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Mathematics 7.

Table of contents
Natural numbers, rational numbers
The forms of rational numbers
Operations with rational numbers
Proportionalities (revision)
Percentages (revision)
Interest
Exponentiation
Operations with exponential terms with the same base
Operations including exponential terms with the same exponent
Prime hunting
Very large numbers
Mixed exercises
Algebraic expressions
Equations, inequalities
Plane geometry I.
Sets, combinatorics
Linear functions, series
Plane geometry II.
Statistics, probability
Spatial geometry
Operations with rational numbers
Revision
Game • Calculate with rational numbers
Make the following cards.
Addition:
\latex{\frac{1}{2} +\frac{1}{3}= \frac{3}{6}+\frac{2}{6}=\frac{5}{6}}
Rational numbers:
\latex{\frac{2}{3}}
\latex{\frac{5}{7}}
\latex{\frac{3}{4}}
\latex{\frac{5}{6}}
\latex{0}
\latex{12}
\latex{3.5}
\latex{1.2}
Division:
signs:
\latex{\times}
\latex{+}
\latex{-}
\latex{-}
\latex{,}
\latex{,}
\latex{,}
\latex{;}
\latex{\frac{2}{3}\div \frac{5}{7}=\frac{2}{3}\times \frac{7}{5}=\frac{14}{15}}
operations:
\latex{\times}
\latex{+}
\latex{-}
\latex{\div }
\latex{, }
\latex{, }
\latex{, }
\latex{.}
\latex{\frac{5}{7}}  reciprocal \latex{\frac{7}{5}}
Flip the cards and shuffle them. Then, start drawing the cards as the shapes indicate it.
Write down the resulting operation and calculate the result. Put the cards back and draw a new series of operations. The winner is the player who gets the largest number when adding the results of four operations.
Multiplication:
\latex{(-3)\times (-4)=12}
\latex{(-3)\times 4=-12}
\latex{\frac{2}{3} \times \frac{5}{7}=\frac{10}{21}}
Multiplying sums and differences
Example 1
Perform the following operations.
  1. \latex{(-2)\times \left(1-\frac{1}{3} \right) }
  1. \latex{-\left(\frac{3}{4}+\frac{1}{2} \right) }
  1. \latex{\left(-\frac{2}{3} \right) \times \left(1+\frac{1}{4}-\frac{1}{2} \right)}
Solution
  1. There are two ways to perform these operations.
  1. First, perform the operations between brackets.
\latex{(-2)\times \left(1-\frac{1}{3} \right)=(-2)\times \left(\frac{3}{3}-\frac{1}{3} \right)=(-2)\times \frac{2}{3}=-\frac{4}{3}.}
  1. You already know that multiplying sums and differences can be performed by multiplying each term separately and then adding or subtracting the products. This is called removing the brackets.
\latex{(-2)\times \left(1-\frac{1}{3} \right)=(-2)\times 1-(-2)\times \frac{1}{3}=-2-\left(-\frac{2}{3} \right)=\\= -2+\frac{2}{3} =-\frac{6}{3}+\frac{2}{3}=-\frac{4}{3} .}
  1. Performing the operations between brackets first:

\latex{-\left(\frac{3}{4}+\frac{1}{2} \right)=-\left(\frac{3}{4}+\frac{2}{4}\right)=-\left(\frac{5}{4} \right)=-\frac{5}{4}.}
 
Removing the brackets first:

\latex{-\left(\frac{3}{4}+\frac{1}{2} \right)=(-1)\times \frac{3}{4}+(-1)\times \frac{1}{2}=-\frac{3}{4}-\frac{1}{2}=-\frac{3}{4}-\frac{2}{4}=-\frac{5}{4}.}
 
  1. Performing the operations between brackets first:
 
\latex{-\left(\frac{3}{4}+\frac{1}{2} \right)=\\=(-1)\times \left(\frac{3}{4}+\frac{1}{2} \right)}
\latex{\left(-\frac{2}{3} \right) \times \left(1+\frac{1}{4}-\frac{1}{2} \right)=\left(-\frac{2}{3} \right)\times \left(\frac{4}{4}+\frac{1}{4}-\frac{2}{4} \right)=\left(-\frac{\bcancel{2 }^{1} }{\cancel{3}^{1} } \right)\times\frac{\cancel{3}^{1} }{\bcancel{4}^{2} }=-\frac{1}{2}.}
 
Removing the brackets first:
\latex{\left(-\frac{2}{3} \right)\times \left(1+\frac{1}{4}-\frac{1}{2} \right)=\left(-\frac{2}{3} \right)\times 1+\left(-\frac{\cancel{2}^{1} }{3} \right)\times \frac{1}{\cancel{4}^{2} }-\left(-\frac{\cancel{2}^{1} }{3} \right)\times \frac{1}{\cancel{2}^{1} }=}

\latex{=-\frac{2}{3}+\left(-\frac{1}{6} \right)-\left(-\frac{1}{3} \right)=-\frac{4}{6}-\frac{1}{6}+\frac{2}{6}=-\frac{\cancel{3}^{1} }{\cancel{6}^{2} }=-\frac{1}{2}.}
Example 2
Perform the following operations.
  1. \latex{\frac{2}{3}-\left(\frac{2}{3}-\frac{1}{5} \right);}
  1. \latex{2-12\times \left(\frac{5}{6}-\frac{1}{3} \right).}
Solution
  1. Performing the operations between brackets first:
\latex{\frac{2}{3}-\left(\frac{2\times 5}{3\times 5}-\frac{1\times 3}{5\times 3} \right)=\frac{2}{3}-\left(\frac{10}{15}-\frac{3}{15} \right)=\frac{2}{3}-\frac{7}{15}=}

\latex{=\frac{2\times 5}{3\times 5}-\frac{7}{15}=\frac{10}{15}-\frac{7}{15}=\frac{3}{15}=\frac{1}{5}.}
Removing brackets first:
\latex{\frac{2}{3}- \left(\frac{2}{3}-\frac{1}{5} \right)=\frac{2}{3}-\frac{2}{3}+\frac{1}{5}=\frac{1}{5}.}
 
This solution required less calculation. Its worth considering which method to choose when solving an exercise.
  1. Performing the operations between brackets first:
\latex{2-12\times \left(\frac{5}{6}-\frac{1}{3} \right)=2-12\times \left(\frac{5}{6}-\frac{1\times 2}{3\times 2} \right)=}
\latex{=2-12\times \left(\frac{5}{6}-\frac{2}{6} \right)=2-\cancel{12}^{2}\times \frac{3}{\cancel{6}}=2-6=-4.}
removing brackets
Removing the brackets first:
\latex{2-\cancel{12}^{2}\times \frac{5}{\cancel{6}}+(-\cancel{12}^{4} ) \times \left(-\frac{1}{\cancel{3}} \right)=2-10+4=-4.}
Example 3
Perform the following operations.
  1. \latex{1-\frac{1}{3}\times \frac{2}{5}-0,1;}
  1. \latex{\left(1-\frac{1}{3} \right)\times \frac{2}{5}-0,1;}
  1. \latex{1-\frac{1}{3}\times \left(\frac{2}{5}-0,1 \right);}
  1. \latex{\left(1-\frac{1}{3} \right)\times \left(\frac{2}{5}-0,1 \right).}
Solution
The difference between these operations is that, due to the brackets, the order of operations is different.
It is better to use the fraction form because of \latex{\frac{1}{3}.}
  1. \latex{1-\frac{1}{3}\times \frac{2}{5}-\frac{1}{10}=1-\frac{2}{15}-\frac{1}{10}=\frac{30-4-3}{30}=\frac{23}{30};}
  1. \latex{\left(1-\frac{1}{3} \right)\times \frac{2}{5}-\frac{1}{10}=\frac{2}{3}\times \frac{2}{5}-\frac{1}{10}=\frac{4}{15}-\frac{1}{10}=\frac{8-3}{30}=\frac{5}{30};}
  1. \latex{1-\frac{1}{3}\times \left(\frac{2}{5}-\frac{1}{10} \right)=1-\frac{1}{3}\times \frac{4-1}{10}=1-\frac{1}{3}\times \frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}=\frac{27}{30};}
  1. \latex{\left(1-\frac{1}{3} \right)\times \left(\frac{2}{5}-\frac{1}{10} \right)=\frac{2}{3}\times \frac{3}{10}=\frac{6}{30}.}
\latex{0,1=\frac{1}{10}}

\latex{1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}}

\latex{\frac{2}{5} -\frac{1}{10}=\frac{4-1}{10} =\frac{3}{10}}

 
Example 4
Which number is greater: \latex{A=\frac{\frac{1}{2} }{3},}  or  \latex{B=\frac{1}{\frac{2}{3} }?}
Solution
Compound fractions:
\latex{A=\frac{\frac{1}{2} }{3}}
\latex{B=\frac{1}{\frac{2}{3} }}
main fraction bar
In the case of compound fractions, the order of operations is determined by the main fraction bar. The main fraction bar is always written at the same height as the equals sign.
\latex{A=\frac{\frac{1}{2} }{3}=\frac{1}{2}\div 3=\frac{1}{2\times 3}=\frac{1}{6};}     \latex{B=\frac{1}{\frac{2}{3} }=1\div \frac{2}{3}=1\times \frac{3}{2}=\frac{3}{2}.}
Since \latex{\frac{3}{2}=\frac{9}{6},}  \latex{A\lt B.}
Note that depending on where the main fraction bar is placed, the position of the brackets may change:
\latex{A=\frac{\frac{1}{2} }{3}=(1\div 2)\div 3,}  while  \latex{B=\frac{1}{\frac{2}{3} }=1\div (2\div 3) .}
During division, changing the position of the brackets results in different quotients.

\latex{(1\div 2)\div 3}\latex{\neq}

\latex{\neq}\latex{1\div (2\div 3)}

Example 5
Simplify the compound fraction \latex{1+\frac{4}{1-\frac{1}{3} }.}
Solution
The numerator of the fraction is \latex{ 4 }, and the denominator is  \latex{1-\frac{1}{3}=\frac{2}{3}.}
 
The fraction can be simplified in the following way:
\latex{1+\frac{4}{1-\frac{1}{3} }=1+\frac{4}{\frac{2}{3} }=1+4\div \frac{2}{3}=1+4\times \frac{3}{2}=1+2\times 3=1+6=7.}
Exercises
{{exercise_number}}. Perform the following operations.
  1. \latex{1+2\times \left(\frac{1}{2}-4 \right) }
  1. \latex{4-3\times (1-2\times 3)}
  1. \latex{3,5-(4\times 3-9)\times \frac{1}{2}}
  1. \latex{1-3\times \left(\frac{1}{3}+\frac{5}{6} \right) }
  1. \latex{\frac{1}{2}+10\times \left(0,2\frac{1}{4} \right)}
  1. \latex{-3\times \left(\frac{1}{9}+\frac{1}{27} \right)+\frac{1}{3}}
{{exercise_number}}. Perform the following operations.
  1. \latex{\frac{2}{3}+\left(\frac{1}{2}-\frac{2}{5} \right)}
  1. \latex{\frac{2}{3} -\left(\frac{1}{2}-\frac{2}{5} \right)}
  1. \latex{\frac{2}{3} -\left(\frac{1}{2}+\frac{2}{5} \right)}
  1. \latex{\frac{2}{5} -0,1\times \left(10+\frac{200}{2} \right)}
  1. \latex{(-5)\times \left(\frac{1}{2}-\frac{2}{8} \right)-\frac{1}{8}}
  1. \latex{\left(\frac{1}{2}-0,4 \right) \times \left(0,4-\frac{1}{10} \right)}
{{exercise_number}}. Perform the following operations.
  1. \latex{\frac{5}{9}\times (-18)+\frac{1}{3} }
  1. \latex{\left(-\frac{4}{9}+\frac{2}{3} \right)\times \left(-\frac{1}{2} \right) }
  1. \latex{\left[-\frac{5}{3}-\left(-\frac{3}{4} \right) \right]\div \left(-\frac{2}{9} \right)}
  1. \latex{\frac{4}{9}\times \left(\frac{3}{2}-\frac{4}{5} \right) \div \left(-\frac{7}{5} \right)}
{{exercise_number}}. Moro Bear really likes honey. He eats honey for three \latex{ days }, a quarter of a honey pot each \latex{ day }. On every fourth \latex{ day }, he eats raspberries. On 'raspberry \latex{ days }', he refills two-thirds of a pot with honey. This morning, he has a pot completely full of honey left.
  1. How much honey will be left in the pot four \latex{ days } later?
  1. How much honey will be left in the pot eight \latex{ days } later?
  1. How many \latex{ days } later will the honey not be enough for Moro Bear?
{{exercise_number}}. The king commissioned the construction of a new castle. During the day, \latex{\frac{1}{30}} of the castle is built; however, during the night, \latex{\frac{1}{3}} of the part built on the given day collapses. When will the construction of the castle be completed?
{{exercise_number}}. You can use the following numbers and signs:
\latex{2}
\latex{1}
\latex{\frac{1}{3} }
\latex{- }
\latex{+ }
\latex{\times }
\latex{( }
\latex{) }
Use them to make a series of operations and calculate the result. Find several solutions.
{{exercise_number}}. Which one is greater?
  1. \latex{A=\frac{\frac{2}{3} }{5}}  or  \latex{B=\frac{2}{\frac{3}{5} } }
  1. \latex{A=\frac{\frac{5}{3} }{5}}  or  \latex{B=\frac{5}{\frac{3}{5} }}
  1. \latex{A=\frac{\frac{5}{3} }{2}}  or  \latex{B=\frac{5}{\frac{3}{2} }}
  1. \latex{A=\frac{-\frac{2}{3} }{5}}  or  \latex{B=\frac{2}{-\frac{3}{5} } }
{{exercise_number}}. Observe the following fractions. Calculate their values.
  1. \latex{\frac{\left(\frac{1}{3}-0.25+1 \right)\div \left(\frac{1}{3}-0.25+1 \right) }{\left(0.5-1\right)\div \left(0.751-25\right) }}
  1. \latex{\frac{\left(\frac{1}{3}-0.25+1.8 \right)\div \left(\frac{1}{3}-0.25+1.8 \right) }{\left(0.25-1\right)\div \left(0.51- 0.25\right) }}
  1. \latex{\frac{\left(\frac{2}{3}-0.5+2 \right)\div \left(\frac{1}{3}-0.25+1\right) }{\left(0.5-1\right)\div \left(0.751-0.25\right) }}
  1. \latex{\frac{\left(\frac{1}{3}-0.25+1 \right)\div \left(\frac{1}{3}-0.25+1\right) }{\left(-1.2\right)\div \left(0.75-1- 0.25\right) }}
{{exercise_number}}. Write the correct signs between the numbers to make the equalities true.
  1. \latex{\frac{1}{2}\bigcirc \frac{1}{3}=\frac{5}{6}}
  1. \latex{\frac{3}{5}\triangle\frac{12}{6}=0.3}
  1. \latex{0.7\triangledown\frac{1}{2}=\frac{1}{5}}
  1. \latex{\frac{13}{70} \Box\frac{39}{10}=\frac{1}{21}}
{{exercise_number}}. Simplify the following compound fractions.
  1. \latex{\frac{\frac{1}{2}-\frac{1}{3}}{5}}
  1. \latex{\frac{2-\frac{2}{3}}{\frac{2}{3}-2}}
  1. \latex{1+\frac{1}{3+\frac{4}{5} }}
  1. \latex{\frac{1+\large{\frac{1}{1+\frac{1}{2}} } }{2}}
{{exercise_number}}. What distance did you cover if, after completing \latex{\frac{2}{3}} of the road, you still had \latex{\frac{2}{3}} of a \latex{ kilometre } left?
{{exercise_number}}. Carl participated in a cycling race. After the race, he told his parents:
"One-third of the competitors finished before me, while half of them finished after me. There were no ties."

What place did Carl finish in?
{{exercise_number}}. Substitute the number cards    1      2     3   into the compound fraction \latex{\frac{\Box}{\Box}} to get the
  1. greatest;
  1. smallest possible number.
Quiz
Write signs between the following fractions and use brackets to make the equalities correct.
  1. \latex{\frac{1}{2}\div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} =\frac{1}{2} ;}
  1. \latex{\frac{1}{2}\div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} =1;}
  1. \latex{\frac{1}{2}\div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{2} =0.}
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