 Jim has$2 (Lois has $8 more dollars than Lois) If you plug in the numbers for each choice y = 12 x = 8 Only A is equal to 2 (12 - 8)/2 = 2 Therefore, the answer is A. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11518 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: Lois has x dollars more than Jim has, and together they have [#permalink] Show Tags 06 Dec 2017, 14:19 1 This post received KUDOS Expert's post Hi All, We're told that Lois has X dollars MORE than Jim has and together they have a TOTAL of Y dollars. We're asked for the number of dollars that Jim has. This question can be solved by TESTing VALUES. IF.... Lois has 5 dollars and Jim has 2 dollars, then X = 3 and Y = 7. So we're looking for an answer that equals 2 when we plug in X = 3 and Y = 7... There's only one answer that matches: Final Answer: [Reveal] Spoiler: A GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Lois has x dollars more than Jim has, and together they have   [#permalink] 06 Dec 2017, 14:19
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