Instruction

1

Let us examine how to solve common tasks.

The solution of problems on velocity we need to know a few formulas and be able to correctly write the equation.

Formula for the solution :

S=V*t formula is the way;

V=S/t formula speed;

t =S/V time formula, where S is distance, V - velocity, t - time.

For example, let us consider how to solve tasks of this type.

Condition: the Truck on the way from city a to city B spent 1.5 hours. The second truck has spent 1.2 hours. The speed of the second car more than 15 km/h than the speed of the first. To find the distance between two cities.

Solution: For convenience, use the following table. In it, specify what is known by the condition:

1 auto 2 auto

S X X

V X/1.5 X/1,2

t 1,5 1,2

For X take what you need to find, i.e. the distance. When setting up the equation to be careful, note that all values were in the same dimension (time - in hours, speed in km/h). By the speed of the 2nd car for more speeds 1 to 15 km/h, i.e., V1 - V2=15. Knowing this, we form and solve the equation:

X/1,2 X/1,5=15

1.5 X - 1.2 X - 27=0

0.3 X=27

X=90(km) - distance between cities.

Answer: the Distance between the towns is 90 km.

The solution of problems on velocity we need to know a few formulas and be able to correctly write the equation.

Formula for the solution :

S=V*t formula is the way;

V=S/t formula speed;

t =S/V time formula, where S is distance, V - velocity, t - time.

For example, let us consider how to solve tasks of this type.

Condition: the Truck on the way from city a to city B spent 1.5 hours. The second truck has spent 1.2 hours. The speed of the second car more than 15 km/h than the speed of the first. To find the distance between two cities.

Solution: For convenience, use the following table. In it, specify what is known by the condition:

1 auto 2 auto

S X X

V X/1.5 X/1,2

t 1,5 1,2

For X take what you need to find, i.e. the distance. When setting up the equation to be careful, note that all values were in the same dimension (time - in hours, speed in km/h). By the speed of the 2nd car for more speeds 1 to 15 km/h, i.e., V1 - V2=15. Knowing this, we form and solve the equation:

X/1,2 X/1,5=15

1.5 X - 1.2 X - 27=0

0.3 X=27

X=90(km) - distance between cities.

Answer: the Distance between the towns is 90 km.

2

In solving problems on motion on the water " you must know that there are several types of velocities: the velocity (V) speed current (Vпо tech.), speed against the current (V [PR]. tech.), flow rate (Vтеч.).

Remember the following formula:

Vпо tech=VC+Vтеч.

V [PR]. tech.=VC-Vтеч.

V [PR]. tech=tech Vпо. - 2Vтеч.

Vпо tech.=V [PR]. tech+2Vтеч.

VC=(Vпо tech.+V [PR] tech.)/2 or VC=Vпо tech.+Vтеч.

Vтеч.=(Vпо tech. - V [PR]. tech)/2

For example, let us consider how to solve them.

Condition: boat Speed the flow of 21.8 km/h and upstream 17.2 km/h to Find a private speed boat and speed of river flow.

Solution: According to the formula: VC=(Vпо tech.+V [PR] tech.)/2 and Vтеч.=(Vпо tech. - V [PR]. tech)/2, we find:

Vтеч = (21,8 - 17,2)/2=4,6\2=2,3 (km/h)

VC = V [PR] tech.+Vтеч=17,2+2,3=19,5 (km/h)

Answer: Vc=19,5 (km/h), Vтеч=2,3 (km/h).

Remember the following formula:

Vпо tech=VC+Vтеч.

V [PR]. tech.=VC-Vтеч.

V [PR]. tech=tech Vпо. - 2Vтеч.

Vпо tech.=V [PR]. tech+2Vтеч.

VC=(Vпо tech.+V [PR] tech.)/2 or VC=Vпо tech.+Vтеч.

Vтеч.=(Vпо tech. - V [PR]. tech)/2

For example, let us consider how to solve them.

Condition: boat Speed the flow of 21.8 km/h and upstream 17.2 km/h to Find a private speed boat and speed of river flow.

Solution: According to the formula: VC=(Vпо tech.+V [PR] tech.)/2 and Vтеч.=(Vпо tech. - V [PR]. tech)/2, we find:

Vтеч = (21,8 - 17,2)/2=4,6\2=2,3 (km/h)

VC = V [PR] tech.+Vтеч=17,2+2,3=19,5 (km/h)

Answer: Vc=19,5 (km/h), Vтеч=2,3 (km/h).

3

Objectives to compare quantities

Condition: Lot of 9 bricks 20 kg more than the weight of one brick. Find the mass of one brick.

Solution: let X (K), then the mass of 9 bricks 9X (kg). From the condition it follows that:

9X - X=20

8x=20

X=2,5

Answer: the Weight of one brick is 2.5 kg.

Condition: Lot of 9 bricks 20 kg more than the weight of one brick. Find the mass of one brick.

Solution: let X (K), then the mass of 9 bricks 9X (kg). From the condition it follows that:

9X - X=20

8x=20

X=2,5

Answer: the Weight of one brick is 2.5 kg.

4

Task on fractions. The main rule when dealing with such this type of tasks: to find a fraction of the number, this number multiplied by this fraction.

Condition: the Camper was on the road 3 days. The first day was it? all the way, in the second the remaining 5/9 of the way, and on the third day - the last 16 km to Find the entire path of the tourist.

Solution: Let the entire path of the tourist is equal to X (km). That first day was it? x (km) second day - 5/9(x -?) = 5/9*3/4x = 5/12x. Since the third day it was 16 km, then:

1/4x+5/12x+16=x

1/4x+5/12x-x= - 16

- 1/3x=-16

X=- 16:(-1/3)

X=48

Answer: All the way a tourist is 48 km away.

Condition: the Camper was on the road 3 days. The first day was it? all the way, in the second the remaining 5/9 of the way, and on the third day - the last 16 km to Find the entire path of the tourist.

Solution: Let the entire path of the tourist is equal to X (km). That first day was it? x (km) second day - 5/9(x -?) = 5/9*3/4x = 5/12x. Since the third day it was 16 km, then:

1/4x+5/12x+16=x

1/4x+5/12x-x= - 16

- 1/3x=-16

X=- 16:(-1/3)

X=48

Answer: All the way a tourist is 48 km away.

Useful advice

To easily solve problems, we must learn to translate them into the language of numbers”, using some tricks. Drawing tables and diagrams as possible helps to understand the problem statement, the relationship of the variables. Also facilitates the process of generating the equations. Of course, you have to know the necessary formulas.

# Advice 2: How to solve problems with improper fractions

A fraction is a mathematical notation of simple rational numbers. It is a number that consists of one or more fractions of a unit can be either in decimal and ordinary form. Today operations convert fractions are of great importance not only in mathematics but also in other fields of knowledge.

Instruction

1

Typically, the most common fractions are wrong, and in this case they require specific action on the part of someone who decides examples and

**tasks**with this fraction.2

Take the tutorial to the task. Please read condition, reading it several times and go to solution. Let's see which fractions are to be solved action. This can be incorrect, correct or decimal fraction. Put the correct fractions in the wrong, but remember that to record all response actions will have to run back, already converting improper fraction to the correct. At common fraction the number above the fraction bar (the numerator) is always greater than the number below the line is denominator. To make the translation from proper fractions to improper follow the next steps.

3

Multiply the denominator by the integer and add the result to the numerator. For example, if a fraction of 2 integers 7/9, 9 must be multiplied by 2 and then add 7 to 18 - the end result will be 25/9.

4

Perform all necessary actions according to the task (addition, subtraction, division, multiplication) using the transformed fraction.Take your answer, it will need to be present in the fractions. To do this, divide the numerator by the denominator. For example, if you have to take the number 25/9 in a proper fraction, divide 25 by 9. As 25 9 not evenly divided, the answer is 2 and as many as seven (numerator) ninth (denominator). Now obtained the proper fraction, where the numerator more than the denominator and a part of it.

5

Record the response

**task**correct fraction. Check your action, if it requires the condition to do**task**or teacher.