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

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

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

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New post 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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New post 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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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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

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

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

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New post 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 :P
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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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New post 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 :P


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

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New post 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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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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New post 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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Re: Lois has x dollars more than Jim has, and together they have [#permalink]

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New post 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.
Re: Lois has x dollars more than Jim has, and together they have   [#permalink] 13 Apr 2017, 11:29
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