Respuesta :
Answer:
C) I and III only
Step-by-step explanation:
Let full pool is denoted by O
Days Hose x takes to fill pool O = a
Pool filled in one day x = O/a
Days Hose y takes to fill pool O = b
Pool filled in one day y = O/b
Days Hose z takes to fill pool O = c
Pool filled in one day z = O/c
It is given that
a>b>c
[tex]a>b>c>d\\\implies x<y<z<(x+y+z)\\[/tex]
Days if if x+y+z fill the pool together = d
1 day if x+y+z fill the pool together [tex]=O(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{O}{d}---(1)[/tex]
I) d < c
d are days when hose x, y, z are used together where as c are days when only z is used so number of days when three hoses are used together must be less than c when only z hose is used. So d < c
III) [tex]\frac{c}{3}<d<\frac{a}{3}[/tex]
Using (1)
[tex]\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad(a>b>c)\\(ab+bc+ca)<3ab\\\\d=\frac{abc}{ab+bc+ca}>\frac{abc}{3ab}\\\\d>\frac{c}{3}[/tex]
Similarly
[tex]\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad a>b>c\\(ab+bc+ca)>3bc\\\\d=\frac{abc}{ab+bc+ca}<\frac{abc}{3bc}\\\\d<\frac{a}{3}[/tex]
So,
[tex]\frac{c}{3}<d<\frac{a}{3}[/tex]
