It takes machine A 'x' hours to manufacture a deck of cards that

Folks,

I have seen the replies of experts. However, I have one query on this question.

Solution:

Work to be performed =100 decks

Rate * time = work
100/x * x = 100
Rate 1: 100/x

Similarly Rate 2 : 100/y

Then why posters have taken the rates as 1/x and 1/y.

Rgds,
TGC!

It takes machine A x hours to manufacture ONE deck --> the rate of A = (job)/(time) = 1/x decks per hour;
It takes machine B y hours to manufacture ONE deck --> the rate of B = (job)/(time) = 1/y decks per hour.

Hope it's clear.

Attached is a visual that should help. Bundle's solution is the most elegant, but if you can't pull that off (which many test-takers can't), then this illustrates an admittedly more work-intensive second option.

...and here is a visual version of Bunuel 's explanation.

Bunuel why is the rate (y+x / yx). Isnt that time? Work rule is 1/r + 1/s = 1/h so doing 1/x + 1/y actually delivers time not rate?

Time is a reciprocal of rate:
1/r + 1/s = 1/h

(s + r)/(rs) = 1/h

h = rs/(r+s).

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

\(time*speed=distance\) <--> \(time*rate=job \ done\). For example when we are told that a man can do a certain job in 3 hours we can write: \(3*rate=1\) --> \(rate=\frac{1}{3}\) job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then \(5*(2*rate)=1\) --> so rate of 1 printer is \(rate=\frac{1}{10}\) job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then \(3*(2*rate)=12\) --> so rate of 1 printer is \(rate=2\) pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is \(rate_a=\frac{job}{time}=\frac{1}{2}\) job/hour and B's rate is \(rate_b=\frac{job}{time}=\frac{1}{3}\) job/hour. Combined rate of A and B working simultaneously would be \(rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) job/hour, which means that they will complete \(\frac{5}{6}\) job in one hour working together.

3. For multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that \(t_1\) and \(t_2\) are the respective individual times needed for \(A\) and \(B\) workers (pumps, ...) to complete the job, then time needed for \(A\) and \(B\) working simultaneously to complete the job equals to \(T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}\) hours, which is reciprocal of the sum of their respective rates (\(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\)).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

\(T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}\) hours.



Hope this helps

We can let the rate of machine A = 1/x and the rate of machine B = 1/y. We are given that machine A operates for z hours, so machine A completes (1/x)(z) = z/x decks of cards when operating alone. Thus, there will be 100 - z/x decks left to complete when machines A and B work together.

We can let n = the number of hours that machines A and B work together to complete (100 - z/x) decks and we can create the following equation:

(1/x + 1/y)(n) = 100 - z/x

Multiplying the entire equation by xy, we have:

(y + x)(n) = 100xy - zy

n = y(100x - z)/(y + x)

Answer: B

Hi All,

This question can be solved by TESTing VALUES.

Machine A takes X hours to make a deck of cards.
Machine B takes Y hours to make a deck of cards.

Since the answers are suitably complex-looking, let's choose really small, easy numbers to work with:

X = 1
Y = 2

So…
Machine A takes 1 hour to make a deck of cards
Machine B takes 2 hours to make a deck of cards
When both machines work together for 2 total hours, 3 decks of cards are made.

The question goes on to state that Machine A will work alone for Z hours, then be joined by Machine B until 100 decks are made.

Z = 1

In that first hour, Machine A will produce 1 deck of cards, leaving 99 decks to go. Since the two machines together can produce 3 decks every 2 hours, the remaining 99 decks will take…
2 hours x 33 sets = 66 hours.

We're looking for the answer that equals 66 when we plug in X=1, Y=2 and Z=1 into the answer choices.

While it "looks like" there's a lot of math to be done, most of the answers are way too small to be 66 (and it shouldn't take too long to figure that out). Only one answer equals 66

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi,

I solved this problem with a different approach but failed to get the correct answer. Could anyone please tell me the mistake that I am doing?

Here is my approach:

Total units = 100

Let x = 2 hrs
So, in 1 hr A will produce 50 units

Similarly, y= 4 hrs
So, in 1 hr B will produce 25 units

Now suppose z= 1 hr
that means A will produce 50 units in 1 hr

Remaining = 50 units
So, combined rate is 50/25+ 50
= 50/75 hrs

Let x=2, implying that A's time to produce 1 deck = 2 hours.
Let y=3, implying that B's time to produce 1 deck = 3 hours.

Machine A operates alone for z hours.
Let A produce ALL 100 DECKS.
Since A takes 2 hours per deck, z = (2 hours per deck)(100 decks) = 200 hours.

How long will the two machines operate simultaneously?
Since A produces all 100 decks -- implying that the two machines work together for 0 hours -- the numerator for the correct answer must yield a value of 0 when x=2, y=3 and z=200.
Only B works:
y(100x – z) = 3(100*2 - 200) = 0

Hi aarushisingla,

Based on your post, it appears that you lost track of the 'units' that you were referring to.

If you set X = 2, then that means that it takes Machine A 2 HOURS to produce 1 deck. By extension, in 1 HOUR, Machine A would produce 1/2 of a deck (not 50 decks). The same issue occurs again with Machine B: if Y = 4, then it takes Machine B 4 HOURS to produce 1 deck - and in 1 HOUR, Machine B would produce 1/4 of a desk (not 25 decks).

If you want to proceed in that way, then you now know A's HOURLY RATE (re: 1/2 a deck per hour) and the COMBINED RATE of the two machines (1/2 + 1/4 = 3/4 of a deck each hour). This makes the overall math a bit more difficult than it needs to be, but you could still get the correct answer with it.

GMAT assassins aren't born, they're made,
Rich

VeritasKarishma Bunuel

If i was to plug in values, does that mean machine A takes "x" hours and machine B takes "y" hours for 1 deck of cards ?

I thought the question was not clear how many decks of cards (1 or more ?) machine A takes "x" hours for or B takes "y" hours for.

Hi jabhatta2,

The first sentence refers to the number of hours for each Machine to "....manufacture A deck of cards...", so those rates are relative to producing 1 deck.

GMAT assassins aren't born, they're made,
Rich

Bunuel KarishmaB EMPOWERgmatRichC

Hi All,

I am taking x = 6 and Y = 3 and Z = 300 which is not giving me correct solution. But if I am plugging x = 3 and Y = 6 and Z = 300 which is giving me 0 and that value is correct. How to identify which between x and y should be greater if I am plugging values.

Variables in the question that are repeated in the answer choices is an invitation to Plug In. An underutilized Plugging In trick is to try to end up with zero at the end.
x=1
y=doesn't matter...I'll use 2
z=100
A makes all the decks on its own.
A and B work together for zero hours to finish.

Let's check the answer choices (we need a numerator equal to zero; denominator is irrelevant):

A. \(\frac{200 – 100}{x + y}\) Not zero. Eliminate.

B. \(\frac{2(100 – 100)}{x + y}\) Keep it.

C. \(\frac{200(1 – 100)}{x + y}\) Not zero. Eliminate.

D. \(\frac{1 + 100}{100xy – z}\) Not zero. Eliminate.

E. \(\frac{1 + 2 – 100}{100xy}\) Not zero. Eliminate.

Answer choice B.


ThatDudeKnowsPluggingIn

Hi himgkp1989,

Since you have not shown your 'work', it's not clear what the issue might be with your approach through your first TEST case. Using those numbers though, we would have:

X = 6
Y = 3
Z = 300

Machine A takes 6 hours to complete a deck of cards and Machine B takes 3 hours to complete a deck of cards.

Machine A works alone for 300 hours, so it completes 300/6 = 50 decks of cards. So now we have 100 - 50 = 50 more decks of cards to create.

From this point, every 6 HOURS will give us 3 more finished decks (1 from Machine A and 2 from Machine B). Thus, the remaining 50 decks will take 6(50/3) = 300/3 = 100 hours) to create. The answer choices are all written as fractions, but we're ultimately looking for the one that equals 100 when X=6, Y=3 and Z=300. There's only one that matches.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

himgkp1989

Nothing wrong with your approach. The answer does not depend on which is greater.

A makes a deck in x = 6 hrs,
B makes a deck in y = 3 hrs,
A works for z = 300 hrs and makes 300/6 = 50 decks.

So together, they need to make another 50 decks (to make total 100)

Combined rate = 1/6 + 1/3 = 1/2
so together they will take 2 hrs to make a deck. So to make 50 decks, they need 100 hrs of work together.

When you plug these values in (B), you get 100.

Though, when a question involves so many variables, I would not suggest you to plug in numbers. The probability of a mistake increases multifold.