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PS Problem from GMATPrep

Expert replies
Source: — Problem Solving |

my answer

by g2000 » Sun Jan 21, 2007 3:44 pm
40 = Circumference of square + perimeter of square

We are to find the sum of areas of both.
Area of circle = pi * r^2
Area of Square = m^2 where m is the one side length of square

4*m = 40 - (2 * pi * r)
m = ( 40 - (2 * pi * r) ) / 4
m = (10 - ((pi * r ) / 2 ) )

Total area enclosed by both shapes
= area of circle + area of square
= (pi * r ^ 2) + (10 - (( pi * r)/ 2)) ^ 2

Answer is the last choice
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by Tsadori » Mon Jan 22, 2007 2:15 pm
OA: E

Thanks :-)
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by BTGmoderatorRO » Fri Oct 06, 2017 6:09 pm
let say the piece of wire was divided into two equally,
(1) circumference of the circle is 2Ï€r = 20
r = 10/π, area = πr^2 = π (10/π)^2
area of circle = 100 / π

(2) Perimeter of square = 4 * side = 20
side = 20/4 = 5, area = 5^2 = 25
the total area is 100/Ï€ + 25
Now, let us substitute the value of r= 10/ π in the answer choices and see which one gives 100/π + 25

In option A, πr^2 = π (10/π)^2 = 100/π wrong
in option B, πr^2 + 10 = 100/π +10 wrong
in option C, πr^2 + (1/4) π^2 r^2 = (100/π) + 1/4 * π^2 * (10/π)^2
= 100/Ï€ + 25 correct
in option D, πr^2 + (40 - 2πr)^2 = 100/π [40 - 2π (10/π)]^2
=100/Ï€ + (40-20)^2
100/Ï€ + 400 wrong
in option E, πr^2 + (10 - 1/2 * πr^2) = 100/π + [10- 1/2π (10/π)^2]
=100/Ï€ + (10-5)^2
100/Ï€ + 25 correct

option C and E seem to be the two correct answer to this question
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PS Problem from GMATPrep

by Brent@GMATPrepNow » Fri Oct 06, 2017 6:24 pm
A thin piece of 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in term of r?

A)πr²
B)πr² + 10
C)π)r² + 1/4(π² * r²)
D)πr² + (40 - 2[pi] * r)²
E)πr² + (10 - 1/2[πr)²
One approach is to plug in a value for r and see what the output should be.

Let's say r = 0. That is, the radius of the circle = 0
This means, we use the entire 40-meter length of wire to create the square.
So, the 4 sides of this square will have length 10, which means the area = 100

So, when r = 0, the total area = 100

We'll now plug r = 0 into the 5 answer choices and see which one yields an output of 100

A) π(0²) = 0 NOPE
B) π(0²) + 10 = 10 NOPE
C) π(0²) + 1/4(π² * 0²) = 0 NOPE
D) π(0²) + (40 - 2π0)² = 1600 NOPE
E) π(0²) + (10 - 1/2[π](0))² = 100 PERFECT!

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: PS Problem from GMATPrep

by Scott@TargetTestPrep » Fri Jan 24, 2020 3:09 pm
Tsadori wrote: ↑
Sun Jan 21, 2007 2:15 pm
SOURCE: GMATPrep
We are given that a thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius R, and the other is used to form a square.

Since the circumference of a circle with radius R is 2πR, the length of wire used to form the circle is 2πR. Thus, we have (40 – 2πR) left over to form the square. In other words, the perimeter of the square is (40 – 2πR). However, since we need to calculate the total area of the circular and the square regions, we need to determine the side of the square in terms of R. Since the perimeter of the square is (40 – 2πR), the side of the square is:

side = (40 – 2πR)/4

side = 10 – (1/2)πR

Now we can determine the areas of the circle and the square.

Area of circle = πR2

Area of square = side^2 = (10 – (1/2)πR)^2

Thus, the combined area of the circle and square is πR2 + (10 – (1/2)πR)2.

Answer: E

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