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M21-33

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Math Expert
Joined: 02 Sep 2009
Posts: 51101

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16 Sep 2014, 00:15
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Difficulty:

35% (medium)

Question Stats:

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

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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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51101

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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:

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

_________________
Intern
Joined: 08 Mar 2014
Posts: 47
Location: United States
GMAT Date: 12-30-2014
GPA: 3.3

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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:

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

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

Regards & Thanks
Math Expert
Joined: 02 Aug 2009
Posts: 7101

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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:

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

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
_________________

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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Intern
Joined: 08 Mar 2014
Posts: 47
Location: United States
GMAT Date: 12-30-2014
GPA: 3.3

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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:

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

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.
Math Expert
Joined: 02 Aug 2009
Posts: 7101

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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
_________________

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

GMAT online Tutor

Intern
Joined: 08 Mar 2014
Posts: 47
Location: United States
GMAT Date: 12-30-2014
GPA: 3.3

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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

Thanks
Senior Manager
Joined: 31 Mar 2016
Posts: 385
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

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02 Aug 2016, 03:47
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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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:

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

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
Senior Manager
Joined: 08 Jun 2015
Posts: 436
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

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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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" The few , the fearless "

Intern
Joined: 07 Oct 2014
Posts: 3

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18 Dec 2017, 19:22
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 $$Area=\frac{1}{2}*Base*Height=7.5$$.

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

Re: M21-33 &nbs [#permalink] 18 Dec 2017, 19:22
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