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805+ (Hard)|   Algebra|      
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chetan86
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For a particular model of moving truck, rental agency A charges a dail Updated on: 22 Oct 2014, 07:15
Originally posted by chetan86 on 22 Oct 2014, 04:48.
Last edited by Bunuel on 22 Oct 2014, 07:15, edited 1 time in total.
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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) \(\frac{m-p}{2(q-n)}\)
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For a particular model of moving truck, rental agency A charges a dail 22 Oct 2014, 07:24
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chetan86
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) \(\frac{m-p}{2(q-n)}\)
Show more

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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For a particular model of moving truck, rental agency A charges a dail 22 Oct 2014, 10:41
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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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For a particular model of moving truck, rental agency A charges a dail 24 Oct 2014, 00:08
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Bunuel
chetan86
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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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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For a particular model of moving truck, rental agency A charges a dail 24 Oct 2014, 13:40
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Bunuel
chetan86
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.
Show more

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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For a particular model of moving truck, rental agency A charges a dail 25 Oct 2014, 06:11
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Bunuel
chetan86
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)
Show more

Because x is already the total number of miles driven in two days.
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For a particular model of moving truck, rental agency A charges a dail 18 Jul 2016, 02:00
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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) \(\frac{m-p}{2(q-n)}\)

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.
Show more

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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For a particular model of moving truck, rental agency A charges a dail 18 Jul 2016, 03:23
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Bunuel
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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) \(\frac{m-p}{2(q-n)}\)

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
Show more

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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For a particular model of moving truck, rental agency A charges a dail 18 Jul 2016, 03:25
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Bunuel
chetan86
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
Show more

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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For a particular model of moving truck, rental agency A charges a dail 18 Jul 2016, 03:46
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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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For a particular model of moving truck, rental agency A charges a dail 21 Sep 2017, 14:48
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chetan86
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) \(\frac{m-p}{2(q-n)}\)
Show more

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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For a particular model of moving truck, rental agency A charges a dail 24 Nov 2018, 05:02
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Bunuel , chetan2u

Can you please explain why we need to divide the cents in 100?
I understand that n/100 gives dollars per mile, but why can't we leave it as cents? (i.e., why do we need to say p cents is 0.0n when its already mentioned that n is cents?)

Thank you in advanced
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For a particular model of moving truck, rental agency A charges a dail 24 Nov 2018, 05:13
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Eladt
Bunuel , chetan2u

Can you please explain why we need to divide the cents in 100?
I understand that n/100 gives dollars per mile, but why can't we leave it as cents? (i.e., why do we need to say p cents is 0.0n when its already mentioned that n is cents?)

Thank you in advanced
Show more


Hi..

When we are finding the cost one way it is some dollars as a constant and some cents per mile..
So, if you have to add both, we have to get them into same units, either dollars or cents...

Say you take 1 hour for first few miles and then takes 30 minutes to complete the rest ..
Total cannot be 1+30, it will be 1+ (30/60) as we have to convert both into same units
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For a particular model of moving truck, rental agency A charges a dail 24 Nov 2018, 05:17
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Eladt
Bunuel , chetan2u

Can you please explain why we need to divide the cents in 100?
I understand that n/100 gives dollars per mile, but why can't we leave it as cents? (i.e., why do we need to say p cents is 0.0n when its already mentioned that n is cents?)

Thank you in advanced


Hi..

When we are finding the cost one way it is some dollars as a constant and some cents per mile..
So, if you have to add both, we have to get them into same units, either dollars or cents...

Say you take 1 hour for first few miles and then takes 30 minutes to complete the rest ..
Total cannot be 1+30, it will be 1+ (30/60) as we have to convert both into same units
Show more


Oh boy...
you're right, totally missed that.

Thank you very much chetan2u !
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For a particular model of moving truck, rental agency A charges a dail 10 May 2020, 11:29
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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) 100(m−p)/q−n

(B) 200(p−m)/n−q

(C) 50(m−p)/q−n

(D) 2(p−m)/n−q

(E) m−p/2(q−n)
____________________________________
2 days on the roads
let d be a mile that a driver driven

rental agency A = 2m + 0.01n * d
rental agency B = 2p + 0.01q * d
A = B
therefore,
2m+0.01nd = 2p + 0.01qd
2m-2p = 0.01qd - 0.01nd
2(m-p) = 0.01d(q-n)
200(m-p) = d(q-n)
200(m-p)/(q-n) = d
but we can't find this one among choices.
assuming that choice (B) looks like our solution above, we try to see if they are the same

take "-" out
-200 (p-m) / q-n
multiply - both numerator and denominator
200(p-m) / n-q
now our solution looks like the answer choice B. (B) 200(p−m)/n−q

therefore the answer is B
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For a particular model of moving truck, rental agency A charges a dail 03 Jan 2021, 11:14
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Since none of the variables is determined in the problem, their values can be selected at random.

Remember that m and p are expressed as dollars and n and q are expressed as cents. Let m = 10, n = 200, p = 20, q = 100. Then agency A charges $10 per day plus $2 per mile, and agency B charges $20 per day plus $1 per mile. (Notice that the agency with the higher daily fee must offer a lower mileage rate; otherwise, the prices will never be the same.)

For a two-day rental, agency A charges $20 plus $2/mile, and agency B charges $40 plus $1/mile. To solve algebraically, let x be the number of miles driven:

20 + 2x = 40 + 1x
x = 20

20 is the target value. Find a match in the answer choices:

(A) 100(–10)/(–100) = 10 miles.
(B) 200(10)/(100) = 20 miles.
(C) 50(–10)/(–100) = 5 miles.
(D) 2(10)/(100) = 0.2 mile.
(E) –10/(2× –100) = 0.05 mile.

Only choice B gives the desired figure of 20 miles.

This question can also be answered algebraically. One cent is equal to 0.01 dollars. Therefore, rental agency A charges $m per day plus $0.01n per mile, and rental agency B charges $p per day plus $0.01q per mile.

Let x be the number of miles at which the fees are equal, for two days’ rental:

2m + (0.01n)x = 2p + (0.01q)x
2m – 2p = 0.01qx – 0.01nx
2(m – p) = 0.01x(q – n)
x=200(p - m)/(n - q)


The correct answer is B.
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For a particular model of moving truck, rental agency A charges a dail 21 Aug 2023, 14:54
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A) 2M + NX/100
B) 2P + QX/100

* you multiply the daily fee for A and B by 2 cz the driver plans to rent for 2 days and divide N and Q by 100 cz those figures are in cents, so I like to treat cents like percentages in my mind and just divide them by 100 when needed to be converted into $

2M + NX/100 = 2P + QX/100

NX/100 - QX/100 = 2P - 2M

X(N-Q)/100 = 2(P-M)

X(N-Q) = 200(P-M)
X= 200(P-M)/N-Q
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For a particular model of moving truck, rental agency A charges a dail 21 Aug 2023, 23:06
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Bunuel
If we consider x miles per day then for 2 days 2x miles.

We get option A as the answer.

Why is it a mistake. What are we missing here.

Silly doubt.
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For a particular model of moving truck, rental agency A charges a dail 21 Aug 2023, 23:44
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Nabneet
Bunuel
If we consider x miles per day then for 2 days 2x miles.

We get option A as the answer.

Why is it a mistake. What are we missing here.

Silly doubt.
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Nabneet - The question doesn't want us to find the number of miles per day the driver must drive for the two rental agencies’ total charges to be equal. The question asks us to find the number of miles this driver must drive across both days. Also, we don't know if the driver drives an equal number of miles on both days, hence the assumption that if the driver drives x miles on the first day, he will drive 2x miles across both days is sort of not correct.

Quote:
"...the number of miles this driver must drive for the two rental agencies’ total charges to be equal..."
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Let's take your example into consideration

The driver drives \(2x\) miles across both the days -

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

\(200m + n(2x) = 200p + q(2x)\)

\(2x(n-q) = 200(p-m)\)

\(2x = \frac{200(p-m)}{n-q}\)

Note, we cannot divide by 2 on both sides, as we need the value of 2x (the number of miles driven across both days) and not x (the number of miles driven per day).

That's Option B
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For a particular model of moving truck, rental agency A charges a dail 02 Apr 2026, 05:12
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