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A thin piece of wire 40 metres long...

This topic has 3 expert replies and 2 member replies
shubh425 Junior | Next Rank: 30 Posts Default Avatar
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19 Jan 2016
Posted:
20 messages

A thin piece of wire 40 metres long...

Post Sun Jun 26, 2016 10:30 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Please explain,which approach will better, whether to use algebra or to plugin numbers??

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    800_or_bust Master | Next Rank: 500 Posts Default Avatar
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    Post Sun Jun 26, 2016 10:37 am
    shubh425 wrote:
    Please explain,which approach will better, whether to use algebra or to plugin numbers??

    Given that the circle has a radius of r units, its circumference must be 2(pi)r. And the remaining wire makes up the perimeter of the square, i.e. 40 - 2(pi)r. Since a square consists of four equal sides, we can determine the length of a side by dividing the perimeter by 4. Thus, each side is 10 - (pi)r/2. And the area of the square is given by (10 - (pi)r/2)^2. The area of the circle is (pi)r^2. And the total area is (pi)r^2 + (10 - (pi)r/2)^2, which is answer choice E.

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    800_or_bust Master | Next Rank: 500 Posts Default Avatar
    Joined
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    Posted:
    199 messages
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    4 members
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    Test Date:
    7/9/2016
    GMAT Score:
    780
    Post Sun Jun 26, 2016 10:47 am
    shubh425 wrote:
    Please explain,which approach will better, whether to use algebra or to plugin numbers??

    Would definitely rely on an algebraic approach. You would need to do a lot of math if you chose numbers. Because you'd have to compare your result with each answer choice. The algebra is not too terrible on this one. I know sometimes it gets pretty bad, especially with compounding interest and things of that sort.

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    Post Sun Jun 26, 2016 11:19 am
    Quote:
    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)(pi)r²
    B)(pi)r² + 10
    C)(pi)r² + 1/4([pi]² * r²)
    D)(pi)r² + (40 - 2[pi] * r)²
    E)(pi)r² + (10 - 1/2[pi] * r)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) (pi)(0²) = 0 NOPE
    B) (pi)(0²) + 10 = 10 NOPE
    C) (pi)(0²) + 1/4([pi]² * 0²) = 0 NOPE
    D) (pi)(0²) + (40 - 2[pi]0)² = 1600 NOPE
    E) (pi)(0²) + (10 - 1/2[pi](0))² = 100 PERFECT!

    Answer: E

    Cheers,
    Brent

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    Post Sun Jun 26, 2016 11:19 am
    Quote:
    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)(pi)r²
    B)(pi)r² + 10
    C)(pi)r² + 1/4([pi]² * r²)
    D)(pi)r² + (40 - 2[pi] * r)²
    E)(pi)r² + (10 - 1/2[pi] * r)²
    Here's an algebraic approach:

    Since r is the radius of the circle, the area of the circle will be (pi)r²

    If r is the radius of the circle, the length of wire used for this circle will equal its circumference which is 2(pi)r

    So, the length of wire to be used for the square must equal 40 - 2(pi)r

    In other words, the perimeter of the square will be 40 - 2(pi)r

    Since squares have 4 equal sides, the length of each side of the square will be [40 - 2(pi)r]/4, which simplifies to be 10 - (pi)r/2

    If each side of the square has length 10 - (pi)r/2, the area of the square will be [10 - (pi)r/2

    So, the total area will equal (pi)r² + [10 - (pi)r/2]², which is the same as E

    Cheers,
    Brent

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    Post Sun Jun 26, 2016 11:19 am
    We demonstrated the two methods (Algebraic and Input-Output) for solving a question type I call Variables in the Answer Choices.
    If you'd like more information on these approaches, we have some videos:
    - Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat-word-problems/video/933
    - Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat-word-problems/video/934
    - Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat-word-problems/video/935

    Here are some more questions to practice with:
    - http://www.beatthegmat.com/what-mistake-am-i-making-in-this-substitution-last-sunday-a-t276293.html
    - http://www.beatthegmat.com/car-dealer-sales-t274136.html
    - http://www.beatthegmat.com/ps-rate-times-t276107.html
    - http://www.beatthegmat.com/to-find-the-least-perfect-square-t273338.html
    - http://www.beatthegmat.com/y-dollars-from-the-rental-venture-t273031.html
    - http://www.beatthegmat.com/meters-in-x-minutes-t273854.html
    - http://www.beatthegmat.com/a-better-explanation-please-t279396.html


    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

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