Answer:
The difference is D. 90 min
Explanation:
The distance that Linda covers (DL) in terms of time in hours (t) is:
![DL=2mph*t](https://img.qammunity.org/2020/formulas/mathematics/high-school/e0mywbqigpmdp2w7t0eajxjdkrp3mhzp2r.png)
and the distance that Tom covers (DT) in terms of time in hours (t) is:
![DT=6mph*(t-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/7nzqtvk5n49e5ol9h4cxcow3to2j5bg9dy.png)
We need to subtract 1 hour from Tom's time because he started walking an hour later than Linda.
When Tom has covered the same distance that Linda has, DT=DL, so:
![2mph*t=6mph(t-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/cdwt00ctjwdaputv9ylt06ba1d6n09uq04.png)
solving for t, we will find the time Tom needs to cover the same distance as Linda:
![2t=6t-6](https://img.qammunity.org/2020/formulas/mathematics/high-school/xh0f1i8zupiypji38vuexlzf32yh2fkldb.png)
![2t-6t=-6](https://img.qammunity.org/2020/formulas/mathematics/high-school/pk66zite3l5i66k87ugbejsltu1xrh2228.png)
![t=(6)/(4)=1.5h](https://img.qammunity.org/2020/formulas/mathematics/high-school/iplsfjexsok1id5329uw85y0mnu90n1m16.png)
Now, when Tom has covered twice the distance than Linda, DT=2DL, so:
![6mph(t-1)=2*2mph*t](https://img.qammunity.org/2020/formulas/mathematics/high-school/xer6905iuac9udw2z370bvdle0tdrlrjxf.png)
Again, solving for t:
![6t-6=4t](https://img.qammunity.org/2020/formulas/mathematics/high-school/yy929xmmuyuojjh4hepzg6msxpno5vla1g.png)
![6t-4t=6](https://img.qammunity.org/2020/formulas/mathematics/high-school/tuyf6f7wyknh8axnqzuc0zaggxl8ztxzaf.png)
![t=(6)/(2)=3h](https://img.qammunity.org/2020/formulas/mathematics/high-school/ry631qe84qzg7yse4nox2ma8ylb8liony3.png)
So, the difference between those two times is:
![3h-1.5h=1.5h=90min](https://img.qammunity.org/2020/formulas/mathematics/high-school/3tgdlhz3trwz7xk3rjqjhu44s3xxziequs.png)