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# Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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Updated on: 15 Jan 2020, 08:32
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55% (hard)

Question Stats:

69% (02:27) correct 31% (02:26) wrong based on 67 sessions

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[GMAT math practice question]

Point $$P$$ is on the line $$y = 2x + a.$$ The x-coordinate of $$P$$ is $$4$$. The area of the shaded quadrilateral is $$24$$. What is the value of $$a$$?

A. $$1$$

B. $$\frac{4}{3}$$

C. $$2$$

D. $$\frac{7}{6}$$

E. $$3$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 15 Jan 2020, 00:26. Last edited by MathRevolution on 15 Jan 2020, 08:32, edited 1 time in total. Intern Joined: 08 Jul 2017 Posts: 19 Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a [#permalink] ### Show Tags 15 Jan 2020, 04:03 1 The question misses the figure. Request you to please revise the question. _________________ GMAT Instructor Intern Joined: 16 Oct 2018 Posts: 30 Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a [#permalink] ### Show Tags 16 Jan 2020, 02:33 Hi, can anyone help with the solution? VP Status: Learning Joined: 20 Dec 2015 Posts: 1077 Location: India Concentration: Operations, Marketing GMAT 1: 670 Q48 V36 GRE 1: Q157 V157 GPA: 3.4 WE: Engineering (Manufacturing) Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a [#permalink] ### Show Tags 16 Jan 2020, 08:17 yashwardhan wrote: Hi, can anyone help with the solution? Hi, Calculate the area of rectangle and triangle and equate the sum of areas to the area to the quadrilateral. Area of rectangle = 4*a Area of triangle is = 1/2*4*(8+a-a)= 16 4*a + 16 = 24 4*a = 8 a = 2 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8564 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a [#permalink] ### Show Tags 17 Jan 2020, 00:07 => When we substitute $$x$$ with $$4$$, we have the point $$P(4, a+8).$$ The area of the trapezoid is $$(\frac{1}{2})(a + a + 8)*4 = 2(2a + 8) = 24$$ and we have $$2a * 8 = 12$$ or $$a = 2.$$ Therefore, C is the answer. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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25 Jan 2020, 00:38
MathRevolution wrote:
=>

When we substitute $$x$$ with $$4$$, we have the point $$P(4, a+8).$$

The area of the trapezoid is $$(\frac{1}{2})(a + a + 8)*4 = 2(2a + 8) = 24$$ and we have $$2a * 8 = 12$$ or $$a = 2.$$

Area of Trapezoid = 1/2 (sum of parallel sides * height)

Can you please explain how sum of parallel sides is a+(a+8) and height is 4?
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Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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25 Jan 2020, 10:50
How come the sum of parallel sides is a + a + 8? I understand a+8 but how come a?
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Re: Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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25 Jan 2020, 11:22
(a+a+8)*4/2=24
a=2

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Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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25 Jan 2020, 12:05
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nandeta wrote:
How come the sum of parallel sides is a + a + 8? I understand a+8 but how come a?

One of the parallel side is along the Y-axis where the side is from the origin to the Y-intercept (x=0) of the line y=2x+a and so the length of this side is calculated from 0 to a which is a.

And as you know the length of the other parallel side is a+8

So the sum of the lengths of parallel sides is a+a+8
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Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a  [#permalink]

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25 Jan 2020, 12:17
MathRevolution wrote:
[GMAT math practice question]

Point $$P$$ is on the line $$y = 2x + a.$$ The x-coordinate of $$P$$ is $$4$$. The area of the shaded quadrilateral is $$24$$. What is the value of $$a$$?

A. $$1$$

B. $$\frac{4}{3}$$

C. $$2$$

D. $$\frac{7}{6}$$

E. $$3$$

y coordinate of P when x=4 is
y=2(4)+a
y=8+a

So, P=(4,8+a)

Split the total area into

1. A rectangle with length $$4$$ and breadth $$a$$

2. A right triangle with base $$4$$ and height $$(8+a)-a = 8$$

Area of the rectangle = $$4a$$
Area of the triangle = $$\frac{1}{2}*4*8$$=$$16$$

Total Area = $$4a+16$$ = $$24$$

$$4a=24-16$$
$$4a=8$$
$$a=2$$

Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a   [#permalink] 25 Jan 2020, 12:17
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