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If these alterations decreased the area of the rectangle by 12 percent, then the value of p is 20.
In the given question,
A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent.
We have to find if these alterations decreased the area of the rectangle by 12 percent, then what is the value of p.
As we know that the area of rectangle is the multiple of length and width.
A = l×b
where A = Area of the rectangle
l = length of the rectangle
b = width of the rectangle
The area before increasing the value is
A = L×B
Since the length the increased by 10%. So the length of the new rectangle is
L(n) = L+L×10%
We can write 10% as
L(n) = L+L×10/100
L(n) = L+L×0.10
L(n) = L+0.10L
L(n) = 1.10L
From the question, the width is decreased by p%.
So W(n) = W+W×p%
W(n) = W(1− p/100)
W(n) = W(100−p/100)
Area of rectangle is decreased by 12%.
A(n) = A−A×12%
A(n) = A−A×12/100
A(n) = A−0.12A
A(n) = 0.88A...................Equation 1
We know that
A=L×W
Now putting the value of A in the Equation 1
A(n) = 0.88(L×W)...................Equation 2
The value of A(n) = L(n)×W(n)
L(n) = length of new rectangle
W(n) = width of new rectangle
Now putting the value of L(n) and W(n)
A(n) = 1.10L×W(100−p/100)
Now putting the value of A(n) in the Equation 2
1.10L×W(100−p/100) = 0.88(L×W)
Divide by L and W on both side
1.10L×W(100−p/100)/LW = 0.88(L×W)/LW
Simplifying
1.10×(100−p/100) = 0.88
Multiply by 100 on both side
1.10×(100−p/100)×100 = 0.88×100
1.10×(100−p) = 88
Using distributive property to simplify the bracket
1.10×100−1.10×p = 88
Simplifying
110−1.10p = 88
Subtract 110 on both side
110−1.10p−110=88−110
−1.10p=−22
Divide by −1.10 on both side
−1.10/−1.10 p=−22/−1.10
p = 20
Hence, if these alterations decreased the area of the rectangle by 12 percent, then the value of p is 20.
To learn more about the area of rectangle link is here
brainly.com/question/8663941
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