Question

I’m stuck with my hone work and I really need some help

Answer 1

Given the time taken by the boys in a relay race.

First, make the denominator of the given time of all boys to 12.

Time took by James

`\begin{gathered} =(2)/(3) \\ =(2)/(3)\cdot(4)/(4) \\ =(8)/(12) \end{gathered}`

Time took by Gilbert = 11/12

Time took by Matthew

`\begin{gathered} =(5)/(6) \\ =(5)/(6)\cdot(2)/(2) \\ =(10)/(12) \end{gathered}`

Time took by SImon = 7/12

1. In the given time, fastest boy is Gilbert (time with highest numerator) and the slowst boy is Simon ( time with lowest numerator).

The difference between the fastest boy's time and the lowest boy's time

`\begin{gathered} =(11)/(12)-(7)/(12) \\ =(11-7)/(12) \\ =(4)/(12) \\ =(1)/(3) \end{gathered}`

2. Time taken by the boys for the relay race equals the sum of time each boy taken. So. time taken for the relay race

`\begin{gathered} =(8)/(12)+(11)/(12)+(10)/(12)+(7)/(12) \\ =(8+11+10+7)/(12) \\ =(36)/(12) \\ \end{gathered}`

School record time for the relay race

`\begin{gathered} =2(7)/(12) \\ =(2\cdot12+7)/(12) \\ =(31)/(12) \end{gathered}`

Since 36 > 31, so the boys taken more time than the school's record.

Time slower were the boys than school record

`\begin{gathered} =(36)/(12)-(31)/(12) \\ =(36-31)/(12) \\ =(5)/(12) \end{gathered}`

The boys are 5/12 units slower than the school's record.

3. Given that:

Gallons of water the boys have to share that day

`\begin{gathered} =4(5)/(6) \\ =(4\cdot6+5)/(6) \\ =(29)/(6) \end{gathered}`

Gallons of water James drank

`\begin{gathered} =1(1)/(3) \\ =(3\cdot1+1)/(3) \\ =(4)/(3) \\ =(8)/(6) \end{gathered}`

Gallons of water Gilbert drank = 5/6

Gallons of water Matthew drank

`\begin{gathered} =1(1)/(2) \\ =(1\cdot2+1)/(2) \\ =(3)/(2) \\ =(9)/(6) \end{gathered}`

Gallons of water Simon drank

`\begin{gathered} =(2)/(3) \\ =(2\cdot2)/(3\cdot2) \\ =(4)/(6) \end{gathered}`

Total gallons of water the boys drank

`\begin{gathered} =(8)/(6)+(5)/(6)+(9)/(6)+(4)/(6) \\ =(8+5+9+4)/(6) \\ =(26)/(6) \end{gathered}`

Gallons of water left at the end of the day

= Gallons of water to share by the boys that day - Total gallons of water the boys drank

`\begin{gathered} =(29)/(6)-(26)/(6) \\ =(29-26)/(6) \\ =(3)/(6) \\ =(1)/(2) \end{gathered}`

Hence, 1/2 gallons of water left at the end of the day.