Advice 1: How to solve the problem with parts

One of the most interesting problems in mathematics are problems". They are of three types: determination of one value over another, the definition of two quantities by the sum of these values, the definition of two variables using the difference data values. To the solution process was as easy as possible, you should of course know the material. Examples of how to solve problems of this type.
How to solve the problem with parts
Instruction
1
Condition 1. The novel caught on the river 2.4 kg bass. 4 pieces he gave it to my sister Lena, part 3 – brother Sergei, and one part left. How many kg of fish have got each of the kids?
Solution: Denote the mass of one particle by X (K), then the mass of three parts – 3X (kg), mass in four parts – 4 (kg). It is known that there were 2.4 kg, we set up and solve the equation:

X + 3X + 4X =2,4

8X = 2,4

X = 0,3 (kg) bass was a novel.

1) 3*0,3 = 0,9 (kg) – fish gave Sergei.

2) 4*0,3 = 1,2 (kg) – bass got a sister Lena.

Answer: 1.2 kg, 0.9 kg, 0.3 kg.
2
The next version will also look at the example:

Condition 2. To prepare the pear compote need water, pear and sugar, the weight of which should be proportional to the numbers 4,3, and 2 respectively. How much should you take of each component ( by weight) to prepare 13.5 kg of juice?
Solution: Let for cooking stewed fruit requires a (kg) water b (kg) pears, c (kg) of sugar.

Then a/4=b/3=C/2. Take each of the relations for x. Then a/4=X, b/3=X/2 = X. it follows that a = 4X, b = 3X, c = 2X.
According to the problem, a + b + c =13,5 (kg). From this it follows that

4 + 3 + 2 =13,5

9X = 13.5 cm

X = 1,5

1) 4*1,5 = 6 (kg) water;

2) 3*1,5 = 4,5 kg – pears;

3) 2*1,5 = 3 (kg) – sugar.

Answer: 6, 4.5 and 3 kg.
3
The following type of decision task in part" - finding a fraction of the number and the number from the fraction. When solving problems of this type it is necessary to remember two rules:

1. To find a fraction of a certain number, this number multiplied by this fraction.

2. To find all the number for a given value of its fractions, the given value must be divided by a fraction.
For example, analyze such problems. Condition 3: Find the value of X if 3 / 5th of this number is 30.

Place the solution in the form of the equation:

In accordance with the rule, have

3/5X = 30

X = 30:3/5

X = 50.
4
Condition 4: Find the area of the garden, if you know that 0,7 dug up the entire garden, and left to dig 5400 m2?

Solution:

Take the whole garden for one (1). Then,

1). 1 – 0,7 = 0,3 – not dug up part of the garden;

2). 5400:0.3 mm = 18000(m2) is the area of the garden.

Response: 18000 m2.
Consider one more example.

Condition 5: the Traveler was in transit for 3 days. On the first day he прошел1/ 4 part of the way, the second – 5/9 the rest of the way, on the last day he walked the remaining 16 km away. you Must find all the way of the traveler.

Solution: Take the entire path for X (km). Then, on the first day he walked 1/ 4(km), the second – 5/9(X – 1/ 4) = 5/9*3/4X = 5/12X. Knowing that on the third day he walked 16 km, then:

1/4 + 5/12 + 16=X

1/4+5/12 X=-16

-1/3=-16

X=-16 :(-1/3)

X=48

Answer: the Entire path of the traveler is equal to 48 km.
5
Condition 6: Bought 60 buckets, with a 5-liter was 2 times more than 10 liter. How many parts have to 5литров bucket, a bucket of 10 liters, all the buckets? How many bought a 5 litre and 10 litre buckets?

Let buckets 10 litre is 1 part, then a 5 litre be 2 parts.

1) 1 + 2 = 3 (parts) — account for all buckets;

2) 60:3 = 20 (buckets.) — accounts for 1 piece;

3) 20·2 = 40 (buckets) — you have 2 part (five-liter bucket).
6
Condition 7: homework (algebra, physics and geometry) Roma spent 90 minutes. In physics he spent 3/4 of that time spent on algebra and geometry for 10 minutes less than on physics. How much time Roma spent on each item separately.

Solution: Let x (min) spent on algebra. Then 3/4x (min) went to physics, and geometry elapsed (3/4x – 10) minutes.

Knowing that all the lessons he spent 90 minutes, we set up and solve the equation:

X+3/4x+3/4x-10=90

5/2=100

X=100:5/2

X=40 (min) – left on algebra;

3/4*40=30(min) - physics;

30-10=20 (min) - geometry.

Response: 40 min, 30 min,20 min.
Useful advice
When solving problems in part need to learn to take the suitable value for 1 piece. Learn how to find out how many parts you have on the other values, their sum or difference.

Advice 2 : How to solve the problem with parts

One of the most interesting problems in mathematics are problems". They are of three types: determination of one value over another, the definition of two quantities by the sum of these values, the definition of two variables using the difference data values. To the solution process was as easy as possible, you should of course know the material. Examples of how to solve problems of this type.
How to solve the problem with parts
Instruction
1
Condition 1. The novel caught on the river 2.4 kg bass. 4 pieces he gave it to my sister Lena, part 3 – brother Sergei, and one part left. How many kg of fish have got each of the kids?
Solution: Denote the mass of one particle by X (K), then the mass of three parts – 3X (kg), mass in four parts – 4 (kg). It is known that there were 2.4 kg, we set up and solve the equation:

X + 3X + 4X =2,4

8X = 2,4

X = 0,3 (kg) bass was a novel.

1) 3*0,3 = 0,9 (kg) – fish gave Sergei.

2) 4*0,3 = 1,2 (kg) – bass got a sister Lena.

Answer: 1.2 kg, 0.9 kg, 0.3 kg.
2
The next version will also look at the example:

Condition 2. To prepare the pear compote need water, pear and sugar, the weight of which should be proportional to the numbers 4,3, and 2 respectively. How much should you take of each component ( by weight) to prepare 13.5 kg of juice?
Solution: Let for cooking stewed fruit requires a (kg) water b (kg) pears, c (kg) of sugar.

Then a/4=b/3=C/2. Take each of the relations for x. Then a/4=X, b/3=X/2 = X. it follows that a = 4X, b = 3X, c = 2X.
According to the problem, a + b + c =13,5 (kg). From this it follows that

4 + 3 + 2 =13,5

9X = 13.5 cm

X = 1,5

1) 4*1,5 = 6 (kg) water;

2) 3*1,5 = 4,5 kg – pears;

3) 2*1,5 = 3 (kg) – sugar.

Answer: 6, 4.5 and 3 kg.
3
The following type of decision task in part" - finding a fraction of the number and the number from the fraction. When solving problems of this type it is necessary to remember two rules:

1. To find a fraction of a certain number, this number multiplied by this fraction.

2. To find all the number for a given value of its fractions, the given value must be divided by a fraction.
For example, analyze such problems. Condition 3: Find the value of X if 3 / 5th of this number is 30.

Place the solution in the form of the equation:

In accordance with the rule, have

3/5X = 30

X = 30:3/5

X = 50.
4
Condition 4: Find the area of the garden, if you know that 0,7 dug up the entire garden, and left to dig 5400 m2?

Solution:

Take the whole garden for one (1). Then,

1). 1 – 0,7 = 0,3 – not dug up part of the garden;

2). 5400:0.3 mm = 18000(m2) is the area of the garden.

Response: 18000 m2.
Consider one more example.

Condition 5: the Traveler was in transit for 3 days. On the first day he прошел1/ 4 part of the way, the second – 5/9 the rest of the way, on the last day he walked the remaining 16 km away. you Must find all the way of the traveler.

Solution: Take the entire path for X (km). Then, on the first day he walked 1/ 4(km), the second – 5/9(X – 1/ 4) = 5/9*3/4X = 5/12X. Knowing that on the third day he walked 16 km, then:

1/4 + 5/12 + 16=X

1/4+5/12 X=-16

-1/3=-16

X=-16 :(-1/3)

X=48

Answer: the Entire path of the traveler is equal to 48 km.
5
Condition 6: Bought 60 buckets, with a 5-liter was 2 times more than 10 liter. How many parts have to 5литров bucket, a bucket of 10 liters, all the buckets? How many bought a 5 litre and 10 litre buckets?

Let buckets 10 litre is 1 part, then a 5 litre be 2 parts.

1) 1 + 2 = 3 (parts) — account for all buckets;

2) 60:3 = 20 (buckets.) — accounts for 1 piece;

3) 20·2 = 40 (buckets) — you have 2 part (five-liter bucket).
6
Condition 7: homework (algebra, physics and geometry) Roma spent 90 minutes. In physics he spent 3/4 of that time spent on algebra and geometry for 10 minutes less than on physics. How much time Roma spent on each item separately.

Solution: Let x (min) spent on algebra. Then 3/4x (min) went to physics, and geometry elapsed (3/4x – 10) minutes.

Knowing that all the lessons he spent 90 minutes, we set up and solve the equation:

X+3/4x+3/4x-10=90

5/2=100

X=100:5/2

X=40 (min) – left on algebra;

3/4*40=30(min) - physics;

30-10=20 (min) - geometry.

Response: 40 min, 30 min,20 min.
Useful advice
When solving problems in part need to learn to take the suitable value for 1 piece. Learn how to find out how many parts you have on the other values, their sum or difference.
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