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28 May 2019, 21:35
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C
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72% (02:14) correct
28%
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A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?
A. 10
B. 12
C. 15
D. 18
E. 20
A. 10
B. 12
C. 15
D. 18
E. 20
wizmaverick
Joined: 22 May 2019
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Schools: Darden '23 (A) ISB'22 (A) IESE'23 (D) Haas '23 (WL) Anderson '23 (I) Tepper '23 (A$) McCombs '23 (A$) Fuqua '23 (D)
GMAT 1: 760 Q50 V42
Schools: Darden '23 (A) ISB'22 (A) IESE'23 (D) Haas '23 (WL) Anderson '23 (I) Tepper '23 (A$) McCombs '23 (A$) Fuqua '23 (D)
GMAT 1: 760 Q50 V42
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28 May 2019, 22:02
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The answer should be C IMO
The series reduced to an Arithmetic Progression with the first term as 1/120 and difference of 1/120.
The sum of the Arithmetic Progression should equal 1
On solving the quadratic equation obtained, you get two values of n : 15 and -16
As the number of workers can't be negative, the answer should be 15
Posted from my mobile device
The series reduced to an Arithmetic Progression with the first term as 1/120 and difference of 1/120.
The sum of the Arithmetic Progression should equal 1
On solving the quadratic equation obtained, you get two values of n : 15 and -16
As the number of workers can't be negative, the answer should be 15
Posted from my mobile device
28 May 2019, 22:48
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Time taken by a person to complete 1 job = 120
Work done by a person in 1 day = 1/120
Work done on 2nd day by 2 persons of same efficiency = 1/120 + 2/120 = 3/120
Work done on 3rd day by 3 persons . of same efficiency = 1/120 + 2/120 + 3/120 = 6/120
Work done on nth day by n persons of same efficiency = (1+2+3+4....... + n)/120
1 Job is completed on nth day. So the work done will be equal to 1:
(1+2+3+4....... + n)/120 = 1
n(n+1)/2 = 120 ( Sum of 1st n natural numbers = n(n+1)/2 )
Substituting the values from the options, we'll get n=15
The correct answer is C
Work done by a person in 1 day = 1/120
Work done on 2nd day by 2 persons of same efficiency = 1/120 + 2/120 = 3/120
Work done on 3rd day by 3 persons . of same efficiency = 1/120 + 2/120 + 3/120 = 6/120
Work done on nth day by n persons of same efficiency = (1+2+3+4....... + n)/120
1 Job is completed on nth day. So the work done will be equal to 1:
(1+2+3+4....... + n)/120 = 1
n(n+1)/2 = 120 ( Sum of 1st n natural numbers = n(n+1)/2 )
Substituting the values from the options, we'll get n=15
The correct answer is C
