In the course of mathematics, various equations and problems are sure to be found, but for many they cause difficulties. The thing is that it is necessary to work out and automate these processes. How to learn to solve problems in mathematics, to understand them, you will learn in this article.
Simple tasks
Let's start with the easiest. To get the right answer to the problem, you need to understand its essence, so you need to train on the simplest examples for elementary school. How to learn how to solve math problems, we will describe to you in this section on specific examples.
Example 1: Vanya and Dima were fishing together, but Dima didn’t bite well. What is the catch of the guys? Dima caught 18 fish less than the whole catch, one of the guys had 14 fish less than the other.
This example is taken from a fourth year math course. To solve a problem, it is necessary to understand its essence, the exact question, what in the end must be found. This example is solved in two simple steps:
18-14 = 4 (fish) - caught Dima;
18 + 4 = 22 (fish) - the guys caught.
Now you can safely record the answer. We recall the main question. What is the total catch? Answer: 22 fish.
Example 2:
A sparrow and an eagle fly, it is known that a sparrow flew fourteen kilometers in two hours, and an eagle flew 210 kilometers in three hours. How many times is the speed of the eagle greater.
We draw attention to the fact that in this example two questions, recording the result, do not forget to indicate two answers.
We pass to the solution. In this problem, you need to know the formula: S = V * T. She is probably known to many.
Decision:
14/2 = 7 (km / h) - sparrow speed;
210/3 = 70 (km / h) - the speed of the eagle;
70/7 = 10 - so many times the speed of an eagle exceeds the speed of a sparrow;
70-7 = 63 (km / h) - how much is the sparrow's speed less than the speed of the eagle.
We write down the answer: 10 times the speed of the eagle exceeds the speed of the sparrow; at 63 km / h the eagle is faster than a sparrow.
More difficult level
How to learn how to solve math problems using tables? Everything is very simple! As a rule, tables are used to simplify and systematize conditions. To understand the essence of this method, we will analyze an example.
Before you is a bookcase with two shelves, on the first three times more books than on the second. If you remove eight books from the first shelf and put 32 on the second, then they will be equally divided. Answer the question: how many books were originally on each shelf?
How to learn how to solve text problems in mathematics, now we will clearly demonstrate everything. To simplify the perception of the conditions, we compile a table.
Condition | 1 shelf | 2 shelf |
| It was | 3x | x |
| Has become | 3x-8 | x + 32 |
Now we can make the equation:
3x-8 = x + 32;
3x-x = 32 + 8;
2x = 40;
x = 20 (books) - was on the second shelf;
20 * 3 = 60 (books) - was on the first shelf.
Answer: 60; 20.
Here is a good example of solving an equation problem using an auxiliary table. It greatly simplifies the perception.
Logics
In the course of mathematics there are more complex tasks. How to learn how to solve logical problems in mathematics, we will consider in this section. To get started, read the condition, it consists of several points:
- Before us is a sheet with numbers from 1 to 2009.
- We crossed out all the odd numbers.
- The numbers standing in odd places were deleted from the remaining ones.
- The last action was performed until only one number remained.
Question: What number is left crossed out?
How to quickly learn how to solve math problems in logic? To begin with, we are in no hurry to write all these numbers and cross out one at a time, believe me, this is a very long and stupid lesson. The task of this type is not difficult to solve in several actions. We suggest reflecting on the solution together.
Decision progress
Let's assume which numbers remain after the first action. If we exclude all the odd ones, then there remain: 2, 4, 6, 8, ..., 2008. Note that they are all multiples of two.
We remove numbers in odd places. What is left with us? 4, 8, 12, ..., 2008. We notice that all of them are a multiple of four (that is, they are divisible by four by a remainder).
Next, remove the numbers on odd places. As a result, we have a number series: 8, 16, 24, ..., 2008. You have probably guessed that they are all multiples of eight.
It is easy to guess about our next actions. Next, leave the numbers a multiple of 16, then 32, then 64, 128, 256.
When we get to the numbers that are multiples of 512, we have only three numbers left: 512, 1024, 1536. The next step is to leave the number that is a multiple of 1024, it is one in our list: 1024.
As you can see, the task is solved elementarily, without much effort and the mass of time spent.
Olympiad
At school, there is such a thing as an olympiad. Children with special skills go there. How to learn how to solve the olympiad problems in mathematics, and what they are, we will consider further.
It’s worth starting from a lower level, further complicating it. To develop skills for solving olympiad problems, we offer examples.
Olympics, Grade 5. Example.
Nine pigs live on our farm; they eat twenty-seven bags of feed in three days. A neighbor farmer asked to leave five of his pigs for five days. How much feed do five pigs need for five days?
Olympics, Grade 6. Example.
A large eagle flies three meters in one second, and an eagle flies one meter in half a second. They simultaneously started from one peak to another. How long will an adult eagle have to wait for its cub if the distance between the peaks is 240 meters?
Solutions
In the last section, we examined two simple olympiad problems for the fifth and sixth grade. How to learn how to solve problems in mathematics of the olympiad level, we suggest considering right now.
Let's start with the fifth grade. What do we need to get started? To find out how many bags nine piglets eat in one day, for this we will make the simplest calculation: 27: 3 = 9. We found the number of bags for nine piglets per day.
Now we calculate how many bags are needed for one pig for one day: 9: 9 = 1. We recall what was said in the condition, the neighbor left five pigs for five days, therefore, we need 5 * 5 = 25 (bags of feed). Answer: 25 bags.
The solution to the problem for the sixth grade:
240: 3 = 80 seconds an adult eagle flew;
the eaglet flies two meters in 1 second, therefore: 80 * 2 = 160 meters the eaglet flies in 80 seconds;
240-180 = 80 meters will remain to fly an eaglet when an adult eagle has already landed on a rock;
80: 2 = 40 seconds, the eagle will still need to fly to the adult eagle.
Answer: 40 seconds.