Question 1065847: A small factory planned to produce a batch of men’s shirts in 8 days. But producing 10 more shirts per day than planned, they finished the task one day early. How many shirts per day was the factory planning to produce.
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy: Answer by josgarithmetic(39617) (Show Source): Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
I solved this problem under this link couple of days ago
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1065526.html
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1065526.html
Here, for your convenience, I am repeating the solution:
Rate x Time = Output
The original scenario was
R*8 = N (where R is "How many shirts per day was the factory planning to produce", N is the entire output).
The rate increase scenario is
(R+10)*7 = N
In both scenarios, the output N is the same. So set them equal to each other:
R*8 = (R+10)*7
8R = 7R+70
R = 70.
Answer. How many shirts per day was the factory planning to produce? 70.
Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
A small factory planned to produce a batch of men’s shirts in 8 days. But producing 10 more shirts per day than planned, they finished the task one day early. How many shirts per day was the factory planning to produce.
Let original number of shirts to produce, per day, be S
Then number of shirts to be produced in 8 days = 8S
Producing 10 more shirts, per day, results in S + 10 shirts, per day
Finishing 1 day early means that it took 7 (8 – 1) days to produce 8S shirts
We then get: 7(S + 10) = 8S
7S + 70 = 8S
70 = 8S – 7S
S, or number it’d planned to produce =
|
|
|