Respuesta :
Answer:
81.25%
Step-by-step explanation:
Given:
A fun size snickers have earlier dimensions 2 inches by 1/2 inches by 1 inches.
Today a fun size snickers has dimensions 1 1/2 inches by 1/4 inches by 1/2 inches.
Question asked:
What is the percent decrease in the volume of the snickers ?
Solution:
Old dimensions of snooker = 2 inches by [tex]\frac{1}{2}[/tex] inches by 1 inches
Volume of old snooker = [tex]length\times breadth \times height[/tex]
[tex]=2\times\frac{1}{2} \times1=\frac{2}{2} =1\ cubic \ inches[/tex]
New dimensions of snooker = [tex]1\frac{1}{2}[/tex] inches by [tex]\frac{1}{4}[/tex] inches by [tex]\frac{1}{2}[/tex]
Volume of new snooker = [tex]length\times breadth\ height[/tex]
[tex]=1\frac{1}{2}\times\frac{1}{4} \times\frac{1}{2} \\ \\ =\frac{3}{2} \times\frac{1}{4} \times\frac{1}{2}=\frac{3}{16} \ cubic\ inches[/tex]
[tex]Percent\ changed=\frac{New\ volume-old\ volume}{old\ volume} \times100[/tex]
[tex]=\frac{3}{16} -1\div1\times100\\ \\ =\frac{3-16}{16} \times100\\ \\ =\frac{-13}{16} \times100\\ \\ =\frac{-1300}{16} \\ \\ =-81.25\%[/tex]
Here negative sign shows that new volume is decreased.
Therefore, 81.25% decrease in the volume of the snickers.
