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Several math problems in the summer homework in the second volume of the eighth grade.
1. Let the speed x of the ship in still water be the downstream speed x+3 and the upstream speed x-3.

66/(x+3)=48/(x-3) x= 19

2. It takes x y days for Class A and Class B to complete the task respectively, so Class A and Class B complete the task 1/x 1/y respectively every day.

6/x+6/y= 1 x= 18

4/x+4/y+6/x= 1 gives y=9.

Another explanation is that "Class B was transferred from another task, and it took Class A six days to finish planting".

The number of days required for Class A to complete the task alone = 2 times the number of days required for Class B to complete the task alone.

It takes1(6+6 * 2) =18 days for Class A to complete the task alone.

It takes 18/2=9 days for Class B to complete the task alone.

Judging from "the efficacy is 1.5 times of the original", the time before adopting the new technology is 1.5 times of the later one.

Therefore, before adopting new technology, it takes1.5 *10/(1.5-1) = 30 hours, 1200/30=40 to complete the task.

It takes 30-10=20 hours to complete the task after adopting the new technology, and the processing capacity per hour is 1200/20=60.

4. If the cost is X, the price is (1+p%)x. If the price is reduced by D%, the new price is (1+p%)x *( 1-D%).

There is also "no loss", that is, (1+p%) x * (1-d%) > =x so p & gt=(x*D%)/( 1-D%)