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

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

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New post 18 Feb 2015, 04:18
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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


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

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New post 18 Feb 2015, 06:11
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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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New post 18 Feb 2015, 06:19
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  [#permalink]

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New post 18 Feb 2015, 06:32
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  [#permalink]

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New post 18 Feb 2015, 08:51
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

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

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New post 18 Feb 2015, 22:51
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  [#permalink]

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New post 22 Feb 2015, 12:17
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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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 10 Jul 2016, 21:17
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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New post 11 Jul 2016, 01:26
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  [#permalink]

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