Gmat_mission wrote: ↑

Thu Dec 17, 2020 12:56 pm

Every day, Tom walks from his home to his office, via the same route, covering \(s\) feet at a speed of \(x\) feet per minute. Today he took a different route and ended up walking \(10\%\) more than he usually does, at a speed of \(x\) meters per minute. What is the percentage change in the time he took today compared to the time he takes on a usual day? (1 feet = 0.3 meter)

A. \(10\%\) decrease

B. \(10\%\) increase

C. \(67\%\) decrease

D. \(67\%\) increase

E. \(200\%\) decrease

Answer: C

Solution:

Let’s let m = the time, in minutes, that it takes Tom to walk s feet at a rate of x ft/min. Thus, we have the following distance equation for Tom’s usual day:

x * m = s

m = s/x

Tom’s rate today - x meters/min - is equivalent to x/0.3 ft/min, and we use this value in his distance equation for today. Letting n = the time, in minutes, that it takes him to walk 10% more distance than normal (1.1s feet) at a rate of x/0.3 ft/min, we have today’s equation as:

x/0.3 * n = 1.1s

n = 1.1s / (x/0.3)

n = 0.33s/x

To determine the percent change in the time it takes him today, compared to his usual time, we use the percent change formula (New - Old)/Old * 100:

(0.33s/x - s/x) / (s/x) *100

(0.33s - s)/s *100

(0.33 - 1)/1 *100 = - 67 percent

Thus, today, Tom took 67% less time than on a usual day.

Answer: C