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Jennifer can buy watches at a price of B dollars per watch, which she
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18 Feb 2015, 03:18
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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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18 Feb 2015, 05:11
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
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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 Answer: Option
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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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 Then Total profit, T = 28*11 = $308. OK, now Substitute T = 308 = 11*28, N = 11, and B = 7, in options 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! Answer: Option
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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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18 Feb 2015, 07:51
Using smart numbers: Jennifer Purchases watches for $5: B=5 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:



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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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 \(x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}\)
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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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 SOLUTIONAlgebraic 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/gmatpract ... QEkpA.dpuf
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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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
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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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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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 Answer: Option 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?



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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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 ?



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Re: Jennifer can buy watches at a price of B dollars per watch, which she
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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
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23 Oct 2018, 11:28
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 dave13Unfortunately, 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
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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) Answer: A
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