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Car X left Town T (RATE PROBLEM, Need Help)

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Car X left Town T (RATE PROBLEM, Need Help)

by mkbigmoz » Tue Apr 07, 2009 8:27 am

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Im completely confused with the answer...I know I must be overlooking something.



Car X left Town T traveling at an average speed of 40 miles per hour. Car Y left Town T 18 minutes after car X left Town T, and car Y traveled at an average speed of 54 miles per hour. When car Y had traveled for z minutes, car Y had traveled 23 miles more that car X had from the time that car X left Town T. What is the value of z ?



A 81
B 90
C 105
D 125
E 150

[spoiler]
E 150

If the answer is E, wouldn't car Y have traveled 135 miles further than car X?[/spoiler]
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by moutar » Tue Apr 07, 2009 8:47 am
distance = speed x time

Car X travels at 40 mph for (z+18)mins = (z+18)/60 hours
Distance = 40(z+18)/60

Car Y travels at 54mph for z mins = z/60 hours
Distance = 54z/60

Car Y travels 23 miles more than Car X so:

(40(z+18)/60) + 23= 54z/60

40(z+18) + 23*60 = 54z (multiply through by 60)

40z + 720 + 1380 = 54z (multiplying out)

14z = 2100 (collecting like terms)

z = 150 (E)
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by Brent@GMATPrepNow » Fri Dec 28, 2018 9:19 am
mkbigmoz wrote:Car X left Town T traveling at an average speed of 40 miles per hour. Car Y left Town T 18 minutes after car X left Town T, and car Y traveled at an average speed of 54 miles per hour. When car Y had traveled for z minutes, car Y had traveled 23 miles more that car X had from the time that car X left Town T. What is the value of z ?
A 81
B 90
C 105
D 125
E 150
IMPORTANT: Let's use HOURS throughout our solution, and then convert hours to minutes at the end.

When car Y had traveled for z HOURS, car Y had traveled 23 miles more that car X had from the time that car X left Town T.
z = car Y's travel time in HOURS
So, z + 3/10 = car X's travel time in HOURS
[18 minutes = 3/10 HOURS, and car X traveled for 18 minutes before car Y started moving]

We can write: (car Y's distance traveled) - (car X's distance traveled) = 23 miles

distance = (speed)(time)
So, we get: (54)(z) - (40)(z + 3/10) = 23 miles
Expand to get: 54z - 40z - 12 = 23
Simplify: 14z - 12 = 23
So: 14z = 35
Solve: z = 35/14 = 5/2 = 2.5 HOURS
= 150 MINUTES

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by jaetpu » Thu Jan 03, 2019 10:57 pm
mkbigmoz wrote:Im completely confused with the answer...I know I must be overlooking something.



Car X left Town T traveling at an average speed of 40 miles per hour. Car Y left Town T 18 minutes after car X left Town T, and car Y traveled at an average speed of 54 miles per hour. When car Y had traveled for z minutes, car Y had traveled 23 miles more that car X had from the time that car X left Town T. What is the value of z ?



A 81
B 90
C 105
D 125
E 150

[spoiler]
E 150

If the answer is E, wouldn't car Y have traveled 135 miles further than car X?[/spoiler]
Let's make this into a simple D=RT problem. Since car X had been traveling for 18 minutes before car Y left, car X must be (18/60)hr*40mph = 12 miles ahead. Later, we learn that car Y had traveled 23 miles more than X. To sum it up, Car Y needs to catch up 35 miles to be 23 ahead of X.

Car Y travels 14mph faster than Car X. So 35=14mph * t. 35/14 = 2.5 hrs = t. 2.5 hrs in minutes is 150.
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by Scott@TargetTestPrep » Fri Feb 01, 2019 5:42 pm
mkbigmoz wrote:Im completely confused with the answer...I know I must be overlooking something.

Car X left Town T traveling at an average speed of 40 miles per hour. Car Y left Town T 18 minutes after car X left Town T, and car Y traveled at an average speed of 54 miles per hour. When car Y had traveled for z minutes, car Y had traveled 23 miles more that car X had from the time that car X left Town T. What is the value of z ?



A 81
B 90
C 105
D 125
E 150
We are given that Car X traveled at a rate of 40 mph and Car Y at a rate of 54 mph.

Since car Y left 18 minutes after car X, car Y traveled for z minutes, or z/60 hours, and car X for (z + 18) minutes or (z + 18)/60 hours; thus:

40[(z + 18)/60] + 23 = 54(z/60)

Multiplying by 60, we have:

40(z + 18) + 1380 = 54z

40z + 720 + 1380 = 54z

2100 = 14z

z = 2100/14 = 150

Answer: E

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