BTGModeratorVI wrote: ↑

Tue Mar 03, 2020 6:38 am

If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?

A) (10nm)/(3n+2m)

B) (3n+2m)/10(n+m)

C) (2n+3m)/(5nm)

D) 10(n+m)/(2n+3m)

E) 5(n+m)/(2n+3m)

Answer: A
Source: Official Guide

Rate of Car A = 2/n mph; Rate of Car B = 3/m mph

Average of their rates = (2/n + 3/m)/2 = (2m + 3n)/2mn mph

Thus, time taken by Car C to travel 5 miles at the speed of (2m + 3n)/2mn mph
= 5*[(2m + 3n)/2mn] = (10nm)/(3n+2m) hours

The correct answer: A

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Tue Mar 03, 2020 6:38 am

If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?


A) (10nm)/(3n+2m)

B) (3n+2m)/10(n+m)

C) (2n+3m)/(5nm)

D) 10(n+m)/(2n+3m)

E) 5(n+m)/(2n+3m)

Answer: A
Source: Official Guide

If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?

Speed = distance/time
Speed of car A = 2/n
Speed of car B = 3/m

Speed of car C = (2/n + 3/m)/2
= (2m/mn + 3n/mn)/2
= [(2m + 3n)/mn]/(2/1)
= [(2m + 3n)/mn](1/2)
= (2m + 3n)/2mn

Which of the following expresses the time it would take Car C to travel 5 miles?
time = distance/speed
Car C's time = 5/[((2m + 3n)/2mn)]
= 5[2mn/(2m + 3n)]
= 10mn/(2m + 3n)

Answer: A

Cheers,
Brent

BTGModeratorVI wrote: ↑

Tue Mar 03, 2020 6:38 am

If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?


A) (10nm)/(3n+2m)

B) (3n+2m)/10(n+m)

C) (2n+3m)/(5nm)

D) 10(n+m)/(2n+3m)

E) 5(n+m)/(2n+3m)

Answer: A
Source: Official Guide

The average of the rates of Cars A and B is:

(2/n + 3/m)/2

[2m/(mn) + 3n/(mn)]/2

[(2m + 3n)/(mn)]/2

(2m + 3n)/(2mn)

Since time = distance/rate, it would take Car C the following amount of time to travel 5 miles:

5/[(2m + 3n)/(2mn)] = 10mn/(2m + 3n)

Answer: A