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

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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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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.
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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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New post 15 Jan 2020, 04:03
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The question misses the figure.
Request you to please revise the question.
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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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New post 16 Jan 2020, 02:33
Hi, can anyone help with the solution?
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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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New post 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
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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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New post 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
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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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New post 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 \(a = 2.\)

Therefore, C is the answer.
Answer: C


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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New post 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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New post 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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New post 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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New post 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 = \(4a+16\) = \(24\)

\(4a=24-16\)
\(4a=8\)
\(a=2\)

Answer is (C)
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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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