One of the most interesting tasks in mathematics are tasks "on a part". They are three types: determination of one size through another, determination of two sizes through the sum of these sizes, determination of two sizes through the difference of these sizes. In order that process of the decision became the easiest, it is necessary to know material, of course. On examples we will consider how to solve problems of this kind.

## Instruction

1. Condition 1. The novel caught 2.4 kg of perches on the small river. It gave 4 parts to the sister Lena, 3 parts – to the brother Seryozha, and kept one part. How many kg of perches were received by each of children? Decision: Designate the mass of one part through X (kg), then the mass of three parts – 3X (kg), and the mass of four parts – 4X (kg). It is known that all there were 2.4 kg, we will work out and we will solve the equation: X + 3X + 4X = 2.48X = 2.4X = 0.3 (kg) – perches the Novel.1) 3*0.3 = received 0.9 (kg) – fishes gave to Seryozha.2) 4*0.3 = 1.2 (kg) – perches the sister Lena received. Answer: 1.2 kg, 0.9 kg, 0.3 kg.

2. We will sort the following option on an example too: Condition 2. For preparation of pear compote water, pears and sugar which mass has to be proportional to numbers 4.3 and 2 respectively is necessary. How many it is necessary to take each component (on weight) to prepare 13.5 kg of compote? Decision: Let preparation of compote require a (kg) of water, b (kg) of pears, c (kg) of sugar. Then a/4=b/3= with / 2. Let's take each of the relations for H. Togd of a/4= ò, b/3= ò, with / 2 = X. From this it follows that a = 4X, b = 3X, c = 2H.Po to a statement of the problem, a + b + c of =13.5 (kg). From this follows, chto4kh + 3X + 2X = 13.59X = 13.5X = 1.51) 4*1.5 = 6 (kg) there are waters; 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 the solution of tasks "on a part" - on finding of fraction from number and number from fraction. At the solution of tasks of this kind it is necessary to remember two rules: 1. To find fraction from a certain number, it is necessary to increase this number by this fraction.2. To find all number on a preset value of its fraction, it is necessary to divide this value into fraction. On an example we will sort such tasks. Condition 3: To find value X if 3/5 parts of this number are equal to 30. Let's issue the decision in the form of the equation: According to the rule, imeem3/5X = 30X = 30:3/5X = 50.

4. Condition 4: To find the area of a kitchen garden if it is known what was dug up by 0.7 all kitchen gardens, and it was necessary to dig up 5400 sq.m? Decision: Let's take all kitchen garden for unit (1). Then, 1). 1 – 0.7 = 0.3 - not dug up part of a kitchen garden; 2). 5400:0.3 = 18000 (sq.m) – the area of all kitchen garden. Answer: 18000 sq.m. Let's review one more example. Condition 5: The traveler was in way 3 days. In the first day it proshel1/4 a part of a way, in the second – 5/9 the remained ways, in the last day it passed the remained 16 km. It is necessary to find all way of the traveler. Decision: Let's take all way for X (km). Then, in the first day it passed 1 / 4X (km), in the second – 5/9 (X – 1/4X) = 5/9*3/4X = 5/12X. Knowing that in the third day it passed 16 km: 1/4X + 5/12 + 16=X1/4X+5/12-X = - 16-1/3X = - 16X =-16 :(-1/3)X =48otvet: All way of the traveler is equal to 48 km.

5. Condition 6: Bought 60 buckets, and 5-liter it was twice more, than 10-liter. How many parts are the share of buckets of 5 liters, of buckets of 10 liters, of all buckets? How many did buy 5-liter and 10-liter buckets? Let buckets 10-liter make 1 part, then 5-liter make 2 part.1) 1 + 2 = 3 (parts) — it is the share of all buckets; 2) 60:3 = 20 (buckets.) — it is the share of 1 part; 3) 20 · 2 = 40 (buckets) — are the share of 2 parts (5 liter buckets).

6. Condition 7: Spent for doing homework (algebra, physics and geometry) of Rohm 90 minutes. It spent 3/4 those for physics time that spent for algebra, and on geometry is 10 min. less, than on physics. What is the time Roma spent for each subject separately. Decision: Let x (min.) it spent for algebra. Then 3/4kh (min.) left on physics, and for geometry it is spent (3/4kh – 10) minutes. Knowing that it spent for all lessons 90 minutes, we will work out and we will solve the equation: X +3/4kh +3/4kh-10=905/2kh = 100X=100:5/2X=40 (min.) – left on algebra; 3/4*40=30 (min.) - on physics; 30-10=20 (min.) - on geometry. Answer: 40 min., 30 min., 20 min.