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

GMAT assassins aren't born, they're made,
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Lois has x dollars more than Jim has, and together they have [#permalink]
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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 = number of dollars that Jim has

Lois has x dollars more than Jim has
So, J + x = number of dollars that Lois has

Together they have a total of y dollars
So, (Jim's $) + (Lois' $) = y
Or: J + (J+x) = y
Simplify: 2J + x = y

Which of the following represents the number of dollars that Jim has?
Solve 2J + x = y for J
Subtract x from both sides to get: 2J = y - x
Divide both sides by 2 to get: (y-x)/2

Answer: A

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 28 Apr 2018, 08:23.
Last edited by BrentGMATPrepNow on 19 Jan 2020, 13:47, edited 1 time in total.
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Re: Lois has x dollars more than Jim has, and together they have [#permalink]
Based on the question it is given :

1. L = J + x
2.L + J = y

By simply substituting L for J + x in the second equation,
J + J+ x = y => 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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