Rates are just ratios. Here are a four problems exploring rates. Remember: no calculator!
1) Someone on a skateboard is traveling 12 miles per hour. How many feet does she travel in 10 seconds? (1 mile = 5280 feet)

(A) 60
(B) 88
(C) 120
(D) 176
(E) 264
2) At 12:00 noon, a machine, operating at a fixed rate, starts processing a large set of identical items. At 1:45 p.m., the twentyfirst item has just been processed, and 15 have not yet been processed. At what time will all 36 items be processed?

(A) 2:25 pm
(B) 3:00 pm
(C) 3:27 pm
(D) 4:13 pm
(E) 5:15 pm
3) An importer wants to purchase N high quality cameras from Germany and sell them in Japan. The cost in Germany of each camera is E euros. He will sell them in Japan at Y yen, which will bring in a profit, given that the exchange rate is C yen per euro. Given the exchange rate of D US dollars per euro, which of the following represents his profit in dollars?
4) Machine A and machine B process the same work at different rates. Machine C processes work as fast as Machines A & B combined. Machine D processes work three times as fast as Machine C; Machine D’s work rate is also exactly four times Machine B’s rate. Assume all four machines work at fixed unchanging rates. If Machine A works alone on a job, it takes 5 hours and 40 minutes. If all four machines work together on the same job simultaneously, how long will it take all of them to complete it?

(A) 8 minutes
(B) 17 minutes
(C) 35 minutes
(D) 1 hour and 15 minutes
(E) 1 hours and 35 minutes
Solutions will come to these at the end of the article.
Rates
Rates are ratios, that is, fractions. Any fraction with different units in the numerator and in the denominator is a rate: miles per hour, $ per pound, grams per cubic centimeter, etc. Many rate questions can be solves by setting up an equation of the form fraction = fraction; such an equation is called a proportion. In setting up this proportion, we have to make sure that units match: the same units in the two numerators, and the same units in the two denominators. Here are a couple pertinent blogs:
b) work rate
Summary
If you understand the rules of fractions and the concept of work rate, there’s nothing about rates you can’t understand. If anything I have said here is unclear, let us know in the comments section.
1) The speed is 12 mph. To change this to feet/second, we need to multiply by (5280 ft/mile), to cancel miles, and to multiply by (1 hour/3600 second) to cancel seconds.
So, in 10 seconds, the skateboarder moves 176 feet. Answer = (D)
2) At 1:45, that is, 105 minutes after starting, the machine has completed 21/36 = 7/12 of the job. Let T be the whole time in minutes. For the total time, set up a proportion
Remember, with proportions, we can cancel a common factor in the two numerators; cancel the factor of 7.
Now, crossmultiply, and use the doubling & halving shortcut for multiplying.
T = 15*12 = 30*6 = 180
Now, 180 minutes = 3 hours, so the task finishes 3 hours later, at 3 p.m.
Answer = (B)
3) All the other currencies are related to euros, so we should focus on getting everything to euros and then changing it all at once to dollars.
Remember that profit = revenue – cost. For one camera, cost is E euros. The revenue is Y yen: let’s change that to euros, so that we can express cost, revenue, and profit all in euros.
We have an exchange rate of C yen/euro, with yen in the numerator and euros in the denominator. If we were to multiply this, we could cancel euros and wind up with yen. We don’t want that. We want to cancel yen and wind up with euros, so we need to divide by C. Y/C is the revenue in euros.
This means that (Y/C – E) is the profit in euros of one camera.
Now, the other exchange rate is D dollars/euro, with dollars in the numerator and euros in the denominator. If we multiply this, we cancel euros and get dollars. That’s exactly what we want. Thus, D(Y/C – E) is the profit, in dollars, of one camera.
Now, just multiply by the number of cameras: ND(Y/C – E)
Answer = (C)
4) Let A, B, C, and D be the rates of Machines A, B, C, and D respectively. We know that
(i) C = A + B
(ii) D = 3C
(iii) D = 4B
Starting with (ii) and (iii), equate the two expressions equal to D, and then substitute in the expression from (i) equal to C.
4B = 3C = 3(A + B) = 3A + 3B
B = 3A
Then, C = A + 3A = 4A, and D = 3*(4A) = 12A
The combined rate,
A + B + C + D = A + 3A + 4A + 12A = 20A
Since the combined rate is 20 times faster than Machine A alone, the combined time should be divided by 20.
Machine A alone takes 5 hr 40 min, or 340 minutes for the whole job. Divide this by 20:
340/20 = 17
The combination of the four machines will take 17 minutes to complete the job.
Answer = (B)
Hey Mike,
I know this post is a bit dated, but for anyone looking for a different approach for number 2, instead of using proportions (which after looking at the solution seems a bit more intuitive), I just used w=rt, and solved for rate.
We know t is going to be 7/4, and at the time the work done is 21 thingamajigs.
So we get have 21 = r*(7/4)
r= (4/7) * 21
r=3
Since already we converted time to hours, we can simply solve for the amount of time to finish 36 thingamajigs.
36 = 12*t
t=36/12
t=3
3 is in hours, and our answer would be 3 hours later from 3pm. Hope that helped someone.
Thanks, Mike. These were useful. Can you please clarify if rate should be written (let’s say Machine A’s rate) as 1 unit/340 mins or 340 mins/1 unit? I know both ways will work so long as we are consistent with other rates, but which is the technically more accurate version?
Regards,
Aviram
Dear Aviram,
I’m happy to respond. 🙂 Actually, as long as you are consistent, it doesn’t matter at all mathematically. In the real world, by convention we tend to write the rate of almost anything as (some units)/time, so time in the denominator is much more common, but in terms of pure mathematics & GMAT problem solving, it doesn’t matter in the least.
Mike 🙂
4) Let A, B, C, and D be the rates of Machines A, B, C, and D respectively. We know that
(i) C = A + B
(ii) D = 3C
(iii) D = 4B
4B = 3C = 3(A + B) = 3A + 3B
B = 3A
Question : where did B = 3A come from?
Also where did 4B = 3C = 3(A + B) = 3A + 3B come from?
Dear Sagnik,
I’m happy to explain. 🙂 One of the statements is D = 3C. Since C = A + B, we can substitute that in for C — D = 3C = 3(A + B) = 3A + 3B.
We also have another statement that says D = 4B. Well, if D = 4B and D also equals 3A + 3B, we can set those equal:
4B = 3A + 3B
Once we have that, we just subtract 3B from both sides:
B = 3A
Does all this make sense?
Mike 🙂