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If the price of coffee, p, doubles every week for 4 weeks

Expert replies
by BTGmoderatorDC » Fri Nov 17, 2017 4:55 am
If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p

How will i start solving this kind of problem?

OA E
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Source: — Problem Solving |

by GMATGuruNY » Fri Nov 17, 2017 5:48 am
lheiannie07 wrote:If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p
Let p=1.

Since the price doubles every week for a total of 4 weeks, we get:
Price after the first week = 2.
Price after the second week = 4.
Price after the third week = 8.
Price after the fourth week = 16.

Since the price triples the next week, we get:
Price after the fifth week = 3*16 = 48.

Since the price quadruples the next week, we get:
Price after the sixth week = 4*48 = 192.

The question stem asks for the price at the end of the 6-week period.
Thus, the target value is 192.
Now plug p=1 into the answer choices to see which yields the target value of 192.
Only E works:
(12)(2�)(p) = (12)(16)(1) = 192.

The correct answer is E.
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by DavidG@VeritasPrep » Fri Nov 17, 2017 11:32 am
lheiannie07 wrote:If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p

How will i start solving this kind of problem?

OA E
And there's always good old-fashioned algebra.

Start: p
After one week: 2*p
After two weeks: 2*2*p
After three weeks: 2*2*2*p
After four weeks: 2*2*2*2*p = $$2^4*p$$

Next week if triples (so it's multiplied by 3): $$3*2^4*p$$
The last week it quadruples (so it's multiplied by 4) $$4*3*2^4*p = 12 * 2^4 * p$$

The answer is E
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by EconomistGMATTutor » Sat Nov 18, 2017 3:19 pm
If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p

How will i start solving this kind of problem?

OA E
Hi lheiannie07,
Let's take a look at your question.

The price of the coffee is p. It doubles every week for 4 weeks, therefore,
After the first week the price of coffee = 2p
After the second week the price of coffee = 2(2p) = 2^2 p
After the third week the price of coffee = 2(2^2 p) =2^3 p
After the third week the price of coffee= 2(2^3 p) = 2^4 p

A week later the price of coffee triples, therefore the total price of coffee after a week later = 3(2^4 p)

and finally quadruples a week after that, therefore,
The final price of the coffee = 4 * 3(2^4 p) = 12 (2^4 p)

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Scott@TargetTestPrep » Wed Dec 13, 2017 12:36 pm
lheiannie07 wrote:If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p
If the initial price of coffee is p, after week one it's 2p, after week two it’s 4p, after week 3 it’s 8p, and after week 4 it’s 16p.

Since the price triples a week later, the new price is (3)(16p), and since it quadruples the following week, it's (4)(3)(16p) = 12*2^4*p .

Answer: E
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by BTGmoderatorDC » Wed Jan 10, 2018 9:49 pm
Scott@TargetTestPrep wrote:
lheiannie07 wrote:If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p
If the initial price of coffee is p, after week one it's 2p, after week two it’s 4p, after week 3 it’s 8p, and after week 4 it’s 16p.

Since the price triples a week later, the new price is (3)(16p), and since it quadruples the following week, it's (4)(3)(16p) = 12*2^4*p .

Answer: E
Thanks a lot!
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by BTGmoderatorDC » Wed Jan 10, 2018 9:50 pm
EconomistGMATTutor wrote:
If the price of coffee, p, doubles every week for 4 weeks, triples a week later, and finally quadruples a week after that, which of the following represents the price of coffee at the end of period, in terms of p?

A. 15p
B. 7*8*p
C. 7*2^4*p
D. 3*4*8*p
E. 12*2^4*p

How will i start solving this kind of problem?

OA E
Hi lheiannie07,
Let's take a look at your question.

The price of the coffee is p. It doubles every week for 4 weeks, therefore,
After the first week the price of coffee = 2p
After the second week the price of coffee = 2(2p) = 2^2 p
After the third week the price of coffee = 2(2^2 p) =2^3 p
After the third week the price of coffee= 2(2^3 p) = 2^4 p

A week later the price of coffee triples, therefore the total price of coffee after a week later = 3(2^4 p)

and finally quadruples a week after that, therefore,
The final price of the coffee = 4 * 3(2^4 p) = 12 (2^4 p)

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
Thanks a lot!
Join the discussion

by [email protected] » Thu Jan 11, 2018 11:16 am
Hi lheiannie07,

We're told that the price of coffee, P, DOUBLES every week for 4 weeks, TRIPlES a week later, and finally QUADRUPLES a week after that. We're asked which of the following represents the price of coffee at the end of the period, in terms of P. This question can be solved by TESTing VALUES.

IF.... starting price = $1
Week 1 price = $2
Week 2 price = $4
Week 3 price = $8
Week 4 price = $16
Week 5 price = $48
Week 6 price = $192

So we're looking for an answer that equals 192 when P=1. There's only one answer that matches...

Final Answer: E

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