Find \(m\) from the relation \(\dfrac{m}{2}=\dfrac{3}{n}\), where \(n\ne0\).
Solution
Using the fundamental property of a proportion, \(m\cdot n=3\cdot2\). Hence \(m=\dfrac{6}{n}\). Answer: Б.
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2
Task
One correct answer
Calculate: \(\dfrac{2^6\cdot5^6}{10^4}\).
Solution
\(\dfrac{2^6\cdot5^6}{10^4}=\dfrac{(2\cdot5)^6}{10^4}=\dfrac{10^6}{10^4}=10^{6-4}=10^2\). Answer: Б.
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3
Task
One correct answer
The remainder when the natural number \(k\) is divided by 5 is 2. Find the remainder when \(k+21\) is divided by 5.
Solution
Let \(k=5p+2\). Then \(k+21=5p+23=5(p+4)+3\). Therefore the remainder is 3. Answer: Г.
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4
Task
One correct answer
Choose the correct inequality.
Solution
With equal denominators, the fraction with the larger numerator is larger; with equal numerators, the fraction with the smaller denominator is larger. For option Г, \(\dfrac{5}{6}=\dfrac{25}{30}\) and \(\dfrac{4}{5}=\dfrac{24}{30}\), so \(\dfrac{5}{6}>\dfrac{4}{5}\). Answer: Г.
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5
Task
One correct answer
An empty pool holding \(x\text{ m}^3\) of water is completely filled in 5 hours at a constant rate. Which formula gives the amount of water \(V\) (in \(\text{m}^3\)) after 2 hours?
Solution
In 5 hours the pool receives \(x\text{ m}^3\), so in 1 hour it receives \(\dfrac{x}{5}\text{ m}^3\), and in 2 hours \(\dfrac{2x}{5}\text{ m}^3\). Thus \(V=\dfrac{2x}{5}\). Answer: Г.
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6
Task
One correct answer
Write \(2^{15}\), \(4^{10}\), and \(10^5\) in increasing order.
Solution
Write the numbers with the same exponent: \(2^{15}=(2^3)^5=8^5\), \(4^{10}=(4^2)^5=16^5\), and \(10^5=10^5\). Since \(8<10<16\), \(2^{15}<10^5<4^{10}\). Answer: Б.
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7
Task
One correct answer
A teacher distributed 72 notebooks among the students of a class. Each student received the same number of notebooks. Which of the following numbers could be the number of students in the class?
Solution
The number of students must divide 72 exactly. Of the given numbers, only \(9\) divides 72 with no remainder: \(72:9=8\). Answer: Б.
\(\dfrac59\cdot0.3=\dfrac59\cdot\dfrac3{10}=\dfrac{15}{90}=\dfrac16\). Answer: А.
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10
Task
Matching
Match each statement about a fraction (1–4) with the fraction (А–Д) for which the statement is true.
А
\(\dfrac57\) Б
\(\dfrac{13}{27}\) В
\(\dfrac{41}{10}\) Г
\(\dfrac7{10}\) Д
\(\dfrac{34}{51}\)
Solution
А: \(\dfrac57\) is the reciprocal of \(1\dfrac25=\dfrac75\), so 4–А. Б: \(\dfrac{13}{27}<\dfrac12\), so 3–Б. В: \(\dfrac{41}{10}>1\), so it is an improper fraction and 2–В. Д: \(\dfrac{34}{51}\) reduces by 17, so 1–Д. Answer: 1–Д, 2–В, 3–Б, 4–А.
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11
Task
Matching
Match each number (1–4) with the set (А–Д) to which it belongs.
А
set of even natural numbers Б
set of integers that are not natural numbers В
set of rational numbers that are not integers Г
set of irrational numbers Д
set of prime numbers
Solution
\(-8\) is an integer but not a natural number; \(23\) is prime; \(\sqrt{16}=4\) is an even natural number; \(1.7\) is rational but not an integer. Answer: 1–Б, 2–Д, 3–А, 4–В.
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12
Task
One correct answer
Choose the set of all values of \(a\) for which \(|a^3-a^2|=a^3-a^2\).
Solution
The equality \(|x|=x\) holds exactly when \(x\ge0\). Hence \(a^3-a^2=a^2(a-1)\ge0\), so \(a=0\) or \(a\ge1\). Therefore \(a\in\{0\}\cup[1;+\infty)\). Answer: Б.
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13
Task
One correct answer
Calculate \(1001^2-999^2\).
Solution
Using the difference of squares: \(1001^2-999^2=(1001-999)(1001+999)=2\cdot2000=4000\). Answer: Д.
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14
Task
One correct answer
Calculate \(\dfrac13+\dfrac23\cdot\dfrac58\).
Solution
\(\dfrac13+\dfrac23\cdot\dfrac58=\dfrac13+\dfrac5{12}=\dfrac4{12}+\dfrac5{12}=\dfrac9{12}=\dfrac34\). Answer: А.
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15
Task
One correct answer
Oksana has a certain number of nuts. When she divides them into groups of 5, 2 nuts remain; when she divides them into groups of 3, none remain. Which of the following numbers COULD be the number of nuts Oksana had?
Solution
The number must be divisible by 3 and leave remainder 2 when divided by 5. For 57, \(57:3=19\) and \(57=5\cdot11+2\). Answer: В.
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16
Task
Matching
Match each number (1–4) with the set (А–Д) to which it belongs.
А
set of natural numbers Б
set of composite numbers В
set of integers that are not natural numbers Г
set of fractional numbers Д
set of irrational numbers
Solution
\(3.4\) is a fractional number; \(\sqrt8\) is irrational; \(\dfrac{10}{2}=5\) is natural; \(-13\) is an integer but not a natural number. Answer: 1–Г, 2–Д, 3–А, 4–В.
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17
Task
One correct answer
Write \(0.351\) in scientific notation.
Solution
Scientific notation has the form \(a\cdot10^n\), where \(1\le a<10\). Thus \(0.351=3.51\cdot10^{-1}\). Answer: А.
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18
Task
One correct answer
\(|2-\sqrt5|+|2+\sqrt5|=\)
Solution
Since \(2<\sqrt5\), \(|2-\sqrt5|=\sqrt5-2\) and \(|2+\sqrt5|=2+\sqrt5\). Therefore \(|2-\sqrt5|+|2+\sqrt5|=2\sqrt5\). Answer: Б.
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19
Task
Matching
One number is chosen at random from all natural numbers greater than 9 and less than 20. Match each event (1–4) with its probability (А–Д).
А
0 Б
0.2 В
0.3 Г
0.4 Д
1
Solution
The possible numbers are 10, 11, …, 19 — 10 numbers in total. The primes are 11, 13, 17, and 19, so \(P=4/10=0.4\). Every number is two-digit, so \(P=1\). None is a divisor of 5, so \(P=0\). The digit sum is divisible by 3 for 12, 15, and 18, so \(P=3/10=0.3\). Answer: 1–Г, 2–Д, 3–А, 4–В.
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20
Task
Matching
Match each question (1–4) with its correct answer (А–Д).
А
8 Б
16 В
17 Г
27 Д
56
Solution
\(16=4^2\), so 1–Б. The number 17 is prime, so 2–В. The number 8 is a divisor of 8, so 3–А. \(56=7\cdot8\) is a multiple of 7, so 4–Д. Answer: 1–Б, 2–В, 3–А, 4–Д.
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21
Task
One correct answer
Arrange the numbers \(\dfrac19;\ 0.1;\ 0.11\) in increasing order.
Solution
We have \(0.1=0.100\), \(0.11=0.110\), and \(\dfrac19=0.111\ldots\). Therefore \(0.1<0.11<\dfrac19\). Answer: Б.
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22
Task
One correct answer
Calculate \((0.3)^2\cdot10^4\).
Solution
\((0.3)^2\cdot10^4=0.09\cdot10000=900\). Answer: Б.
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23
Task
One correct answer
Calculate \(3\dfrac5{12}+\dfrac78\).
Solution
Use the common denominator 24: \(\dfrac5{12}=\dfrac{10}{24}\) and \(\dfrac78=\dfrac{21}{24}\). Thus \(3\dfrac5{12}+\dfrac78=3+\dfrac{31}{24}=4\dfrac7{24}\). Answer: Д.
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24
Task
One correct answer
Which of the given digits should replace the asterisk in 257* so that the resulting number is divisible by 3?
Solution
A number is divisible by 3 when the sum of its digits is divisible by 3. Here \(2+5+7=14\). Only 7 gives \(14+7=21\), which is divisible by 3. Answer: Г.
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25
Task
One correct answer
Calculate \(\dfrac{\frac23}{4}+0.5\).
Solution
\(\dfrac{\frac23}{4}+0.5=\dfrac23\cdot\dfrac14+\dfrac12=\dfrac16+\dfrac12=\dfrac46=\dfrac23\). Answer: А.
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26
Task
Matching
Match each numerical expression (1–4) with its value (А–Д).
In a garden, pears and apples were collected in separate boxes. The number of boxes with apples is to the number of boxes with pears as 7:3. Which of the given numbers could be the total number of boxes with apples and pears?
Solution
The ratio has \(7+3=10\) total parts, so the total number of boxes must be a multiple of 10. Among the given numbers, 80 satisfies this condition. Answer: Г.
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30
Task
Matching
Match each expression (1–4) with the statement about its value (А–Д) when \(a=15\).
А
less than 20 Б
is a prime number В
is even Г
is divisible by 3 Д
is divisible by 5
Solution
For \(a=15\): \(\dfrac73a=35\), so it is divisible by 5; \(2a-1=29\) is prime; \(a^2+12a+36=(a+6)^2=21^2=441\) is divisible by 3; \(a^2+13^2=225+169=394\) is even. Answer: 1–Д, 2–Б, 3–Г, 4–В.
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31
Task
One correct answer
The number \(a\) is 5 times as large as the positive number \(b\). Then \(a=\)
Solution
If \(a\) is 5 times as large as \(b\), then \(a=5b\). Answer: Г.
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32
Task
Matching
Match the numerical expressions (1–4) with their values (А–Д).
А
64 000 Б
27 000 В
24 000 Г
22 500 Д
14 400
Solution
1. \(2003^2-1997^2=(2003-1997)(2003+1997)=6\cdot4000=24000\), so 1–В. 2. Since \(132=2\cdot66\), \(186^2-186\cdot132+66^2=(186-66)^2=120^2=14400\), so 2–Д. 3. Since \(104=2\cdot52\), \(98^2+98\cdot104+52^2=(98+52)^2=150^2=22500\), so 3–Г. 4. Since \(51=3\cdot17\) and \(141=3\cdot47\), the expression equals \((47-17)^3=30^3=27000\), so 4–Б. Answer: 1–В, 2–Д, 3–Г, 4–Б.
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33
Task
One correct answer
Calculate \(\dfrac13\cdot5.8+\dfrac13\cdot8.3\).
Solution
Factor out \(\dfrac13\): \(\dfrac13\cdot5.8+\dfrac13\cdot8.3=\dfrac13(5.8+8.3)=\dfrac{14.1}{3}=4.7\). Answer: В.
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34
Task
One correct answer
Calculate \(27^{\frac23}-16^{\frac14}\).
Solution
\(27^{\frac23}=(3^3)^{\frac23}=3^2=9\), and \(16^{\frac14}=(2^4)^{\frac14}=2\). Therefore \(9-2=7\). Answer: Д.
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35
Task
One correct answer
At a shop, 6 identical notebooks and several pens at 3 UAH each were bought. Which of the following numbers could represent the total cost of the purchase (in UAH)?
Solution
The cost of 6 identical notebooks is a multiple of 6, and the cost of the pens is a multiple of 3. Thus the total cost is divisible by 3. Of the given numbers, only 24 is divisible by 3. Answer: Г.
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36
Task
One correct answer
If \(m=n-1\), then \(7-m=\)
Solution
Substitute \(m=n-1\): \(7-m=7-(n-1)=7-n+1=8-n\). Answer: В.
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37
Task
One correct answer
If the numbers \(x\) and \(y\) satisfy \(2y+4=x\), then \(y=\)
Solution
From \(2y+4=x\), we get \(2y=x-4\), so \(y=\dfrac{x-4}{2}\). Answer: В.
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38
Task
One correct answer
If \(a\le-7\), then \(\left|\dfrac{a^2-49}{a+7}\right|=\)
Solution
The expression is defined only for \(a\ne-7\). Together with \(a\le-7\), this means \(a<-7\). Then \(\left|\dfrac{a^2-49}{a+7}\right|=\left|\dfrac{(a-7)(a+7)}{a+7}\right|=|a-7|\). Since \(a-7<0\), \(|a-7|=7-a\). Answer: А.
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39
Task
One correct answer
\(-2xy^2-(3xy^2-2x^2y)=\)
Solution
Remove the parentheses, changing the signs inside: \(-2xy^2-(3xy^2-2x^2y)=-2xy^2-3xy^2+2x^2y=-5xy^2+2x^2y\). Answer: А.
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40
Task
One correct answer
If \(a<2\), then \(1+|a-2|=\)
Solution
Since \(a<2\), we have \(a-2<0\), so \(|a-2|=-(a-2)=2-a\). Therefore \(1+|a-2|=1+2-a=3-a\). Answer: Д.
