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M21-33

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M21-33  [#permalink]

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New post 16 Sep 2014, 00:15
1
5
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

79% (01:34) correct 21% (02:21) wrong based on 98 sessions

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Re M21-33  [#permalink]

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New post 16 Sep 2014, 00:15
1
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:

Image

The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A
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Re: M21-33  [#permalink]

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New post 20 Jun 2015, 21:16
Bunuel wrote:
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:

Image

The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A


Hello,
Can you please tell that how you got the vertices of the triangle from the given equations.

Regards & Thanks
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Re: M21-33  [#permalink]

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New post 20 Jun 2015, 22:16
3
1
vik09 wrote:
Bunuel wrote:
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:

Image

The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A


Hello,
Can you please tell that how you got the vertices of the triangle from the given equations.

Regards & Thanks


Hi,
you have three equations which are equqtion of line..
when you solve for x and y between any two of them, these value of x and y will give you intersection of the two lines, which would be same as the vertices..
you will hAVE three sets of two equations so three vertices..
just an example
y=5−x, 2y=3x, and y=0..

say y=5-x and y=0..
here x=5 and y=0.. this is one of the vertices

similarly 2y=3x and y=0..
here x=0 and y=0..

and the third can be found from
y=5−x, 2y=3x
hope it is clear
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M21-33  [#permalink]

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New post 20 Jun 2015, 23:01
chetan2u wrote:
vik09 wrote:
Bunuel wrote:
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:

Image

The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A


Hello,
Can you please tell that how you got the vertices of the triangle from the given equations.

Regards & Thanks


Hi,
you have three equations which are equqtion of line..
when you solve for x and y between any two of them, these value of x and y will give you intersection of the two lines, which would be same as the vertices..
you will hAVE three sets of two equations so three vertices..
just an example
y=5−x, 2y=3x, and y=0..

say y=5-x and y=0..
here x=5 and y=0.. this is one of the vertices

similarly 2y=3x and y=0..
here x=0 and y=0..

and the third can be found from
y=5−x, 2y=3x
hope it is clear


Hello,
Thanks a lot for your reply. I want to ask one thing that every time when we will solve for x and y from two equations the values that we get for x and y will be the intersection of two points ??. The other point is clear that they will act as vertices.

Sorry if my question is lame ... actually I struggle a lot in Co-Geo apart from P&C.
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Re: M21-33  [#permalink]

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New post 20 Jun 2015, 23:16
1
@vik09 wrote:

Hello,
Thanks a lot for your reply. I want to ask one thing that every time when we will solve for x and y from two equations the values that we get for x and y will be the intersection of two points ??. The other point is clear that they will act as vertices.

Sorry if my question is lame ... actually I struggle a lot in Co-Geo apart from P&C.


Hi,
look what does each equation of line tells us ..
it has two coord x and y , and by taking different values of x and y that fit into that equation and putting them into the graph we have a line..
So when we have two such lines , we will have two sets of values of x and y with only one set of values of x and y common to two sets and that is the point where these two lines intersect... of course we will have no set of value common incase of parallel lines and their slope will be same
hope it was helpful
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M21-33  [#permalink]

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New post 21 Jun 2015, 02:14
chetan2u wrote:
@vik09 wrote:

Hello,
Thanks a lot for your reply. I want to ask one thing that every time when we will solve for x and y from two equations the values that we get for x and y will be the intersection of two points ??. The other point is clear that they will act as vertices.

Sorry if my question is lame ... actually I struggle a lot in Co-Geo apart from P&C.


Hi,
look what does each equation of line tells us ..
it has two coord x and y , and by taking different values of x and y that fit into that equation and putting them into the graph we have a line..
So when we have two such lines , we will have two sets of values of x and y with only one set of values of x and y common to two sets and that is the point where these two lines intersect... of course we will have no set of value common incase of parallel lines and their slope will be same
hope it was helpful


Thanks :)
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Re M21-33  [#permalink]

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New post 02 Aug 2016, 03:47
I think this is a high-quality question and I agree with explanation.
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M21-33  [#permalink]

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New post 22 Feb 2017, 14:58
Bunuel wrote:
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:

Image

The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A


A 30 second shortcut is to find the determinant of a matrix and halve it. So...

|1 1 5|
|-3 2 0|
|0 1 0|

gives 1*(2*0-1*0) - 1*(-3*0-0*0) + 5*(-3*1--2*0) = -15
area is abs(-15)/2 = 15/2
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Re: M21-33  [#permalink]

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New post 18 Dec 2017, 02:20
+1 for A. Draw a graph an get the answer. The area of the triangle is (1/2)*(b)*(h). The sides are x-axis, line x+y=5 and y=(3/2)x. The base is 5 units and height is 3. The area is 15/2 or 7.5 units. The answer is hence option A.
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Re: M21-33  [#permalink]

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New post 18 Dec 2017, 19:22
mbadude2017 wrote:
Bunuel wrote:
Official Solution:

What is the area of the triangle formed by lines \(y = 5-x\), \(2y = 3x\), and \(y = 0\)?

A. 7.5
B. 8.0
C. 9.0
D. 10.5
E. 15.0


Look at the diagram below:



The vertices of the triangle are \((0, 0)\), \((5, 0)\) and \((2, 3)\). So, \(Base=5\), \(Height=3\) and \(Area=\frac{1}{2}*Base*Height=7.5\).


Answer: A


A 30 second shortcut is to find the determinant of a matrix and halve it. So...

|1 1 5|
|-3 2 0|
|0 1 0|

gives 1*(2*0-1*0) - 1*(-3*0-0*0) + 5*(-3*1--2*0) = -15
area is abs(-15)/2 = 15/2


Can you please explain the shortcut please? thanks
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Re: M21-33 &nbs [#permalink] 18 Dec 2017, 19:22
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