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Taxi driver charges

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by tritrantran » Thu Dec 11, 2008 8:44 pm
If a taxi driver charges x cents for the first quarter-mile of a trip and (x/5) cents for each additional quarter-mile, what is the charge, in cents, for a trip whose distance in miles is the whole number y?

A) (x+xy)/125

B) (4x + 4xy)/5

C) (4x+xy)/500

D) (4x+xy)/5

E) xy/25

OA B

Two question:
1) What am I doing wrong?
2) What are the other methods that would be faster and more efficient?


Total cost = [cost of 1st quarter mile] + [cost of rest]

= x(1/4) + (x/5)*4*[y-(1/4)]

= x/4 + 4x/5*[y-(1/4)]

= x/4 + 4xy/5 - x/5

= 5x/20 + 16xy/20 - 4x/20

= (x+ 16xy)/20

???
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Source: — Problem Solving |

by weena82 » Thu Dec 11, 2008 9:08 pm
Hi,

Check your process against mine:

Total distance y mile = 4y quarter-mile

Total expense for the first quarter-mile = x cents
Total expense for the following quarter-mile = (x/5)(4y-1)

Total expense in cents = x + (x/5)(4y-1)

Solve this equation,

(5x + 4xy – x)/5

you’ll get (4x+4xy)/5
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by tritrantran » Thu Dec 11, 2008 9:15 pm
Ah, thanks! I found my mistake...I multipled x by (1/4) when I should just multiplied x by 1.

So eqtn should have been...

x*1 + (x/5)(4)*[y - (1/4)]
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by 720dreaming » Thu Dec 11, 2008 10:25 pm
You can plug in numbers. I've still yet to decide if plugging in makes more sense. Using algebra is quicker, but if you make a mistake you will spend more time trying to fix it or go back to plugging in numbers.
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by logitech » Thu Dec 11, 2008 10:47 pm
The question says that Y needs to be a whole number so:


for y = 1

x + 3x/5 = 8x/5

Which options gives us this when y=1 ?

4x+4y/5
LGTCH
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by tritrantran » Fri Dec 12, 2008 8:22 am
logitech wrote:The question says that Y needs to be a whole number so:


for y = 1

x + 3x/5 = 8x/5

Which options gives us this when y=1 ?

4x+4y/5
I think this is the winner for the fastest and most efficient way to solve the problem.

Thanks logitech!
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by james » Mon Oct 27, 2014 12:38 am
if i plugin x=5; y=8
1st Quarter: 5 cents; distance = 2
2nd Quarter: 1 cent; distance = 2
3rd Quarter: 1 cent; distance = 2
4th Quarter: 1 cent; distance = 2

Total charges = 5+1+1+1 = 8
Total distance covered = 2*4 = 8

If i check my answer with B(4x + 4xy)/5 = (20+160)/5 = 56cents

Pls i must be missing some logic here; somebody assist.


Thanks
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by GMATGuruNY » Mon Oct 27, 2014 3:20 am
tritrantran wrote:If a taxi driver charges x cents for the first quarter-mile of a trip and (x/5) cents for each additional quarter-mile, what is the charge, in cents, for a trip whose distance in miles is the whole number y?

A) (x+xy)/125

B) (4x + 4xy)/5

C) (4x+xy)/500

D) (4x+xy)/5

E) xy/25
Let x=5.
Cost for the first quarter-mile = x = 5.
Cost for each additional quarter-mile = x/5 = 1.
Let y=1, implying that the total distance traveled = 1 mile.
Total cost to travel 1 mile = cost for the first quarter-mile + cost for 3 additional quarter-miles = 5 + 3(1) = 8. This is our target.

Now plug x=5 and y=1 into the answers to see which yields our target of 8.
Only B works:
(4x + 4xy)/5 = (4*5 + 4*5*1)/5 = 8.

The correct answer is B.
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by GMATGuruNY » Mon Oct 27, 2014 3:38 am
james wrote:if i plugin x=5; y=8
1st Quarter: 5 cents; distance = 2
2nd Quarter: 1 cent; distance = 2
3rd Quarter: 1 cent; distance = 2
4th Quarter: 1 cent; distance = 2
The portion in red does not reflect what is described in the prompt.
x is the cost not for 1/4 of the entire distance but for the FIRST 1/4 MILE.
y is the cost not for each additional 1/4 of the distance for but EACH ADDITIONAL 1/4 MILE.
Total charges = 5+1+1+1 = 8
Total distance covered = 2*4 = 8

If i check my answer with B(4x + 4xy)/5 = (20+160)/5 = 56 cents.

Pls i must be missing some logic here; somebody assist.


Thanks
The portion in red miscalculates the total charge for a distance of 8 miles.
The value in blue should be 36.

If x=5, then the cost for the first quarter-mile = 5 cents, and the cost for each additional quarter-mile = x/5 = 5/5 = 1 cent.
If y=8 miles -- since 8*4 = 32 -- the total distance is composed of 32 quarter-miles.
Thus:
Total charge for 8 miles = 5 cents for the first quarter-mile + 1 cent for each of the 31 additional quarter-miles = 5 + 1*31 = 36. This is our target.

Plugging x=5 and y=8 into the OA, we get:
(4x + 4xy)/5 = (4*5 + 4*5*8)/5 = 36.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by Matt@VeritasPrep » Mon Oct 27, 2014 10:15 am
It also doesn't hurt to do this algebraically.

If the trip is y miles long, it must also be 4y miles long.

The first 1/4 mile is x¢, so the other (4y-1) quarter miles are each (x/5)¢.

Hence our trip costs x + (4y-1)(x/5), or (4x + 4xy)/5 cents.
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by james » Mon Oct 27, 2014 1:19 pm
GMATGuruNY wrote:
james wrote:if i plugin x=5; y=8
1st Quarter: 5 cents; distance = 2
2nd Quarter: 1 cent; distance = 2
3rd Quarter: 1 cent; distance = 2
4th Quarter: 1 cent; distance = 2
The portion in red does not reflect what is described in the prompt.
x is the cost not for 1/4 of the entire distance but for the FIRST 1/4 MILE.
y is the cost not for each additional 1/4 of the distance for but EACH ADDITIONAL 1/4 MILE.
Total charges = 5+1+1+1 = 8
Total distance covered = 2*4 = 8

If i check my answer with B(4x + 4xy)/5 = (20+160)/5 = 56 cents.

Pls i must be missing some logic here; somebody assist.


Thanks
The portion in red miscalculates the total charge for a distance of 8 miles.
The value in blue should be 36.

If x=5, then the cost for the first quarter-mile = 5 cents, and the cost for each additional quarter-mile = x/5 = 5/5 = 1 cent.
If y=8 miles -- since 8*4 = 32 -- the total distance is composed of 32 quarter-miles.
Thus:
Total charge for 8 miles = 5 cents for the first quarter-mile + 1 cent for each of the 31 additional quarter-miles = 5 + 1*31 = 36. This is our target.

Plugging x=5 and y=8 into the OA, we get:
(4x + 4xy)/5 = (4*5 + 4*5*8)/5 = 36.
i got the jist...not 1/4 of d total distance BUT a 1/4-mile (or 0.25 mile ONLY)
So, if y=1, there are 4 quarter miles...left with x/5 * 3 cents
if y=2, there are 8 quarter miles...left with x/5*7
if y=3, there will be 12 quarter miles....left with x/5*11...etc, etc.

Thanks GmatGuru
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by Jeff@TargetTestPrep » Tue Jan 02, 2018 10:17 am
tritrantran wrote:If a taxi driver charges x cents for the first quarter-mile of a trip and (x/5) cents for each additional quarter-mile, what is the charge, in cents, for a trip whose distance in miles is the whole number y?

A) (x+xy)/125

B) (4x + 4xy)/5

C) (4x+xy)/500

D) (4x+xy)/5

E) xy/25
Since 1 mile has 4 quarter-miles, a trip of y miles has 4y quarter-miles, and after the first quarter-mile that charges x cents, there are 4y - 1 additional quarter-miles that charge x/5 cents each. Thus, the total cost is:

x + (x/5)(4y - 1)

5x/5 + (4xy - x)/5

(4x + 4xy)/5

Answer: B

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