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In rectangle ABCD, the length is 60 percent of the width.
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28 Jan 2020, 10:54
28 Jan 2020, 10:54
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42% (01:20) correct
57% (01:21) wrong based on 35 sessions
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In rectangle ABCD, the length is 60 percent of the width.
Quantity A |
Quantity B |
\(\frac{1}{10}\) area of rectangle |
The length of the rectangle |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: In rectangle ABCD, the length is 60 percent of the width.
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03 Feb 2020, 10:38
03 Feb 2020, 10:38
2
Plug in numbers:
case i)
L = 6 and w = 10 --> Area = 6*10 =60
1/10 *60 = 6 = length --> both equal
case ii)
L = 12 and w = 20
12*20 = 240
1/10 *240 = 24 > 12 --> A > B
therefore the relationship cannot be determined --> D
case i)
L = 6 and w = 10 --> Area = 6*10 =60
1/10 *60 = 6 = length --> both equal
case ii)
L = 12 and w = 20
12*20 = 240
1/10 *240 = 24 > 12 --> A > B
therefore the relationship cannot be determined --> D
Re: In rectangle ABCD, the length is 60 percent of the width.
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17 Jul 2021, 04:32
17 Jul 2021, 04:32
1
Giving that rectangle has length that is .6*width.
asking which bigger.
A) Area of Rectangle/ 10
B) Length of Rectangle ".6w".
A) (.6w^2)/10
B) .6w
multiply be 10 each we get
.6w^2 (A)
6w (b)
divide by w
(A) 0.6w, (B)6
now if w=10 the quantities are equal
however if it is 100 a is bigger then
the answer is D
asking which bigger.
A) Area of Rectangle/ 10
B) Length of Rectangle ".6w".
A) (.6w^2)/10
B) .6w
multiply be 10 each we get
.6w^2 (A)
6w (b)
divide by w
(A) 0.6w, (B)6
now if w=10 the quantities are equal
however if it is 100 a is bigger then
the answer is D
