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Rates & Work Problems - Manhattan Gmat

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by shanice » Thu Apr 19, 2012 8:30 am
Please help me with the following problems:-

1)Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at
a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour.If
Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather
walked when the two meet?

a)7 miles
b)8 miles
c)9 miles
d)10 miles
e)12 miles

Answer is B - 8 miles.


2)If Lucy walks to work at a rate of 4 miles per hour, but she walks home by the same route at a rate
of 6 miles per hour, what is Lucy's average walking rate for the round trip?

Answer is 4.8 miles per hour.


Thank you in advance.
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by aneesh.kg » Thu Apr 19, 2012 9:14 am
Solution 1:
Stacy's speed = 6mph
Heather's speed = 5 mph
If Heather travels for t hours, then Stacy travels for (t hours + 24 minutes) or (t + 0.4) hours.
Total distance between them = Distance traveled by Heather + Distance traveled by Stacy
D = s1*t1 + s2*t2
20 = 5t + 6(t + 0.4)
20 = 11t + 2.4
17.6 = 11t
t = 1.6 hours

Heather's distance = her speed * time taken by her = 5*t = 5*1.6 = 8 miles
Last edited by aneesh.kg on Thu Apr 19, 2012 9:18 am, edited 1 time in total.
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by aneesh.kg » Thu Apr 19, 2012 9:17 am
Solution 2:
Let the distance from her home to work be 'D'
Average Speed = (Total distance)/(Total Time)
= (Distance from home to work + Distance from work to home)/(Time taken from home to work + Time taken from work to home)
= (D + D)/(D/s1 + D/s2)
where
s1 = 4mph
s2 = 6mph

Therefore,
Average Speed = (2D)/(D/4 + D/6)
= 2/(1/4 + 1/6)
= 2*4*6/(4 + 6)
= 4.8 mph
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by shanice » Thu Apr 19, 2012 9:48 am
Thank you, Aneesh. I understand it now.
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by Anurag@Gurome » Thu Apr 19, 2012 5:21 pm
1)Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when the two meet?

a)7 miles
b)8 miles
c)9 miles
d)10 miles
e)12 miles

Answer is B - 8 miles.
Stacy's speed = 6 mi/hr
Heather's speed = 5 mi/hr

Distance covered by Stacy in 24 mins = (24/60) * 6 = 2.4 miles
Since both are walking in opposite directions, so sum of their speeds = 6 + 5 = 11 mi/hr
Distance to be covered = 20 - 2.4 = 17.6
Time = Distance/Speed = 17.6/11 = 1.6 hrs
So, Heather will cover = 5 * 1.6 = 8 miles

The correct answer is B.
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by tomada » Fri Apr 20, 2012 4:05 pm
For the 2nd problem, I observed that both 4 and 6 (her walking speeds to and from work, respectively) factor evenly into 12, so I chose 12 miles as the one-way distance.

Her walking speed to work is 4 mph. With a distance of 12 miles, this would require 3 hours of walking.
Her walking speed from work is 6 mph. With the same distance of 12 miles, this would require 2 hours of walking.

Combined distance = 24 miles.
Combined time = 5 hours

24/5 = 4.8
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by GMATGuruNY » Fri Apr 20, 2012 7:19 pm
shanice wrote:Please help me with the following problems:-

1)Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at
a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour.If
Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather
walked when the two meet?

a)7 miles
b)8 miles
c)9 miles
d)10 miles
e)12 miles
Distance traveled by Stacy in 2/5 of an hour = r*t = 6*(2/5) = 12/5 miles.
Remaining distance = 20 - 12/5 = 88/5 miles.

Stacy and Heather work together to cover the remaining distance.
Since Stacy travels at 6 miles per hour and Heather travels at 5 miles per hour, Heather travels 5 of every 11 miles that the two travel together.
Thus, Heather travels 5/11 of the remaining distance:
(5/11)(88/5) = 8.

The correct answer is B.
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