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Bunuel
Type A employees are 100% more productive than type B employees and type B employees are 100% more productive than type C employees. If 30 type A employees need to work for six hours a day, for 50 days to complete a project, how long will it take a team of 80 type B employees and 80 type C employees to complete the same project if they work ten hours a day?

A. 10
B. 15
C. 20
D. 30
E. 60
Solution:

  • Productivity can be equivalent to the Rate of work
  • "Type A employees are 100% more productive than type B" means we can write \(R_A=R_B(1+\frac{100}{100})⇒R_A=2R_B⇒R_B=\frac{R_A}{2}\)
  • "Type B employees are 100% more productive than type C" means we can write \(R_B=R_C(1+\frac{100}{100})⇒R_B=2R_C⇒R_C=\frac{R_A}{4}\)
  • Now, we can make a table for \(W=nRT\) as follows:
    Attachment:
    WRT3.png
    WRT3.png [ 6.93 KiB | Viewed 11389 times ]
  • Let us assume 80 type B and 80 type C employees are working 10 hours a day for n days:
    Attachment:
    WRT4.png
    WRT4.png [ 20.46 KiB | Viewed 11260 times ]
  • Thus, we can say \(800nR_B+800nR_C=9000R_A\)
    \(⇒800n(R_B+R_C)=9000R_A\)
    \(⇒8n(\frac{R_A}{2}+\frac{R_A}{4})=90R_A\)
    \(⇒\frac{24n}{4}=90\)
    \(⇒n=15\)

Hence the right answer is Option B
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Let the productivity of Type C be x.
Then, Type B is 2x and Type A is 4x.

Work done by 30 Type A employees in 6 hours per day for 50 days:
30 employees × 4x × 6 hours × 50 days = 36000x

Combined rate of 80 Type B and 80 Type C employees:
(80 × 2x) + (80 × x) = 240x per hour

Let the number of days they work be d.
Then, total work they do in 10 hours per day for d days = 240x × 10 × d = 2400x × d

Equating both works:
36000x = 2400x × d
d = 15
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Bunuel
Type A employees are 100% more productive than type B employees and type B employees are 100% more productive than type C employees. If 30 type A employees need to work for six hours a day, for 50 days to complete a project, how long will it take a team of 80 type B employees and 80 type C employees to complete the same project if they work ten hours a day?

A. 10
B. 15
C. 20
D. 30
E. 60
This one was quite challenging for me. Here is the approach I experimented with:
1. Setup a "multiplicative ratio table"
rb = Rate of Type B Employees
Employee Type# of Employees*(rate)*(time)=(work)
C80(1/2)*(rb)(10 hr/day)*(t day)
B80(rb)(10 hr/day)*(t day)
A30(2)*(rb)(6 hr/day)*(50 day)1 project

2. Work of (B + C) = (Work of A) = (1 project)
(80)*(rb)*(10)*(t) + (80)*(1/2)*(rb)*(10)*(t) = (30)*(2)*(rb)*(6)*(50)
i) Divide both sides by common terms (A pair of 10s and rb)
(8)*(t) + (4)*(t) = (3)*(2)*(6)*(5)
ii) Solve for t
t*(8+4) = (3)*(12)*(5)
t = 3*5 = 15

ANSWER: B

As an exercise in translating the multiplicative ratio table to English and back to math:
i) MRT -> English
A has three-eights the employees of B or C.
A's rate is twice B's or four times C's.
A works three-fifths (6/10) the hours per day as B or C.
A completes the same amount of work as B or C in fifty days.

ii) English -> Math
For "B or C" statements, we can [factor] out these values (given a multiplicative relationship)
Work of A = Work of B + Work of C
(3)*(2*rb)*(3)*(50) = [(8)*(rb)*(5)*(t)]*(1 + (1/2))
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