Last visit was: 22 Dec 2024, 11:38 It is currently 22 Dec 2024, 11:38

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 30461
Own Kudos [?]: 36818 [5]
Given Kudos: 26100
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12234 [0]
Given Kudos: 136
Send PM
General Discussion
Senior Manager
Senior Manager
Joined: 03 Dec 2020
Posts: 440
Own Kudos [?]: 61 [0]
Given Kudos: 68
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 30461
Own Kudos [?]: 36818 [2]
Given Kudos: 26100
Send PM
Re: A manufacturer makes umbrellas at the cost of c dollars per umbrella, [#permalink]
1
Expert Reply
1
Bookmarks
GreenlightTestPrep wrote:
This is a tough one to use the INPUT-OUTPUT approach, but here is goes:

Let c = $2 (it cost $2 to make each umbrella)
Let x = 10 (we make 10 umbrellas)
Let r = $5 (the retail price is $5 per umbrella)
Let b = $0 (the below-cost sale price is $0 per umbrella)

So, the manufacturer made 10 umbrellas at the cost of $2 per umbrella. So the total cost = $20
We need a 100% profit. So, we must earn $40 in revenue. In other words, we must sell 8 umbrellas at $5 each.
This means we can "sell" 2 umbrellas at the below-cost sale price of $0 each.

At this point, we must plug c = 2, x = 10, r = 5 and b = 0 into each expression and see which one yields an OUTPUT of 2

A. \(\frac{b(2c-r)}{(x-r)}\) = 0 ELIMINATE A

B. \(\frac{2x(c-r)}{(b-r)}\) = 12 ELIMINATE B

C. \(\frac{x(2c-r)}{(b-r)}\) = 2 KEEP C

D. \(\frac{2b(c-r)}{(x-r)}\) = 0 ELIMINATE D

E. \(\frac{2(xc-r)}{(x-r)}\) = 6 ELIMINATE E

Answer : C


Thank you sir for the explanation

Regards
Senior Manager
Senior Manager
Joined: 03 Dec 2020
Posts: 440
Own Kudos [?]: 61 [0]
Given Kudos: 68
Send PM
Re: A manufacturer makes umbrellas at the cost of c dollars per umbrella, [#permalink]
i solved it later, but it requires lot of time.
any simle approach?
avatar
Intern
Intern
Joined: 08 Jun 2021
Posts: 1
Own Kudos [?]: 1 [1]
Given Kudos: 0
Send PM
A manufacturer makes umbrellas at the cost of c dollars per umbrella, [#permalink]
1
This one took me a while unfortunately so I probably would have skipped it on the actual test. I apologize in advance for being unfamiliar on the formatting for this forum.

Cost of manufacturing = c

Retail price = r

Below price = b

total quantity sold = x

Total Profit = 100%

The key to this question is starting with some solid variables and a fundamental definition what profit is.

Lets say the quantity sold for retail price is q1 and the quantity sold for below price is q2. The costs for both remain the same.

q1 + q2 must equal the total quantity x. I.e. x = q1 + q2

100% profit = You make 200% return whatever you spent, but lose 100% due to cost (doubling your investment).

100% profit = 2 x (Cost) - 1 x (Cost) = Cost = cx

Total Profit = Total Revenue - Total Cost
= (r-c)(q1) + (b-c)(q2)

Set Total profit = 100% profit

cx = (r-c)(q1) + (b-c)(q2)

Using substitution, q1 = x - q2. When you plug into this formula and set equal to the cost, you get the final equation

(r-c)(x-q2) + (b-c)(q2) = cx

solving for q2, the number of umbrellas sold at below cost, gives you x(2c-r)/(b-r)
Prep Club for GRE Bot
A manufacturer makes umbrellas at the cost of c dollars per umbrella, [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne