At a loading dock, each worker on the night crew loaded 3/4 : GMAT Problem Solving (PS)
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26 Mar 2012, 02:55
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1. At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew, If the night crew has 4/5 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load?

A. 1/2
B. 2/5
C. 3/5
D. 4/5
E. 5/8

2. As x increases from 165 to 166, which of the following must increse?

I. 2x-5
II. 1-1/x
III. 1/(x^2-x)

A. I only
B. III only
C. I and II
D. I and III
E. II and III

OPEN DISCUSSION OF THE SECOND QUESTION IS HERE: as-x-increases-from-165-to-166-which-of-the-following-must-144732.html

Last edited by Bunuel on 18 Feb 2013, 03:30, edited 2 times in total.
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26 Mar 2012, 03:14
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gayathriz wrote:
1. At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew, If the night crew has 4/5 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load?

A. 1/2

B. 2/5

C. 3/5

D. 4/5

E. 5/8

2. As x Increases from 165 to 166,which of the following must increse?

I. 2x-5

II. 1-1/x

III. 1/square of x - x

A) I only

B) III only

C) I and II

D) I and III

E) II and III

Q #1:
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
A. 1/2
B. 2/5
C. 3/5
D. 4/5
E. 5/8

Pick some smart numbers.
Let the day crew has 5 workers and each day crew worker loads 4 boxes, so total boxes loaded by the day crew is 5*4=20;

Then the night crew would have 5*4/5=4 workers and each night crew worker would load 4*3/4=3 boxes, so total boxes loaded by the night crew would be 4*3=12;

Fraction of day crew is 20/(20+12)=20/32=5/8.

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26 Mar 2012, 03:22
Q#2:
As x increases from 165 to 166, which of the following must increse?

I. 2x-5
II. 1-1/x
III. 1/(x^2-x)

A. I only
B. III only
C. I and II
D. I and III
E. II and III

I. $$2x - 5$$ --> $$x$$ increases from 165 to 166 --> $$2x$$ increases --> $$2x-5$$ increases. Correct.

II. $$1-\frac{1}{x}$$ --> $$x$$ increases from 165 to 166 --> $$\frac{1}{x}$$ decreases --> $$1-\frac{1}{x}$$ increases. Correct.

At this point you can stop and even not consider III, since only answer choice to have both I and II is C.

Still if interested:
III. $$\frac{1}{x^2-x}$$ --> $$x$$ increases from 165 to 166 --> $$x^2$$ increases more than $$x$$ --> $$x^2-x$$ increases --> $$\frac{1}{x^2-x}$$ decreases.

P.S. Pleas post ONE QUESTION PER TOPIC.
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22 Oct 2012, 07:16
Sir Bunuel,
Could you please provide an alternate soln to 1 without picking smart numbers?
Regards,
Sach
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24 Oct 2012, 04:13
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sachindia wrote:
Sir Bunuel,
Could you please provide an alternate soln to 1 without picking smart numbers?
Regards,
Sach

total boxes loaded = numbers of workers * boxes loaded per worker

For day shift,
let numbers of workers be x and boxes per worker be y

then for night,
number of workers = 4x/5
boxes per worker = 3y/4

total boxes loaded by day crew: xy
total boxes loaded by night crew: 4x/5 * 3y/4 = 3xy/5

So, ratio of boxes loaded by day crew to total boxes loaded = xy/(xy+3xy/5) = 1/(1+3/5) = 5/8

Cheers,
CJ
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16 Feb 2013, 08:56
x^2-x= x(x-1)
when x increase, both x and x-1 increase so the result increase

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16 Feb 2013, 10:15
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Sachin9 wrote:
Sir Bunuel,
Could you please provide an alternate soln to 1 without picking smart numbers?
Regards,
Sach

make a grid like this

------------------------------------------------------Night---------------day

No. of workers------------------------------------(4/5)y---------------y

No. of boxes loaded by each worker ----------(3/4)x-----------------x

Total No of boxes loaded ----------------------(4/5)(3/4)xy-------xy

Now we are asked ratio of TOTAL NO OF BOX LOADED DAY / (TOTAL NO OF BOX LOADED DAY + TOTAL NO OF BOX LOADED NIGHT)

XY/(XY+(3/5)XY) ----------> (XY/1) / (8XY/5) --------> 5/8 ----------> E

I) in 2x-5 if x increase by 1 then equation will be 2(x+1)-5. If x is positive the equation will increase by 2 and if x is negative the equation will decrease by 2. We are given in ques that x increased from 165 to 166 that means x is positive so the equation will increase by 2. Eliminate B and E

II) 1-(1/x) now this equation will increase if 1/x decrease and 1/x will decrease if x increase. x is increased so this equation is also sure to increase. Eliminate A. Now Option D will also get eliminated because it does not contain II, so from here we do not need to proceed to check III. The correct choice is C. However for the study part we will check III

III) 1/(X^2-x) -------> 1/x(x-1) -------> we know as denominator increased the fraction decreases. in this case with the increase in x the denominator increases and the fraction decreases. so this equation will not increase with an increase in X.

Regards,

Abhijit
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16 Feb 2013, 21:02
1. We have to pick a no. for eg 20 (LCM of 4 and 5) and form a rough matrix:
Night crew Day crew
Boxes 3/4*20=15 20
Team size 4/5*20=16 20 Proportion of boxes loaded by day crew= 400/640=5/8 =Option E
Boxes per crew 15*16=240 400

2. We have plug in 165 and 166 in I, as it satisfies the question eliminate option C and E and check by plugging in the values in II and III. we will find option C is the answer.
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18 Feb 2013, 03:30

OPEN DISCUSSION OF THE SECOND QUESTION IS HERE: as-x-increases-from-165-to-166-which-of-the-following-must-144732.html

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