A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these altercations decreased the area of the rectangle by 12 percent, what is the value of p?
Original area =L*W
New area = 1.1*(1-p/100)LW
1.1*(1-p/100) = 0.88
1-p/100 = 0.8
p = 20%
A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent.
If these altercations decreased the area of the rectangle by 12 percent,
what is the value of p?
1.1l (xw) = 0.88lw
1.1x = 0.88
x = 0.88/1.1 = 0.8
1.1*L * x*w = 0.88a
1L * 1w =1a
1.1L * [(100 – p)/100]w = 0.88Lw
110 – 1.1p = 88
p = 22/1.1 = 20 %
10% increase is a factor of 1.1
12% decrease is a factor of 0.88
(1.1L)(xW) = LW(0.88)
x = 0.8
Comments
Original area =L*W
New area = 1.1*(1-p/100)LW
1.1*(1-p/100) = 0.88
1-p/100 = 0.8
p = 20%
A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent.
If these altercations decreased the area of the rectangle by 12 percent,
what is the value of p?
1.1l (xw) = 0.88lw
1.1x = 0.88
x = 0.88/1.1 = 0.8
p = 20%
1.1*L * x*w = 0.88a
1L * 1w =1a
1.1L * [(100 – p)/100]w = 0.88Lw
110 – 1.1p = 88
p = 22/1.1 = 20 %
10% increase is a factor of 1.1
12% decrease is a factor of 0.88
(1.1L)(xW) = LW(0.88)
1.1x = 0.88
x = 0.8
p = 20%