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#104 in OG 13th edition

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#104 in OG 13th edition

by cfleck » Mon Feb 02, 2015 4:08 pm
Can someone please explain to me the estimation technique used in the answer to this OG question?

If a square mirror has a 20-inch diagonal what is the closest approximate perimeter of the mirror, in inches?
A- 40
B- 60
C- 80
D- 100
E- 120

I solved for each side, but I didn't understand the OG's explanation for how to estimate to get 60 as an answer.

Thanks in advance!
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by GMATGuruNY » Mon Feb 02, 2015 4:19 pm
cfleck wrote:Can someone please explain to me the estimation technique used in the answer to this OG question?

If a square mirror has a 20-inch diagonal what is the closest approximate perimeter of the mirror, in inches?
A- 40
B- 60
C- 80
D- 100
E- 120
The diagonal of a square with side s = s√2.
Every test-taker should know that √2 ≈ 1.4.

We can PLUG IN THE ANSWERS, which represent the perimeter of the mirror.
When the correct answer is plugged in, the diagonal ≈ 20.

Answer choice D: 100
Since p=100, s=25.
Too big:
The side of the square cannot be longer than the diagonal.
Since the correct answer must be smaller, eliminate D and E.

Answer choice B: 60
Since p=60, s=15.
Diagonal = 15√2 ≈ (15)(1.4) = 21.
Success!

The correct answer is B.
Last edited by GMATGuruNY on Mon Feb 02, 2015 4:42 pm, edited 2 times in total.
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by Brent@GMATPrepNow » Mon Feb 02, 2015 4:21 pm
cfleck wrote: If a square mirror has a 20-inch diagonal what is the closest approximate perimeter of the mirror, in inches?
A- 40
B- 60
C- 80
D- 100
E- 120
Let x = the length of each side of the square.

If we draw a diagonal, we create a RIGHT TRIANGLE with a hypotenuse (diagonal) with length 20
Since, we have a RIGHT TRIANGLE we can apply the Pythagorean Theorem.
The two legs have length x, so we can write: x² + x² = 20²
Simplify to get: 2x² = 400
Divide both sides by 2 to get: x² = 200
This means x = √200

NOTE: the questions asks for an APPROXIMATION and the answer choices are quite SPREAD APART.
So, we won't wasted time finding a precise value for √200

Notice that √196 = 14 and √225 = 15
Since 200 is BETWEEN 196 and 225, we can conclude that √200 is BETWEEN 14 and 15.
So, √200 = 14.something
In other words, x = 14.something

Since x = the length of each side of the square, the PERIMETER = (4)(14.something)
ASIDE: (4)(14) = 56, so (4)(14.something) = a bit more than 56

Choose B [it's the best answer]

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by ceilidh.erickson » Sun Feb 08, 2015 12:53 pm
With PS questions, it's worthwhile to take stock of the answer choices and try to eliminate a few before solving. Since the question asks us for the perimeter of the square, ask yourself what the side length of the square would be for each of the answer choices:
A) 40 -> 10
B) 60 -> 15
C) 80 -> 20
D) 100 -> 25
E) 120 -> 30

Since we know that the diagonal of a square can't be equal to or less than the side length, we can eliminate C, D, and E.

Now test the other two answer choices. If the side length were 10, the diagonal would be 10√2. Even if you didn't know that √2 ≈ 1.4, you know that it's significantly less than 2, which would be needed to get 10*2=20.

The answer must be B.
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by DavidG@VeritasPrep » Sun Feb 08, 2015 1:59 pm
Another fun little shortcut:

Anytime you're going from the 45 side of a 45:45:90 triangle to the 90 side, you're multiplying by root 2, or about 1.4.

Anytime you're going from the 90 side to the 45 side, you're multiplying by (root 2)/2, or about .7.

If hypotenuse is 20, each side is 20 * .7 = 14. 14*4 = 56. Closest is 60.
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by Matt@VeritasPrep » Sun Feb 08, 2015 10:15 pm
Here's one last quick algebraic approach.

Suppose each side of the square has length x. We know that the diagonal of the square has length x√2, so x√2 = 20 and x = 20/√2. We want the perimeter, which is 4x, or 4 * (20/√2), or 80/√2.

Now we'll simplify. 80 = 40 * 2 = 40 * √2 * √2, so

80/√2 = (40*√2*√2)/√2 = 40*√2

40√2 = 40 * (≈1.41) = ≈56

So anything close to 56 should do, and the answers are spread out enough that 60 is the best choice.
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