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The length of a rectangular canvas is increased by x percent
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25 Dec 2018, 16:43
25 Dec 2018, 16:43
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The length of a rectangular canvas is increased by x percent, and the width of the canvas is decreased by x percent.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The area of the new canvas if x = 20 |
The area of the new canvas if x = 40 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The length of a rectangular canvas is increased by x percent
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07 Jan 2019, 00:14
07 Jan 2019, 00:14
1
Let the length of the canvas be l and
Let the width of the canvas be w then,
Quantity A:
If the area is altered by x percent where x = 20 then new area would be
l(1.2) * w(0.8)
Quantity B:
If the area is altered by x percent where x = 40 then new area would be
l(1.4)* w(0.6)
We can cancel l and w from both sides to get:
Quantity A as 1.2 * 0.8
Quantity B as 1.4 * 0.6
use the onscreen calculator to get,
Quantity A = 0.96
Quantity B = 0.84
Let the width of the canvas be w then,
Quantity A:
If the area is altered by x percent where x = 20 then new area would be
l(1.2) * w(0.8)
Quantity B:
If the area is altered by x percent where x = 40 then new area would be
l(1.4)* w(0.6)
We can cancel l and w from both sides to get:
Quantity A as 1.2 * 0.8
Quantity B as 1.4 * 0.6
use the onscreen calculator to get,
Quantity A = 0.96
Quantity B = 0.84
Re: The length of a rectangular canvas is increased by x percent
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07 Jan 2019, 13:40
07 Jan 2019, 13:40
Answer: A
old_length = a
old_width = b
new_length = a + a*x/100 = a(1+x/100)
new_width = b - b*x/100 = b(1- x/100)
A: The area of the new canvas if x=20: a(1+20/100) * b(1- 20/100) = a(1+0.2) * b(1- 0.2) = ab*(1.2)*(0.8) = 0.96*ab
B: The area of the new canvas if x=20: a(1+40/100) * b(1- 40/100) = a(1+0.4) * b(1- 0.4) = ab*(1.4)*(0.6) = 0.84*ab
So A is bigger than B.
old_length = a
old_width = b
new_length = a + a*x/100 = a(1+x/100)
new_width = b - b*x/100 = b(1- x/100)
A: The area of the new canvas if x=20: a(1+20/100) * b(1- 20/100) = a(1+0.2) * b(1- 0.2) = ab*(1.2)*(0.8) = 0.96*ab
B: The area of the new canvas if x=20: a(1+40/100) * b(1- 40/100) = a(1+0.4) * b(1- 0.4) = ab*(1.4)*(0.6) = 0.84*ab
So A is bigger than B.
