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H: 14
V: 6
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Difficulty:
45%
(medium)
Question Stats:
68% (02:04) correct
32%
(02:27)
wrong
based on 1016
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In the standard (x,y) coordinate plane, the graph of \(|3x| + |7y| = 21\) is a parallelogram with 2 vertices on the x-axis and 2 vertices on the y-axis. Let H and V be the lengths, in coordinate units, of the horizontal and vertical diagonals, respectively, of this parallelogram.
Select for H and for V values that are consistent with the given information. Make only two selections, one in each column.
DI -3.png [ 45.75 KiB | Viewed 10345 times ]
Select for H and for V values that are consistent with the given information. Make only two selections, one in each column.
Attachment:
DI -3.png [ 45.75 KiB | Viewed 10345 times ]
| H | V | |
| 4 | ||
| 6 | ||
| 8 | ||
| 10 | ||
| 14 | ||
| 20 |
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Official Answer
H: 14
V: 6
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19 Jan 2024, 05:52
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yikes000
The question tells us that the points make a parallelogram.
Even the solution of equation will give two intersection on x axis at equal distance from origin and similarly 2 points on y axis at equal distance from origin.
One of the diagonals will be formed by y-axis intersection and the other by x-axis intersection.
Let us solve it
\(|3x| + |7y| = 21\) Take y as 0 to find intersection at y-axis. |3x|=21 or x=7 and -7.
Horizontal Diagonal = 7-(-7)=14
\(|3x| + |7y| = 21\) Take x as 0 to find intersection at x-axis. |7y|=21 or y=3 and -3.
Vertical Diagonal = 3-(-3)=6
sukoon9334, you are correct with your method.
General Discussion
sukoon9334
Joined: 22 Oct 2019
Last visit: 01 Feb 2024
Posts: 29
Given Kudos: 70
Location: India
Schools: HEC Montreal '21
GMAT 1: 660 Q50 V30
19 Jan 2024, 01:37
Kudos
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It's an 80% inequalities question and 20% coordinate planes.
|3x|+|7y|=21 can be written as |7y|=21-|3x|
Now, since |7y| is always zero or positive, hence, 21 - |3x| will also always be zero or positive.
This means that 0≤ |3x|≤ 21
Case 1: 3x ≤ 21
=> x≤7
Case 2: -3x≤21
=> x≥-7
Case 3: 0≤ |3x|
=> x=0
Analysis I:
|7y|=0 for both extremes i.e. x=7 and x=-7
This implies, y=0
So, our two coordinate points of the parallelogram are (7,0) and (-7,0)
Analysis II:
|7y|=21 for x=0
This creates two more cases:
Case 3: 7y≤ 21
=> y≤ 3
Case 4: -7y≤ 21
=> y≥ -3
So, our other two coordinates for the parallelogram are (0,3) and (0,-3)
Now, when we plot these 4 coordinates on the coordinate plane, we get a shape whose diagonals are the x-axis and y-axis itself.
V=distance from (0,3) to (0,-3) = 6 units
H= distance from (7,0) to (-7,0) = 14 units
This might seem daunting. I felt the same when I saw it and got it incorrect in the mock test but when I sat down later, I got this one in one shot. All the best.
Bunuel approve this, please.
|3x|+|7y|=21 can be written as |7y|=21-|3x|
Now, since |7y| is always zero or positive, hence, 21 - |3x| will also always be zero or positive.
This means that 0≤ |3x|≤ 21
Case 1: 3x ≤ 21
=> x≤7
Case 2: -3x≤21
=> x≥-7
Case 3: 0≤ |3x|
=> x=0
Analysis I:
|7y|=0 for both extremes i.e. x=7 and x=-7
This implies, y=0
So, our two coordinate points of the parallelogram are (7,0) and (-7,0)
Analysis II:
|7y|=21 for x=0
This creates two more cases:
Case 3: 7y≤ 21
=> y≤ 3
Case 4: -7y≤ 21
=> y≥ -3
So, our other two coordinates for the parallelogram are (0,3) and (0,-3)
Now, when we plot these 4 coordinates on the coordinate plane, we get a shape whose diagonals are the x-axis and y-axis itself.
V=distance from (0,3) to (0,-3) = 6 units
H= distance from (7,0) to (-7,0) = 14 units
This might seem daunting. I felt the same when I saw it and got it incorrect in the mock test but when I sat down later, I got this one in one shot. All the best.
Bunuel approve this, please.
