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Mathematics desk and chair
1) implicit quantification: the labor cost of repairing each set of tables and chairs is fixed, that is, the total amount of money is fixed.

Solution: If Party A uses X days and Party B uses x-20 days (according to the number of days x the amount of money per day = the total amount of money).

40x= 120X(x-20)

40x= 120x-2400

80x=2400

x=30

B: 30-20= 10 (days)

Without the maintenance price of each set of tables and chairs, we can know how many sets of tables and chairs Party A and Party B maintain every day.

2) Scheme 1:

Nail repair cost: 40X30= 1200 (Yuan) Quality Supervision:10x30 = 300 * * Cost: 1200+300= 1500 (Yuan).

Option 2:

Second maintenance requirements:120x10 =1200 (yuan) Quality supervision:10x10 =100 * * Cost:1200+.

Option 3:

1, daily workload of Party A and Party B * * *:1/30+10 = 2/15.

2. Days required for Party A and Party B to cooperate together: 1 divided by 2/ 15=7.5 (days).

A: 40X7.5=300 yuan; B: 120X7.5=900 yuan; Quality Supervision: 10X7.5=75 yuan.

Supply and demand price: 300+900+75= 1275 (yuan)

1500 > 1300 > 1275

Option 3 saves money and time.

Note: Personally, I think each set of tables and chairs is short of repair price.