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Does the rectangular mirror have an area greater than \(10 \,cm^2\)?
1) The perimeter of the mirror is \(24 \,cm\).
2)The diagonal of the mirror is less than \(11 \,cm\).
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Does the rectangular mirror have an area greater than \(10 \,cm^2\)?
This topic has expert replies
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deloitte247
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$$Check\ if\ the\ area\ of\ the\ rec\tan gular\ mirror\ >10cm^2$$
Statement 1: The perimeter of the mirror is 24cm
Let length = L, and Breadth = B
Perimeter = 2 (L+B)
$$2\left(L+B\right)=24$$
$$\left(L+B\right)=\frac{24}{2}=12$$
The exact value of L and B are unknow, hence, statement 1 is NOT SUFFICIENT.
Statement 2: The diagonal of the mirror is less than 11 cm.
With a given diagonal, the rectangular mirror can be treated as 2 right-angle triangles where hypotenuse = diagonal, and the other two sides = length and breadth respectively.
Using the Pythagoras theorem=>
$$\left(diagonal\right)^2>L^2+B^2$$
$$11^2>L^2+B^2$$
$$i.e\ L^2+B^2<11^2$$
Therefore, the exact value of L and B are still unknown, hence, statement 2 is NOT SUFFICIENT.
Combining both statements together:
(L + B) = 12.......equation 1
$$L^2+B^2<121...........equation\ 2$$
Square rooting equation 1
$$\left(L+B\right)^2=12^{^2}=144.........\ equation\ 3$$
$$L^2+2LB+B^2=144.......equation\ 4$$
Subtracting equation 2 from equation 4
$$L^2+2LB+B^2=144\ -\ L^2+B^2<121=\frac{2LB}{2}<\frac{23}{2}$$
Therefore, LB<11.5 but given that diagonal was less than 11, any lesser value of the diagonal
will give a larger Area that will correlate with the fact that area LB is less than 11.5 so Area is
greater than.
Both statements together ARE SUFFICIENT
Answer = C
Statement 1: The perimeter of the mirror is 24cm
Let length = L, and Breadth = B
Perimeter = 2 (L+B)
$$2\left(L+B\right)=24$$
$$\left(L+B\right)=\frac{24}{2}=12$$
The exact value of L and B are unknow, hence, statement 1 is NOT SUFFICIENT.
Statement 2: The diagonal of the mirror is less than 11 cm.
With a given diagonal, the rectangular mirror can be treated as 2 right-angle triangles where hypotenuse = diagonal, and the other two sides = length and breadth respectively.
Using the Pythagoras theorem=>
$$\left(diagonal\right)^2>L^2+B^2$$
$$11^2>L^2+B^2$$
$$i.e\ L^2+B^2<11^2$$
Therefore, the exact value of L and B are still unknown, hence, statement 2 is NOT SUFFICIENT.
Combining both statements together:
(L + B) = 12.......equation 1
$$L^2+B^2<121...........equation\ 2$$
Square rooting equation 1
$$\left(L+B\right)^2=12^{^2}=144.........\ equation\ 3$$
$$L^2+2LB+B^2=144.......equation\ 4$$
Subtracting equation 2 from equation 4
$$L^2+2LB+B^2=144\ -\ L^2+B^2<121=\frac{2LB}{2}<\frac{23}{2}$$
Therefore, LB<11.5 but given that diagonal was less than 11, any lesser value of the diagonal
will give a larger Area that will correlate with the fact that area LB is less than 11.5 so Area is
greater than.
Both statements together ARE SUFFICIENT
Answer = C
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Jay@ManhattanReview
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Say the lengths of two legs are a and b cm, respectively; thus, the hypotenuse = √(a^2 + b^2) cm
Area of the mirror = ab/2
We have to determine whether ab/2 < 10 or ab < 20.
Let's take each statement one by one.
1) The perimeter of the mirror is \(24 \,cm\).
=> 2(a + b) = 24 => a + b = 12
Case 1: Say a = 1; thus, b = 11. Thus, ab = 1*11 = 11 < 20. The answer is yes.
Case 2: Say a = 6; thus, b = 6. Thus, ab = 6*6 = 36 > 20. The answer is no.
No unique answer. Insufficient.
2)The diagonal of the mirror is less than \(11 \,cm\).
=> √(a^2 + b^2) < 11
a^2 + b^2 < 121
Case 1: Say a = 2; thus, b = 10. Thus, (a^2 + b^2) = (2^2 + 10^2) = 104 < 121. Thus, ab = 2*10 = 20 = 20. The answer is no.
Case 2: Say a = 6; thus, b = 6. Thus, (a^2 + b^2) = (6^2 + 6^2) = 72 < 121. Thus, ab = 6*6 = 36 > 20. The answer is no.
No unique answer. Insufficient.
(1) and (2) together
We know that a + b = 12 => (a + b)^2 = 144 => a^2 + b^2 + 2ab = 144 => 2ab = 144 – (a^2 + b^2)
Again, we know that a^2 + b^2 < 121, from 2ab = 144 – (a^2 + b^2), we can write that 2ab > 144 – 121
Or, 2ab > 23 => ab > 11.5. The answer is no. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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