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Sub 505 (Easy)|   Algebra|                           
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Bunuel
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x+y = 80 ---(1)
x+z = 160---(2)
z+y= 120---(3)
Subtract equation 1 from 2 & we get--> z-y = 80----(4)
Add equation (4) & (3) we get--> 2z= 200
z=100
Answer D
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x + y = 80 ......(1)
y + z = 120 .....(2)
x + z = 160 ......(3)

From (2) above, z=160-y .....Substitute value of z in (3)

==> x-y = 40 ....(4)

Solve (1) and (4), to get x = 60
==> y = 20
==> z = 100

Thus, number of green marbles in Jar R = 100 (Ans = D)
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SOLUTION


In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

We need to find the value of \(z\), while given that:

\(x+y=80\);
\(y+z=120\);
\(x+z=160\).

Sum these 3 equations: \(2x+2y+2z=360\) --> reduce by 2: \(x+y+z=180\) --> since we know that \(x+y=80\), then \(80+z=180\) --> \(z=100\).

Answer: D.

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\(x + y = 80\) eq 1
\(y + z = 120\) eq 2
\(x + z = 160\) eq 3
______________
\(2x + 2y + 2z = 360 --> x + y+ z = 180\) eq 4

Combine eq 4 and eq 1:

\(80 + z = 180 --> z = 100\)

Answer: D
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Bunuel
SOLUTION


In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

We need to find the value of \(z\), while given that:

\(x+y=80\);
\(y+z=120\);
\(x+z=160\).

Sum these 3 equations: \(2x+2y+2z=360\) --> reduce by 2: \(x+y+z=180\) --> since we know that \(x+y=80\), then \(80+z=180\) --> \(z=100\).

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

Are we always allowed to sum the 3 equations? Do we need to have some commonalities to be able to sum the equations?
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X+Y =80 --------- (1)

Y+Z =120 -------- (2)

X+Z =160 -------- (3)

Subtract (3) -(1)

we get

Z - Y = 80 -----------(4)

ADD (4) and (2) equations,
Z=100;

Hence D.
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russ9
Bunuel
SOLUTION


In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

We need to find the value of \(z\), while given that:

\(x+y=80\);
\(y+z=120\);
\(x+z=160\).

Sum these 3 equations: \(2x+2y+2z=360\) --> reduce by 2: \(x+y+z=180\) --> since we know that \(x+y=80\), then \(80+z=180\) --> \(z=100\).

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

Are we always allowed to sum the 3 equations? Do we need to have some commonalities to be able to sum the equations?

Yes, we can sum/subtract/multiply equations. I think you are mixing equations with inequalities, for which there are specific rules.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.
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Bunuel
russ9
Bunuel
SOLUTION


In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

We need to find the value of \(z\), while given that:

\(x+y=80\);
\(y+z=120\);
\(x+z=160\).

Sum these 3 equations: \(2x+2y+2z=360\) --> reduce by 2: \(x+y+z=180\) --> since we know that \(x+y=80\), then \(80+z=180\) --> \(z=100\).

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

Are we always allowed to sum the 3 equations? Do we need to have some commonalities to be able to sum the equations?

Yes, we can sum/subtract/multiply equations. I think you are mixing equations with inequalities, for which there are specific rules.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.

Thanks for clarifying.

Just to confirm one of your comments above -- "Yes, we can sum/subtract/multiply equations." -- would this be valid for the problem even if one of the equations didn't have any common variables. What I mean is, if the equations read:

\(x+y=80\);
\(a+b=120\);
\(x+z=160\).

Can we still add the 3?
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russ9
Bunuel
russ9

Are we always allowed to sum the 3 equations? Do we need to have some commonalities to be able to sum the equations?

Yes, we can sum/subtract/multiply equations. I think you are mixing equations with inequalities, for which there are specific rules.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.

Thanks for clarifying.

Just to confirm one of your comments above -- "Yes, we can sum/subtract/multiply equations." -- would this be valid for the problem even if one of the equations didn't have any common variables. What I mean is, if the equations read:

\(x+y=80\);
\(a+b=120\);
\(x+z=160\).

Can we still add the 3?
_______________________
Yes.
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The table provided in this question was a boon for me. See how below.....
Attachment:
Table.png
Table.png [ 45.65 KiB | Viewed 33758 times ]

1. Replaced y with (80-x). The equation remains intact on "Jar P" row

2. Copied (80-x) in "Jar Q" row.

These 2 steps directly eliminates x & y

3. Adding rows "Jar Q" & "Jar R"

2z+80 = 280

z = 100

Answer = D
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From the given table we could make equations like
Equation 1. Given x+y=80 ----> x=80-y
Equation2. Given y+z=120
Equation3. Given x+z = 160
Substituting the value of x in Equation 3 from Equation 1
Equation 4. (80-y)+z=160 ----> -y+z = 80
Adding Equation 2 and Equation 4
2z=200 ---> Z=100.
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just substract from the first equation the 2nd and the 3rd you'll get --> x+y-y-z-x-z = -200 ->x,y cancel out and Z=100 (D)
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Bunuel

In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

We can create three equations from the information presented in the table.

Equation 1: x + y = 80

Equation 2: y + z = 120

Equation 3: x + z = 160

These equations present us with a good opportunity to use the “combination method” to solve multiple equations. Here, we can add equations together. Because we need the number of green marbles in jar R, we need to determine the value of z.

We can start by multiplying equation 1 by -1.

-1(x + y = 80) = -x – y = -80

Next, we add this equation to equation 2. So we have:

-x - y = -80 + (z + y = 120) = z – x = 40

Now we can add z – x = 40 to equation 3. So we have:

–x + z = 40 + (x + z = 160)

2z = 200

z = 100

Answer D.
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Bunuel

In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

Practice Questions
Question: 55
Page: 159
Difficulty: 600

Attachment:
Table.png


Simply add the three given details

x+y=80
y+z=120-
x+z=160

2(x+y+z) = 360
x+y+z = 180

Now We want green marbles in Jar R. That is we want z.
x+y+z=180
x+y= 80; this means z = 100
Option D is the answer.
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Hi All,

Complex-"looking" prompts can often hide hidden patterns that will help you to reduce the amount of work you need to do to answer the question.

The question asks us for the number of green marbles in Jar R, so we're asked "what is the value of Z?"

In the table, you might notice that each variable shows up EXACTLY TWICE, so if we add all of the equations together, we have…

2X + 2Y + 2Z = 360

So…

X + Y + Z = 180

Notice in Jar P….X + Y = 80

Combining these two equations gives us the value of Z…..100

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Bunuel

In the table above, what is the number of green marbles in Jar R ?

(A) 70
(B) 80
(C) 90
(D) 100
(E) 110

Practice Questions
Question: 55
Page: 159
Difficulty: 600

Attachment:
Table.png

Adding 3 equations
x+y+z=180
x+y=80
z=100

IMO D

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