Relative Range Distance Rate Time GMAT, GRE Problem

The distance between Mercury and Earth changes due to the orbits of the planets. When Mercury is at its closest point to Earth, it is 48 million miles away. When Mercury is at its furthest point from Earth, it is 138 million miles away. For a science project, Ruby calculates the maximum and minimum amount of time it would take to travel from Earth to Mercury in a spacecraft traveling 55 miles per hour. Approximately what are the times, in days?

(A) 3,636 and 10,454
(B) 14,545 and 41,818
(C) 36,364 and 104,545
(D) 87,272 and 250,909
(E) 872,727 and 2,509,091

Following the solution will be a refresher on rate time distance problems.

What we know:

• When Mercury is at its closest point to Earth the distance is 48 million miles
• When Mercury is at its furthest point from Earth the distance is 138 million miles
• The speed of spacecraft is 55 miles per hour

What we need to find:

• The maximum and minimum time the spacecraft will take to travel from Earth to Mercury

Closest distance = 48 million miles = 48000000 miles

• Therefore, the minimum time required = 4800000055hours=872727hours=87272724days=36363days

Furthest distance = 138 million miles = 138000000 miles

• Therefore, the maximum time required = 13800000055hours=2509090hours=250909024days=104545days

C.

Distance word problems can almost all be solved using the d = rt formula, where d stands for distance, r stands for the average rate of speed, and t stands for time.

Make sure that the units for time and distance are the same units for the rate. If they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. If they use yards or something else that doesn't match most of the units, convert everything to the units that the answer choices are written in.

Let's look at an example of a distance question and use a chart to parse out the parts of d=rt.

A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

	d	r	t
part 1	d	105	t
part 2	555 – d	115	5 – t
total	555	---	5



Using d = rt, the first row gives me d = 105t, and the second row gives us

555 – d = 115(5 – t)

Since the two distances add up to 555, add the two distance expressions, and set their sum equal to the given total

555 = 105t + 115(5 – t)

555 = 105t + 575 – 115t
555 = 575 – 10t
–20 = –10t
2 = t




According to the grid, t stands for the time spent on the first part of the trip, so the answer is the plane flew for 2 hours at 105 mph and 3 hours at 115 mph.

If you're at a loss for what you're looking for, simply write d=rt as below and cover up the term you want to find. The equation is what remains visible.