C can complete the work alone in 60 days. A and B take 40% and 75% more time than C. The work was started by A and B, and C worked with A on every third day, In how many day the work will be completed.
Answers
Answer:
48 ⅘
Step-by-step explanation:
C an complete work in = 60 days
A takes 40% more time than C, So time taken by A = (40/100)*60 + 60
= 84 days
B takes 75% more time than C, So time taken by B = 75% of 60 + 60
= (75/100) * 60 + 60
= 105 days
(Note: To make calculation easy take total work as LCM of time taken by A, B and C)
So,
Person Days Rate Total work
A 84 20 1680
B 105 16 1680
C 60 28 1680
For 3 days:
First day A worked alone + second day A worked alone + third day all 3 worked
20 + 20 +(20+16+28)
= 104 unit of work in 3 days
No of days = Total work / Unit in 3 days
= 1680 / 104 unit/3days (in integers)
= 16 *3 (as rate is per 3 days)
= 48 days
now
in 48 days work done = 1664
Remaining work = 1680 -1664 = 16 unit
On 49 day A will work Alone
So,
20 unit work in 1 day
16 unit work in = (1/20)*16
= 16/20
= 4/5
So total time is 48 ⅘
Given: C can complete the work alone in 60 days.
A and B take 40% and 75% more time than C.
The work was started by A and B, and C worked with A on every third day, In how many day the work will be completed.
To Find :
Solution:
C can complete the work alone in 60 days.
A can complete work in 60 + (40/100)60 = 84 days
B can complete work in 60 + (75/100)60 = 105 days
Work Done in 3 Days =
3 * (1/84) + 1/105 + 1/60
= (15 + 4 + 7)/420
=26/420
= 13/210
Work done in 16 * 3 = 48 days
= 16 * 13/210
= 208/210
Work left = 2/210 = 1/105
1/105 Work done by A in (1/105)/(1/84) = 4/5 day
Hence work will be completed in 48 4/5 Days
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