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# Jennifer can buy watches at a price of B dollars per watch, which she

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Math Expert
Joined: 02 Sep 2009
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Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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18 Feb 2015, 03:18
2
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00:00

Difficulty:

45% (medium)

Question Stats:

66% (02:14) correct 34% (02:39) wrong based on 151 sessions

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Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Kudos for a correct solution.

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Joined: 15 Aug 2013
Posts: 53
Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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18 Feb 2015, 05:11
2
1
Cost price of N watches = BN (cost of each watch, B * total nuber of watches, N)
Total Profit = T
But this profilt is also equal to BN (x/100), where x is the profit percentage
Hence, BN (x/100) =T
or x = 100T/ (NB)
Hence (A).
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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18 Feb 2015, 05:19
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Let's say Jennifer marks up by x%

Total Profit = Marked up price of N watches - Cost of N watches

i.e. T = NB[1+(x/100)] - NB
i.e. (T) / NB = [(x/100)]
i.e. x = 100(T)/NB

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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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18 Feb 2015, 05:32
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Let's say The price of each watch, i.e. B = 7,

Let's say Mark up = 400%==> 400% of 7 is 28, and the $7 watches are marked up$28,

then the selling price per watch = $35. The profit per watch is$28, and so

if she sells N = 11 watches

Markup of 100%; now selling for $10: x=10 She sells 5 watches: N=5 Her total profit will be (N*x)-(B*N) (5*10)-(5*5)= 25 T=25 Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100. B=5 T=25 N=5 Answer option: SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1820 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 21:51 Answer = A. 100T/(NB) Cost price per watch = b Selling price per watch = $$\frac{t}{n}$$ Let x = percent of the markup from her buy price to her sell price $$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$ _________________ Kindly press "+1 Kudos" to appreciate Math Expert Joined: 02 Sep 2009 Posts: 53063 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 22 Feb 2015, 11:17 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB) Answer = (A) That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers. Numerical Solution: The price of each watch, I will pick B = 7, a prime number. For the percent increase, I will make this easy. Picking 100% is too easy, and too predictable, so I am going to pick 400% — 400% of 7 is 28, and the$7 watches are marked up $25, then the selling price is$35. The profit per watch is $28, and so if she sells N = 11 watches (another prime number), that would be a profit of T = 28*11 =$308. Leaving T in unmultiplied form will make it easier to cancel.

OK, now we will plug in T = 308 = 11*28, N = 11, and B = 7, and hope to get 400 as our answer.

(A) 100T/(NB) = 100*11*28/(11*4) = 100*28/7 = 110*4 = 400 = works!
(B) TB/(100N) = 11*28*7/(100*11) = 28*7/100 = doesn’t work!
(C) 100TN/B = 100*11*28*11/7 = 100*11*4*11 = doesn’t work!
(D) ((T/N) – B)/(100B) = [(11*28/11) – 7]/(7*100) = (28 – 7)/(7*100) = 21/(7*100) = 3/100 = doesn’t work
(E) 100(T – NB)/N = 100(11*28 – 11*7)/7 = 100*11*21/7 = 100*11*3 = doesn’t work!

We were lucky here. With one choice of numbers, we were able to eliminate four answer choices, leaving (A) as the only possible answer.

- See more at: http://magoosh.com/gmat/2014/gmat-pract ... QEkpA.dpuf
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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10 Jul 2016, 20:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Kudos for a correct solution.

Cost price of one watch = B
Cost price of N watches = B*N

Total profit on N watches = T

Markup percentage = Profit/Cost Price *100 = T/BN * 100 = 100T/BN

Correct Option: A
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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11 Jul 2016, 00:26
let B be cost price of one watch
CP of N watches = NB
Profit = T = MP - NB (where MP= marked price)
profit % = (profit/CP)*100
= (T/NB)*100 =A
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Joined: 14 Aug 2012
Posts: 90
Location: United States
GMAT 1: 620 Q43 V33
GMAT 2: 690 Q47 V38
Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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22 Oct 2018, 16:33
GMATinsight wrote:
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Let's say Jennifer marks up by x%

Total Profit = Marked up price of N watches - Cost of N watches

i.e. T = NB[1+(x/100)] - NB
i.e. (T) / NB = [(x/100)]
i.e. x = 100(T)/NB

Hi, it looks like there are 2 NB's here - the one attached to the parantheses gets divided but what happened to the -NB on the far right?
VP
Joined: 09 Mar 2016
Posts: 1284
Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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23 Oct 2018, 08:34
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Kudos for a correct solution.

let B be the cost +mark up price 2+1 = $3 T = total profit =$10t
N = total number of items sold

$$\frac{10t}{3b*n}$$

Hence, A

pushpitkc is my solution correct ?
Manager
Joined: 14 Jun 2018
Posts: 223
Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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23 Oct 2018, 09:47
Cost Price = B
No of items sold = N
Profit after selling N items = T ; Profit per item = $$\frac{T}{N}$$
Let "k" be the markup percentage , therefore selling price = $$\frac{(100+K)}{100} * B$$

We know ,
Selling Price - Cost Price = Profit
=> $$\frac{(100+K)}{100} * B - B = \frac{T}{N}$$
=> $$K = 100 (\frac{T}{NB} + \frac{B}{B}) - 100$$
=> $$K = 100(\frac{T}{NB})$$
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Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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23 Oct 2018, 11:28
1
dave13 wrote:
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Kudos for a correct solution.

let B be the cost +mark up price 2+1 = $3 T = total profit =$10t
N = total number of items sold

$$\frac{10t}{3b*n}$$

Hence, A

pushpitkc is my solution correct ?

Hey dave13

Unfortunately, this is wrong. I am not quite clear what you trying to do here

If you are assigning values B = $5(cost per watch). She marks up the watches and sells for a total profit of T =$1
Assuming there is a N = 1 watch and she keeps a profit of 20%(which is 0.2*$5 =$1)

Now, working backward, trying to calculate the percentage by substituting the values
of B,N, and T, we will get

A. 100T/(NB) = 100/5 = 20 - Our solution is Option A($$\frac{100T}{NB}$$)

We don't need to substitute the values in the other 4 answer options as we have a match!

Hope that helps.
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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25 Oct 2018, 08:19
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

The profit (or markup) per watch is T/N. So the markup is:

(T/N)/B x 100 = 100T/(NB)

percent of the purchase price.

Alternate Solution:

Jennifer’s profit per watch is T/N. Using the formula for profit, we can compute Jennifer’s sell price:

profit = sell price – buy price

T/N = sell price – B

T/N + B = sell price

We now use the percent markup formula:

% markup = (Sell price – Buy price) / Buy price x 100

% markup = (T/N + B – B) / B x 100

% markup = T/N / B x 100

% markup = T/BN x 100

% markup = 100T / (BN)

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Re: Jennifer can buy watches at a price of B dollars per watch, which she   [#permalink] 25 Oct 2018, 08:19
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