For each trip, a taxi company charges a fixed fee of $2.00

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For each trip, a taxi company charges a fixed fee of $2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ 0.75r/2]
B. 2.00 + 0.75[ r/2 ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]

OA:E

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by [email protected] » Mon Dec 19, 2016 7:16 am
Hi NandishSS,

This question can be solved by TESTing Values (although since the prompt discusses 'fractions of a 1/2 mile' iit's 'hinting' that we should be thinking about non-integers...

Let's TEST R = 2.1

According to the prompt, for a 2.1-mile trip, the taxi company will charge $2 + $0.75(5) = $5.75

When we plug R = 2.1 into the answers, and we follow the definition of [x], we get....

Answer A: 2 + 1 = 3 NOT a match
Answer B: 2 + 1.5 = 3.5 NOT a match
Answer C: 2 + 2.25 = 4.25 NOT a match
Answer D: 2 + 4 = 6 NOT a match
Answer E: 2 + 3.75 = 5.75 This IS a match

Final Answer: E

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by GMATGuruNY » Mon Dec 19, 2016 7:23 am
NandishSS wrote:For each trip, a taxi company charges a fixed fee of $2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ 0.75r/2]
B. 2.00 + 0.75[ r/2 ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]
Let r = 3/4 mile.
Base charge = 2.
Since 75 cents is charged for each 1/2 mile or fraction of 1/2 mile, we get:
Charge for the first 1/2 mile = 0.75.
Charge for the next 1/4 mile = 0.75.
Total charge = 2 + 0.75 + 0.75 = 3.5. This is our target.

Now plug r = 3/4 into the answers to see which yield our target of 3.5.

A: 2.00 + [0.75r/2] = 2 + [(3/4)(3/8)] = 2 + [9/32] = 2 + (least integer greater than or equal to 9/32) = 2 + 1 = 3.
Eliminate A.

B: 2.00 + 0.75[ r/2 ] = 2 + (3/4)[3/8] = 2 + (3/4)(least integer greater than or equal to 3/8) = 2 + (3/4)(1) = 2.75.
Eliminate B.

C: 2.00 + 0.75[ r ] = 2 + (3/4)[3/4] = 2 + (3/4)(least integer greater than or equal to 3/4) = 2 + (3/4)(1) = 2.75.
Eliminate C.

D: 2.00 + [1.5r ] = 2 + [(3/2)(3/4)] = 2 + [9/8] = 2 + (least integer greater than or equal to 9/8) = 2 + 2 = 4.
Eliminate D.

The correct answer is E.

E: 2.00 + 0.75 [ 2r] = 2 + (3/4)[(2)(3/4)] = 2 + (3/4)[6/4] = 2 + (3/4)(least integer greater than or equal to 6/4) = 2 + (3/4)(2) = 3.5.
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by NandishSS » Mon Dec 19, 2016 9:24 am
GMATGuruNY wrote:
NandishSS wrote:For each trip, a taxi company charges a fixed fee of $2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ 0.75r/2]
B. 2.00 + 0.75[ r/2 ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]
Let r = 3/4 mile.
Base charge = 2.
Since 75 cents is charged for each 1/2 mile or fraction of 1/2 mile, we get:
Charge for the first 1/2 mile = 0.75.
Charge for the next 1/4 mile = 0.75.
Total charge = 2 + 0.75 + 0.75 = 3.5. This is our target.

Now plug r = 3/4 into the answers to see which yield our target of 3.5.

A: 2.00 + [0.75r/2] = 2 + [(3/4)(3/8)] = 2 + [9/32] = 2 + (least integer greater than or equal to 9/32) = 2 + 1 = 3.
Eliminate A.

B: 2.00 + 0.75[ r/2 ] = 2 + (3/4)[3/8] = 2 + (3/4)(least integer greater than or equal to 3/8) = 2 + (3/4)(1) = 2.75.
Eliminate B.

C: 2.00 + 0.75[ r ] = 2 + (3/4)[3/4] = 2 + (3/4)(least integer greater than or equal to 3/4) = 2 + (3/4)(1) = 2.75.
Eliminate C.

D: 2.00 + [1.5r ] = 2 + [(3/2)(3/4)] = 2 + [9/8] = 2 + (least integer greater than or equal to 9/8) = 2 + 2 = 4.
Eliminate D.

The correct answer is E.

E: 2.00 + 0.75 [ 2r] = 2 + (3/4)[(2)(3/4)] = 2 + (3/4)[6/4] = 2 + (3/4)(least integer greater than or equal to 6/4) = 2 + (3/4)(2) = 3.5.
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HI Guru & Rich,

You guys are awesome!!! And I'm really grateful to you guys.

One quick question, is it feasible to using TESTing method in these kind of problems?

Thanks
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by [email protected] » Mon Dec 19, 2016 8:07 pm
Hi NandishSS,

I'm not exactly clear on what you're asking - can you go into a bit more detail about what you're interested in? Both Mitch and I used variations on the same approach (TESTing VALUES), so it's clearly a viable way to approach the prompt.

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by NandishSS » Mon Dec 19, 2016 9:05 pm
[email protected] wrote:Hi NandishSS,

I'm not exactly clear on what you're asking - can you go into a bit more detail about what you're interested in? Both Mitch and I used variations on the same approach (TESTing VALUES), so it's clearly a viable way to approach the prompt.

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Rich
Hi Rich,

To be more precise is it efficient way to use (TESTing VALUES) in these kind of problems?

Thanks
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by [email protected] » Tue Dec 20, 2016 1:37 pm
Hi NandishSS,

TESTing VALUES is a useful approach (and a relatively easy/efficient one) on many of the Quant questions that you'll face on the GMAT. You can even use it on certain IR questions and rare CR questions. The key to maximizing your use of this Tactic is to practice it repeatedly. If you're not comfortable TESTing VALUES before you get to Test Day, then there's no way that you'll be able to effectively use it when you really need it.

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by DavidG@VeritasPrep » Tue Dec 20, 2016 2:32 pm
NandishSS wrote:For each trip, a taxi company charges a fixed fee of $2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ 0.75r/2]
B. 2.00 + 0.75[ r/2 ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]

OA:E

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And there's always good old-fashioned algebra. We know there are 2 half-miles in each mile, right? So a trip of r miles will contain 2r half-miles. If it costs $.75 for each half-mile, that's .75 * 2r on top of the $2 fixed cost, for a total of 2 + .75 * 2r. Slap some brackets on the '2r' and the answer is E
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by gmatstudent2017 » Mon Sep 04, 2017 1:13 pm
Hi Mitch and Rich,

I had actually used numbers to solve this problem and ended up changing my answer from E to D. The reason being, the first number I picked for r was 1, which means the total cost will be $2 + (2*.75) since there is a $2 fixed fee and 1 mile consists of 2 half miles. This gave me a total cost of $3.50. Both choices D and E resulted in $3.50, so I picked another number for r. I then picked 2 for r, which gave me a final cost of $2 + (4*.75) = $5. Answer choice E doesn't result in $5, but D does that is why I ended up picking D. I'm not sure where I went wrong.

Thank you in advance!

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by Admin1 » Mon Sep 04, 2017 6:18 pm
Solution:
Firstly, it is important to know the meaning of the expression, [x], which is the least integer greater than or equal to x. For example, [2.3] = 3.

Now, in order to know how many half-miles there are in r miles, we have
r / (1/2) = 2 r.
But, a fraction of a half-mile will still be counted as one half-mile in terms of money. Hence, we need to have, [2 r], in order to include that fraction.

And since the charge for this half-mile and fraction of half-mile is $ 0.75 and the fixed price is $ 2.00, we therefore have

2.00 + 0.75 [2 r].

The answer is E.

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by DavidG@VeritasPrep » Tue Sep 05, 2017 7:34 am
gmatstudent2017 wrote:Hi Mitch and Rich,

I had actually used numbers to solve this problem and ended up changing my answer from E to D. The reason being, the first number I picked for r was 1, which means the total cost will be $2 + (2*.75) since there is a $2 fixed fee and 1 mile consists of 2 half miles. This gave me a total cost of $3.50. Both choices D and E resulted in $3.50, so I picked another number for r. I then picked 2 for r, which gave me a final cost of $2 + (4*.75) = $5. Answer choice E doesn't result in $5, but D does that is why I ended up picking D. I'm not sure where I went wrong.

Thank you in advance!
Answer choice E doesn't result in $5;

E would also give you $5 if r = 2. They key is to pick a non-integer value for r, such as 2.5, in which case, E will work and D will not.
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by [email protected] » Tue Sep 05, 2017 10:18 am
Hi gmatstudent2017,

If you take a good look at how Answers D and E are 'designed', you'll notice that they provide the SAME result IF you choose an integer value for R. In your example, BOTH answers total $5 when you TEST R = 2.

The wording of the prompt heavily implies that we need to consider a NON-integer to get to the correct answer, which is why I chose R = 2.1 in my explanation.

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