Arrcoding to the context, 4 balls must be put into 4 different tubes that each tube must have a ball. So the free balls are: \(7-4=3\left(balls\right)\)

There are: \(6.3=18\left(ways\right)\) divided by 1 free ball/tube.

There are: \(4.3.3=36\left(ways\right)\) divided by 2 free ball/tube.

There are: \(4.1=4\left(ways\right)\)divided by 3 free ball/tube.

There are total: \(18+36+4=58\left(ways\right)\)

ANSWER: 58 ways

I'll just answer this, who thinks it's correct will .

100 + 100 = 200

Arrcoding to the context, \(\dfrac{x}{2}=\dfrac{y}{3}\)

=>\(x.3=y.2\)   => \(\dfrac{3}{2}x=y\)  => \(x< y\)

We have 5 cases that satisfy\(x< y\)   and \(x+y=12\) . They are: 1+11 ; 2+10 ; 3+9 ; 4+8 ; 5+7. (We avoid\(-x\) and \(y\)   because \(-x\ne y\) ) In that 5 cases, no case are suitable for this text => NO ANSWER.

Answer: NO ANSWER

There are 2 vertices in the left triangle that can connect to 2 vertices in the right triangle. So there are \(2.2=4\left(ways\right)\)

ANSWER: 4 ways.

The total weight of 9 pearls are:

\(12+2+3+4+5+6+7+8+9=56\left(g\right)\)

The total weight of 4 rings are:

\(17+13+7+5=42\left(g\right)\)

The weight of the unused pearl is:

\(56-42=14\left(g\right)\)

Because there are no \(14g\) pearl => NO ANSWER.

Answer: NO ANSWER.

(This question is only enable to answer if it's weight is changed lower with 1 pearl in the distance of \(14-2=12\left(g\right)\))

6 dollars ($6) = 600 cents, 1 nickel = 5 cents, 1 dime = 10 cents.

There were: \(600:\left(5+10\right)=40nickels\)

ANSWER: 40 nickels

We call:

a - the youngest brother

b - the 2nd young brother

c - the third young brother

d - the oldest brother

We have:

a + b + c + d = 1200 that \(\left(d-300\right)+\left(d-200\right)+\left(d-100\right)+d=1200\)

<=> \(\left(d-300\right)+d=\left(d-200\right)+\left(d-100\right)=1200:2=600\)

=> \(d=450\) 

=> \(d-100=350=c\)

=>\(d-200=250=b\)

=>\(d-300=150=a\)

Arrcoding to the solution, the youngest brother (a) gets 150.

ANSWER: 150

Arrcoding to the context, the line AB = 7cm and \(\left|AP-PB\right|=5cm\)

=> AP-PB = 5cm or AP-PB = -5cm.

1st situation: AP-PB = 5cm.

=> AP = \(5+\left(7-5\right):2=6\left(cm\right)\) /// PB = \(7-6=1\left(cm\right)\)

=> The position of P is 6cm from A and 1cm from B.

2nd situation: AP-PB = -5cm

=> PB = \(\left|-5+\left(7-5\right):2\right|=6\left(cm\right)\) /// AP = \(7-6=1\left(cm\right)\)

ANSWER: The position of P = 6cm from A and 1cm from B or 1cm from A and 6cm from B

The answer is C: 19 logs.

There were: \(72-53=19\left(logs\right)\)

\(\left(3.x-2^4\right).7^3=2.7^4\)

\(\left(3.x-2^4\right)=\left(2.7^4\right):7^3\)

\(3.x-2^4=2.7\)

\(3.x-16=14\)

\(3.x=14+16\)

\(3.x=30\)

\(x=30:3\)

\(x=10\)

1 middle square = \(1:9=\dfrac{1}{9}\)(area)

1 black little square = \(\dfrac{1}{9}:4=\dfrac{1}{36}\)(area)

ANSWER: \(\dfrac{1}{36}area\)

1. The smaller number is \(248:2-1=123\). 

2. The least natural number is \(0\)

4. There different is: 40

(Explaination) We have the smallest 3 digit number is 100.

The smaller number = 100 : (3+7) . 3 = 30

The bigger number = 100 - 30 = 70

Their difference = 70 - 30 = 40.

The answer is A: \(2011^1\) and C: \(1\cdot2011\)  There are two possible answers in this question.

The value of A= \(2011^1=2011\).

The value of B=\(1^{2011}=1\).

The value of C=\(1\cdot2011=2011\)

The value of D= \(1:2011=\dfrac{1}{2011}\)

The answer is C: 19

There are: 33 bricks of white, 3 bricks of blue (33:11=3), 14 bricks of red (50-33-3=14)

There are: 33-14=19 fewer red bricks than white ones.

The answer is C: 1

\(\dfrac{2011\cdot2,011}{201,1\cdot20,11}=\dfrac{4044,121}{4044,121}\)= 1

How many cubes are there ?

The answer is A: 5050

- Total of \(1+2+3+...+100\) = \(\left(1+100\right)\cdot100:2\) = \(5050\)

There are 3 squares can be drawn.

The answer is C: 2 (girls)

- There are: \(9+13=22\) children in the class.

- There are: \(22:2=11\) children have got a cold.

- There are at least:\(11-9=2\) girls have got a cold.