[Math Revolution GMAT math practice question]
What is the perimeter of a rectangle?
1) The square of the diagonal is 52.
2) The area of the rectangle is 24.
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What is the perimeter of a rectangle?
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Max@Math Revolution
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Brent@GMATPrepNow
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Target question: What is the perimeter of a rectangle?Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
What is the perimeter of a rectangle?
1) The square of the diagonal is 52.
2) The area of the rectangle is 24.
This is a good candidate for rephrasing the target question.
Let x = the length of the rectangle's base
Let y = the length of the rectangle's height
So, the perimeter of the rectangle = 2x + 2y
REPHRASED target question: What is the value of 2x + 2y?
Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
Statement 1: The square of the diagonal is 52
In other words, (length of diagonal)² = 52
We can create a RIGHT triangle with the base, height and diagonal.
As such, we can apply the Pythagorean Theorem to write: x² + y² = 52
Is this information (x² + y² = 52) enough to determine the value of 2x + 2y?
NO.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 4 and y = 6. Notice that x² + y² = 4² + 6² = 52. In this case, the answer to the REPHRASED target question is 2x + 2y = 8 + 12 = 20
Case b: x = √51 and y = 1. Notice that x² + y² = (√51)² + 1² = 52. In this case, the answer to the REPHRASED target question is 2x + 2y = 2√51 + 2
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: The area of the rectangle is 24
In other words, xy = 24
Is this information enough to determine the value of 2x + 2y?
NO.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 4 and y = 6. Notice that xy = (4)(6) = 24. In this case, the answer to the REPHRASED target question is 2x + 2y = 8 + 12 = 20
Case b: x = 2 and y = 12. Notice that xy = (2)(12) = 24. In this case, the answer to the REPHRASED target question is 2x + 2y = 4 + 24 = 28
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that x² + y² = 52
Statement 2 tells us that xy = 24, which also means 2xy = 48
Add the two red equations to get: x² + 2xy + y² = 100
Factor the left side to get: (x + y)² = 100
This means that: x + y = 10
Multiply both sides by 2 to get: 2x + 2y = 20
Perfect!!
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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fskilnik@GMATH
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\[? = {\text{perim}}\left( {{\text{rectangle}}} \right)\]Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
What is the perimeter of a rectangle?
1) The square of the diagonal is 52.
2) The area of the rectangle is 24.
Excellent opportunity to GEOMETRICALLY BIFURCATE each statement alone, as presented below:
\[\left( 1 \right)\,\,\,{\text{dia}}{{\text{g}}^{\,{\text{2}}}} = 52\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{diag}}\,\, > \,\,0} \,\,\,{\text{diag}}\,\,{\text{unique}}\,\,\,{\text{but}}\,\,\,{\text{INSUFF}}.\]
\[\left( 2 \right)\,\,\,{\text{area}} = 24\,\,\,\,\, \Rightarrow \,\,\,\,{\text{INSUFF}}{\text{.}}\]
Let L and W be the length and width of our focused-rectangle. Hence:
\[? = {\text{2}}\left( {L + W} \right)\]
\[\left( {1 + 2} \right)\,\,\left\{ \begin{gathered}
{L^2} + {W^2} = 52 \hfill \\
2LW = 2 \cdot 24\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,{\left( {L + W} \right)^2} = 52 + 48 = 100\]
\[{\left( {L + W} \right)^2} = 100\,\,\,\,\mathop \Rightarrow \limits^{L + W\,\, > \,\,0} \,\,\,\,L + W\,\,\,{\text{unique}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 2\left( {L + W} \right)\,\,\,\,{\text{unique}}\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Forget conventional ways of solving math qAnswer: CAnswer: Cuestions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
When we apply VA method to geometry, we need to count the number of variables. For a rectangle, we have two variables for the length and the width of the rectangle. Let x and y be the length of the width of the rectangle, respectively.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have x^2+y^2 = 52 by Pythagoras' theorem, and Area = xy = 24.
So, (x+y)^2 = x^2+2xy + y^2 = (x^2+y^2) +2xy = 52 + 48 = 100.
Therefore, x+y = 10 and we can calculate the perimeter.
Both conditions (together) are sufficient.
Therefore, C is the answer.
Answer: C
Forget conventional ways of solving math qAnswer: CAnswer: Cuestions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
When we apply VA method to geometry, we need to count the number of variables. For a rectangle, we have two variables for the length and the width of the rectangle. Let x and y be the length of the width of the rectangle, respectively.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have x^2+y^2 = 52 by Pythagoras' theorem, and Area = xy = 24.
So, (x+y)^2 = x^2+2xy + y^2 = (x^2+y^2) +2xy = 52 + 48 = 100.
Therefore, x+y = 10 and we can calculate the perimeter.
Both conditions (together) are sufficient.
Therefore, C is the answer.
Answer: C
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