Answer:
Three full scoring drill can be completed in
hrs with
hrs still remaining in end of practice session.
Solution:
Given that
Time remaining before end of soccer practice in hours =
![(5)/(8)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vxddhg228csxjy2evrh0lqiufb2suv2tlh.png)
Time required for 1 full scoring drill =
![(1)/(6)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1hfsyqujpiw8kz1nwo0zeow75fj0n2oyv6.png)
Need to identify how many drills can be complete before practice is over.
For sake of simplicity lets make denominators of both the numbers same.
![(5)/(8)=(5)/(8) * (3)/(3)=(15)/(24)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/10gcd1338cpx4w0vefzg568rqfb1izdtlv.png)
![(1)/(6)=(1)/(6) * (4)/(4)=(4)/(24)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a4so6r6j3tn01r8h78clok9x0i0cri3hji.png)
is time for 1 drill
Lets say
![(4)/(24) = d](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tqp4dpbekpp78tjzht1mjvpixa5cj6au2m.png)
Time remaining =
![(15)/(24) = (4 + 4 + 4+ 3)/(24) = (4)/(24) + (4)/(24) + (4)/(24) + (3)/(24) = d + d +d + (3)/(24)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zmsrh7w9i2viat4diur3swfww92yra758q.png)
![(15)/(24) = 3d + (3)/(24)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xul0gxeiaslr1m41xgd685wahlzik6cyhg.png)
3d means three drills will be completed. Fourth drill will not complete because
is not sufficient to complete a drill.
Hence mathematical representation of given situation is
![(5)/(8) = 3d + (1)/(8)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4kvkvl4sllqfct2yw8czqgli043xtqshe3.png)
Where d is the time required to complete one drill equal to
of hour.
So three full scoring drill can be completed in
hrs with
hrs still remaining in end of practice session.