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Difficulty:
95%
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Question Stats:
34% (02:33) correct
66%
(02:32)
wrong
based on 112
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Three labourers worked together for 30 days, in the course of work, all of them remained absent for few days. One of them was absent for 10 days more than the second labourer and the third labourer did one-third of the total work. How many days more than the third labourer was the first one absent?
(A) 4
(B) 5
(C) 6
(D) 7
(E) Cannot be determined
Are You Up For the Challenge: 700 Level Questions
(A) 4
(B) 5
(C) 6
(D) 7
(E) Cannot be determined
Are You Up For the Challenge: 700 Level Questions
18 Jan 2022, 10:11
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Let k be the efficiency of A, B and C
And working days are x-10, x, y
According to question
(x-10)k + (x)k = 2/3
(y)k = 1/3
k =1/3y
substituting the value of k in above equation
(x-10)1/3y + (x)1/3y = 2/3
x-5 = y
y-x = 5
And working days are x-10, x, y
According to question
(x-10)k + (x)k = 2/3
(y)k = 1/3
k =1/3y
substituting the value of k in above equation
(x-10)1/3y + (x)1/3y = 2/3
x-5 = y
y-x = 5
General Discussion
20 Jun 2024, 06:06
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Set a1 and a3 as days absent for workers 1 & 3 respectively.
Worker 3 did 1/3 of the work and with R = rate of work:
1/3 = (30-a3)R so
R = 1/(3(30-a3))
Since worker 2 was absent 10 fewer days than 1, his time worked was:
30-a1+10 = 40-a1
Therefore, 2/3 of the work was done by workers 1&2:
2/3 = R(30-a1+40-a1) =
R(70-2a1)
Therefore R also equals:
1/(3(35-a1)
The denominators of the two expressions for R above must be equal:
30-a3 = 35-a1 and
a1-a3 = 5
Posted from my mobile device
Worker 3 did 1/3 of the work and with R = rate of work:
1/3 = (30-a3)R so
R = 1/(3(30-a3))
Since worker 2 was absent 10 fewer days than 1, his time worked was:
30-a1+10 = 40-a1
Therefore, 2/3 of the work was done by workers 1&2:
2/3 = R(30-a1+40-a1) =
R(70-2a1)
Therefore R also equals:
1/(3(35-a1)
The denominators of the two expressions for R above must be equal:
30-a3 = 35-a1 and
a1-a3 = 5
Posted from my mobile device
