Answer:
[tex]66\dfrac{2}{3}\%[/tex]
Step-by-step explanation:
Given,
The number of cartons = k,
Time taken by machine a = 8 hours,
So, the number of cartons made by machine a in one hour
= [tex]\frac{\text{Cartons in 8 hours}}{8}[/tex]
= [tex]\frac{k}{8}[/tex]
Time taken by machine b = 4 hours ,
So, the number of cartons made by machine b in one hour
= [tex]\frac{k}{4}[/tex]
Total cartons made in 1 hour = [tex]\frac{k}{8}+\frac{k}{4}[/tex]
[tex]=\frac{k+2k}{8}[/tex]
[tex]=\frac{3k}{8}[/tex]
∵ for the whole number value of k,
[tex]\frac{k}{4}>\frac{k}{8}[/tex]
i.e. machine b is faster,
Also, the percent of the cartons were sealed by the machine b
[tex]=\frac{\text{Cartons made in 1 hour by machine b}}{\text{Total cartons}}\times 100[/tex]
[tex]=\frac{\frac{k}{4}}{\frac{3k}{8}}\times 100[/tex]
[tex]=\frac{2}{3}\times 100[/tex]
[tex]=\frac{200}{3}[/tex]
[tex]=66\dfrac{2}{3}\%[/tex]
Hence, OPTION D is correct.
