Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.
Since rate times time = distance, we can set the equation as 45 t = 30 (1- t), 75t = 30 so
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.
Solution II:
Let D be the distance from Esther's home to work.
\(\frac{\Large{D}}{\Large{45}}\) + \(\frac{\Large{D}}{\Large{30}}\) = 1 (hour)
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90
D = 18 miles
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90
D = 18 miles
Solution III:
The rate ratio between driving to work and returning home is 45 : 30 or 3 : 2.
Since rate and time are inversely related (rt = d), the time ratio between the two is 2 : 3.
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles
Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.
Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)
Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel.
Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance.
So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.
Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed?
Solution I:
To find average speed, you use total distance over total time it takes Sally to drive.
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours
to drive 84 * 2 = 168 miles.
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph
Solution II:
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time? How far away is her office ?
Solution I:
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up
the following equation:
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ; 25 = 20t ; t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles
To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph
Solution II :
Again, have you noticed that if both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles
Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.
Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)
 |
Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel.
Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance.
So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.
Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed?
Solution I:
To find average speed, you use total distance over total time it takes Sally to drive.
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours
to drive 84 * 2 = 168 miles.
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph
Solution II:
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time? How far away is her office ?
Solution I:
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up
the following equation:
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ; 25 = 20t ; t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles
To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph
Solution II :
Again, have you noticed that if both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
