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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Jennifer can buy watches at a price of B dollars per watch, which she  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50544 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 03:18 2 7 00:00 Difficulty: 45% (medium) Question Stats: 70% (01:48) correct 30% (02:08) wrong based on 141 sessions ### HideShow timer Statistics 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. _________________ Manager Joined: 15 Aug 2013 Posts: 54 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 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). CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 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 _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 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!

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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, 07:51
1
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

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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, 21:51

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  [#permalink]

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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 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)

That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.

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: 179 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 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 => $$\frac{(100+K)}{100} * B - B = \frac{T}{N}$$ => $$K = 100 (\frac{T}{NB} + \frac{B}{B}) - 100$$ => $$K = 100(\frac{T}{NB})$$ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3301 Location: India GPA: 3.12 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 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 &nbs [#permalink] 25 Oct 2018, 08:19
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