OG 138

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sun Jun 08, 2008 5:12 pm

OG 138

by ssraf » Tue Jun 10, 2008 12:24 pm
On 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 was loaded by the day crew?

A)1/2
B)2/5
C)3/5
D)4/5
E)5/8
Source: — Problem Solving |

Master | Next Rank: 500 Posts
Posts: 192
Joined: Tue Apr 29, 2008 7:41 am
Thanked: 14 times

by aatech » Tue Jun 10, 2008 12:48 pm
Ans 3/5

Suppose there are d no of day workers and they loaded x boxes/worker so total no of boxes loaded by day worker = dx

There are 4d/5 night workers and they loaded 3x/4 boxes/worker.. total no of boxes loaded by night worker = (4d/5)*(3x/4) = 3dx/5

ratio = (3dx/5 )/dx = 3/5

Junior | Next Rank: 30 Posts
Posts: 14
Joined: Mon Jun 02, 2008 1:38 pm
Thanked: 3 times

by llewellyn27 » Tue Jun 10, 2008 1:58 pm
Is the Answer 5/8

Assume Days had 20 boxes/ worker and had 10 workers.

Total boxes by Days = 200

Therefore Nights loaded 3/4 * 20 = 15 boxes per worker
And
Nights had 4/5 * 10 = 8 workers

Total boxes by nights = 120

Fraction of boxes loaded by Days as part of whole = Part / whole
= 200/320 = 5/8

Please give some feedback

Master | Next Rank: 500 Posts
Posts: 151
Joined: Tue May 13, 2008 7:04 pm
Location: dallas,tx usa
Thanked: 6 times

yeah i think answer is 'E'

by ektamatta » Tue Jun 10, 2008 2:36 pm
yeah i think answer is 'E'

This is how i solve the problem...by plug in

Assumed day crew loaded = 20 night crew loaded 3/4 of 20=15

then, assumed day crew workers 15 and night crew worker= 4/5 of 15=12

Now according to the question
Night crew= 15@12=180
Day crew = 20@15 =300
When we combine both the crew together we will get 300 +180=480
In question ..we have to find fraction of both the crew by day crew so
300/480=5/8answer

Master | Next Rank: 500 Posts
Posts: 171
Joined: Sat May 03, 2008 3:51 pm
Thanked: 8 times

by wawatan » Tue Jun 10, 2008 3:11 pm
this question is confusing and the best way is to plug in numbers. first you have to break the question up by parts. i think the easiest way is to find how many day crew members vs. night crew members. Lets say the day crew has
20 members. Well night crew (4/5)* 20= 16.

night crew =16
day crew =20

each night crew member loaded 3 boxes
each day crew member loaded 4 boxes.

20*4= 80= boxes total for day crew
16*3= 48= boxes total for night crew

boxes total for day crew/ boxes total for 2 crews = 80/128= 5/8

Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sun Jun 08, 2008 5:12 pm

by ssraf » Tue Jun 10, 2008 5:34 pm
thanks all! OA is 5/8

Senior | Next Rank: 100 Posts
Posts: 33
Joined: Thu Apr 16, 2009 6:58 am
Location: Boston
Thanked: 2 times

by bynddrvn » Wed Aug 25, 2010 4:04 pm
I just found this solution to the Official Guide 12ed problem solving question #138, he uses the table method to solve the problem. Very easy to understand; although, if people keep making the math look so easy we will probably start to see some Physics II questions on the exam. Physics II hurt my brain :(

https://www.youtube.com/watch?v=qwmMtKO3Rn4