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If the length of a rectangle were increased by 20 percent and the widt
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18 Feb 2021, 04:17
18 Feb 2021, 04:17
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If the length of a rectangle were increased by 20 percent and the width were decreased by 20 percent, what is the ratio of the original area to the new area?
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: If the length of a rectangle were increased by 20 percent and the widt
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18 Feb 2021, 06:26
18 Feb 2021, 06:26
1
Carcass wrote:
If the length of a rectangle were increased by 20 percent and the width were decreased by 20 percent, what is the ratio of the original area to the new area?
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
Let the dimensions of the ORIGINAL rectangle be 10 by 10 [Yes, a square is a type of rectangle]
So, the area of the ORIGINAL rectangle = (base)(height) = (10)(10) = 100
If the length of a rectangle were increased by 20 percent and the width were decreased by 20 percent...
So, the NEW rectangle has dimensions 120 by 80
So, the area of the NEW rectangle = (base)(height) = (12)(8) = 96
...what is the ratio of the original area to the new area?
The ratio = 100 : 96
Divide both sides by 4 to get the equivalent ratio: 25 : 24
Answer: E
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Status:Founder & Quant Trainer
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Re: If the length of a rectangle were increased by 20 percent and the widt
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18 Feb 2021, 06:28
18 Feb 2021, 06:28
Carcass wrote:
If the length of a rectangle were increased by 20 percent and the width were decreased by 20 percent, what is the ratio of the original area to the new area?
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
A. 4:3
B. 5:4
C. 10:9
D. 15:13
E. 25:24
Approach 1:
Let;
Original Length be L
Original Width be W
Area = LW
Then,
New Length = 1.2L
New Width = 0.8W
Area = 0.96LW
Ratio = \(\frac{LW}{0.96LW} = \frac{100}{96} = \frac{50}{48} = \frac{25}{24}
\)
Hence, option E
Approach 2: Assume the values
Original Length = 100
Original Width = 100
Area = 10,000
Then,
New Length = 120
New Width = 80
Area = 9,600
Ratio = \(\frac{10,000}{9,600LW} = \frac{100}{96} = \frac{50}{48} = \frac{25}{24}
\)
