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# If the width of the rectangle above is increased by 50 percent and the

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Math Expert
Joined: 02 Sep 2009
Posts: 42652

Kudos [?]: 135984 [0], given: 12719

If the width of the rectangle above is increased by 50 percent and the [#permalink]

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04 Oct 2017, 10:17
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(N/A)

Question Stats:

96% (00:24) correct 4% (00:00) wrong based on 23 sessions

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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

[Reveal] Spoiler:
Attachment:

2017-10-04_1124.png [ 1.53 KiB | Viewed 442 times ]
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135984 [0], given: 12719

Manager
Status: Preparing
Joined: 05 May 2016
Posts: 62

Kudos [?]: 12 [0], given: 150

Location: India
Concentration: International Business, Finance
Re: If the width of the rectangle above is increased by 50 percent and the [#permalink]

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04 Oct 2017, 11:05
Bunuel wrote:

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

[Reveal] Spoiler:
Attachment:
2017-10-04_1124.png

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

Kudos [?]: 12 [0], given: 150

VP
Joined: 22 May 2016
Posts: 1140

Kudos [?]: 408 [0], given: 648

If the width of the rectangle above is increased by 50 percent and the [#permalink]

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04 Oct 2017, 14:25
Bunuel wrote:

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

[Reveal] Spoiler:
Attachment:
2017-10-04_1124.png

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

Kudos [?]: 408 [0], given: 648

If the width of the rectangle above is increased by 50 percent and the   [#permalink] 04 Oct 2017, 14:25
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# If the width of the rectangle above is increased by 50 percent and the

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