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04 Oct 2017, 11:17
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
94% (00:44) correct
6%
(00:09)
wrong
based on 43
sessions
History
Date
Time
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Not Attempted Yet
If the width of the rectangle above is increased by 50 percent and the length remains the same, by what percent will the area of the rectangle be increased?
(A) 50%
(B) 100%
(C) 150°
(D) 200°
(E) cannot be determined from the information given
Attachment:
2017-10-04_1124.png [ 1.53 KiB | Viewed 1870 times ]
04 Oct 2017, 12:05
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Bunuel
New area will be 1.5xy where x is width and y is length.
Old area is xy.
(1.5xy-xy)/xy*100=50%
Answer should be A
04 Oct 2017, 15:25
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Bunuel
Assume x = 10 and y = 4. (Choose easily multiplied numbers. For y, choose a number that, when multiplied by 1.5 or \(\frac{3}{2}\), yields an integer.)
Original area = (L * W) = 40
Width increases by 50 percent. New width is \((4)*\frac{3}{2} = 6\)
New area = (L * W) = 10 * 6 = 60
Percent increase in area =
\(\frac{New - Old}{Old} * 100\)
\(\frac{60 - 40}{40} =(\frac{20}{40}) * 100\) = 50 percent
ANSWER A
Algebra might be faster (not by much). Typically, both lengths increase. There are more multipliers, often squared. Here, just one length increases:
Original area = (y)(x)
New area = (1.5y)(x)
Percent increase:
\(\frac{1.5yx - 1yx}{yx} =\\
\frac{.5}{1} * 100\) = 50 percent
