Urgent.... Grade 8 Semester 2 Math Application Questions

4+1 2=9 2 hours Cycling for the same distance takes 9 2 hours longer than riding a car.

Knowing that the ratio of the speed of the bicycle to the speed of the car is 2:5, the time ratio is 5:29 2 5 (5-2)=15 2 hours of cycling time.

9 2 2 (5-2) = 3 hours of travel time.

125 15 2 = 50 3 km Cycling.

125 3=125 3 km By car.

Equation: Let the cycling speed be x km/h and the riding speed be 5x 2 km/h.

125-4x) x-125 (5x 2)=1 2 solution: x=50 3

5x/2 =125/3

Solution: Let the speed of the bicycle be x meters hour, then the speed of the car is meter hour, and the time of the car is t.

4x+(t+

x= 50/3

t=3 car speed (50 3) x5 2 = 250 6 = 125 3 The question is wrong, it is impossible to ride for several hours at 125 meters.

Let the time of the car travel be th, and the speed of the bicycle is xm h

Then there is the equation: 5 2xt = 125(1).

The solution yields x=25m h

So the speed of the bicycle is 25m h and the speed of the car is.

set, the speed of the passenger is x

The cyclist speed is 2 5 x=

then, 125 x+4+1 2=125

solution, x=125 3 (meter hours).

m-hours) In addition, the answer to this question is not logical, and it seems that 125 meters should be changed to 125 kilometers in the original question.

Let the speed of the car be a, the speed of the bicycle be b, and the equation of the column is a b=5 2; 125 b=125 a+, and the system of equations gives a=125 3; b=50/3

Let the time taken by the car be t and the speed is v, then it can be seen from the title that the time of the bicycle is t+ and the speed is 2 5v, and the equation is as follows:

v*t=125

2/5v*(t+

Just solve it.

If the bicycle speed x is set to the car speed of 5 2x

The column equation is 125 x-4+1 2=125 (5 2x).

Then solve the equation.

The bicycle speed is 50 3 meters per hour, and the car speed is 125 3 meters per hour. In addition, the distance between A and B should be kilometers!

1.Solution: If each Type A car has x seats, then each Type B car has (X+5) seats, because the rental of Type A cars is called by a number of blockades, just full, then Type A cars are used for 240 X, and if Type B cars are rented, use [(240 x)-2] cars, X+5)[(240 x)-2]-10=240

The solution is x1=-30 (rounded).

x2=20x+5=25

So each Type A has 20 seats and each Type B has 25 seats.

2.Solution: From the system of equations, in the two sets of solutions, x1=y2, x2=y1, that is, a1=b2, a2=y1, so the value of a1a2-b1b2 is 0

1.If the A-type car has x seats, and (X>0), then the B-type car has (x+5) seats, it can be seen from the question that all teachers and students should be seated, and if the A car is rented, it is required.

240 x cars, if you rent a B car, you need (240 x)-2 cars, at this time, [

240/x)-2 ]×

x+5)=240+10, simplifying the equation gets: (1200 x)-2x=20, that is, x 2+10x-600=0, and solving the equation gives x=20. So.

Type A has 20 seats and Type B has 25 seats.

2.It can be seen from the question.

When b=0, solve the system of equations to get x1=a1=a, y1=b1=0;x2=a2=0,y2=b2=a；So.

a1a2-b1b2=0；

When b≠0, then x,y are not 0, and the state simplification equation system obtains a noisy message: y=b x,x+b x=a; So x 2-ax+b=0;So x1x2=b,x1+x2=a;

In the same way, y1y2=b, y1+y2=a; So.

a1a2-b1b2=0

In summary, a1a2-b1b2=0.

A typed a manuscript that takes 4 hours and B needs to be completed in 5 hours.

A and B play together for 4 hours = A 4 + B 4 = B 5 + B 4 = B 9, so B shares 9 + 6 = 15 hours.

Total workload = B 15 = 3 * B 5 = 3 * A 4 = A 12 hours A uses.

Solution, it takes X hours for A to type this manuscript alone, and it takes Y hours for B to type this manuscript alone.

According to the title, yes.

4/x=5/y

4/x+5/y+6/y=1

Solution: x=12, y=15

Use parentheses yourself.

A typed a 4-hour manuscript and B needed 5 hours to complete, so A's 1-hour manuscript = 5 4 B's 1-hour manuscript.

B alone to type 4 5 4 + 4 + 6 = 15 hours.

If it takes x hours for a separate draft, then x 15 = 4 5, x = 12

Suppose A's velocity is x, then B's velocity is.

The time it takes for A to complete the course is: 80 x, and the time for B to complete the course is 80 (as you know: 80 x-20-60=80 (.)

Solution: x = 1 3 (km min) = 20 km h B is the speed of km h.

Let the velocity of A be x km h, then the velocity of B is km h(80-x) x-80

Multiply both sides at the same time to get:

80-x)* x

The solution is x=20km h

So if the velocity of A is 20 km h, then the velocity of B is 30 km h

The speed of A x

80/x=80/(

x=40 is the root of the square.

Solution: Let the speed of A be x kilometers per hour, then the speed of B is kilometers per hour.

80÷x－80÷

x＝20………The speed of A.

kilometers per hour.........B's velocity.

Let the first speed be x, then the second speed is.

80/x-80/

A x = 20 B.

Solution: If there are x children, then the play has 3x+4, and the last person has 3 toys by the relation 0

The number of toys for the last 1 person Total number of toys The number of toys for all people in front of it, the column formula 0 3x+4 4(x 1) 3

Solution, 5 x 8, since the number of people is an integer, so x = 6,7,8

When x=6, the number of toys is 22;

When x=7, the number of toys is 25;

When x=8, the number of toys is 28;

Answer: 6 children, 22 toys; 7 children, 25 toys; 8 children, 28 toys

Solution: If there are x people, then the game has: (3x+4), according to the meaning of the topic:

3x+4) 4=(x) So: 1, when 3x+4=4(x-1)+1, the solution: x=7, 3x+4=25,2, when 3x+4=4(x-1)+2, the solution: x=6, 3x+4=22, to sum up: human: 7, toy: 25; Or people: 6, toys: 22 A: omitted.

There are more answers to this question for the following reasons:"The last person got less than three"The last person in this sentence can get 2 pieces or one piece can also make 0 pieces, all of which meet the conditions of less than three:

1, get two pieces, that is, less than two pieces are not enough, (4 + 2) (4-3) = 6 (people) 2, get one, that is, less than three pieces are not enough, (4 + 3) (4-3) = 7 (people) 3, get 0 pieces, that is, less than 4 pieces are not enough, (4 + 4) (4-3) = 8 (people).

Solution: Let the number of people be x and the toy y

From the title, it can be seen that the number of toys of the last person may be 0,1,2(1) 3x+4=y

4(x-1)+0=y

The solution is x=8 and y=28

2)①3x+4=y

4(x-1)+1=y

The solution is x=7 and y=25

3)①3x+4=y

4(x-1)+2=y

The solution is x=6 and y=22

Solution: Let the velocity of B be x h, then.

30/x－30/(x＋2)＝

x 10 that fits the topic

The velocity of A is 12 h, and the velocity of B is 10 h.

Solution: Set the flower bed on one side to grow xm, then.

x(25－2x)＝72

The solution is in line with the topic's x 9

The long side of the flower bed is 9 m long, and the short side is 8 m long.

1>.Let the velocity of A be x, then the equation obtained from the problem B velocity is (x-2): 30 (x-2) -30 x = get: x=12, or x=-10 (rounded).

x-2=10

2>.If the length of the two equal sides of the flower bed is x, then the length of the other side is (25-2x), and (25-2x)<10 is x >

The equation is: x*(25-2x)=72

Find: x=8, or x=rounded).

25-2x=9

1.Let B be x kilometers per hour, then A is (x+2) kilometers per hour, and we get 30 x = 30 (x+2)+

The solution yields x1 = 10 and x2 = -12 (rounded).

A: So B is 10 km/h, and A is 10+2=12km/h.

2.Let one side be y meters (the side against the wall), then the other side is (25-2y) meters, and y(25-2y) = 72

The solution yields y1=8, y2=

Note: The 10 meter old wall is useless here).

1.Let A's velocity be x km/h, then B's velocity is x-2 km/h, which is 30 x+ from the title

2.Let the length of the two sides be x,y

Simultaneous xy=72 and 2x+y=25, and finally the solution is x=8, y=9, so the two adjacent sides are 8 meters and 9 meters long.

A speed x, B is x-2000

30 divided by x-30 divided by (x-2) = half an hour) by yourself.

Let this two-digit dozens be x;

Because the number in the tens is 2 greater than the number in the single digit

Then the number of the single digit is x-2 (understand it here) and the product of the number in the ten digit and the number in the single digit is smaller than this two-digit number40 (1) The product of the number in the ten digit and the number in the single digit = x*(x-2) = x 2-2x(2) Two digits = 10x+(x-2) (understand here, the number of tens should be multiplied by 10) Yes.

10x+(x-2)-x*（x-2）=40

11x-2-x^2+2x-40=0

x^2+13x-42=0

x^2-13x+42=0

x-6)(x-7)=0 - This is cross multiplication, have you learned?

If not, you can check it out.

x1 = 6 - the single digit is 4, then the double digit is 64x2 = 7 - the single digit is 5, then the double digit is 75, so the number is 64 or 75

Let this two-digit dozens be x;

10x+(x-2)-x×（x-2）=40

11x-2-x2+2x-40=0

x2+13x-42=0

x2-13x+42=0

x-6)(x-7)=0

x = 6 or 7, so the number is 64 or 75

Is it necessary to set a binary quadratic equation, the product of the number in the ten digit and the number in the single digit is 40 smaller than the two digits, then the single digit can only be 4 or 5, so it is 75 or 64.

Let the ten digits be x, then the single digit is x-2, and the equation x*(x-2)=10*x+x-2-40

Solve this equation x=6 or x=7, and when x=6 the number is 64 and when x=7 the number is 75

The above is solved, but it is not solved.

Equation thinking. As long as you see this kind of problem, you can solve an equation.

Let the digit x on the ten

Then from the title, 10x+(x-2)-x(x-2)=4011x-2-x2+2x-40=0

x2+13x-42=0

x2-13x+42=0

x-6)(x-7)=0

x = 6 or 7, so the number is 64 or 75

x-y=2, 10x+y-x*y=40, so x=y+2, substituting into the equation obtains: 11y+20-y*y-2y=40

Solution: y=4 or y=5. When y=4, x=6; When y=5, y=7Both answers are correct.

1. Let the speed of A be x, then the speed of B is x-2, the time of A is 30 x, and the time of B is 30 (x-2).

The equation 30 x+1 2=30 (x-2) gives x=12, the velocity of A is 12 km h, and the velocity of B is 10 km h

2. Let the velocity of A be x, then the velocity of B is 3x, the time of A, and the time of B are the column equations, so that x=5, the velocity of A is 5km h, and the velocity of B is 15km h

Solution: The first one.

Let B travel x kilometers per hour, then A travels (x+2) kilometers per hour, and there is an equation according to the meaning of the problem.

30 Solve x= Lift.

The second course is 40 minutes late, what does the society mean? Is it 40 minutes later than B or is it another company?