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In the table above, what is the number of green marbles in J [#permalink ]
01 Oct 2012, 04:11
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Re: In the table above, what is the number of green marbles in J [#permalink ]
01 Oct 2012, 04:11
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Re: In the table above, what is the number of green marbles in J [#permalink ]
01 Oct 2012, 04:19
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To solve, I took the sum of the expressions as seen below:
X + Y = 80
Y + Z = 120
X + Z = 160 2X + 2Y + 2Z = 360
Dividing by 2 we get X + Y + Z = 180.
Since we know X + Y = 80 from Jar P, we can deduce that Z = 100. Since Z is the number of green marbles in Jar R we have our solution.
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Re: In the table above, what is the number of green marbles in J [#permalink ]
01 Oct 2012, 19:22
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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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Re: In the table above, what is the number of green marbles in J [#permalink ]
03 Oct 2012, 23:41
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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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Re: In the table above, what is the number of green marbles in J [#permalink ]
04 Oct 2012, 13:20
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Re: In the table above, what is the number of green marbles in J [#permalink ]
05 Oct 2012, 02:19
x+y=80...(1) y+z=120 ==> z=120-y x+z=160 ==> z=120-x...(3) 120-x=160-y ==> x-y=40...(2) sloving (1) & (2) we get x=60 put value of x=60 in eqn (3), 60+z=160 => z=100 Ans
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Re: In the table above, what is the number of green marbles in J [#permalink ]
10 Dec 2012, 03:43
\(x + y = 80\) 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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Re: In the table above, what is the number of green marbles in J [#permalink ]
13 Apr 2014, 07:52
Bunuel wrote:
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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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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Re: In the table above, what is the number of green marbles in J [#permalink ]
13 Apr 2014, 20:16
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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Re: In the table above, what is the number of green marbles in J [#permalink ]
14 Apr 2014, 00:30
russ9 wrote:
Bunuel wrote:
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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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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Re: In the table above, what is the number of green marbles in J [#permalink ]
28 Apr 2014, 20:27
Bunuel wrote:
russ9 wrote:
Bunuel wrote:
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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Re: In the table above, what is the number of green marbles in J [#permalink ]
28 Apr 2014, 23:58
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Re: In the table above, what is the number of green marbles in J [#permalink ]
29 Apr 2014, 06:47
Bunuel wrote:
russ wrote:
Can we still add the 3?
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Yes.
Thanks -- that clarifies a lot!
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Re: In the table above, what is the number of green marbles in J [#permalink ]
29 May 2014, 08:03
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Re: In the table above, what is the number of green marbles in J [#permalink ]
29 May 2014, 08:04
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Re: In the table above, what is the number of green marbles in J [#permalink ]
20 Aug 2014, 22:47
Another approach can be back solving by taking a value from choices for z and finding x and y to see if they make sense per the table.
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Re: In the table above, what is the number of green marbles in J [#permalink ]
09 Sep 2014, 21:05
The table provided in this question was a boon for me. See how below.....
Attachment:
Table.png [ 45.65 KiB | Viewed 4323 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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Re: In the table above, what is the number of green marbles in J [#permalink ]
21 Nov 2014, 02:28
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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Re: In the table above, what is the number of green marbles in J [#permalink ]
22 Jul 2015, 08:52
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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