So, I know how to do this type of tasks but it takes me a lot of time to to them. Is there some quicker way of doing it, please explain? Thank you in advance.
On Thursday, Jeanine drove 20 miles to work, at an average speed of 40 miles per hour. She then drove home along the same route at an average speed of 50mi/h. How many hours did she spend driving to and from work on Thursday?
If you’re running out of time, just look for the answer that corresponds to the time that is slightly below the middle of the two speeds. In this case, 40 miles at less than 45 mph so the answer should be around 50 minutes…
(20 mi)/(40 mph) + (20 mi)/(50 mph) = 1/2 hr + 2/5 hr = 9/10 hr (units are in hours)
If you have no time, you can do as @CHD2013 suggested, but answer choices can sometimes fool you. I wouldn’t be too surprised if 8/9 hours was an answer choice (= (40 mi)/(45 mph)). But because the average speed is slightly less than 45 mph, you would want to pick a slightly larger answer.
@CHD2013 This one was an example of a grid in question. But I have found the solution.
Now I have a big problem with this one, it takes me a lot of time to come to a solution: A boy goes to school at speed of 3 mi/h, and jogs back using the same route at speed of 5 mi/h. If the total time of the trip is 1h, how long was the whole return journey?
@MITer94 Thank you!
@MITer94 Could you help me with the second question too?Thanks 
My suggestion was based on problems like the one below. Clearly, for grid-ins and when time is not a factor, @MITer94 's suggested methodology makes more sense.
A girl rides her bicycle to school at an average speed of 8 mph. She returns to her house using the same
route at an average speed of 12 mph. If the round trip took 1 hour, how many miles is the round trip.
A. 8
B. 9 3/5
C. 10
D. 11 1/5
E. 12
The answer is B, which is the highest number that’s less than the simple average of 8 and 12. Several similar problems have been used in real tests.
@Relax2954 suppose the distance from home to school is D. (it turns out that we don’t actually care what D is)
Writing the equation for time, we have D/(3 mph) + D/(5 mph) = 1 hr. Solving for D, we have 8D/15 = 1 or D = 15/8 mi.
The length of the return trip is D/(5 mph) = 3/8 hours.
A similar but slightly shorter way is to note that the time for the return trip is 3/(3+5) of the total trip since the ratio of the speeds is 5:3. So the return trip takes 3/8 hours.
For grid-in problems, it is good to use the average rate formula.
If r1=first rate, r2=second rate, average rate=
2(r1)(r2)/(r1+r2)
In words: For a round-trip, the average rate is twice the product of the two rates divided by the sum of the two rates
- For the first problem: average rate=(2x40x50)/(50+40)=4000/90=400/9
Then time=distance/rate =40 (9/400)=9/10
Using this method, all the calculations are very easy. You don’t have to add fractions with different denominators.
When you are multiplying fractions, remember to REDUCE FIRST, and MULTIPLY AFTER to keep the numbers small
-
For the boy who goes to school. the average rate is 2(3)(5)/8=15/4
Then distance=rate x time= 15/4 -
For the girl on the bicycle, the average rate is 2(8)(12)/20 =48/5
Then distance=rate x time = 48/5=9 3/5