An extremely difficult 7th grade math problem 1

At 7 o'clock, the hour hand and the minute hand are 210 degrees apart, and the difference is 180 degrees when forming a straight line, so the minute hand can go 30 degrees more than the hour hand, the hour hand can go one minute, and the minute hand will go 6 degrees a minute, so set 7 o'clock x minutes into a straight line Column: 6x— solution x=

Go 120 degrees more at a straight angle Column: 6x—

The speed of the minute hand is 12 times that of the hour hand A catch-up question about angles, seven o'clock is the starting point There are two answers to right angles.

Solution: The minute hand advances 360 60 6 degrees.

The hour hand advances 360 (60*24) 1 4 degrees per minute.

So, the minute hand catches up with 6-1 4 23 4 degrees per minute.

Right angle 90 degrees or 270 degrees.

So 90 (23 4) is about 15 minutes and 36 seconds.

270 (23 4) about minutes 54 seconds.

So, 7:15:36 is a right angle, and 7:46:54 is also a right angle.

7:00 The outer angle of the minute and hour hands is 210 degrees.

The minute hand of one minute goes 6° and the hour hand goes.

210 and 2 11 7:38 and 2 11 in a straight line, 90 7:16 and 4 at 11 o'clock at a straight angle.

38 and 2 11 + 180 11 = 54 and 6 11 7: 54 and 6 11 at a straight angle.

1. Subtract one-part of y=0 from one-part of x, get one-part of x=one-part of y, get y=3x, (

x +xy-y ) x +y ) = 3x 10x = three tenths.

2. Original formula = 1 + 2-25 1 = -22

3. It is obtained from (x+3y) (4x-y)=3.

x+3y=12x-3y, we get 11x=6y, the value of x y is 64/11, x +10x+25=(x+5) is greater than 0, so x-1 is less than 0, so x is less than 1

5, (6x -12x+6) (1-x) = 6 (1-x), (1-x) is greater than 0 and is the divisor of 6 x = one-half, one-third, one-quarter, one-seventh,。。

1) The speed of a pedestrian is equivalent to 1 meter per second in kilometers per hour) 2) The speed of a bicycle.

Let the train xm1*22+x) 22=(3*26+x) 26 (the length of the train is equal).

22*13+13x=3*26*11+11x2x=572

x=286

1 msec.

2 Set the train length x meters.

Cyclist speed = meters and seconds.

1*22+x)/22=(3*26+x)/2622*13+13x=3*26*11+11x2x=572

x=286

Let the number in the single digit be x, then the number in the tenth digit is 2x

The original number is 10*2x+x=21x

The current number is 10x+2x=12x

So 21x-12x=36

x=42x=2*4=8

Therefore, this number is 8*10+4=84

Set x laborers to dig and 120-x laborers to haul soil.

5x=（120-x）×3

x = 45 45 laborers for digging and 75 laborers for transporting soil.

The person who excavates and transports the soil is x,y x+y=120 5x=3y, and the solution is x=75 y=45

Set up excavator x person, 5x = (120-x) * 3

x=45 (digging).

120-45 = 75 (earth).

Set up excavation x transport y

According to the title. x+y=120

5x=3y.

x=45y=75

The first question is equal to n

Question 2: n n+1 - n-1 n = (n-(n-1)) n+1)x(n-1)) 1 (nxn-1).

Set: Y tree planted by three teams.

Solution: First team a.

Second team 2a+8

Third team (2a+8 2)-6

a+2a+8+[(2a+8 2)-6]=yWhen a=100.

y=406 trees.

1.Q: The three teams planted =a+2a+8+1 2(2a+8)-6=4a+6

2.When a = 100 trees, the three teams planted a total of trees = 4a + 6 = 406

Trees planted by the second team: 2a+8

Trees planted by the third team: 1 2 (2a+8)-6

Three teams co-planted: a+(2a+8)+1 2(2a+8)-6=4a+6When a=100, 100*4+6=406

The second planting team = 100 * 2 + 8 = 208

Third planting team = 208 2-6 = 98

Second team: A times 2+8

Third pair: (a times 2+8) 2-6

Total: a + a times 2 + 8 + (a times 2 + 8) 2-6

Team 1: A Team 2: (2a+8) Team 3: [(2a+8) 2 6] trees.