Primary 5 Math Workbook, Primary 5 Math Workbook Page 20 How to Do

Because these oranges are divided into fruit bowls, whether they are 6 or 7 in a plate, they are just finished, that is, the number of these oranges is both a multiple of 6 and a multiple of 7, that is, a common multiple, and because of the requirement to have at least how many oranges? , so find the least common multiple of 6 and 7. Namely:

There are at least 42 of these oranges.

Because there are 2 ropes that are 18 decimeters and 24 decimeters long, it is now necessary to cut them into small segments of the same length, that is, the length of each small segment is both an approximate of 18 and a divisor of 24, that is, the common divisor, namely: 1 decimeter, 2 decimeter, 3 decimeter, 6 decimeter. If the maximum requirement can be a decimeter, it is 6 decimeters.

1.The least common multiple is 6*7=42 answers, and there are at least 42 answers.

2.The greatest common divisor, 18 and 24, is 6, and the answer can be 6 decimeters.

There are at least 42 oranges.

It can be 6 decimeters.

The multiple of 6*7 should be found to be a small common multiple, then.

6 7=42 So it should be at least 42 oranges.

There are two ropes, one 18 decimeters and one 24 decimeters.

Then ask for their factor.

Then they have the same factor: 6

So two pieces of rope, one 18 decimeters and one 24 decimeters, divide them into two small segments of equal length, each of which should be 6 decimeters.

1: At least 6*7=42.

The greatest common factor of 2:18 and 24 is 6, so 18 decimeters can be divided into 3 segments, and 24 decimeters can be divided into 4 segments.

1) Solution: Suppose that there are x people in general.

5x+100=850

x=150A:. . .

2) Solution: Let the daughter be x years old.

6x=33-3

x=5A:. . .

It's better to do it yourself, 5th grade math is not very difficult.

Can I send the question down? If you write I'll change it.

What are you doing in class?

Didn't have long ears. Do it yourself!!

You send us the question, we don't know where we lost the book, and the title is different.

I only have the Beijing Normal University edition, and I still borrowed the textbook from my sister.

Exercise 5 15/4 1 and 3/5 1/8 13/20 3 and 1/4 1/100 1/4 = 3/8 = 4/5 = 100 7 = 3/20 = 4/25 =

2.Horizontally read ah greater than greater than less than equal to less than 3Read horizontally 7/9 1 3/4 2/3 3/5 3/3

Read horizontally one and four fifteenths one-half seven-eighths one-sixth one-twelfth twenty-three thirty-eighths two-nineths of nineths.

Okay, right?

Is it a reflection question of the Sujiao version? The answer is seconds).

What's the version? But you still have to do it yourself, don't rely on others alone!!

What a version! You have to say it! The human-taught version?

This is a travel problem, and to solve the travel problem, you must learn to use a line diagram to represent the route of action. Then solve the equation according to the distance of row A + distance of row B = total distance series. The total distance of this question is 758 + A round-trip 2 hours round-trip, so let's meet Ship A after X hours.

35x+37x=758+35*2

Solution: Let the time of the encounter be x hours.

35x+37x=758+1*35*2

72x=758+70

72x=828

Solution: Let the time of the encounter be x hours.

35x+37x=758+1*35*2 (because the two ships of A and B meet and are always in motion, so the movement time is the same).

72x=758+70

72x=828

x = A: Ship B meets Ship A after hours.

(758+35*2) (35+37)=hours).

Answer: Ship B met Ship A after an hour.

Solution: Let the time of the encounter be x hours.

35x+37x=758+1*35*2

72x=758+70

72x=828x=

Solution: Let ship B meet ship A after x hours.