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a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
a + b + c/2 = 60
–a – b + c/2 = –10
(A) Column A is larger
(B) ...
(C) ...
(D) ...
(E) ...
*This question is included in Nova Math - Diagnostic/Review:
–a – b + c/2 = –10
| Column A | Column B | |
| b | c |
(A) Column A is larger
(B) ...
(C) ...
(D) ...
(E) ...
*This question is included in Nova Math - Diagnostic/Review:
Replies to This Thread: 1 | ----
I don't understand how you got c=50. I keep getting c=100 because the a and b cancel out, then you're left with 2c/4=50 which equals 100 right?
Reply: that's what I get. I guess they wrote the problem down strange or in a way that we're not seeing it correctly.
Admin
Reply 1 of 1
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Posted: 04/14/2015 03:02
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Hi Tony,
c/2 + c/2 = c, not 2c/4, which reduces to c/2.
It appears that Marquita is adding the denominators (bottoms) of the fractions. But when adding fractions that have a common denominator, we write down the common denominator and add the numerators (tops) of the fractions. So, c/2 + c/2 = (c + c)/2 = 2c/2 = c.
Nova Press
c/2 + c/2 = c, not 2c/4, which reduces to c/2.
It appears that Marquita is adding the denominators (bottoms) of the fractions. But when adding fractions that have a common denominator, we write down the common denominator and add the numerators (tops) of the fractions. So, c/2 + c/2 = (c + c)/2 = 2c/2 = c.
Nova Press
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Marquita, thanks for asking. I see where you made a misstep: c/2 + c/2 = 2c/2 = c, not 2c/4. I hope this helps.
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I eventually figured out the answer (with a different way... took too long) but I really don't understand why you would add those two equations together. What would make you do that? And then you multiplied them together, I don't understand why...
Admin
Reply 1 of 1
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Posted: 07/18/2012 18:53
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Nichalla, please recall high school algebra. Solving for equations with unknowns. Adding the equations together allows us to solve for the unknown c, because we can eliminate the other unknowns (a and b, in this case). Multiplying the result is just to further solve for c.
Replies to This Thread: 1 | ----
I got that c=50 but how do u solve for b ?
Admin
Carla, since we have 2 equations with 3 unknowns, we can't really solve for b. What we want to do is express b in terms of a, so that we can deduce whether b or c is larger, or whether we don't really have enough information to decide.
Reply 1 of 2
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Posted: 06/24/2013 18:55
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I solved for c and got 100. Then I plug it back into my equation for a? When I do this I end up with an impossible equation where 120=20 and b cancels out
Reply 2 of 2
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Posted: 07/02/2013 11:13
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Leslie, when you solve for c, you should not get 100, but you should get 50.
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a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
Posted: 07/05/2013 16:48
I dont understand what the columns are referring to...
Admin
Reply 1 of 1
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Posted: 07/05/2013 21:26
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Hi Bethany,
This is a Quantitative Comparison problem. For these problems, you need to determine whether Column A is larger, or Column B is larger, or the columns are equal, or there is not enough information to decide.
Nova Press
This is a Quantitative Comparison problem. For these problems, you need to determine whether Column A is larger, or Column B is larger, or the columns are equal, or there is not enough information to decide.
Nova Press
Replies to This Thread: 1 | ----
a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
Posted: 07/05/2013 21:59
Thanks! I was mostly confused because it was so close together that I couldn't tell what was in each column.
Admin
Reply 1 of 1
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Posted: 07/06/2013 00:38
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Hi Bethany,
Column A contains just the letter b, and Column B contains just the letter c.
Nova Press
Column A contains just the letter b, and Column B contains just the letter c.
Nova Press
Replies to This Thread: 1 | ----
a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
Posted: 08/22/2013 12:08
Can someone help me understand why the following solutions aren't correct? A=35, B=0, C=50. I found c by adding the equations together. Then I found a by making the first equation b= 60-a-c/2 and plugging that equation in for b in the second. Found a=35 from that and used my a and c values to find that b=0
Admin
Reply 1 of 1
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Posted: 08/22/2013 14:39
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Hi Arielle,
When plugging b = 60 - a - c/2 into the second equation -a - b + c/2 = -10, we get
-a - (60 - a - c/2) + c/2 = -10
Distributing the negative sign yields
-a - 60 + a + c/2 + c/2 = -10
Simplifying yields
- 60 + c = -10 (note that the a's cancelled)
Adding 60 to both sides of this equation yields
c = 50
But we have already determined that c = 50. More importantly, we were not able to determine the value of a.
Nova Press
When plugging b = 60 - a - c/2 into the second equation -a - b + c/2 = -10, we get
-a - (60 - a - c/2) + c/2 = -10
Distributing the negative sign yields
-a - 60 + a + c/2 + c/2 = -10
Simplifying yields
- 60 + c = -10 (note that the a's cancelled)
Adding 60 to both sides of this equation yields
c = 50
But we have already determined that c = 50. More importantly, we were not able to determine the value of a.
Nova Press
Replies to This Thread: 1 | ----
a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
Posted: 11/05/2013 13:51
isn't the real, final answer simply that you can not find three unknowns with two equations? why is any algebra required at all?
Reply 1 of 1
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Posted: 11/08/2013 15:21
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Joe, even though we know we can't solve the equations, sometimes expressing b and c in terms of the other variable tells us whether it is larger than the other or not without solving them numerically.
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a + b + c/2 = 60 –a – b + c/2 = –10 ... ...
Posted: 02/14/2014 01:10
Please can I get more explanations on such equations.
If possible another equation where a solution is achieved.
If possible another equation where a solution is achieved.
Admin
Reply 1 of 1
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Posted: 03/27/2014 16:37
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Chukwuka, thanks for using the app. This problem has been explained many times by different contributors. Please look up the discussion thread and try to follow the explanations.