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Eight key answers to interesting math problems in fifth grade of primary school
1.( 1)(294.4- 19.2×6)÷(6+8)

(2) 12.5×0.76×0.4×8×2.5

2. Multiply two numbers (1). If the multiplier is increased by 12, the multiplier remains unchanged and the product is increased by 60; If the multiplicand is constant, the multiplier increases by 12, and the product increases by 144, what is the original product?

(2) June 1990 is Friday, so what day is June 10 in 2000?

3. How many different currencies can hexagonal, dihedral and octagonal form?

4. Now put 12 pieces in the 20 squares in the figure, and put 1 piece at most in each square. It is required that the sum of the number of pieces placed in each row and column is even, and how to place them should be shown on the map.

There is a residential building, and each family subscribes to two different newspapers. Residential buildings subscribe to three newspapers, including 34 China TV newspapers, 30 Beijing Evening News and 22 reference news. So how many * * * subscribe to Beijing Evening News and reference news?

There are three playing cards on the table in a row. We already know:

(1) At least one of the two cards on the right of k is an A.

(2) One of the two cards on the left of A is also A. ..

(3) At least one of the two cards on the left side of the square is a heart.

(4) One of the two cards to the right of the heart is also a heart.

Please write these three cards in order.

7. Arrange even numbers in the table below:

A B C D E

2 4 6 8

16 14 12 10

18 20 22 24

32 30 28 26

……

So, under which letter is the number 1998?

8. Fill in an integer in the box 14 in the figure below. If the sum of the numbers filled in any three adjacent squares is 20, it is known that the fourth square is filled with 9, and 12 square is filled with 7. So, what number should be filled in the eighth square?

9. Divide by natural number 1, 2,3 ...15 into two groups of numbers A and B. Prove that in A or B, the sum of two different numbers must be a complete square number.

10. Cut a piece of paper into 6 pieces, take a few pieces at will, each piece is cut into 6 pieces, then take a few pieces at will, each piece is cut into 6 pieces, and so on. Q: Can you just cut it into 1999 pieces after a limited number of times? Explain why.

Test answer 1

1.( 1)(294.4- 19.2×6)÷(6+8)

= 179.2÷ 14

= 12.8

(2) 12.5×0.76×0.4×8×2.5

=( 12.5×8)×(0.4×2.5)×0.76

= 100× 1×0.76=76

2.

(1) Solution: When two numbers are multiplied, if the multiplicand is increased by 12, the multiplier remains unchanged and the product is increased by 60; If the multiplicand is constant, the multiplier increases by 12, and the product increases by 144, what is the original product?

Let the original title be a× b.

According to the meaning of the question: (a+ 12) × b = a× b+60.

Available: 12× b = 60 b = 5.

Similarly: (b+ 12) × a = a× b+ 144.

So:12× a =144a =12.

\ Original product is: 12× 5 = 60.

(2) Solution: June 1990 is Friday, so what day is June 2000 10?

There are 365 days in a year, 10 plus leap year 1992,1996,2000 plus days in June, July, August and September, 10/0, * *.

3650+3+30+3 1+3 1+30+ 1

=3776

3776÷7=539……3

1990 June 1 Friday, so June 2000/kloc-0 June 1 is Sunday.

3. How many different currencies can hexagonal, dihedral and octagonal form?

All the money * * * has 9 yuan 60 cents.

The smallest denomination is a dime, and there are six of them. Together with the pentagon, they can form all the whole coins of a dime, a dime, a dime and a dollar coin. So you can form all the whole corners from ten cents to nine yuan and sixty cents, and ***96 different kinds of money.

4. Now put 12 pieces in the 20 squares in the figure, and put 1 piece at most in each square. It is required that the sum of the number of pieces placed in each row and column is even, and how to place them should be shown on the map.

Chart (○) stands for chess):

The answer is not unique.

There is a residential building, and each family subscribes to two different newspapers. Residential buildings subscribe to three newspapers, including 34 China TV newspapers, 30 Beijing Evening News and 22 reference news. So how many * * * subscribe to Beijing Evening News and reference news?

Solution: Each family subscribes to 2 different newspapers, while * * * subscribes.

34+30+22 = 86 (copies)

So, * * * has 43.

China TV has 34 subscriptions, so this newspaper has 9 subscriptions.

Those who don't subscribe to China TV News must subscribe to Beijing Evening News and reference news.

So there are nine subscriptions to Beijing Evening News and reference news.

There are three playing cards on the table in a row. We already know:

(1) At least one of the two cards on the right of k is an A.

(2) One of the two cards on the left of A is also A. ..

(3) At least one of the two cards on the left side of the square is a heart.

(4) One of the two cards to the right of the heart is also a heart.

Please write these three cards in order.

Solution: Let the three cards on the table be A, B and C, and there are two cards on the right side of the condition (1)k, then A must be K, and at least one of B and C is A.

According to the condition (2), there is a to the left of A, so it is inevitable that both B and C are A. ..

Similarly, it can be inferred from (4) that A is the heart. From (3), C is a cube, and from (4), B is a heart.

\ The order of the three cards is: K of hearts, A of hearts and A of diamonds.

7. Arrange even numbers in the table below:

A B C D E

2 4 6 8

16 14 12 10

18 20 22 24

32 30 28 26

……

So, under which letter is the number 1998?

Solution: As can be seen from the chart, even numbers are arranged in sequence, and every eight even numbers are arranged in the order of columns B, C, D, E, D, C, B and A. ..

Looking at column A again, the ranking order obtained by column E is cyclical with 16.

1998÷ 16= 124…… 14

Therefore, 1998 and 14 are listed in column B.

8. Fill in an integer in the box 14 in the figure below. If the sum of the numbers filled in any three adjacent squares is 20, it is known that the fourth square is filled with 9, and 12 square is filled with 7. So, what number should be filled in the eighth square?

Solution: Let A, B, C and D be numbers in any four consecutive squares.

a+b+c=20=b+c+d

\a=d

Then, the numbers in 1, 4, 7, 10, 13 are the same, and they are all 9.

Similarly, the numbers in the boxes 3, 6, 9 and 12 are all 7.

Then, the numbers in the boxes 2, 5, 8, 1 1 4 are the same, and they should all be:

20-9-7=4

9. Divide by natural number 1, 2,3 ...15 into two groups of numbers A and B. Prove that in A or B, the sum of two different numbers must be a complete square number.

Solution: Assuming that the sum of two different numbers of group A and group B is a complete square number, we can explain that this is impossible.

Let's set 1 in group A.

1+3=4= , 1+ 15= 16=

\ 3 15 are all in group B.

3+6=9=

6 must be in group a.

6+ 10= 16=

It is concluded that 10 should be in group B. At this time, the sum of two numbers in group B is a complete square number.

10+ 15=25=

Therefore, in Group A or Group B, the sum of two different numbers must be a complete square number.

10. Cut a piece of paper into 6 pieces, take a few pieces at will, each piece is cut into 6 pieces, then take a few pieces at will, each piece is cut into 6 pieces, and so on. Q: Can you just cut it into 1999 pieces after a limited number of times? Explain why.

Solution: suppose that after cutting into 6 pieces, the pieces are taken out for the first time, and each piece is cut into 6 pieces, so there are 5 more pieces. At this time, * * * has:

6+5 = 1+5+5

= 5 (+ 1)+ 1 (block)

The second time, I took out a few pieces from it, cut each piece into 6 pieces and added 5 pieces. At this time, * * * has been

6+5 +5

= 5 (++ 1)+ 1 (block)

And so on, the nth block, cut into 6 pieces * * * has been.

5 (++...++ 1)+ 1 (block)

So after each cut, the total number of papers is a natural number (5k+ 1) (that is, 1 divided by 5).

1999÷5=399……4

So it is impossible to get 1999 sheets of paper.

1. There are 9 trees. We need to plant 10 rows, with 3 trees in each row. Please help us.

According to the meaning of the question, there are 3 trees in each row, and it seems that it takes 30 trees to plant 10 row. However, there are only nine trees now. So at least plant some trees (mathematically called key points) at the intersection of several rows. To this end, we can design six three points (three lines intersect) and three four points (four lines intersect).

A tree is eight meters high. A person climbs four meters every minute and drops three meters. How many minutes does it take him to reach the top of the tree?

(8-4)/(4-3)+ 1=5

Grandpa said to Xiaojun, "I am seven times your age now, six times your age in a few years, five times, four times, three times and two times your age in a few years." How old are Grandpa and Xiaojun now?

Grandpa said to Xiaojun, "I am seven times your age now."

So grandpa's age now is a multiple of 7.

Consider a multiple of 7 within 100, as follows

7 14 2 1 28 35 42 49 56 63 70 77 84 9 1 98

Because this is a practical problem.

Grandpa's age is five figures: 56 63 70 77 84.

Then the corresponding age of Xiaojun is 8 9101112.

Grandpa, it's been x years, six times bigger than Xiaojun.

The column equation (8+x)*6=56+x is not an integer, so the answer that Xiaojun is 8 years old is excluded.

The column equation (9+x)*6=63+x is not an integer, so Xiaojun is 9 years old.

The column equation (10+x)*6=70+x is solved with x=2, and the answer of Xiaojun 10 can be considered.

The column equation (1 1+x)*6=84+x is not an integer, so the answer of Xiaojun 1 1 year is excluded.

In fact, as long as grandpa's age is MINUS 6 times that of Xiaojun, 10 times, the conditions will be met.

So now we have the answer. Xiaojun 10 years old, grandpa is 70 years old.

Then we will verify the known conditions.

Grandpa, who has been for X years, is five times as old as Xiaojun.

Solve the column equation (10+x)*5=70+x to get x=5.

Grandpa, it's been x years, four times bigger than Xiaojun.

Solve the column equation (10+x)*4=70+x to get x= 10.

Grandpa, it's been x years, three times bigger than Xiaojun.

Solve the column equation (10+x)*3=70+x to get x=20.

Grandpa, it's been x years, twice as big as Xiaojun.

Solve the column equation (10+x)*2=70+x to get x=50.

1.( 1)(294.4- 19.2×6)÷(6+8)

(2) 12.5×0.76×0.4×8×2.5

2. Multiply two numbers (1). If the multiplier is increased by 12, the multiplier remains unchanged and the product is increased by 60; If the multiplicand is constant, the multiplier increases by 12, and the product increases by 144, what is the original product?

(2) June 1990 is Friday, so what day is June 10 in 2000?

3. How many different currencies can hexagonal, dihedral and octagonal form?

4. Now put 12 pieces in the 20 squares in the figure, and put 1 piece at most in each square. It is required that the sum of the number of pieces placed in each row and column is even, and how to place them should be shown on the map.

There is a residential building, and each family subscribes to two different newspapers. Residential buildings subscribe to three newspapers, including 34 China TV newspapers, 30 Beijing Evening News and 22 reference news. So how many * * * subscribe to Beijing Evening News and reference news?

There are three playing cards on the table in a row. We already know:

(1) At least one of the two cards on the right of k is an A.

(2) One of the two cards on the left of A is also A. ..

(3) At least one of the two cards on the left side of the square is a heart.

(4) One of the two cards to the right of the heart is also a heart.

Please write these three cards in order.

7. Arrange even numbers in the table below:

A B C D E

2 4 6 8

16 14 12 10

18 20 22 24

32 30 28 26

……

So, under which letter is the number 1998?

8. Fill in an integer in the box 14 in the figure below. If the sum of the numbers filled in any three adjacent squares is 20, it is known that the fourth square is filled with 9, and 12 square is filled with 7. So, what number should be filled in the eighth square?

9. Divide by natural number 1, 2,3 ...15 into two groups of numbers A and B. Prove that in A or B, the sum of two different numbers must be a complete square number.

10. Cut a piece of paper into 6 pieces, take a few pieces at will, each piece is cut into 6 pieces, then take a few pieces at will, each piece is cut into 6 pieces, and so on. Q: Can you just cut it into 1999 pieces after a limited number of times? Explain why.

Test answer 1

1.( 1)(294.4- 19.2×6)÷(6+8)

= 179.2÷ 14

= 12.8

(2) 12.5×0.76×0.4×8×2.5

=( 12.5×8)×(0.4×2.5)×0.76

= 100× 1×0.76=76

2.

(1) Solution: When two numbers are multiplied, if the multiplicand is increased by 12, the multiplier remains unchanged and the product is increased by 60; If the multiplicand is constant, the multiplier increases by 12, and the product increases by 144, what is the original product?

Let the original title be a× b.

According to the meaning of the question: (a+ 12) × b = a× b+60.

Available: 12× b = 60 b = 5.

Similarly: (b+ 12) × a = a× b+ 144.

So:12× a =144a =12.

\ Original product is: 12× 5 = 60.

(2) Solution: June 1990 is Friday, so what day is June 2000 10?

There are 365 days in a year, 10 plus leap year 1992,1996,2000 plus days in June, July, August and September, 10/0, * *.

3650+3+30+3 1+3 1+30+ 1

=3776

3776÷7=539……3

1990 June 1 Friday, so June 2000/kloc-0 June 1 is Sunday.

3. How many different currencies can hexagonal, dihedral and octagonal form?

All the money * * * has 9 yuan 60 cents.

The smallest denomination is a dime, and there are six of them. Together with the pentagon, they can form all the whole coins of a dime, a dime, a dime and a dollar coin. So you can form all the whole corners from ten cents to nine yuan and sixty cents, and ***96 different kinds of money.

4. Now put 12 pieces in the 20 squares in the figure, and put 1 piece at most in each square. It is required that the sum of the number of pieces placed in each row and column is even, and how to place them should be shown on the map.

Chart (○) stands for chess):

The answer is not unique.

There is a residential building, and each family subscribes to two different newspapers. Residential buildings subscribe to three newspapers, including 34 China TV newspapers, 30 Beijing Evening News and 22 reference news. So how many * * * subscribe to Beijing Evening News and reference news?

Solution: Each family subscribes to 2 different newspapers, while * * * subscribes.

34+30+22 = 86 (copies)

So, * * * has 43.

China TV has 34 subscriptions, so this newspaper has 9 subscriptions.

Those who don't subscribe to China TV News must subscribe to Beijing Evening News and reference news.

So there are nine subscriptions to Beijing Evening News and reference news.

There are three playing cards on the table in a row. We already know:

(1) At least one of the two cards on the right of k is an A.

(2) One of the two cards on the left of A is also A. ..

(3) At least one of the two cards on the left side of the square is a heart.

(4) One of the two cards to the right of the heart is also a heart.

Please write these three cards in order.

Solution: Let the three cards on the table be A, B and C, and there are two cards on the right side of the condition (1)k, then A must be K, and at least one of B and C is A.

According to the condition (2), there is a to the left of A, so it is inevitable that both B and C are A. ..

Similarly, it can be inferred from (4) that A is the heart. From (3), C is a cube, and from (4), B is a heart.

\ The order of the three cards is: K of hearts, A of hearts and A of diamonds.

7. Arrange even numbers in the table below:

A B C D E

2 4 6 8

16 14 12 10

18 20 22 24

32 30 28 26

……

So, under which letter is the number 1998?

Solution: As can be seen from the chart, even numbers are arranged in sequence, and every eight even numbers are arranged in the order of columns B, C, D, E, D, C, B and A. ..

Looking at column A again, the ranking order obtained by column E is cyclical with 16.

1998÷ 16= 124…… 14

Therefore, 1998 and 14 are listed in column B.

8. Fill in an integer in the box 14 in the figure below. If the sum of the numbers filled in any three adjacent squares is 20, it is known that the fourth square is filled with 9, and 12 square is filled with 7. So, what number should be filled in the eighth square?

Solution: Let A, B, C and D be numbers in any four consecutive squares.

a+b+c=20=b+c+d

\a=d

Then, the numbers in 1, 4, 7, 10, 13 are the same, and they are all 9.

Similarly, the numbers in the boxes 3, 6, 9 and 12 are all 7.

Then, the numbers in the boxes 2, 5, 8, 1 1 4 are the same, and they should all be:

20-9-7=4

9. Divide by natural number 1, 2,3 ...15 into two groups of numbers A and B. Prove that in A or B, the sum of two different numbers must be a complete square number.

Solution: Assuming that the sum of two different numbers of group A and group B is a complete square number, we can explain that this is impossible.

Let's set 1 in group A.

1+3=4= , 1+ 15= 16=

\ 3 15 are all in group B.

3+6=9=

6 must be in group a.

6+ 10= 16=

It is concluded that 10 should be in group B. At this time, the sum of two numbers in group B is a complete square number.

10+ 15=25=

Therefore, in Group A or Group B, the sum of two different numbers must be a complete square number.

10. Cut a piece of paper into 6 pieces, take a few pieces at will, each piece is cut into 6 pieces, then take a few pieces at will, each piece is cut into 6 pieces, and so on. Q: Can you just cut it into 1999 pieces after a limited number of times? Explain why.

Solution: suppose that after cutting into 6 pieces, the pieces are taken out for the first time, and each piece is cut into 6 pieces, so there are 5 more pieces. At this time, * * * has:

6+5 = 1+5+5

= 5 (+ 1)+ 1 (block)

The second time, I took out a few pieces from it, cut each piece into 6 pieces and added 5 pieces. At this time, * * * has been

6+5 +5

= 5 (++ 1)+ 1 (block)

And so on, the nth block, cut into 6 pieces * * * has been.

5 (++...++ 1)+ 1 (block)

So after each cutting, the total number of papers is the natural number of (5k+ 1) (that is, divided by 5, 1).

1999÷5=399……4

So it is impossible to get 1999 sheets of paper.

1. There are 9 trees. We need to plant 10 rows, with 3 trees in each row. Please help us.

According to the meaning of the question, there are 3 trees in each row, and it seems that it takes 30 trees to plant 10 row. However, there are only nine trees now. So at least plant some trees (mathematically called key points) at the intersection of several rows. To this end, we can design six three points (three lines intersect) and three four points (four lines intersect).

A tree is eight meters high. A person climbs four meters every minute and drops three meters. How many minutes does it take him to reach the top of the tree?

(8-4)/(4-3)+ 1=5

Grandpa said to Xiaojun, "I am seven times your age now, six times your age in a few years, five times, four times, three times and two times your age in a few years." How old are Grandpa and Xiaojun now?

Grandpa said to Xiaojun, "I am seven times your age now."

So grandpa's age now is a multiple of 7.

Consider a multiple of 7 within 100, as follows

7 14 2 1 28 35 42 49 56 63 70 77 84 9 1 98

Because this is a practical problem.

Grandpa's age is five figures: 56 63 70 77 84.

Then the corresponding age of Xiaojun is 8 9101112.

Grandpa, it's been x years, six times bigger than Xiaojun.

The column equation (8+x)*6=56+x is not an integer, so the answer that Xiaojun is 8 years old is excluded.

The column equation (9+x)*6=63+x is not an integer, so Xiaojun is 9 years old.

The column equation (10+x)*6=70+x is solved with x=2, and the answer of Xiaojun 10 can be considered.

The column equation (1 1+x)*6=84+x is not an integer, so the answer of Xiaojun 1 1 year is excluded.

In fact, as long as grandpa's age is MINUS 6 times that of Xiaojun, 10 times, the conditions will be met.

So now we have the answer. Xiaojun 10 years old, grandpa is 70 years old.

Then we will verify the known conditions.

Grandpa, who has been for X years, is five times as old as Xiaojun.

Solve the column equation (10+x)*5=70+x to get x=5.

Grandpa, it's been x years, four times bigger than Xiaojun.

Solve the column equation (10+x)*4=70+x to get x= 10.

Grandpa, it's been x years, three times bigger than Xiaojun.

Solve the column equation (10+x)*3=70+x to get x=20.

Grandpa, it's been x years, twice as big as Xiaojun.

Solve the column equation (10+x)*2=70+x to get x=50.

Final answer

Grandpa is now 70 years old and Xiaojun 10 years old.

Grandpa's age is six times that of Xiaojun in two years.

Grandpa is five times as old as Xiaojun in five years.

After 10, grandpa's age is four times that of Xiaojun.

Twenty years later, grandpa is three times as old as Xiaojun.

Grandpa is twice as old as Xiaojun at 50.

There are still a few that can't be pulled down. See for yourself.

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