Children three years old

Children three years of age are beginning to recognize numbers and count things they love. Help keep the natural curiosity of the child, daring her to master numerical concepts. These kinds of games can help your child at the beginning of kindergarten.

When you are walking, count your steps aloud with your child. Tell steps. Count the things you put in your shopping basket. Count the plates at the dinner table. Tell candy in her hand, and count again after eating each. Get ten magnets and a clean cookie sheet. Hold the paper so that your child can not see the one hand, and beat some magnets in it, one at a time. Make him guess how many are in the sheet and then tell for sure. Repeat. Count in ascending order, descending fast, slow, high and low. You will never be boring for your three year old.

Data and dominoes Roll the dice and count the dots. Use big data, small data, color data - every change is a new game. After counting the points count the same number of objects to extend the activity. To really liven up the game, get two dice, one for you and one for your child. Roll the dice. The one with the most points wins! The same games can be played with dominoes brought them a box or bag, or turning instead of rolling them.

Multiplying with your fingers. For children from 3 years of age. Show your children, grandchildren, nephews!

Did you know that you can use your fingers to perform multiplications between numbers 6-10? Therefore, it is necessary to identify the fingers as follows:

For example, to calculate 8x9, abuts finger equivalent to 8 to 9 by the equivalent finger on the other hand, as shown below.

The result will be a two-digit number, where the tens digit is equal to the sum of fingers that are below (including those in contact), and the digit of the units will be equal to the multiplication of fingers that are above. The following figure illustrates the multiplication.

More about discovering ages

The solution is the following:

AGES TWO PEOPLE THERE WERE 8 YEARS IN REASON OF 8 TO 11; NOW IS THE REASON OF 4 TO 5. WHAT IS THE AGE OLDER CURRENTLY?

We call y the age of the youngest person.

We call it x age of the older person.

The problem is that now (currently) ages are in the ratio of 4 to 5. Then:

y / x = 4/5 (equation 1)

The problem is that for 8 years ages were in the ratio of 8 to 11. So:

(8-y) / (x-8) = 8/11 (Equation 2)

Isolating y in equation 1:

4x = y / 5

Putting this value of y in equation 2 we have:

((4x / 5) -8) / (x-8) = 8/11

(4x / 5) = 8/11 -8. (X-8)

Making mmc on both sides we have:

(4x-40) / 5 = (8x-64) / 11

11. (4x-40) = 5 (8x-64)

440 = 44x-40x-320

44x-40x = 440-320

4x = 120

x = 30

So the age of the oldest person is 30 years !!!

Mais questões sobre idades

AS IDADES DE DUAS PESSOAS HÁ 8 ANOS ESTAVAM NA RAZÃO DE 8 PARA 11; AGORA ESTÃO NA RAZÃO DE 4 PARA 5. QUAL É A IDADE DA MAIS VELHA ATUALMENTE?

A solução é a seguinte:

Chamaremos de y a idade da pessoa mais nova.

Chamaremos de x a idade da pessoa mais velha.

O problema diz que agora (atualmente) as idades estão na razão de 4 para 5. Então:

y/x = 4/5 (equação 1)

O problema diz que há 8 anos as idades estavam na razão de 8 para 11. Então:

(y-8)/(x-8) = 8/11 (equação 2)

Isolando y na equação 1:

y = 4x/5

Colocando esse valor de y na equação 2 temos:

((4x/5)-8)/(x-8) = 8/11

(4x/5)-8 = 8/11.(x-8)

Fazendo o mmc dos dois lados temos:

(4x-40) / 5 = (8x-64) / 11

11.(4x-40) = 5.(8x-64)

44x-440 = 40x-320

44x-40x = 440-320

4x = 120

x= 30

Portanto a idade da pessoa mais velha é 30 anos!!! 

Combinatorial analysis

A CAR COMPORTA TWO PASSENGERS IN BANK OF FRONT AND BACK ON THREE BANK. CALCULATE THE NUMBER OF DIFFERENT ALTERNATIVES FOR THE AUTOMOBILE fill USING 7 PEOPLE, SO SUCH PEOPLE NEVER MIND A PLACE IN FRONT SEATS

THE PROBLEM IS RESOLVED AS FOLLOWS:

There are 7 people, and one can never go in the front seat.

Let's call this person of John, for example.

So let's first calculate the number of ways to fill the car WITHOUT John, using the other six only:

As we have 6 people and 5 people in the car then we calculate the array of 6 elements, taken 5-5:

A6,5 = 720

Now let's calculate the number of ways to fill the car WITH John.

We know that John may not be in the front seats, so it should be in one of three banks back.

Then we fix the John in one of the rear seats (4 places left over then in the car), and then calculate the number of ways to put the other 6 people in those four places, that is, an array of six elements, taken 4-4:

A6,4 = 360

The John may be in any of the three rear seats, so we should multiply that result by 3:

3 x A6,4 = 3 x 360 = 1080

The total number of ways to fill the car is the sum of the two arrangements (COM John and John SEM).

So the total number is 720 + 1080 = 1800 ways !!!

Anásise Combinatória

UM AUTOMÓVEL COMPORTA DOIS PASSAGEIROS NO BANCO DA FRENTE E TRÊS NO BANCO DE TRÁS. CALCULE O NÚMERO DE ALTERNATIVAS DISTINTAS PARA LOTAR O AUTOMÓVEL UTILIZANDO 7 PESSOAS, DE MODO QUE UMA DESSAS PESSOAS NUNCA OCUPE UM LUGAR NOS BANCOS DA FRENTE.

Solução:

O PROBLEMA SE RESOLVE DA SEGUINTE MANEIRA:

São 7 pessoas, sendo que uma nunca pode ir num banco da frente.

Vamos chamar essa pessoa de João, por exemplo.

Então primeiro vamos calcular o número de maneiras de lotar o automóvel SEM o João, usando apenas as outras seis pessoas:

Como temos 6 pessoas e 5 lugares no carro então calculamos o arranjo de 6 elementos, tomados 5 a 5:

A_6,5= 720

Agora vamos calcular o número de maneiras de lotar o automóvel COM o João.

Sabemos que o João não pode estar nos bancos da frente, portanto ele deve estar em um dos três bancos de trás.

Então fixamos o João em um dos lugares traseiros (então sobram 4 lugares no carro), e depois calculamos o número de maneiras de colocar as outras 6 pessoas nesses 4 lugares, ou seja, um arranjo de 6 elementos, tomados 4 a 4:

A_6,4= 360

O João pode estar em qualquer um dos três bancos de trás, portanto devemos multiplicar esse resultado por 3:

3 x A_6,4= 3 x 360 = 1080

O número total de maneiras de lotar o automóvel é a soma dos dois arranjos (COM João e SEM João).

Portanto número total é 720+1080 = 1800 maneiras!!! 

Discovering ages

I HAVE THE AGE OF DOUBLE WHAT you had WHEN I WAS YOUR AGE. If thou MY AGE, THE SUM OF OUR AGE WILL BE 45 YEARS. WHAT ARE OUR AGES ???


SOLUTION: Only read after trying to solve!

You had an age which we call the x and today HAS an age which we will call y.

I HAVE twice the age that you had when I was your current age y (double x), ie I HAVE 2x years.

THEN:

You had x and y now has.
I HAD y and I now have 2x.

So we have:

y = 2x-x-y

2y = 3x

x = (2/3) * y

SO substituting the value of x, we have:

You had (2/3) * ye now has y.Eu HAD ye now have (4/3) * y.

Now pay attention to the second sentence:

If thou MY AGE, THE SUM OF OUR AGE WILL BE 45 YEARS.

You have y, and to have my age, which is (4/3) * y, must be added to your age y more (1/3) * y.

Adding y + (1/3) * y you will have my age, that is, you will have (4/3) * y.

As we add (1/3) * y to their age, we must add to my well, ie:

Now I have (4/3) * y + (1/3) * y, then I have (5/3) * y.

The sum of our ages should be equal to 45:

(4/3) * y + (5/3) * y = 45

(9/3) * y = 45

3y = 45

y = 15

Earlier we found that x = (2/3) * y thus x = (2/3) * 15, then x = 10.

FINALLY: WHAT ARE OUR AGES ???

AS SAID EARLIER, YOUR CURRENT AGE IS y, THAT IS 15 YEARS.

AND MY AGE IS 2x, IE 2.10, THAT IS EQUAL TO 20 YEARS.

SO THE AGES ARE 20 AND 15 YEARS !!!