Dan drove home from college at an average rate of 60 miles per hour.

Question

Dan drove home from college at an average rate of 60 miles per hour. On his trip back to college, his rate was 10 miles per hour slower and the trip took him one hour longer than the drive home. How far is Dan’s home from the college?

(A) 65 miles
(B) 100 miles
(C) 200 miles
(D) 280 miles
(E) 300 miles

Source: Master GMAT

(A) 65 miles
(B) 100 miles
(C) 200 miles
(D) 280 miles
(E) 300 miles

Source:  Master GMAT

rajatchopra1994 · Answer

60x=50(x+1)
10x=50
x=5

Distance = 60*5 = 300miles

IMO-E

GMATWhizTeam · Answer

Solution 
 • Dan’s speed while going to home from college = 60 miles/hour
• While coming back to college from home, his speed was 10 miles per hour slower.
 o So, Dan’s speed while coming back to college from home  = 60-10 = 50  miles/hour 
• Let’s say Dan took  t  hours to complete his trip from college to home.
 o Then, time taken by Dan to complete his trip back to college  = t+1  hours 
• Let us assume that Dan’s home is at a  d  miles distance from his college.
 o Then,  60*t = 50*(t+1) = d 
o Or,  60*t = 50*t + 50 
o Or,  t = 5  hours 
• Hence,  d = 60*5 = 300   miles 
Thus, the correct answer is   Option E  .

Alternately, 
 • Let us assume that the Dan’s home is d miles away from his college. 
 • Dan’s speed on his trip to home  = 60  miles/hour
• Since on his trip back to college his speed was 10 miles per hour slower,
 o So, his speed while going back to college from home  = 60-10 = 50  miles/hour 
• It is given that on his trip back to college, he took one hour more than his trip to home.
 o So,  \frac{d}{50} – \frac{d}{60 }=1 
o Or,  \frac{6d – 5d}{300} =1 
o Or,  6d – 5d =300 
o Or,  d = 300  miles.  
Thus, the correct answer is   Option E.

EgmatQuantExpert · Answer

Solution

Given  
 • Dan drove home from college at an average rate of 60 miles per hour.
• On his trip back to college, his rate was 10 miles per hour slower
 o And, the trip took him one hour longer than the drive home.

To find  
 • Distance of Dan’s home from the college

Approach and Working out

Method -1 
 • Let T1 is time when Dan goes college at 60 miles per hours and,
• T2 is the time when he comes back from the college at 50 miles per hours(10 miles per hours less)
• And, D is distance of Dan’s home from college
• While coming back he takes 1 hour more
 o So, T2 -T1 =1
o T2 = D/50 and T1 = D/60
o D/50 -D/60 =1
o 10D=50*60
o D =300 Miles

Hence, option E is the correct answer.

Correct Answer: Option E

EgmatQuantExpert · Answer

Method -2

Distance = Speed * Time or Time = Distance/Speed
Hence, speed and time have inverse relation.

Now, we will use this concept to solve this question.
 • Speed reduced from 60 miles per hour to 50 miles per hour.
 o So, speed become 5/6 times of itself in the return journey.
o Now, speed and time has inverse relation.
o So, time will become 6/5 times of itself for the return journey.
o T -6/5T =1
o 1/5T = 1
o T = 5 Hours 
• Hence, Dan takes 5 hours to go to school from his home at 60 miles per hours.
• Therefore, D = 5*60 = 300 miles

Correct Answer: Option E

EgmatQuantExpert · Answer

Method -3

Distance = Speed * Time 
Let time take while going to college from home is T.

Hence, time taken to come back to home is T+1 hours.

• In both the cases, distance is same.
• Hence, 60*T = 50 * (T+1)
 o T =5
o Distance = 300 miles

Correct Answer: Option E

Dan drove home from college at an average rate of 60 miles per hour.

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Dan drove home from college at an average rate of 60 miles per hour.

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