Question 1011261
Do we have to write linear equations?
 
WITH LINEAR EQUATIONS:
{{{x}}}= the family's speed (which for the purpose of this problem is constant), in miles per hour, and
{{{y}}}= total distance the family travels to grandma's house, in miles.
When Billy asks "Are we there yet?",
the family has been traveling for {{{1.5hours}}} (from 1:00pm to 2:30pm).
They have traveled {{{1.5x}}} miles, and their distance (in miles) to grandma's house is
{{{y-1.5x=70}}} .
When Jane asks "Are we there yet?",
the family has been traveling for {{{1.75hours}}} (from 1:00pm to 2:45pm).
They have traveled {{{1.75x}}} miles, and their distance (in miles) to grandma's house is
{{{y-1.75x=55}}} .
We need to solve the system
{{{system(y-1.5x=70,y-1.75x=55)}}}-->{{{system(y-1.5x=70,y-1.5x-(y-1.75x)=70-55)}}}-->{{{system(y-1.5x=70,y-1.5x-y+1.75x)=15)}}}-->{{{system(y-1.5x=70,0.25=15)}}}-->{{{system(y-1.5x=70,x=15/0.25=60)}}}-->{{{system(y-1.5*60=70,x=60)}}}-->{{{system(y-90=70,x=60)}}}-->{{{system(y=70+90=highlight(160),x=60)}}} .
So the family lives {{{highlight(160)}}} miles from grandma's house,
and at {{{60}}} mph it would take them
{{{160miles/"60 mph"=about2&2/3}}}{{{hours=2hours+40minutes}}} ,
so they will arrive at grandma's house at {{{3:40}}}{{{pm}}} .
 
THE FIFTH GRADER SOLUTION:
IN the {{{45-30=15}}} minutes between Billy's question and Jane's question
the family had traveled {{{70-55=15}}} miles,
so they are traveling 1 mile per minute.
When Billy asked, at 2:30pm, they were {{{70}}} miles from Grandma's house.
At {{{1}}} mile per minute, they will cover that distance in {{{70}}} minutes,
so they will arrive at grandma's house {{{70-minutes=1hour+10minutes}}} after 2:30pm.
They will arrive at {{{2:30}}}{{{pm+1:10=3:40}}}{{{pm}}} .