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11 Jun 2020, 00:01
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A group of 12 workers of equal productivity start working on a project. After every three days they are joined by two additional workers and they complete the project in 120 days. How many days would it have taken to complete the project if only the core group of 12 workers had worked throughout?
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
11 Jun 2020, 00:14
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effatara
A group of 12 workers of equal productivity start working on a project. After every three days they are joined by two additional workers and they complete the project in 120 days. How many days would it have taken to complete the project if only the core group of 12 workers had worked throughout?
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
Let, productivity of each worker in 1 day = 1 unit
Total work done by workers in 120 days
120 days i.e 120/3 = 40 cycles of 3 days
PROPERTY:
Sum of n terms of an Arithmetic progression with first term a and common difference d \(= (\frac{n}{2})*[2a+(n-1)*d]\)
Sum of n terms of an Arithmetic progression with first term a and common difference d \(= (\frac{n}{2})*[2a+(n-1)*d]\)
Total word units in 120 days\( = 3*(12+14+16+18.....+40terms) = 3*(\frac{40}{2})*[2*12+(40-1)*2] \)
This work when done by only 12 workers throughout will take days = \((\frac{1}{12})\)*\(3*(\frac{40}{2})*[2*12+(40-1)*2]\)
This work when done by only 12 workers throughout will take days \(= (5)*[2*12+(40-1)*2] = 510\)
Answer: Option D
11 Jun 2020, 00:20
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effatara
A group of 12 workers of equal productivity start working on a project. After every three days they are joined by two additional workers and they complete the project in 120 days. How many days would it have taken to complete the project if only the core group of 12 workers had worked throughout?
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
(A) 480
(B) 570
(C) 450
(D) 510
(E) 390
SO first 3 days - 12 work, or 12*3 men days
next 3 days - 12+2 work or 14*3 men days
For final 3 days - 12+(40-1)*2 or 90*3 men days..
Total men days - \(12*3+14*3+...90*3 = 3(12+14+...90)=3*\frac{(12+90)}{2}*40\) , that is 3* average of extreme terms of the AP*number of terms.
\(3*\frac{(12+90)}{2}*40=3*51*40\)
If 12 work on each day, number of days = \(3*51*40/12=51*10=510\)
D
