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In the table above, what is the number of green marbles in J

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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


[Reveal] Spoiler:
Attachment:
Table.png
Table.png [ 40.01 KiB | Viewed 13089 times ]
[Reveal] Spoiler: OA

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Re: In the table above, what is the number of green marbles in J [#permalink]

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SOLUTION

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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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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 01 Oct 2012, 05:19
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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.

[Reveal] Spoiler:
Answer is D, 100

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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 01 Oct 2012, 20:22
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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]

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New post 04 Oct 2012, 00:41
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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]

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New post 04 Oct 2012, 14:20
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SOLUTION

Image
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.
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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 05 Oct 2012, 03:19
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]

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New post 10 Dec 2012, 04:43
\(x + y = 80\) 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]

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New post 13 Apr 2014, 08:52
Bunuel wrote:
SOLUTION

Image
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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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]

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New post 13 Apr 2014, 21:16
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]

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New post 14 Apr 2014, 01:30
russ9 wrote:
Bunuel wrote:
SOLUTION

Image
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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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 28 Apr 2014, 21:27
Bunuel wrote:
russ9 wrote:
Bunuel wrote:
SOLUTION

Image
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]

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New post 29 Apr 2014, 00:58
russ9 wrote:
Bunuel wrote:
russ9 wrote:
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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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

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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 29 Apr 2014, 07:47
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]

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New post 20 Aug 2014, 23:47
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]

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New post 09 Sep 2014, 22:05
The table provided in this question was a boon for me. See how below.....
Attachment:
Table.png
Table.png [ 45.65 KiB | Viewed 8366 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]

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New post 21 Nov 2014, 03:28
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]

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New post 22 Jul 2015, 09:52
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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Re: In the table above, what is the number of green marbles in J [#permalink]

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New post 22 Jul 2015, 09:56
x+y = 80 (i)
y+z = 120 (ii)
x+z = 160 (iii)

from (i) and (ii):
z-x = 40
z+x = 160

so, z = 100, x = 60, y = 20.

Ans (D).
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In the table above, what is the number of green marbles in J [#permalink]

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New post 30 May 2016, 05:32
Bunuel wrote:
Image
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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