SOLUTION: Three masons build of wall. Mason A builds 7.0m/day , B builds 6.0m/day , and C builds5.0 m/day . Mason B works twice as many days as A , and C works half as many days a

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Question 1055486: Three masons build of wall. Mason A builds 7.0m/day , B builds 6.0m/day , and C builds5.0 m/day . Mason B works twice as many days as A , and C works half as many days as A and B combined. How many days did each work?
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52908) About Me  (Show Source):
You can put this solution on YOUR website!
.
Three masons build of wall. Mason A builds 7.0m/day , B builds 6.0m/day , and C builds 5.0 m/day.
Mason B works twice as many days as A , and C works half as many days as A and B combined. How many days did each work?
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First part of the condition (first line) is not relevant to the question and, therefore, doesn't bring the useful info at all.

The second part and the formulation in whole has no sufficient info to answer the question.


Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!
Three masons build of wall. Mason A builds 7.0m/day , B builds 6.0m/day , and C builds5.0 m/day . Mason B works twice as many days as A , and C works half as many days as A and B combined. How many days did each work?
The masons built 318 m of wall. This was OMITTED.

Let number of days A worked be D
Then number of days B worked = 2D
Number of days C worked =
Amounts built by A, B, and C are: 7D, 6(2D), or 12D, and matrix%281%2C3%2C+5%283D%2F2%29%2C+or%2C+15D%2F2%29, respectively
Since a 318-meter wall was built, we get: 7D+%2B+12D+%2B+15D%2F2+=+318
19D+%2B+15D%2F2+=+318
38D + 15D = 636 ------- Multiplying by LCD, 2
53D = 636
D, or number of days A worked = 636%2F53, or highlight_green%2812%29
Number of days B worked: 2(12), or highlight_green%2824%29
Number of days C worked: 3%2812%29%2F2, or highlight_green%2818%29