Emile ran 6 kilometers yesterday at constant speed. This mo

Question

Emile ran 6 kilometers yesterday at constant speed. This morning, he ran 4 kilometers at a speed that was 1/5 greater yesterday’s speed and then reduced his speed by 4/9 for the remaining 2 kilometers. If today’s run was 1 minute and 15 seconds longer than yesterday’s, what was his speed in kilometers per hour yesterday?

A) 10 km/h
B) 12 km/h
C) 12.5 km/h
D) 15 km/h
E) 16 km/h

A) 10 km/h
B) 12 km/h
C) 12.5 km/h
D) 15 km/h
E) 16 km/h

MarkGMATMentor · Answer

Work / rate and distance / rate problems can often be much more manageable if you use a table and limit the number of variables.

We're looking for Emile's speed yesterday, so let's set that as r. This morning, at first his speed was 1/5 greater, so his speed was:
 \frac{6}{5}r

Then, his reduced his speed by 4/9, so his new speed was 5/9 of  \frac{6}{5}r :
 \frac{5}{9}(\frac{6}{5}r)=\frac{2}{3}r

Let's put what we know into a distance / rate table using the formula Distance = Rate * Time:
    
 D 
 = 
 R 
 T 
 
  yesterday 
 6 
 = 
 r

today fast 
 4 
 = 
  \frac{6}{5}r

today slow 
 2 
 = 
  \frac{2}{3}r

Now, rather than adding additional variables for the times, let's solve for each one in terms of r and put them into the table:
    
 D 
 = 
 R 
 T 
 
  yesterday 
 6 
 = 
 r 
  \frac{6}{r}  
 
  today fast 
 4 
 = 
  \frac{6}{5}r  
  \frac{10}{3r}  
 
  today slow 
 2 
 = 
  \frac{2}{3}r  
  \frac{3}{r}

Finally, since we know that today's run was 1 minute 15 seconds longer than yesterday's, we can build an equation and solve for the rate. But we're looking for kilometers per  hour.  1 minute 15 seconds is equal to 75 seconds. Perform the conversion to hours:
 75 seconds*\frac{1 minute}{60 seconds}*\frac{1 hour}{60 minutes}=\frac{5}{4*60} hour = \frac{1}{48} hour

Then, since today's run was 1/48 hour longer than yesterday's, we have:
 \frac{10}{3r}+\frac{3}{r}=\frac{6}{r}+\frac{1}{48}

\frac{1}{3r}=\frac{1}{48}

r = 16

The answer is E.

A table is often an effective tool on distance / rate and work / rate problems. Remember also to limit the number of variables and align the units.

paragw · Answer

Let yesterday's speed be 30x
T1 = 6/30x = 1/5x

Now for today
4km @ 36x kmph & 2km @ 20x kmph
T2 = 4/36x + 2/20x = 1/9x + 1/10x = 19/90x

Now, T2 - T1 = 75sec
19/90x - 1/5x = 75/3600 = 1/48
1/90x = 1/48
30x = 16

Hence, 16kmph

	D	=	R	T
yesterday	6	=	r	\(\frac{6}{r}\)
today fast	4	=	\(\frac{6}{5}r\)	\(\frac{10}{3r}\)
today slow	2	=	\(\frac{2}{3}r\)	\(\frac{3}{r}\)

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