problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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4.056. Find the fifth term of an increasing geometric progression, given that its first term is equal to $7-3 \sqrt{5}$ and that each of its terms, starting from the second, is equal to the difference of the two adjacent terms. | Solution.
We have:
$$
\left\{\begin{array}{l}
b_{1}=7-3 \sqrt{5} \\
b_{2}=b_{3}-b_{1}, \\
|q|>1
\end{array}\right.
$$
$q_{1}=\frac{1-\sqrt{5}}{2}$ does not fit, as $\left|q_{1}\right|<1 ; q_{2}=\frac{1+\sqrt{5}}{2}$.
From this,
$$
b_{5}=b_{1} q^{4}=\frac{(7-3 \sqrt{5})(1+\sqrt{5})^{4}}{16}=\frac{(17-3 \sqrt{5}) 8(... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.061. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x}+\ldots+\frac{1}{x}=3
$$
where $x$ is a positive integer. | ## Solution.
Multiplying both sides of the equation by $x$, we have
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1, a_{n}=1$,
$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{1-x+1}{-1}+1=x-1 . \text { Since } S_{n}=\frac{a_{1}+a_{n... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.063. Given two geometric progressions consisting of the same number of terms. The first term and the common ratio of the first progression are 20 and $3 / 4$, respectively, while the first term and the common ratio of the second progression are 4 and $2 / 3$, respectively. If the terms of these progressions with the ... | ## Solution.
Let $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ be the terms of two geometric progressions, where $a_{1}=20, q_{1}=\frac{3}{4}; b_{1}=4, q_{2}=\frac{2}{3}$ and $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\ldots=$ $=158.75$. We find $a_{2}=a_{1} q_{1}=20 \cdot \frac{3}{4}=15, a_{3}=a_{1} q_{1}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
(Note: The equation is the same in both languages, so the translation is identical to the original text.) | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq 2 .\end{array}\right.$
After bringing all terms of the equation to a common denominator, we have
$$
\begin{aligned}
& \frac{\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)+2(x+1)(x-2)}{(x+1)(x-2)}=0 \Leftrightarrow \\
& \Leftrightarro... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$. | ## Solution.
We have:
$$
\begin{aligned}
& \left(x^{2}+2 x+1\right)(x+2)+\left(x^{2}-2 x+1\right)(x-2)-12=0 \Leftrightarrow \\
& \Leftrightarrow x^{3}+2 x^{2}+2 x^{2}+4 x+x+2+x^{3}-2 x^{2}-2 x^{2}+4 x+x-2-12=0 \\
& 2 x^{3}+10 x-12=0, x^{3}+5 x-6=0
\end{aligned}
$$
The last equation can be rewritten as:
$x^{3}+5 x-5... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.146. $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq-2, \\ x \neq-3, \\ x \neq-4 .\end{array}\right.$
$$
\begin{aligned}
& \text { From the condition } \frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}-1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(x-1)(x-4)(x-2)(x-3)-(x+1)(x+4)(x+2)(x+3)}{(x+... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.158. $\sqrt[3]{9-\sqrt{x+1}}+\sqrt[3]{7+\sqrt{x+1}}=4$. | ## Solution.
Domain of definition: $x+1 \geq 0$ or $x \geq-1$.
Let $\sqrt{x+1}=y \geq 0$. The equation in terms of $y$ becomes $\sqrt[3]{9-y}+\sqrt[3]{7+y}=4$. Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 9-y+\sqrt[3]{(9-y)^{2}(7+y)}+3 \sqrt[3]{(9-y)(7+y)^{2}}+7+y=64 \Leftrighta... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$. | ## Solution.
Domain of definition: $x+2 \geq 0, x \geq-2$.
$$
\begin{aligned}
& \text { From the condition } \sqrt{x+2}=\sqrt[3]{3 x+2} \Rightarrow(x+2)^{3}=(3 x+2)^{2} \\
& x^{3}+6 x^{2}+12 x+8=9 x^{2}+12 x+4 \Leftrightarrow x^{3}-3 x^{2}+4=0 \Leftrightarrow \\
& \Leftrightarrow(x+1)\left(x^{2}-x+1\right)-3(x-1)(x+1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.160. $\sqrt{\frac{20+x}{x}}+\sqrt{\frac{20-x}{x}}=\sqrt{6}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{20+x}{x} \geq 0, \\ \frac{20-x}{x} \geq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x+20) \geq 0, \\ x(x-20) \geq 0, \\ x \neq 0,0<x \leq 20 .\end{array}\right.\right.$
By squaring both sides of the equation, we have
$$
\begin{aligned}
& \frac{2... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.164. $\sqrt{x+8+2 \sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$. | Solution.
Let $\sqrt{x+7}=y \geq 0, x+7=y^{2}, x=y^{2}-7$.
With respect to $y$, the equation takes the form
$$
\begin{aligned}
& \sqrt{y^{2}+2 y+1}+\sqrt{y^{2}-y-6}=4, \sqrt{(y+1)^{2}}+\sqrt{y^{2}-y-6}=4 \\
& |y+1|+\sqrt{y^{2}-y-6}=4
\end{aligned}
$$
Since $y \geq 0$, we have $y+1+\sqrt{y^{2}-y-6}=4, \sqrt{y^{2}-y-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$.
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition, we have $\sqrt{x}=2-\frac{2 x+1}{x+2} \Leftrightarrow \sqrt{x}=\frac{2 x+4-2 x-1}{x+2}, \sqrt{x}=\frac{3}{x+2}$.
Squaring both sides of the equation, we get $x=\frac{3}{x^{2}+4 x+4} \Leftrightarrow$ $\Leftrightarrow \frac{x^{3}+4 x^{2}+4 x-9}{x^{2}+4... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.170. $\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^{2}-16}-6$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+4 \geq 0, \\ x-4 \geq 0,\end{array} \Leftrightarrow x \geq 4\right.$.
By squaring both sides of the equation, we get
$$
\begin{aligned}
& \frac{x+4+2 \sqrt{(x+4)(x-4)}+x-4}{4}=\left(x+\sqrt{x^{2}-16}\right)^{2}-12\left(x+\sqrt{x^{2}-16}\right)+36 \\
& 2\le... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.173. $2 \sqrt{5 \sqrt[4]{x+1}+4}-\sqrt{2 \sqrt[4]{x+1}-1}=\sqrt{20 \sqrt[4]{x+1}+5}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ 2 \sqrt[4]{x+1}-1 \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \sqrt[4]{x+1} \geq 1,\end{array} \Leftrightarrow x \geq-\frac{15}{16}\right.\right.$.
Let $\sqrt[4]{x+1}=y$, where $y \geq 0$. The equation in terms of $y$ be... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$.
6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ x \geq 0\end{array} \Leftrightarrow x \geq 0\right.$.
Let $\sqrt{x+1}+\sqrt{x}=y$, where $y \geq 0$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& y^{3}+y^{2}-2=0 \Leftrightarrow y^{3}-1+y^{2}-1=0 \Leftrightarrow \\
& \Leftrightarrow(... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$.
6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$. | ## Solution.
Domain of definition: $x+3 \geq 0, x \geq-3$.
From the condition $\sqrt[3]{x+7}=\sqrt{x+3}$ and raising both sides to the sixth power, we get $(x+7)^{2}=(x+3)^{3} \Leftrightarrow x^{2}+14 x+49=x^{3}+9 x^{2}+27 x+27 \Leftrightarrow$
$\Leftrightarrow x^{3}+8 x^{2}+13 x-22=0$. By checking, we find that $x_... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.250. For what values of $a$ do the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$ have a common root? | Solution.
If for some $a$ and $x$ the left parts of the equations are equal to 0, then they are equal to each other:
$$
\left\{\begin{array}{l}
x^{2}+a x+1=0 \\
x^{2}+x+a=0
\end{array}\right.
$$
Subtracting the second equation of the system from the first, we get
$a x-x+1-a=0,(a-1) x-(a-1)=0,(a-1)(x-1)=0$, from whi... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.160. Simplify the expression $\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m$, given that $m^{2}=a^{2}-b^{2}$. | Solution.
$\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m=\log _{a+b} m+\frac{\log _{a+b} m}{\log _{a+b}(a-b)}-$
$-2 \frac{\log _{a+b} m \cdot \log _{a+b} m}{\log _{a+b}(a-b)}=\log _{a+b} m \cdot\left(1+\frac{1}{\log _{a+b}(a-b)}-\frac{2 \log _{a+b} m}{\log _{a+b}(a-b)}\right)=$
$=\frac{\log _{a+b} ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$.
7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$. | Solution.
Domain of definition: $x>0$.
Rewrite the equation as $2^{2 \log _{5} x}-2 \cdot 2^{\log _{5} x}+\frac{2^{\log _{5} x}}{2}-1=0 \Leftrightarrow$ $\Leftrightarrow 2 \cdot 2^{2 \log _{5} x}-3 \cdot 2^{\log _{5} x}-2=0$. Solving this equation as a quadratic equation in terms of $2^{\log _{5} x}$, we find $2^{\lo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}9-2^{x}>0, \\ 3-x \neq 0,\end{array} \Leftrightarrow 3 \neq x<\log _{2} 9\right.$.
## From the condition
$$
\log _{2}\left(9-2^{x}\right)=3-x \Leftrightarrow 9-2^{x}=2^{3-x} \Leftrightarrow 2^{2 x}-9 \cdot 2^{x}+8=0
$$
Solving it as a quadratic equation in ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.179. $\log _{a^{2}} x^{2}+\log _{a}(x-1)=\log _{a} \log _{\sqrt{5}} 5$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>1, \\ 0<a \neq 1 .\end{array}\right.$
From the condition we have
$$
\log _{a} x+\log _{a}(x-1)=\log _{a} 2 \Rightarrow \log _{a} x(x-1)=\log _{a} 2
$$
from which $x^{2}-x-2=0 \Rightarrow x_{1}=2, x_{2}=-1 ; x_{2}=-1$ does not satisfy the domain of definition... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.186. $\log _{12}\left(4^{3 x}+3 x-9\right)=3 x-x \log _{12} 27$. | ## Solution.
Domain of definition: $4^{3 x}+3 x-9>0$.
Rewrite the equation as
$$
\log _{12}\left(4^{3 x}+3 x-9\right)+\log _{12} 27^{x}=3 x \Rightarrow \log _{12} 27^{x}\left(4^{3 x}+3 x-9\right)=3 x
$$
from which $27^{x}\left(4^{3 x}+3 x-9\right)=12^{3 x} \Leftrightarrow 4^{3 x}+3 x-9=4^{3 x}, 3 x-9=0, x=3$.
Answ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.187. $x^{2} \cdot \log _{x} 27 \cdot \log _{9} x=x+4$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 3, then
$$
\frac{3 x^{2}}{\log _{3} x} \cdot \frac{\log _{3} x}{2}=x+4 \Leftrightarrow 3 x^{2}-2 x-8=0
$$
from which $x_{1}=2, x_{2}=-\frac{4}{3} ; x_{2}=-\frac{4}{3}$ does not satisfy the domain of definition.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.197. $5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition we have
$$
\begin{aligned}
& 5^{\frac{x}{\sqrt{x}+2}} \cdot 5^{-\frac{4}{\sqrt{x}+2}}=5^{3 x-12} \cdot 5^{-2 x+4} \Leftrightarrow 5^{\frac{x}{\sqrt{x}+2}-\frac{4}{\sqrt{x}+2}}=5^{3 x-12-2 x+4} \Leftrightarrow \\
& \Leftrightarrow \frac{x-4}{\sqrt{x}+2... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.200. $\frac{1+2 \log _{9} 2}{\log _{9} x}-1=2 \log _{x} 3 \cdot \log _{9}(12-x)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}0<x<12, \\ x \neq 1 .\end{array}\right.$
Switch to base 3. Then we get
$$
\begin{aligned}
& \frac{1+\frac{2 \log _{3} 2}{\log _{3} 9}}{\frac{\log _{3} x}{\log _{3} 9}}-1=\frac{2}{\log _{3} x} \cdot \frac{\log _{3}(12-x)}{\log _{3} 9} \Leftrightarrow \frac{2+... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.208. $2^{x-1}+2^{x-4}+2^{x-2}=6.5+3.25+1.625+\ldots$ (the expression on the right side is an infinite geometric progression). | Solution.
On the right side, we have the sum of the terms of an infinitely decreasing geometric progression $S$, where $b_{1}=6.5 ; q=\frac{3.25}{6.5}=0.5 \Rightarrow S=\frac{b_{1}}{1-q}=\frac{6.5}{1-0.5}=13$.
Rewrite the equation as $\frac{2^{x}}{2}+\frac{2^{x}}{16}+\frac{2^{x}}{4}=13 \Leftrightarrow \frac{13}{16} \... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.209. $49^{1+\sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}=-7$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.209. $49^{1+\sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}=-7$. | Solution.
Domain of definition: $x \geq 2$.
Rewrite the equation as $49 \cdot 7^{2 \sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}+7=0$. Solving it as a quadratic equation in terms of $7^{\sqrt{x-2}}$, we get $\left(7^{\sqrt{x-2}}\right)=7^{-2}$ or $\left(7^{\sqrt{x-2}}\right)_{2}=7$, from which $(\sqrt{x-2})_{1}=-2$ (no soluti... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.212. $\frac{\log _{4 \sqrt{x}} 2}{\log _{2 x} 2}+\log _{2 x} 2 \cdot \log _{1 / 2} 2 x=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \frac{1}{16}, \\ x \neq \frac{1}{2}, \\ x \neq 1 .\end{array}\right.$
Transition to base 2. We have
$\frac{\frac{\log _{2} 2}{\log _{2} 4 \sqrt{x}}}{\frac{\log _{2} 2}{\log _{2} 2 x}}+\frac{\log _{2} 2}{\log _{2} 2 x} \cdot \frac{\log _{2} 2 x}{\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.214. $9^{x}+6^{x}=2^{2 x+1}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.214. $9^{x}+6^{x}=2^{2 x+1}$. | Solution.
Rewrite the equation as $3^{2 x}+2^{x} \cdot 3^{x}-2 \cdot 2^{2 x}=0$ and divide it by $2^{2 x} \neq 0$. Then $\left(\frac{3}{2}\right)^{2 x}+\left(\frac{3}{2}\right)^{x}-2=0 \Rightarrow\left(\left(\frac{3}{2}\right)^{x}\right)=-2$ (no solutions) or $\left(\left(\frac{3}{2}\right)^{x}\right)_{2}=1 \Rightarro... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.218. $\left(\frac{3}{5}\right)^{2 \log _{9}(x+1)} \cdot\left(\frac{125}{27}\right)^{\log _{27}(x-1)}=\frac{\log _{5} 27}{\log _{5} 243}$. | Solution.
Domain of definition: $x>1$.
From the condition we have
$\left(\frac{3}{5}\right)^{\log _{3}(x+1)} \cdot\left(\frac{3}{5}\right)^{\log _{3}(x-1)}=\frac{3}{5} \Leftrightarrow\left(\frac{3}{5}\right)^{\log _{3}(x+1)+\log _{3}(x-1)}=\frac{3}{5} \Rightarrow$
$\Rightarrow \log _{3}(x+1)+\log _{3}(x-1)=1 \Right... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.219. $5^{1+x^{3}}-5^{1-x^{3}}=24$.
7.219. $5^{1+x^{3}}-5^{1-x^{3}}=24$. | Solution.
We have $5 \cdot 5^{x^{3}}-\frac{5}{5^{x^{3}}}-24=0 \Leftrightarrow 5 \cdot\left(5^{x^{3}}\right)^{2}-24 \cdot 5^{x^{3}}-5=0$. Solving this equation as a quadratic in terms of $5^{x^{3}}$, we get $5^{x^{3}}=-\frac{1}{5}$ (no solutions), or $5^{x^{3}}=5 \Rightarrow x^{3}=1, x=1$.
Answer: 1 . | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.220. $3^{2 x+4}+45 \cdot 6^{x}-9 \cdot 2^{2 x+2}=0$.
7.220. $3^{2 x+4}+45 \cdot 6^{x}-9 \cdot 2^{2 x+2}=0$.
(No change needed as the text is already in English and contains only a mathematical equation.) | Solution.
Rewrite the equation as $81 \cdot 3^{2 x}+45 \cdot 3^{x} \cdot 2^{x}-36 \cdot 2^{x}=0$. Dividing it by $9 \cdot 2^{2 x}$, we get $9 \cdot\left(\frac{3}{2}\right)^{2 x}+5 \cdot\left(\frac{3}{2}\right)^{x}-4=0 \Rightarrow\left(\frac{3}{2}\right)^{x}=-1$ (no solutions), or $\left(\frac{3}{2}\right)^{x}=\left(\f... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.225. $\log _{3 x+7}(5 x+3)+\log _{5 x+3}(3 x+7)=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0-\frac{3}{5}, x \neq-\frac{2}{5}\right.$.
Multiplying the equation by $\log _{3 x+7}(5 x+3) \neq 0$, we get
$$
\begin{aligned}
& \log _{3 x+7}^{2}(5 x+3)-2 \log _{3 x+7}(5 x+3)+1=0 \Leftrightarrow\left(\log _{3 x+7}(5 x+3)-1\right)^{2}=0 \Leftrightarrow \\
& \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.227. $(\lg (x+20)-\lg x) \log _{x} 0.1=-1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+20>0, \\ 0<x \neq 1\end{array}\right.$ or $0<x \neq 1$.
Switch to base 10. We have $(\lg (x+20)-\lg x)\left(-\frac{1}{\lg x}\right)=-1 \Leftrightarrow$ $\Leftrightarrow \lg (x+20)-\lg x=\lg x \Leftrightarrow \lg (x+20)=2 \lg x \Leftrightarrow \lg (x+20)=\lg x^... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.229. $27 \cdot 2^{-3 x}+9 \cdot 2^{x}-2^{3 x}-27 \cdot 2^{-x}=8$. | Solution.
Transform the equation:
$$
\begin{aligned}
& 27+9 \cdot 2^{4 x}-2^{6 x}-27 \cdot 2^{2 x}=8 \cdot 2^{3 x} \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-9 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{2 x}-27=0 \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-2^{4 x}-8 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{x}-27=0... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.230. $\log _{x+1}(x-0.5)=\log _{x-0.5}(x+1)$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x+1 \neq 1, \\ 0<x-0.5 \neq 1\end{array}\right.$ or $0.5<x \neq 1.5$.
Multiplying both sides of the equation by $\log _{x+1}(x-0.5) \neq 0$, we get
$$
\begin{aligned}
& \log _{x+1}^{2}(x-0.5)=1 \Rightarrow \log _{x+1}(x-0.5)=-1 \Rightarrow \\
& \Rightarrow x-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.235. $x \log _{x+1} 5 \cdot \log _{\sqrt[3]{1 / 5}}(x+1)=\frac{x-4}{x}$. | ## Solution.
$$
\text { Domain: }\left\{\begin{array}{l}
0<x+1 \neq 1, \\
x \neq 0
\end{array} \Leftrightarrow-1<x \neq 0\right.
$$
Switch to base 5. We have $\frac{x}{\log _{5}(x+1)} \cdot(-3) \log _{5}(x+1)=\frac{x-4}{x}$,
$-3 x=\frac{x-4}{x}$ when $\log _{5}(x+1) \neq 0$. From here, $3 x^{2}+x-4=0, x_{1}=-\frac{4... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.241. $\frac{\log _{2}\left(x^{3}+3 x^{2}+2 x-1\right)}{\log _{2}\left(x^{3}+2 x^{2}-3 x+5\right)}=\log _{2 x} x+\log _{2 x} 2$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{3}+3 x^{2}+2 x-1>0, \\ 0<x^{3}+2 x^{2}-3 x+5 \neq 1, \\ 0<x \neq \frac{1}{2} .\end{array}\right.$
By the formula for changing the base we have
$$
\begin{aligned}
& \log _{x^{3}+2 x^{2}-3 x+5}\left(x^{3}+3 x^{2}+2 x-1\right)=\log _{2 x} 2 x \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$. | Solution.
Domain: $x^{3}+2 x+1>0$.
From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as
$\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$.
Solving this equation as a quadratic in te... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.243. $5^{x} \cdot \sqrt[x]{8^{x-1}}=500$. | Solution.
Rewrite the equation as $5^{x} \cdot 8^{\frac{x-1}{x}}=500 \Leftrightarrow \frac{5^{x} \cdot 8}{8^{1 / x}}=500 \Leftrightarrow$ $\Leftrightarrow \frac{5^{x} \cdot 2}{8^{1 / x}}=125 \Leftrightarrow 5^{x-3}=2^{3 / x-1} \Rightarrow\left\{\begin{array}{l}x-3=0, \\ \frac{3}{x}-1=0,\end{array}\right.$ from which $x... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.245. $\log _{1+x}\left(2 x^{3}+2 x^{2}-3 x+1\right)=3$. | Solution.
Domain of definition: $\left\{\begin{array}{l}2 x^{3}+2 x^{2}-3 x+1>0, \\ -1<x \neq 0 .\end{array}\right.$
We have
$$
\begin{aligned}
& 2 x^{3}+2 x^{2}-3 x+1=(1+x)^{3} \Leftrightarrow 2 x^{3}+2 x^{2}-3 x+1=1+3 x+3 x^{2}+x^{3} \Leftrightarrow \\
& \Leftrightarrow x^{3}-x^{2}-6 x=0 \Leftrightarrow x\left(x^{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.246. $\log _{2} \sqrt[3]{x}+\sqrt[3]{\log _{2} x}=\frac{4}{3}$. | ## Solution.
Domain of definition: $x>0$.
## From the condition we have
$$
\frac{1}{3} \log _{2} x+\sqrt[3]{\log _{2} x}=\frac{4}{3} \Leftrightarrow \log _{2} x+3 \sqrt[3]{\log _{2} x}-4=0
$$
Let $\sqrt[3]{\log _{2} x}=y$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& y^{3}+3 y-4=0 \Leftrightarrow\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.247. $\sqrt{\log _{5} x}+\sqrt[3]{\log _{5} x}=2$. | Solution.
Domain of definition: $\log _{5} x \geq 0$ or $x \geq 1$.
Rewrite the equation as $\sqrt[6]{\left(\log _{5} x\right)^{3}}+\sqrt[6]{\left(\log _{5} x\right)^{2}}-2=0$. Let $\sqrt[6]{\log _{5} x}=y$. The equation in terms of $y$ becomes
$$
y^{3}+y^{2}-2=0 \Leftrightarrow\left(y^{3}-1\right)+\left(y^{2}-1\rig... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.252. $\frac{2}{15}\left(16^{\log _{9} x+1}-16^{\log _{3} \sqrt{x}}\right)+16^{\log _{3} x}-\log _{\sqrt{5}} 5 \sqrt{5}=0$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as
$$
\begin{aligned}
& \frac{2}{15}\left(16 \cdot 16^{\frac{1}{2} \log _{3} x}-16^{\frac{1}{2} \log _{3} x}\right)+16^{\log _{3} x}-3=0 \Leftrightarrow \\
& \Leftrightarrow 16^{\log _{3} x}+2 \cdot 16^{\frac{\log _{3} x}{2}}-3=0
\end{aligned}
$$
Solvin... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.254. $\log _{2} \sqrt[3]{4}+\log _{8}\left(9^{x+1}-1\right)=1+\log _{8}\left(3^{x+1}+1\right)$. | ## Solution.
Domain of definition: $9^{x+1}-1>0 \Leftrightarrow x>-1$.
Since $\log _{2} \sqrt[3]{4}=\frac{2}{3}$, we have
$$
\begin{aligned}
& \log _{8}\left(9 \cdot 3^{2 x}-1\right)-\log _{8}\left(3 \cdot 3^{x}+1\right)=\frac{1}{3} \Leftrightarrow \log _{8} \frac{9 \cdot 3^{2 x}-1}{3 \cdot 3^{x}+1}=\frac{1}{3} \Lef... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.255. $25^{\log _{4} x}-5^{\log _{16} x^{2}+1}=\log _{\sqrt{3}} 9 \sqrt{3}-25^{\log _{16} x}$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as
$5^{\log _{2} x}-5 \cdot 5^{\frac{1}{2} \log _{2} x}=5-5^{\frac{1}{2} \log _{2} x} \Leftrightarrow 5^{\log _{2} x}-4 \cdot 5^{\frac{1}{2} \log _{2} x}-5=0$.
Solving this equation as a quadratic equation in terms of $5^{\frac{1}{\log _{2} x}}$, we get... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.111. Find the non-negative integer values of $x$ that satisfy the inequality $\frac{x+3}{x^{2}-4}-\frac{1}{x+2}<\frac{2 x}{2 x-x^{2}}$. | Solution.
Rewrite the inequality as $\frac{x+3}{x^{2}-4}-\frac{1}{x+2}+\frac{2 x}{x(x-2)}<0 \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow \frac{x+3}{(x-2)(x+2)}-\frac{1}{x+2}+\frac{2}{x-2}<0 \text { for } x \neq 0: \Leftrightarrow\left\{\begin{array}{l}
\frac{2 x+9}{(x-2)(x+2)}<0, \\
x \neq 0
\end{array} \Lef... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
10.199. The bases of the trapezoid are 4 and 16 cm. Find the radii of the inscribed and circumscribed circles of the trapezoid, given that these circles exist. | Solution.
In trapezoid $ABCD$ (Fig. 10.9) $BC \| AD, BC=4$ cm, $AD=16$ cm. Since a circle is circumscribed around the given trapezoid, then $AB=CD$. Since a circle can be inscribed in the given trapezoid, then $AD+BC=AB+CD=2AB; AB=\frac{AD+BC}{2}=10$ cm. $BK$ is the height of the trapezoid. Then $AK=\frac{AD-BC}{2}=6$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.206. What integers represent the sides of an isosceles triangle if the radius of the inscribed circle is $3 / 2 \mathrm{~cm}$, and the radius of the circumscribed circle is $25 / 8$ cm | Solution.
In $\triangle ABC$, $AB = BC$, $BD$ is the altitude, $O_1$ is the center of the inscribed circle, and $O_2$ is the center of the circumscribed circle (see Fig. 10.16). The radii of these circles are $r = \frac{3}{2}$ cm and $R = \frac{25}{8}$ cm, respectively. $E$ is the point of intersection of the ray $BD$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.208. Perpendiculars are drawn from the vertex of the acute angle of a rhombus to the lines containing the sides of the rhombus to which this vertex does not belong. The length of each perpendicular is 3 cm, and the distance between their bases is $3 \sqrt{3}$ cm. Calculate the lengths of the diagonals of the rhombus... | Solution.
Since $\triangle A E F$ is isosceles (Fig. 10.18), the bisector $A M$ is perpendicular to $E F$ and lies on the diagonal of the rhombus. We find $A M^{2}=A F^{2}-M F^{2}=9-\frac{27}{4}=\frac{9}{4}$, i.e., $A M=\frac{3}{2}$ (cm). In $\triangle A C F$, we have $\angle F=90^{\circ}$ and $F M \perp A C \Rightarr... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.211. On the larger leg of a right triangle, a circle is constructed with this leg as its diameter. Determine the radius of this circle if the smaller leg of the triangle is 7.5 cm, and the length of the chord connecting the vertex of the right angle with the point of intersection of the hypotenuse and the circle is ... | Solution.
In $\triangle ABC$ (Fig. 10.21) $\angle ACB=90^{\circ}, BC>AC, AC=7.5$ cm, $N$ is the intersection point of the circle mentioned in the condition and the hypotenuse $AB, CN=6$ cm. $\angle CNB$ is an inscribed angle and subtends the diameter. Therefore, $\angle CNB=90^{\circ}$. From $\triangle ANC \quad (\ang... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.215. In a right-angled triangle, the medians of the legs are $\sqrt{52}$ and $\sqrt{73}$. Find the hypotenuse of the triangle. | ## Solution.
In $\triangle A B C$ (Fig. 10.25) $\angle A C B=90^{\circ}, B P$ and $A E$ are medians, $B P=\sqrt{52}$, $A E=\sqrt{73}$. Let $B C=x, A C=y$. Then $A B=\sqrt{x^{2}+y^{2}}$.
From $\triangle A C E\left(\angle A C E=90^{\circ}\right):$
$$
A C^{2}+C E^{2}=A E^{2} ; y^{2}+\frac{x^{2}}{4}=73
$$
From $\triang... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.216. Two circles, with radii of 4 and 8, intersect at a right angle. Determine the length of their common tangent. | Solution.
Let $A$ be the point of intersection of the circles with centers $O$ and $O_{1}$ (Fig. 10.26), $O A=8 \text{ cm}, O_{1} A=4 \text{ cm}$, $M N$ - their common tangent. According to the problem, the circles intersect at a right angle, so the tangents to them at point $A$ are perpendicular to each other. Theref... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.218. A line parallel to the bases of a trapezoid passes through the point of intersection of its diagonals. Find the length of the segment of this line, enclosed between the lateral sides of the trapezoid, if the bases of the trapezoid are 4 and 12 cm. | ## Solution.
In trapezoid $ABCD$ (Fig. 10.28) $BC \| AD, BC=4$ cm, $AD=12$ cm, $O$ is
Fig. 10.28
 we have: $A B=B C=18, A C=12$. Let $B D=B E=x$. Then $A D=E C=18-x$. The perimeter of trapezoid $A D E C$ is: $P=A C+D E+2 A D=40$, from which $D E=40-(2 A D+A C)=2 x-8$. $\triangle A B C \sim$ $\sim \triangle D B E$, therefore, $\frac{B D}{D E}=\frac{A B... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.229. A parallelogram with sides of 3 cm and 5 cm and a diagonal of 6 cm is inscribed in a triangle. Find the sides of the triangle, given that the diagonals of the parallelogram are parallel to the lateral sides of the triangle, and the smaller of its sides lies on the base of the triangle. | Solution.
Let in $\triangle A B C D K=E F=3 \text{cm}, D E=5 \text{cm}, D F=6$ cm, $D K \| E F$, $O$ - the point of intersection of the diagonals $D F$ and $E K$ of the parallelogram $D E F K$ (Fig. 10.38). Let $E O=x$ cm. Then $D F^{2}+E K^{2}=2\left(D E^{2}+E F^{2}\right)$; $36+4 x^{2}=2(25+9) ; x=2 \sqrt{2}$. OEBF ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.234. A circle with a radius of 13 cm touches two adjacent sides of a square with a side length of $18 \mathrm{~cm}$. Into which two segments does the circle divide each of the other two sides of the square? | ## Solution.
Draw the radius $O K$ (Fig. 10.42); then $K L=\sqrt{K O^{2}-L O^{2}}$. $L O=B E=18-13=5(\mathrm{~cm}) \Rightarrow K L=\sqrt{13^{2}-5^{2}}=12$ (cm), $B K=13-12=1$ (cm). We have found that the side of the square is divided into segments of 1 and 17 cm.
Answer: 1 and 17 cm. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.255. Through point $A$ of a circle with a radius of 10 cm, two mutually perpendicular chords $A B$ and $A C$ are drawn. Calculate the radius of the circle that is tangent to the given circle and the constructed chords, if $A B=16$ cm. | Solution.
$O$ is the center of a circle with a radius of 10 cm (Fig. 10.60), $O_{1}$ is the center of a circle that touches the chords $A B$ and $A C$ and the given circle at points $M, D$, and $K$ respectively. Then $O_{1} D \perp A C, O_{1} M \perp A B$. $\angle B A C$ is an inscribed and right angle. Therefore, $B ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.256. The lengths of two sides of an acute triangle are $\sqrt{13}$ and $\sqrt{10} \mathrm{~cm}$. Find the length of the third side, knowing that this side is equal to the height drawn to it. | Solution.
Let in $\triangle A B C$ (Fig. 10.61) $A B=\sqrt{10}$ cm, $B C=\sqrt{13}$ cm, height $B D=A C$. Since $\triangle A B C$ is an acute-angled triangle, point $D$ lies on the segment $A C$. Let $B D=A C=x$ cm. From $\triangle A D B\left(\angle A D B=90^{\circ}\right)$: $A D=\sqrt{A B^{2}-B D^{2}}=\sqrt{10-x^{2}}... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.259. On the segment $A C$ of length 12 cm, a point $B$ is constructed such that $A B=4 \text{ cm}$. On the segments $A B$ and $A C$ as diameters, semicircles are constructed in the same half-plane with boundary $A C$. Calculate the radius of the circle that is tangent to the constructed semicircles and $A C$. | ## Solution.
Points $O_{1}$ and $O_{2}$ are the centers of semicircles with diameters $A B$ and $A C$ and radii $R_{1}=2$ cm and $R_{2}=6$ cm, respectively (Fig. 10.64). $O_{3}$ is the center of the circle with the unknown radius $x$ cm, $x>0$. Then $O_{1} O_{3}=(x+2)$ cm,
, $\angle ABC = \angle ADC = 90^\circ$, $\angle BAD = 120^\circ$, $\angle BCD = 60^\circ$. $P$ is the intersection point of $AC$ and $BD$, and $AC \perp BD$. Since $\angle ABC = 90^\circ$, $AC$ is the diameter of the given circle. $BP$ is t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.283. The area of a right-angled triangle is $24 \mathrm{~cm}^{2}$, and the hypotenuse is 10 cm. Find the radius of the inscribed circle. | Solution.
Let $a$ and $b$ be the legs of the triangle. We have the system $\left\{\begin{array}{l}a b=48, \\ a^{2}+b^{2}=100,\end{array} \Rightarrow\right.$ $\Rightarrow a^{2}+2 a b+b^{2}=196, a^{2}-2 a b+b^{2}=4, \Rightarrow a+b=14, a-b=2$ and, consequently, $a=8$ (cm), $b=6$ (cm). Since $S=p r$, then $24=12 r, r=2$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.309. The area of an isosceles trapezoid circumscribed about a circle is $32 \mathrm{~cm}^{2}$; the acute angle of the trapezoid is $30^{\circ}$. Determine the sides of the trapezoid. | ## Solution.
Let the height $B H$ of the trapezoid be $h$ (Fig. 10.98). Then $A B=2 h$ (since $\angle A=30^{\circ}$), $B C+A D=4 h$ (since $B C+A D=A B+C D=2 A B$). The area of the trapezoid $S=\frac{(B C+A D) h}{2}=2 h^{2}=32\left(\mathrm{~cm}^{2}\right)$, from which $h=4$ cm. Therefore, $A B=C D=8 \mathrm{~cm}, B C+... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.321. The medians of a triangle are 3, 4, and 5 cm. Find the area of the triangle. | ## Solution.
Let $A N, B M, C P$ be the medians of $\triangle A B C$ (Fig. 10.110), $O$-the point of their intersection, $A N=3 \text{ cm}, B M=4 \text{ cm}, C P=5 \text{ cm}$. On the extension of segment $B M$ beyond point $M$, lay off segment $D M=O M$. Then $D O=B O=$ $=\frac{2}{3} B M=\frac{8}{3}$ cm. Drop a perpe... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.323. The diagonal of an isosceles trapezoid is $10 \mathrm{~cm}$, and the area is $48 \mathrm{~cm}^{2}$. Find the height of the trapezoid. | Solution.
In trapezoid $ABCD$ (Fig. 10.112) $AB=CD, AC=10$ cm, $CE$ is the height. Let $AE=x$ cm, $CE=y$ cm. Since $AE=\frac{AD+BC}{2}$, then by the condition $xy=48$ cm. From $\triangle AEC\left(\angle AEC=90^{\circ}\right): AE^{2}+CE^{2}=AC^{2} ; x^{2}+y^{2}=100$. We solve the system of equations: $\left\{\begin{arr... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.346. A circle with center $O_{1}$ is divided by the diameter $A B$ into two semicircles. In one of them, two new semicircles are constructed, based on $O_{1} A$ and $O_{1} B$ as diameters. A circle is inscribed in the curvilinear figure bounded by the contours of these three semicircles. How many times smaller is it... | ## Solution.
Let $O_{2}$ be the midpoint of $A O_{1}$ - the center of one semicircle, $O_{3}$ be the midpoint of $B O_{1}$ - the center of the second semicircle, $O_{4}$ - the center of the circle inscribed in the curvilinear figure mentioned in the problem (Fig. 10.129).
. According to the problem, $\angle B A D=30^{\circ}, A B \cdot h=6$ dm $^{2}, A D \cdot h=12$ d... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.194. In a trapezoid, the smaller base is equal to 2, the adjacent angles are $135^{\circ}$ each. The angle between the diagonals, facing the base, is $150^{\circ}$. Find the area of the trapezoid.
.
In $\triangle B O C: \angle A C B=\angle D B C=15^{\circ}$, then $\angle B A C=180^{\circ}-(\angle A B C+$ $+\angle A C B)=30^{\circ}$.
By ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.223. Point $C$ is located 12 km downstream from point $B$. A fisherman set out from point $A$, located upstream from point $B$, to point $C$. After 4 hours, he arrived at $C$, and the return trip took 6 hours. On another occasion, the fisherman used a motorboat, thereby tripling his own speed relative to the water, ... | Solution.
Let $v_{T}$ km/h be the speed of the current; $v_{,}$ km/h be the speed of the boat; $a$ km/h be the distance between $A$ and $B$. Then, according to the problem,
$$
\left\{\begin{array}{l}
\left(v_{T}+v_{T}\right) \cdot 4=12+a, \\
\left(v_{L}-v_{T}\right) \cdot 6=12+a, \\
a=\left(3 v_{T}+v_{T}\right) \cdot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.225. A certain order takes 3.6 hours longer to complete in workshop No.1 compared to workshop No.2, and 10 hours longer compared to workshop No.3. If under the same conditions, workshops No.1 and No.2 combine to complete the order, the completion time will be the same as in workshop No.3 alone. By how many hours mor... | ## Solution.
Let $x$ hours be the time to complete the order in Workshop No.3. Then in the 1st workshop, this time is $x+10$ hours, and in the 2nd workshop, it is $-x+6.4$ hours.
$$
\text { According to the condition } \frac{1}{x+10}+\frac{1}{x+6.4}=\frac{1}{x} \text {, from which } x=8 \text { hours. }
$$
This mean... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.227. Two workers were assigned a task. The second one started working an hour later than the first. After 3 hours from the time the first one started, they had 9/20 of the entire work left to complete. By the end of the work, it turned out that each had completed half of the entire work. How many hours would each, w... | Solution.
Let the 1st worker can complete the entire job in $x$ hours, the second in $y$ hours. The productivity of the 1st: $\frac{1}{x} ; 2$nd: $\frac{1}{y}$.
$$
\text { According to the condition }\left\{\begin{array}{l}
\frac{1}{x} \cdot 3+\frac{1}{y} \cdot 2=1-\frac{9}{20}, \\
\frac{x}{2}-\frac{y}{2}=1
\end{arra... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.228. Two workers were assigned to manufacture a batch of identical parts. After the first worked for 2 hours and the second for 5 hours, it turned out that they had completed half of the entire work. Working together for another 3 hours, they found that they had 0.05 of the entire work left to complete. In what time... | ## Solution.
Let the 1st worker can complete the entire job in $x$ hours, the second - in $y$ hours.
Productivity of the 1st: $\frac{1}{x}$, the second $-\frac{1}{y}$.
According to the condition $\left\{\begin{array}{l}\frac{1}{x} \cdot 2+\frac{1}{y} \cdot 5=\frac{1}{2}, \\ \frac{1}{2}+\left(\frac{1}{x}+\frac{1}{y}\... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.232. From the railway station to the beach is 4.5 km. A boy and a scheduled bus left the station for the beach simultaneously. After 15 minutes, the boy met the bus returning from the beach, and he managed to walk another $9 / 28$ km from the place of the first meeting with the bus, when he was caught up by the same... | ## Solution.
Let $v_{\text {m }}, v_{\text{a}}$ be the speeds of the boy and the bus, respectively.
According to the problem, we have the system
$\left\{\begin{array}{l}\frac{15}{60} v_{M}+\frac{15-4}{60} v_{\text{a}}=2 \cdot 4.5, \\ \frac{9}{28 v_{M}}=\frac{4}{60}+\frac{2 \frac{15}{60} v_{M}+\frac{9}{28}}{v_{\text{a... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.233. A tourist was returning from vacation on a bicycle. On the first leg of the journey, which was 246 km, he traveled on average 15 km less per day than he did on the last leg of the journey, which was 276 km. He arrived home right on time at the end of his last vacation day. It is also known that it took him one ... | ## Solution.
Let the tourist travel 267 km in $x$ days, and 246 km in $\frac{x}{2}+1$. On the first segment of 246 km, his speed was $v$ km/day, and on the second segment of 276 km, it was $(v+15)$ km/day.
Then $\left\{\begin{array}{l}\left(\frac{x}{2}+1\right) v=246, \\ x(v+15)=276,\end{array}\right.$
from which $x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.251. The hour and minute hands coincide at midnight, and a new day begins. At what hour of this new day will the hour and minute hands coincide again for the first time, assuming that the clock hands move without jumps? | ## Solution.
Let's assume the speed of the minute hand is 1, then the speed of the hour hand is $-\frac{1}{12} \cdot$ The hands will coincide when $\frac{1}{12} \cdot t + 60 = t$, where $t$ is the time of movement of the hands.
Solving the equation, we find $t = \frac{720}{11}$ (min).
Answer: 1 hr $5 \frac{5}{11}$ m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.262. It is known that a freely falling body travels 4.9 m in the first second, and in each subsequent second, it travels 9.8 m more than in the previous one. If two bodies start falling from the same height, one 5 seconds after the other, then after what time will they be 220.5 m apart? | ## Solution.
Let $t \mathrm{c}$ be the required time. In this time, the first body will travel $4.9 t^{2} \mathrm{M}$, and the second body will travel $4.9(t-5)^{2}$ m.
According to the condition, $4.9 t^{2}-4.9(t-5)^{2}=220.5$, from which $t=7 \mathrm{c}$.
Answer: 7 seconds after the first body starts falling.
![]... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.264. To rise in an ordinary elevator to the top floor of an eight-story building (height 33 m) with two 6-second intermediate stops, it takes as much time as it would to rise in an elevator of a high-rise building with one 7-second intermediate stop at the 20th floor (height 81 m). Determine the lifting speed of the... | ## Solution.
Let $v_{1}$ m/s be the speed of a regular elevator ($v_{1} \leq 3.5 \text{ m/s}$), and $v_{2}$ m/s be the speed of the elevator in a high-rise building, $v_{2}=v_{1}+1.5 \text{ m/s}$.
According to the condition,
$$
\frac{33}{v_{1}}+2 \cdot 6=\frac{81}{v_{1}+1.5}+7 \text{, from which } v_{1}=1.5 \text{ m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.269. At the pier, two passengers disembarked from a steamship and headed to the same village. One of them walked the first half of the way at a speed of 5 km/h and the second half at a speed of 4 km/h. The other walked the first half of the time at a speed of 5 km/h and the second half at a speed of 4 km/h, arriving... | Solution.
Let $x$ km be the distance between the pier and the village, $t$ h be the time it took the second passenger to walk to the village.
According to the problem, $\left\{\begin{array}{l}\frac{x}{2 \cdot 5}+\frac{x}{2 \cdot 4}=t+\frac{1}{60}, \\ 5 \cdot \frac{t}{2}+4 \cdot \frac{t}{2}=x,\end{array}\right.$
from... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.273. The ball is rolling perpendicular to the sideline of a football field. Suppose, moving uniformly decelerated, the ball rolled 4 m in the first second, and 0.75 m less in the next second. A football player, initially 10 m away from the ball, started running in the direction of the ball's movement to catch it. Mo... | Solution.
Let $t$ be the desired time. In this time, the ball will roll $\frac{2 \cdot 4-0.75(t-1)}{2} \cdot t \mathrm{M}$, and the footballer will run $\frac{2 \cdot 3.5+0.5(t-1)}{2} \cdot t \mathrm{M}$. (The formula for the sum of $n$ terms of an arithmetic progression was used).
According to the condition,
$$
\fr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.283. Two pedestrians set out towards each other simultaneously from two points, the distance between which is 28 km. If the first pedestrian had not stopped for 1 hour at a distance of 9 km from the starting point, the meeting of the pedestrians would have taken place halfway. After the stop, the first pedestrian in... | Solution.
Since pedestrians could meet halfway, their speeds are equal. Let $V$ be the initial speed of one of the pedestrians.
According to the condition $\frac{9}{V}+\frac{4}{V+1}+1=\frac{28-(9+4)}{V}$, from which $V=3$ (km/h).
Answer: initially, both were walking at the same speed of 3 km/h. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.290. If both pipes are opened simultaneously, the pool will be filled in 2 hours and 24 minutes. In reality, however, only the first pipe was opened initially for $1 / 4$ of the time it takes the second pipe to fill the pool on its own. Then, the second pipe was opened for $1 / 4$ of the time it takes the first pipe... | Solution.
Let's take the entire volume of the pool as 1. Let $x, y$ (liters) per hour leak through the first and second pipe, respectively. According to the conditions, we have the system
... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.293. Two pipes of different cross-sections are connected to a swimming pool. One is a uniform water supply pipe, and the other is a uniform water discharge pipe, with the first pipe filling the pool 2 hours longer than the second pipe empties it. When the pool was filled to $1 / 3$ of its capacity, both pipes were o... | ## Solution.
Let's write down the values of the sought and given quantities in the form of a table:
| Pipe | Time, h | Capacity | Productivity |
| :---: | :---: | :---: | :---: |
| Supply | $x+2$ | 1 | $\frac{1}{x+2}$ |
| Discharge | $x$ | 1 | $\frac{1}{x}$ |
| Both together | 8 | $\frac{1}{3}$ | $\frac{1}{24}$ |
Ac... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.294. Two workers were assigned to manufacture a batch of identical parts; after the first worked for $a$ hours, and the second for $0.6 a$ hours, it turned out that they had completed $5 / n$ of the entire work. After working together for another $0.6 a$ hours, they found that they still had to manufacture $1 / n$ o... | ## Solution.
Let the entire volume of work be 1. Let $x, y$ be the labor productivity (the number of parts produced per hour) of the first and second workers, respectively. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
a x+0.6 a y=\frac{5}{n} \\
a x+0.6 a y+0.6 a(x+y)=1-\frac{1}{n}
\end{a... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.299. Two points rotate uniformly along two concentric circles. One of them completes a full revolution 5 seconds faster than the other, and therefore manages to make two more revolutions per minute. Let the rays directed from the center of the circle to these points coincide at the beginning of the motion. Calculate... | Solution.
Let $V_{1}, V_{2}$ be the rotational speeds of the points, $l$ be the length of the circumference, and $x$ be the length of the arc between the points after 1 second of rotation. Then the angle between the rays will be $\alpha=\frac{x \cdot 360^{\circ}}{l}$ (1). According to the problem, we have the system $... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.300. The smaller arc between points $A$ and $B$, located on a circle, is 150 m. If the points start moving towards each other along the smaller arc, they will meet after $10 \mathrm{c}$, and if they move along the larger arc, the meeting will occur after $14 \mathrm{c}$. Determine the speeds of the points and the le... | ## Solution.
Let $V_{a}$ and $V_{b}$ be the speeds of points $A$ and $B$, and $l$ be the length of the circumference. According to the conditions, we have the system $\left\{\begin{array}{l}10\left(V_{a}+V_{b}\right)=150, \\ 14\left(V_{a}+V_{b}\right)=l-150, \\ \frac{l}{V_{a}}=\frac{90}{V_{b}},\end{array}\right.$
fro... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.302. Two points move uniformly in the same direction along a circle 60 m in circumference. One of them completes a full revolution 5 seconds faster than the other. At the same time, the points coincide every 1 minute. Determine the speeds of the points. | Solution.
Let $V_{1}, V_{2}$ be the speeds of the points. According to the problem, we have the system
$$
\left\{\begin{array}{l}
\frac{60}{V_{2}}-\frac{60}{V_{1}}=5, \\
60\left(V_{1}-V_{2}\right)=60,
\end{array} \text { from which } V_{1}=4(\mathrm{M} / \mathrm{c}), V_{2}=3(\mathrm{M} / \mathrm{c})\right.
$$
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.309. There are two gold and silver alloys. In one alloy, the quantities of these metals are in the ratio $1: 2$, in the other - $2: 3$.
How many grams should be taken from each alloy to obtain 19 g of an alloy in which gold and silver are in the ratio 7:12? | Solution.
Let $x$ (g) be the amount of the first alloy taken, which contains $\frac{1}{3} x$ gold and $\frac{2}{3} x$ silver. According to the problem, we have the system
$$
\left\{\begin{array}{l}
x+y=19 \\
\frac{1}{3} x+\frac{2}{5} y=\frac{7}{12}\left(\frac{2}{3} x+\frac{3}{5} y\right), \text{ from which } x=9(\mat... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.317. One tourist set out at 6 a.m., and the second - towards him at 7 a.m. They met at 8 a.m. and, without stopping, continued their journey. How much time did each of them spend on the entire journey, if the first arrived at the place where the second started 28 minutes later than the second arrived at the place wh... | ## Solution.
Let $V_{1}, V_{2}$ be the speeds of the first and second tourists, respectively, and $x$ be the time it takes for the second tourist to cover the distance that the first tourist covers in 2 hours. According to the problem, we have the system $\left\{\begin{array}{l}x V_{2}=120 V_{1}, \\ V_{2} x+60 V_{2}=1... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.319. A vessel with a capacity of 8 liters is filled with a mixture of oxygen and nitrogen, with oxygen accounting for $16 \%$ of the vessel's capacity. From this vessel, a certain amount of the mixture is released and an equal amount of nitrogen is added, after which the same amount of the mixture is released again ... | Solution.
Initially, the vessel contained $16 \cdot \frac{8}{100}=\frac{32}{25}$ liters of oxygen. The released $x$ liters of the mixture contain $\frac{16 x}{100}=\frac{4 x}{25}$ liters of oxygen. Now, for every 8 liters of the mixture in the vessel, there are $\frac{32-4 x}{25}$ liters of oxygen, which constitutes $... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.320. Impurities constitute $20 \%$ of the total volume of the solution. What is the smallest number of filters through which the solution must be passed so that the final impurity content does not exceed $0.01 \%$, if each filter absorbs $80 \%$ of the impurities? (It is known that $\lg 2 \approx 0.30$.) | Solution.
Impurities constitute $\frac{1}{5}$ of the solution. After the first filtration, $\left(\frac{1}{5}\right)^{2}$ of the impurities will remain, and after the $k$-th filtration, $-\left(\frac{1}{5}\right)^{k+1}$. According to the condition, $\left(\frac{1}{5}\right)^{k+1} \leq 10^{-4} ;-(k+1) \lg 5 \leq-4$, fr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.325. 9000 parts can be manufactured on several new machines of the same design and one machine of the old design, which works twice as slowly as each of the new machines. The old machine can also be replaced by a new machine of the same design as the others. In this second option, each machine would produce 200 fewe... | Solution.
Let $x$ parts be manufactured on one new machine, $\frac{x}{2}$ - on the old one, $n$ - the number of machines. According to the condition, we have the system
$$
\left\{\begin{array}{l}
(n-1) x+\frac{x}{2}=9000, \\
\frac{9000}{n}=x-200, \quad \text { from which } n=5
\end{array}\right.
$$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.330. Two trucks were supposed to transport a certain cargo in 6 hours. The second truck was delayed in the garage, and when it arrived at the loading site, the first truck had already transported $3 / 5$ of the total cargo; the remaining part of the cargo was transported by the second truck, and the entire cargo was... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the weight of cargo transported by each machine in one trip. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
\frac{1}{x+y}=6, \\
\frac{3}{5 x}+\frac{2}{5 y}=12,
\end{array} \text { from which } x=\frac{1}{10}, y=\frac{1}{15}... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.333. A ball falls from a height of 2 m 43 cm and, upon hitting the ground, bounces back up, each time reaching $2 / 3$ of the height from which it falls again. After how many bounces will the ball rise to a height of 32 cm? | Solution.
The numbers expressing the height to which the ball rises form a geometric progression with $b_{1}=243 q$ and $q=\frac{2}{3}$. According to the condition $q_{n}=32=b_{1} q^{n-1}$, from which $n=5$.
Answer: after 5 bounces. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.340. It is known that the difference between the variable quantities $z$ and $y$ is proportional to the quantity $x$, and the difference between the quantities $x$ and $z$ is proportional to the quantity $y$. The coefficient of proportionality is the same and is a positive integer $k$. A certain value of the quantit... | Solution.
From the condition, we have the system $\left\{\begin{array}{l}z-y=k x, \\ x-z=k y, \\ x-y=\frac{3}{5} z\end{array}\right.$
From the first two equations, we express $\frac{x}{z}$ and $\frac{y}{z}$ in terms of $k$, then substitute the obtained expressions for $\frac{x}{z}$ and $\frac{y}{z}$ into the third eq... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.346. The volume of substance A is half the sum of the volumes of substances B and C, and the volume of substance B is 1/5 of the sum of the volumes of substances A and C. Find the ratio of the volume of substance C to the sum of the volumes of substances A and B. | Solution.
Given that $2 V_{A}=V_{B}+V_{C}$ and $5 V_{B}=V_{A}+V_{C}$. Let $V_{A}=x V_{C}$ and $V_{B}=y V_{C}$. Then we get the system $\left\{\begin{array}{l}2 x-y=1, \\ -x+5 y=1,\end{array}\right.$ from which $x=\frac{2}{3}, y=\frac{1}{3}$. Therefore, $\frac{V_{C}}{V_{A}+V_{B}}=\frac{1}{x+y}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.349. From post office $A$ to village $B$, one needs to walk 9 km. The postman walks the entire path there and back, without stopping in the village, in 3 hours and 41 minutes. The road from $A$ to $B$ goes uphill first, then on flat ground, and finally downhill. Over what distance does the road run on flat ground, i... | ## Solution.
Let $x, y, z$ be the lengths of the road uphill, on flat ground, and downhill, respectively. According to the problem, we have the system
from which $y=4$ (km).
Answer: 4 k... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.362. A certain alloy consists of two metals in the ratio $1: 2$, while another contains the same metals in the ratio $2: 3$. How many parts of each alloy should be taken to obtain a third alloy containing the same metals in the ratio $17: 27?$ | Solution.
Let $x$ parts of the first metal and $y$ parts of the second be taken. Then
$$
\frac{1}{3} x+\frac{2}{5} y=\frac{17}{44}(x+y), \text { hence } \frac{y}{x}=\frac{35}{9}
$$
Answer: 9 and 35 parts. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.369. Around the house, lindens and birches are planted, and their total number is more than 14. If the number of lindens is doubled and the number of birches is increased by 18, then the number of birches will be greater than the number of lindens. If, however, the number of birches is doubled without changing the n... | Solution.
Let $x, y$ be the number of lindens and birches, respectively. According to the condition, we have the system of inequalities $\left\{\begin{array}{l}x+y>14, \\ y+18>2 x, \\ x>2 y .\end{array}\right.$
Construct the lines $x+y=14$ (1), $y+18=2 x$ (2), $x=2 y$ (3). From Fig. 13.20, it is clear that the point ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.4 \frac{3 a^{2}+2 a x-x^{2}}{(3 x+a)(a+x)}-2+10 \frac{a x-3 x^{2}}{a^{2}+9 x^{2}}$.
$1.5\left(\left(\frac{x}{y-x}\right)^{-2}-\frac{(x+y)^{2}-4 x y}{x^{2}-x y}\right)^{2} \frac{x^{4}}{x^{2} y^{2}-y^{4}}$. | 1.4 According to formula (1.8), we have $a^{2}-9 x^{2}=(a-3 x)(a+3 x)$. Simplify the fraction $\frac{a x-3 x^{2}}{(a-3 x)(a+3 x)}$ by canceling $a-3 x$, assuming $a \neq 3 x$; we get $\frac{x}{a+3 x}$. Next, using formula (1.15), factor the numerator of the first fraction:
$$
\begin{aligned}
& 3 a^{2}+2 a x-x^{2}=-\le... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.6 \frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}} ; a=7,4, b=\frac{5}{37}$. | 1.6 Perform the following transformations in the numerator and denominator:
$$
\begin{aligned}
& \frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}=\frac{b+a-2 c}{a b} \\
& \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}=\frac{b^{2}+a^{2}+2 a b-4 c^{2}}{a^{2} b^{2}}= \\
& =\frac{(b+a)^{2}-(2 c)^{2}}{a^{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.24 \frac{\sqrt{\frac{a b c+4}{a}+4 \sqrt{\frac{b c}{a}}}}{\sqrt{a b c}+2} ; a=0.04$. | 1.24 Assuming $a b c \neq 0$, we have:
$$
\begin{aligned}
& \frac{a b c+4}{a}+4 \sqrt{\frac{b c}{a}}=b c+\frac{4}{a}+4 \sqrt{\frac{b c}{a}}= \\
& =(\sqrt{b c})^{2}+4 \sqrt{\frac{b c}{a}}+\left(\frac{2}{\sqrt{a}}\right)^{2}=\left(\sqrt{b c}+\frac{2}{\sqrt{a}}\right)^{2} \\
& \frac{\sqrt{\left(\sqrt{b c}+\frac{2}{\sqrt{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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