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For a particular model of moving truck, rental agency A charges a dail

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For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post Updated on: 22 Oct 2014, 06:15
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For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)

Originally posted by chetan86 on 22 Oct 2014, 03:48.
Last edited by Bunuel on 22 Oct 2014, 06:15, edited 1 time in total.
Edited the question.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 22 Oct 2014, 06:24
4
3
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 22 Oct 2014, 09:41
1
This is mostly like a plug and chug. Let the # of days =2 and the # of miles be equal for both drivers. Just remember to divide n and q by 100 to convert from cents to dollars.

You should get 2m+n*miles/100 = 2p + q*miles/100

Solving for miles gets you B.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 23 Oct 2014, 23:08
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Hi Bunuel,
Thanks a lot for your explanation.
Next time I will take care to formulate mathematical expression correctly. Thanks for the link.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 24 Oct 2014, 12:40
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).

Hi Bunuel,
Can you please clarify why we are not multiplying 2 days to cents? I was stuck on this question because I calculated as 2(m + n/100)
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 25 Oct 2014, 05:11
pairakesh10 wrote:
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).

Hi Bunuel,
Can you please clarify why we are not multiplying 2 days to cents? I was stuck on this question because I calculated as 2(m + n/100)


Because x is already the total number of miles driven in two days.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 18 Jul 2016, 01:00
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Hi Bunuel,
Could you please explain why you are taking the number of miles for both the agencies as equal. Couldn't it be possible that -
Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*y (q/100 gives dollars per mile).

And the total miles would be x+y.

Thanks
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 18 Jul 2016, 02:23
gmat730 wrote:
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Hi Bunuel,
Could you please explain why you are taking the number of miles for both the agencies as equal. Couldn't it be possible that -
Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*y (q/100 gives dollars per mile).

And the total miles would be x+y.

Thanks


The question asks: which of the following expressions gives the number of miles (x in our case) this driver must drive for the two rental agencies’ total charges to be equal? So, for what x, are the charges of two agencies equal.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 18 Jul 2016, 02:25
gmat730 wrote:
Bunuel wrote:
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


Let x be the number of miles this driver must drive for the two rental agencies’ total charges to be equal.

Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*x (q/100 gives dollars per mile).

Equate and solve for x:

\(2m + \frac{n}{100}*x=2p + \frac{q}{100}*x\);

\(200m+nx=200p+qx\);

\(x=\frac{200p-200m}{n-q}\).

Answer: B.

Hope it's clear.

P.S. Please read Writing Mathematical Formulas on the Forum. Thank you.


Hi Bunuel,
Could you please explain why you are taking the number of miles for both the agencies as equal. Couldn't it be possible that -
Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*y (q/100 gives dollars per mile).

And the total miles would be x+y.

Thanks


To understand better check similar questions:
salesperson-a-s-compensation-for-any-week-is-360-plus-30977.html
health-insurance-plan-a-requires-the-insured-to-pay-1000-or-106447.html

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 18 Jul 2016, 02:46
Hi Bunuel,
Could you please explain why you are taking the number of miles for both the agencies as equal. Couldn't it be possible that -
Agency A's charges for two days = 2m + n/100*x (n/100 gives dollars per mile).
Agency B's charges for two days = 2p + q/100*y (q/100 gives dollars per mile).

And the total miles would be x+y.

Thanks[/quote]

To understand better check similar questions:
salesperson-a-s-compensation-for-any-week-is-360-plus-30977.html
health-insurance-plan-a-requires-the-insured-to-pay-1000-or-106447.html

Hope it helps.[/quote]

Got it. Thank you. I was thinking about another possibility in which the driver could travel x miles for agency A and y miles for agency B and still get the total charges as equal.
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Re: For a particular model of moving truck, rental agency A charges a dail  [#permalink]

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New post 21 Sep 2017, 13:48
chetan86 wrote:
For a particular model of moving truck, rental agency A charges a daily fee of m dollars, plus n cents per mile. For the same model of truck, rental agency B charges a daily fee of p dollars, plus q cents per mile. If a driver plans to rent this model of truck for two days, which of the following expressions gives the number of miles this driver must drive for the two rental agencies’ total charges to be equal?

(A) \(\frac{100(m-p)}{q-n}\)

(B) \(\frac{200(p-m)}{n-q}\)

(C) \(\frac{50(m-p)}{q-n}\)

(D) \(\frac{2(p-m)}{n-q}\)

(E) \(\frac{m-p}{2(q-n)}\)


We can create the following equation in which z = the number of miles driven. Since the daily fee is in dollars and the mileage fee is in cents, we convert the daily fee to cents. We should remember that m dollars = 100m cents and p dollars = 100p cents.

2(100m) + nz = 2(100p) + qz

200m + nz = 200p + qz

nz - qz = 200p - 200m

z(n - q) = 200(p - m)

z = 200(p - m)/(n - q)

Answer: B
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