SOLUTION: Normally a factory produces 400 radios in x days. IF the factory were to produce 20 more radios each day, then it would take 10 days less to produce 400 radios. find x
Question 1191077: Normally a factory produces 400 radios in x days. IF the factory were to produce 20 more radios each day, then it would take 10 days less to produce 400 radios. find x Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website! .
Normally a factory produces 400 radios in x days.
IF the factory were to produce 20 more radios each day, then it would take 10 days less to produce 400 radios.
find x
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Producing x items per day, it will take days to make 400 items.
Producing (x+20) items each day, it will take days to make 400 items.
The difference - is 10 days, according to the problem.
So we have this "time" equation
- = 10 days. (1)
Dividing both sides by 10 gives
- = 1.
At this point, you may guess the solution MENTALLY: it is x = 20.
Alternatively, you may reduce it to the quadratic equation and solve it formally.
For it, multiply both sides by x*(x+20). You will get
40*(x+20) + 40*x = x*(x+20)
40x + 800 - 40x = x^2 + 20x
x^2 + 20x - 800 = 0
Factor left side
(x-20)*(x+40) = 0
Of the two roots, 20 and -40, we disregard the negative root x= -40 and accept the positive root x= 20. ANSWERCHECK. We will check the starting equation (1)
- = - = 20 - 10 = 10 days. ! Correct !
The other tutor has provided a good detailed response showing a formal algebraic solution.
If a formal algebraic solution is not required, you can find it quickly with a bit of mental arithmetic.
The conditions of the problem require you to find two ways to expression 400 as the product of two numbers...
ab=400
cd=400
... in which the difference between a and c (the numbers of radios produced per day) is 20 and the difference between b and d (the number of days) is 10.
Quick trial and error should easily find
20*20=400
40*10=400
Those show that 400 radios can be produced 20 per day for 20 days, or 40 per day for 10 days. Those numbers satisfy the conditions of the problem: producing 20 more per day requires 10 fewer days.