Instruction
1
Tasks "on the job". For their solution it is necessary to know the definition and formula. Remember the following:

A=R*t formula works;

P=A/t – formula performance.

t=A/P – time formula, where A is work, f - labor productivity, and t is time.

If the problem statement does not specify the work, then it take 1.
2
The examples will examine how one can solve such problems.

Condition. Two workers, working at the same time, dug up the garden for 6 h. First, a worker might perform the same work in 10 hours how many hours the second worker can dig up the garden?

Solution: Take all the work for 1. Then, in accordance with the formula for productivity is P=A/t , 1/10 of the work is done the first working for 1hour. 6/10 he does for 6 hours. Therefore, the second work in 6 hours makes 4/10 ( 1 – 6/10). We determined that productivity of the second worker is equal to 4/10. In a joint operation, according to the problem is 6 hours. For X take what you need to find, i.e. the work of the second worker. Knowing that t=6, P=4/10, then write and solve the equation:

0.4 x=6,

x=6/0,4,

x=15.

Response: the Second worker can plant a garden in 15 hours.
3
Let us consider another example: For filling the container with water, there are three pipes. The first tube for filling the container must time three times less than the second, and 2 hours more than a third. Three pipes operating simultaneously, fill the container for 3h, but the conditions simultaneously can work only two tubes. Determine the minimum cost of filling the container, if the value of 1H of one of the tubes is 230 rubles.

Solution: This problem is convenient to solve using the table.

1). Take all of the work for 1. For X we take the time required for the third pipe. The condition of the first tube need 2 hours more than the third. Then the first tube need (X+2) hours. And the third pipe need 3 times more time than the first, i.e. 3(X+2). Based on the performance formula, we get: 1/(X+2) – the performance of the first trumpet, 1/3(X+2) – a second tube, 1\X – third of the pipe. Put all the data in the table.

Work time,the hour performance

1 tube A=1 t=(X+2) P=1/X+2

2 pipe A=1 t=3(X+2) P=1/3(X+2)

3 pipe A=1 t=X P=1/X

Along A=1 t=3 P=1/3

Knowing that the joint performance is equal to 1/3, we set up and solve the equation:

1/(X+2)+1/3(X+2)+1/X=1/3

1/(X+2)+1/3(X+3)+1/X-1/3=0

3X+X+3X+6-x2-2X=0

5X+6-x2=0

X2-5X-6=0

When solving quadratic equations find the roots. It turns out

X=6(hours) – time to need a third pipe for filling the container.

From this it follows that the time should the first pipe is equal to (6+2)=8 (hours) and the second = 24(hours).

2). From these data we conclude that the minimum time is for the 1 and 3 pipes ,i.e., 14h.

3). Determine the minimum cost of filling the container with two pipes.

230*14=3220(RUB)

Response: 3220 RUB.
4
There are tasks the most difficult, where you must enter several variables.

Condition: the expert and the trainee, working together, have done a certain work in 12 days. If the first specialist performed one half and then the second half finished one Intern, all would have gone to 25 days.

a) Find the time that could spend a specialist to complete all of the work, provided that it will work faster and one Intern.

b) How to divide the workers received for work sharing of 15000 rubles?
1).Let all the work specialist can perform in X days, and Intern for Y days.

Get 1 day specialist performs 1/X operation, and the trainee for 1/Raboty.

2). Knowing that, working together, for all the work it took them 12 days will get:

(1/X+1/Y)=1/12 – ‘this is the first equation.

The condition, working by turns, alone, had spent 25 days will get:

X/2+Y/2=25

X+Y=50

Y=50-X is the second equation.

3) Substitute the second equation into the first gives: (50 - x +x) / (x(x-50)) = 1/12

X2-50X + 600 = 0,x1= 20,x2=30 (then Y=20) that satisfies the condition.