M21-33 : GMAT Club Tests

vik09

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Re: M21-33 [#permalink]

20 Jun 2015, 22:16

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:

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

vik09

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Re: M21-33 [#permalink]

21 Jun 2015, 00:01

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:

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.

L

chetan2u

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Re: M21-33 [#permalink]

21 Jun 2015, 00:16

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Expert Reply

@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

vik09

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Re: M21-33 [#permalink]

21 Jun 2015, 03:14

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

Senthil7

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Re: M21-33 [#permalink]

02 Aug 2016, 04:47

I think this is a high-quality question and I agree with explanation.

B

mbadude2017

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Re: M21-33 [#permalink]

22 Feb 2017, 15:58

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:

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

S

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Re: M21-33 [#permalink]

18 Dec 2017, 03:20

+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.

B

pinguuu

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Re: M21-33 [#permalink]

18 Dec 2017, 20:22

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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

L

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Re: M21-33 [#permalink]

21 Nov 2019, 03:54

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pinguuu wrote:

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

Take each value x, y, constant and form a matrix and divide by 2

B

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Re: M21-33 [#permalink]

14 May 2020, 09:55

Hey!

I struggle in Coordinate Geometry.
So, my question might be dumb.

Aren't there 2 triangles being formed in the figure. The area of once is coming as 7.5 & the other's 5. How do I decide which one to pick

L

Bunuel

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Re: M21-33 [#permalink]

14 May 2020, 10:01

Expert Reply

SpecterLitt wrote:

Hey!

I struggle in Coordinate Geometry.
So, my question might be dumb.

Aren't there 2 triangles being formed in the figure. The area of once is coming as 7.5 & the other's 5. How do I decide which one to pick

Check the image below:

Three lines \(y = 5-x\) (blue line), \(2y = 3x\) (green line), and \(y = 0\) (red line) form only one triangle.

Show Spoiler

Attachment:

Untitled.png [ 11.83 KiB | Viewed 5093 times ]

S

davidbeckham

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Re: M21-33 [#permalink]

08 Aug 2020, 06:03

Hi Bunuel, coordinate geometry is my weak area, I didn't understand how did you derive the coordinates from the equations - could you please explain?

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

L

Bunuel

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Re: M21-33 [#permalink]

08 Aug 2020, 06:25

Expert Reply

davidbeckham wrote:

Hi Bunuel, coordinate geometry is my weak area, I didn't understand how did you derive the coordinates from the equations - could you please explain?

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

24. Coordinate Geometry

Theory
Math: Coordinate Geometry
Beauty of Coordinate geometry
Dealing with Tangents on the GMAT

Questions
DS Questions
PS Questions

For more check:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

gmatclubot

Re: M21-33 [#permalink]

08 Aug 2020, 06:25