Atult718 wrote: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
On problems like these according to a lot of people say it is always better to plug in numbers and work. How would you solve this question by plugging in numbers?
*pick a total for the number of boxes loaded (choose a number that is a multiple of both denominators) - T = 100
*pick variables for number of boxes loaded - B(d) = day crew boxes loaded; B(n) = night crew boxes loaded
*pick variables for number of each crew - D = # of day crew workers; N = number of night crew workers
-->immediately write down what you are looking for to avoid careless error after doing all the calculations: what is D*B(d) / 100???
--> so you ONLY need D*B(d) and none of the other variables involved
* create first equation - B(n) = 3/4*B(d)
* create second equation - N = 4/5*D
*create second equation - D*B(d) + N*B(n) = 100
since this question is asking for a fraction you dont need to know what the real total number of boxes is nor do you need to know what all the individual numbers are; you only need to know what D*B(d) is since you already know the total.
substitute:
D*B(d) + (4/5*D)(3/4*B(d) = 100
D*B(d) + (3/5)*D*B(d) = 100
(8/5) D*B(d) = 100
D*B(d) = 125/2 <-- you have D*B(d) now, but don't forget what we were looking for!
D*B(d)/100 = (125/2) / 100 =
5/8
