Elementary school math problems, 100 points are awarded, and everyone solves 100

3. Use a cuboid with a length and width of 2 cm and a height of 6 cm to process a maximum cone, and find the volume of the cone.

Diameter of the cone: 2 cm.

Height: 6 cm.

Area: 5. The sum of all edges and lengths of a cuboid model is 72dm, and the ratio of length, width, and height is 4:3:2, what is its volume and surface area?

Length: 4 (4+3+2)*72=32dm

Width: 3 (4+3+2)*72=24dm

Height: 2 (4+3+2)*72=16dm

Area: 32 * 24 * 16 = 12288 (square decimeters).

1000 points can't be answered, no picture!

Don't have any diagrams?

Who can do it? Are you trying to make it difficult for us or are you making a joke!

You give me the picture and I can do it.

There are 19 more questions, how can I send you a picture??

If you're a handsome guy, you draw it!!

500 points, it's all done for you, and you have to put the picture on.

Brother, you don't have a picture of the two words "big fool", we are all God? Guess?

Let's use our minds to help him solve the problem.

Send me the diagram and the title to my mailbox = =

Big brother, what about those pictures of yours?

It's so easy and it's called a puzzle.

It's a lot, can't you finish it at all.

2.(Premise that there is no water at B) Let the water at A be the unit 1, then A has 7 9 flowing into B. That is, 1*7 9=7 9, and 4 5 flows into c at b, i.e. 7 9*4 5=28 45

Premise that there is water at b) let the water at a be x and the water at b be y, then ((7 9) x+y) 4 5 can be obtained

1. When the denominator b is the smallest, a+b is equal to (43).

The least common multiple of the solution is 42, so a b must be greater than 6 42, but less than 1 6, to form the same numerator, that is, 6 36, so the denominator can be taken between 36-42, so the minimum is 37, so a + b = 6 + 37 = 42

There is a problem with question 2 and cannot be solved.

1) a+b=6+37 equals (43).

The least common multiple of the solution is 42, so a b must be greater than 6 42, but less than 1 6, to form the same numerator, that is, 6 36, so the denominator can be taken between 36-42, so the minimum is 37, so a + b = 6 + 37 = 42

The ratio of boys to girls in class A is: 6:5, assuming that there are 6x boys and 5x girls;

The ratio of boys to girls in class B is: 5:4, assuming that there are 5y boys and 4y girls;

Then: 6x+5x+5y+4y=11x+9y=109, because: 11x+9y=55+54, so: x=5, y=6, 6x+5y=30+30=60 people.

109=55+54

Class A boys: 55x6 11=30 (person).

Class B boys: 54x5 9=30 (person).

The two classes of boys have a total of 30 + 30 = 60 (people).

Class A is a multiple of 11 and Class B is a multiple of 9:

10 + 11n is a multiple of 9, and n<9 then n=455+54

Class A boys: 55x6 11=30 (person).

Class B boys: 54x5 9=30 (person).

The two classes of boys have a total of 30 + 30 = 60 (people).

There are X boys in class A and y boys in class B.

109-x*11 6=[(109-x*11 6)-y]*9 4 is sorted out to get 55x+54y=545*6

Because the two classes of boys are x+y, x+54 55y=59+5 11 or 55 54x+y=60+5 9

Because (x+54 55y) can only take the value x+y=60

It should be: 109*(6 11+4 9)!

Not necessarily right!

Class A has a total of 55, including 30 males

Class B has a total of 54, including 30 males, and a total of 60

The first prize is 196 yuan.

Let the third prize be x yuan, and the first prize will be 4 times the third prize by the conditions, so the first prize is 4x=308, x=77, so as to get the total number of prizes, 539

Now one first prize, two second prizes, three third prizes, still set the third prize as x The equation is 3 x + 2 (2x) + 4 x = 539, and the solution is x = 49, so the first prize is 4 49 = 196 yuan.

From the first award, the first prize is 308 yuan, and the total prize is 308 + 308 2 + 308 4 = 539

There is a second award, and the first prize is set x round, which can be seen to be x+2 (x 2) + 3 (x 4) = 539

i.e. 11x 4 = 539

x=196

It is known that v A = v B + 5, t A = t B.

Car A arrives at place C on the way half an hour earlier than car B, and when car B arrives at place C, car A arrives at place B.

The time it takes for car A to travel from C to B is hours.

and xcb=20km

v A = XCB

v B = 35 km h

t A = t B.

xac v A = xac v b.

xac/40=xac/

xac=140km

xcb=20km

xab=xac+xcb=160km

A speed 20

A shares 20 5 = 4h

The total distance is 40*4=160km

I don't know how to hi again

Let ac=s ac=s+20 and the speed of car B is v and the speed of car A is v+5

s v=(s+20) (v+5) s=4v because A arrives at place C on the way half an hour earlier than B: (v+5) 2=20 v=35

So s=140 s+20=160

Let the velocity of B be x, then the velocity of A is x+5 , ab=y

y/x-y/(x+5)=1/2

20/(x+5)=1/2

Let the speed of A be x km h and the speed of B be y km h, x-y = 5

20 x = solution x = 40

y=35

2 major questions and 2 small questions. x five-thirds y twenty-five z minus seven.

From the first award of the regret slip, the first prize bonus is 308 yuan, and it can be seen that the total prize money is 308 + 308 2 + 308 4 = 539

There is a second award, and the first prize is set x round, which can be seen to be x+2 (x 2) + 3 (x 4) = 539

i.e. 11x 4 = 539

x=196

The first prize is 196 yuan.

Set the third prize as x yuan, by the conditions, the first prize is 4 times the third prize, so the first prize of the silver cover is 4x = 308, x = 77, so as to get the total number of bonuses, 539

Now a first prize fighter, two celebrations a second prize, three third prizes, still set the third prize for x The formula is 3 x + 2 (2x) + 4 x = 539, and the solution is x = 49, so the first prize is 4 49 = 196 yuan.

The first prize is 196 yuan.

Let the third prize be x yuan, and the first prize will be 4 times that of the third prize, so the first prize is 4x=308, x=77, so as to get the total number of bonuses, 539

Now there is one first prize, two second prizes, and three thirds.

and other prizes, and the third prize is still set as x

Formula. is 3 x + 2 (2x) + 4 x = 539, and the solution is x = 49, so the first prize is 4 49 = 196 yuan.

I'm a teacher. Thank you.

The solution is known by the question, if the evaluation.

Two first, second and third prizes.

The second type of fiber prize is 308 2 = 154 yuan.

The third prize is 154 2 = 77 yuan.

The total amount of lead imitation gold (308 + 154 + 77) x 2 = 1078 yuan.

If one first prize, two second prizes, and three third prizes are awarded.

It is equivalent to 2x2 + 2x2 + 3 = 11 third prizes.

The third prize is 1078 11 = 98 yuan.

The first prize is equivalent to 2x2=4 times the third prize.

98x4=392 yuan.

Cylindrical v = sh conical v = 1 3sh

1. Because s is equal and v is equal, the height of the cone is three times the height of the cylinder, so the height of the cylinder is more than 3 cm, and the reason for the question is three times the same as the first one.

3 is triple as it equals low equal volume.

Let the cylindrical height be x, then the cone height is 3x

x+3x=72

x = 18 cylinder equals cone 18 equals 54

4 questions v = 1 3sh

So h=3v s

s=h=3*314/(

5. Question cuboid v=abc cylinder v=sh

ab=s, so the cuboid is higher than the height of the cylinder.

When the volume and the ground area are equal, the cone is three times as tall as the cylinder, so the cone is three times that of the cuboid.