Current location - Training Enrollment Network - Mathematics courses - Solution to synchronous training of mathematics in the second volume of the sixth grade
Solution to synchronous training of mathematics in the second volume of the sixth grade
A, B and C are distributed on the same road in turn from west to east. A, B, and C start from A, B, and C at the same time, respectively, with A, B heading east and C heading west. B and C meet at a distance of B 18km, A and C meet at B. When A catches up with B at C, C has passed 32km. So, how many kilometers is the distance between AC?

Solution: Assuming that B and C intersect at point D, this problem can be solved by using proportional knowledge.

① First calculate the length of the CD.

When Party B meets Party C, Party B's CD is 18km.

When A catches up with B, line B CD line C 18+32 = 50km.

Then 18: CD = CD: 50, CD×CD= 18×50=30×30.

So the length of the CD is 30 kilometers.

② Calculate the length of AC again.

C line 50+30 = 80km, A line AC.

The length of line C is 32km, and the BC of line A is 30+ 18=48.

So 32: 48 = 80: AC

Therefore, the length of AC is 80 ÷ 32/48 = 120km.

Two people, Party A and Party B, produce a batch of parts. The efficiency ratio of Party A and Party B is 2: 1. Co-shoot for three days, and Party B will shoot alone for the other two days. At this time, Party A produces 14 more parts than Party B. How many parts are there in this batch?

Solution: Take the work efficiency of B as the unit 1.

Then a's work efficiency is 2.

B 2 days to complete 1×2=2.

Otsuichi * * * produces 1×(3+2)=5.

A * * * Output 2×3=6

So the work efficiency B = 14/(6-5)= 14/ day.

A's work efficiency = 14×2 = 28/ day.

A * * * has 28×3+ 14×5= 154 parts.

Or let the work efficiency of Party A and Party B be 2a/ day and A/ day respectively.

2a×3-(3+2)a= 14

6a-5a= 14

a= 14

A * * * has 28×3+ 14×5= 154 parts.