BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Difficult Math Problem #104 - Work

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pm
Thanked: 11 times
Followed by:5 members

Difficult Math Problem #104 - Work

by 800guy » Fri Mar 02, 2007 10:24 am
from diff math doc. oa coming when some people answer with explanations:

Machine A can produce 50 components a day while
machine B only 40. The monthly maintenance cost for
machine A is $1500 while that for machine B is $550. If
each component generates an income of $10 what is the
least number of days per month that the plant has to work
to justify the usage of machine A instead of machine B?

(A) 6
(B) 7
(C) 9
(D) 10
(E) 11
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

by jayhawk2001 » Sat Mar 03, 2007 8:19 am
Let x be the number of days per month required for A to generate more
revenue than B

(50 * 10 * x - 1500) - (40 * 10 * x - 550) > 0

Solving for x
500x - 1500 - 400x + 550 > 0
100x - 950 > 0
or
x > 9.5

So answer should be D (10 days)
Top
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Fri Jan 05, 2007 4:20 am

by slimsohn » Sat Mar 03, 2007 8:21 am
I think that the answer would be 10, (D).

Machine A will generate 50 * $10 = $500/day
Machine B will generate 40* $10 = $400/day

So every day that they both run, machine A makes $100 more revenue. The difference in maintenance cost is $950, so in order to justify using Machine A, it will have to run for at least 10 days which will result in $1000 extra revenue.

That $1000 will offset the difference in maintenance costs plus give added revenue.
Top
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pm
Thanked: 11 times
Followed by:5 members

oa

by 800guy » Mon Mar 05, 2007 12:18 pm
oa:

Let x be the number of days that need to be worked
500x - 1500 > 400x - 550
100x > 950
x > 9.5 (D)
Top
Moderator
Posts: 772
Joined: Wed Aug 30, 2017 6:29 pm
Followed by:6 members

by BTGmoderatorRO » Sun Sep 03, 2017 12:09 pm
The problem can be solved as a word problem

Machine A produces 50 components on a daily basis each at $10
therefore, Machine A will generate 50 * $10 = $500 on a daily basis

Machine B on the other hand produce 40components in a day each at $10
machine B will generate 40 * $10 = $400

monthly maintenance fee of A is $1500
monthly maintenance fee of B is $ 550

Based on their daily productivity, A earn $100 more than B
difference in the maintenance cost is $1500 - $550 = $950

for us to justify the usage of Machine A over B, let us consider the difference in the income earned by A on a daily basis. i.e $10
machine A spend more on maintenance and will only to operate for at least of 10 days which will gives $100 turnover and will cover for the difference in maintenance cost between A and B.
Top
Post new topic Post Reply
• Page 1 of 1