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14 Jun 2021, 06:58
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (02:51) correct
29%
(03:09)
wrong
based on 287
sessions
History
Date
Time
Result
Not Attempted Yet
Two electric cars each completed a 300 mile race. If the faster car completed the race in 3 hours less time than did the slower car, and at an average speed 5 miles per hour greater than that of the slower car, what was the slower car’s average speed for the race?
A) 12 miles per hour
B) 20 miles per hour
C) 24 miles per hour
D) 25 miles per hour
E) 30 miles per hour
A) 12 miles per hour
B) 20 miles per hour
C) 24 miles per hour
D) 25 miles per hour
E) 30 miles per hour
14 Jun 2021, 08:23
Kudos
Bookmarks
Solution:
Slower Car:
Assume the time taken to cover 300 miles = \(t\) hours.
Average speed = \(\frac{300}{t}\) miles/hour.
Faster car:
Assume the time taken to cover 300 miles = \(t-3\) hours.
Average speed = \(\frac{300}{t-3}\) miles/hour.
The question is telling us that faster car is moving at speed 5 miles/hour greater than slower. which means:
\(\frac{300}{t-3} -5 = \frac{300}{t}\)
\(⇒ \frac{300-5(t-3)}{t-3} = \frac{300}{t}\)
\(⇒ 300t+5t^2+15t=300t-900\)
\(⇒ -5t^2+15t+900=0\)
\(⇒ t^2-3t-180=0\)
\(⇒ t^2-15t+12t-180=0\)
\(⇒ t(t-15)+15(t-15)=0\)
\(⇒ t = 15\) or \(-12\).
Time cannot be negative. Thus \(t = 15\) hours.
Thus slower car’s average speed = \(\frac{300}{t}=\frac{300}{15}=20\) miles/hour.
Hence the right answer is Option B.
Slower Car:
Assume the time taken to cover 300 miles = \(t\) hours.
Average speed = \(\frac{300}{t}\) miles/hour.
Faster car:
Assume the time taken to cover 300 miles = \(t-3\) hours.
Average speed = \(\frac{300}{t-3}\) miles/hour.
The question is telling us that faster car is moving at speed 5 miles/hour greater than slower. which means:
\(\frac{300}{t-3} -5 = \frac{300}{t}\)
\(⇒ \frac{300-5(t-3)}{t-3} = \frac{300}{t}\)
\(⇒ 300t+5t^2+15t=300t-900\)
\(⇒ -5t^2+15t+900=0\)
\(⇒ t^2-3t-180=0\)
\(⇒ t^2-15t+12t-180=0\)
\(⇒ t(t-15)+15(t-15)=0\)
\(⇒ t = 15\) or \(-12\).
Time cannot be negative. Thus \(t = 15\) hours.
Thus slower car’s average speed = \(\frac{300}{t}=\frac{300}{15}=20\) miles/hour.
Hence the right answer is Option B.
20 Jun 2021, 19:31
Kudos
Bookmarks
As D=TxR
300=T1 x R1----1) The faster
300=T2 x R2----2) The slower
The equation from the question:
T2-T1=3----> replace T by using R
R1-R2=5
So, 300/R2-300/R1=3 ---- make it easier ---> 100R1-100R2=R1R2
Then, solve the equation and got result
500=R1R2
From the choice, the slower must be 20
300=T1 x R1----1) The faster
300=T2 x R2----2) The slower
The equation from the question:
T2-T1=3----> replace T by using R
R1-R2=5
So, 300/R2-300/R1=3 ---- make it easier ---> 100R1-100R2=R1R2
Then, solve the equation and got result
500=R1R2
From the choice, the slower must be 20
