problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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13.372 A consignment shop has accepted for sale cameras, watches, fountain pens, and receivers for a total of 240 rubles. The sum of the prices of a receiver and a watch is 4 rubles more than the sum of the prices of a camera and a fountain pen, and the sum of the prices of a watch and a fountain pen is 24 rubles less ... | Solution. Let $x$ be the price of a watch, $y$ be the price of a pen, $z$ be the price of a receiver, $t$ be the price of a camera, and $n$ be the number of cameras. Then, according to the problem, we have:
$\left\{\begin{array}{l}t=10 n, \\ y \leq 6, \\ z+x=y+t+4, \\ 3 n y+n x+n z+n t=240, \\ n, y \text { - integers ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,217 |
13.373 A computing machine was given the task to solve several problems sequentially. Registering the time spent on the assignment, it was noticed that the machine spent the same multiple of time less on solving each subsequent problem compared to the previous one. How many problems were proposed and how much time did ... | Solution. Let $n$ be the number of problems, $b$ be the time to solve the first problem, and $q$ be the common ratio of the geometric progression that represents the times to complete the problems. Using the formula for the sum of $k$ terms of a geometric progression, we have:
$$
\begin{aligned}
& \left\{\begin{array}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,218 |
13.374 Three candles have the same length but different thicknesses. The first candle was lit 1 hour earlier than the other two, which were lit simultaneously. At some point, the first and third candles had the same length, and 2 hours after this, the first and second candles had the same length. How many hours does it... | Solution. Let $x$ be the burning time of the first candle, and $l$ be the length of the candles. Then $\frac{l}{x}, \frac{l}{12}, \frac{l}{8}$ are the burning rates of the candles (i.e., the rates at which the lengths of the candles decrease). According to the problem, after $\frac{\frac{l}{x} \cdot 1}{\frac{l}{8}-\fra... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,219 |
13.375 Find a three-digit number, knowing that the number of its tens is the geometric mean of the number of hundreds and units. If in its notation the digits of hundreds and units are swapped and the new number is subtracted from the desired one, then the difference will be equal to 297. | Solution. Let $\overline{x y z}$ be the desired three-digit number. Then $\overline{x y z}-\overline{z y x}=297$ $\Rightarrow(100 x+10 y+z)-(100 z+10 y+x)=297 \Rightarrow x=z+3$.
By the condition, $y^{2}=x z=z(z+3) \Rightarrow y$ is even $\Rightarrow z(z+3) \vdots 4 \Rightarrow$
$\Rightarrow\left[\begin{array}{l}z: 4... | 300or421 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,220 |
13.376 The desired three-digit number ends with the digit 1. If it is erased and then written as the first digit of the number, the new three-digit number obtained will be less than the desired one by $10 a^{\log _{\sqrt{a}} 3}$. Find this number. | Solution. Let $\overline{x y 1}$ be the desired number. Then
$$
\begin{aligned}
& \overline{x y 1}-\overline{1 x y}=10 a^{\log _{\sqrt{a}} 3} \Leftrightarrow(100 x+10 y+1)-(100+10 x+y)=90 \Rightarrow \\
& \Rightarrow 10 x+y=21 \Leftrightarrow \overline{x y}=21 .
\end{aligned}
$$
Answer: 211 . | 211 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,221 |
13.377 The difference of the logarithms of the hundreds and tens digits of a three-digit number is equal to the logarithm of the difference of the same digits, and the sum of the logarithms of the hundreds and tens is equal to the logarithm of the sum of the same digits, increased by 4/3 times. If this three-digit numb... | Solution. Let $\overline{x y z}$ be the desired number. According to the condition, $\log _{a} x-\log _{a} y=\log _{a}(x-y) \Rightarrow x=y(x-y) \Rightarrow x: y \Rightarrow x=n y, n-$ integer. From here $n y=y(n y-y) \Rightarrow n=y(n-1) \Rightarrow n:(n-1)$
But $n$ and $n-1$ are coprime, hence $n-1=1, y=2, x=4$.
Ac... | 421 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,222 |
13.378 In a piece of alloy weighing 6 kg, copper is contained. In another piece of alloy weighing 8 kg, copper is contained in a different percentage than in the first piece. From the first piece, a certain part was separated, and from the second - a part twice as heavy as from the first. Each of the separated parts wa... | Solution. Let $x$ be the mass of the part separated from the first alloy piece, $p(q)$ be the concentration of copper in the first (second) alloy. Then $x+(8-2 x)=8-x$ is the mass of the third alloy, $2 x+(6-x)=6+x$ is the mass of the fourth alloy, $p x+(8-2 x) q$ is the amount of copper in the third alloy, $p(6-x)+2 x... | 2.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,223 |
13.379 The price of a diamond is proportional to the square of its mass. A diamond with a mass of $p$ carats was broken into two parts, after which its value decreased by a factor of $k$. Find the mass of the parts into which the diamond was broken. Prove that the greatest loss in the value of the diamond occurs when b... | Solution. Let $x$ and $p-x$ be the masses of the parts into which the diamond was divided. Then, according to the problem, we have:
$$
\frac{p^{2}}{x^{2}+(p-x)^{2}}=k \Rightarrow 2 k x^{2}-2 p k x+p^{2} k-p^{2} \Rightarrow
$$
$\Rightarrow x=\frac{p k \pm \sqrt{2 k p^{2}-p^{2} k^{2}}}{2 k}$, and $\left\{\begin{array}{... | \frac{pk\\sqrt{2kp^{2}-p^{2}k^{2}}}{2k} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,224 |
13.380 Several kilograms of two grades of goods were purchased: the first grade at 45 rubles and the second grade at 20 rubles, with 1 kg more of the first grade being purchased. The cost of 1 kg of the first grade is a rubles higher than the cost of 1 kg of the second grade. How many kilograms of each grade of goods w... | Solution. Let $x$ be the mass (in kg) of goods of the first grade. Then $\frac{45}{x}, \frac{20}{x-1}$ are the costs per 1 kg of goods of the first and second grades, respectively. Hence, $\frac{45}{x}-\frac{20}{x-1}=a \Rightarrow a x^{2}-x(a+25)+45=0 \Rightarrow x=\frac{a+25 \pm \sqrt{D}}{2 a}, D=(a-5)(a-125)$.
By th... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,225 |
13.381 Coal mined at point A is sold at q rubles per ton, while coal mined at point B is sold at $p\%$ more expensive. Points A and B are connected by a road $s$ km long. In which zone of this road AB are the consumers of coal located, for whom the purchase and delivery of coal from B is cheaper than from A, if the tra... | Solution. Let the coal consumer be located $x$ km from V. Then the costs for purchasing and delivering 1 ton of coal from A are $r(s-x)+q$, and from V $-r x+\frac{q(100+p)}{100}$.
Therefore, the inequality $r(s-x)+q \leq \frac{100 r x+q(100+p)}{100}$ is equivalent to $x \geq \frac{s}{2}-\frac{q p}{200 r}$. From the la... | \frac{}{2}-\frac{}{200r} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,226 |
13.382 Point $P$ is located on the diameter of a circle with radius $R$ between the ends of the diameter $A B$. From this point $P$, three unit masses move in the directions of segments $P A, P B$, and $P C$ such that $P C$ is a half-chord perpendicular to the diameter $\boldsymbol{A} \boldsymbol{B}$. At what distance ... | Solution. Let $x=PA$, then $PB=2R-x, PC^2=PA \cdot PB=x(2R-x)$ (PC is the height of the right triangle $\triangle ABC$). Since the masses of the moving points are unitary and their speeds coincide with the corresponding distances $PA, PB$, and $PC$ (points $A, B, C$ are reached in one unit of time), the total energy ex... | R\\sqrt{2a^2-3R^2},\frac{3}{2}R^2\leq^2<2R^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,227 |
13.383 Several workers complete a job in 14 days. If there were 4 more people and each worked 1 hour longer per day, then the same job would be done in 10 days. If there were still 6 more people and each worked 1 hour longer per day, then this
job would be completed in 7 days. How many workers were there and how many h... | Solution. Let $x$ be the number of workers, $y$ be the duration (in hours) of the workday, and $t$ be the productivity of one worker per hour. Then
$$
\begin{aligned}
& 14 x y t=10(x+4)(y+1) t=7(x+10)(y+2) t \Leftrightarrow \text { (the volume of work remains constant) } \\
& \left\{\begin{array} { l }
{ 1 4 x y = 1 ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,228 |
13.384 Five people perform a certain job. The first, second, and third, working together, can complete the entire job in 327.5 hours; the first, third, and fifth together - in 5 hours; the first, third, and fourth together - in 6 hours; and the second, fourth, and fifth together - in 4 hours. In what time interval will... | Solution. Let $x_{i}$ be the productivity of the $i$-th worker, $y$ be the volume of work. Then, according to the problem,
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=\frac{y}{7.5} \\
x_{1}+x_{3}+x_{5}=\frac{y}{5} \\
x_{1}+x_{3}+x_{4}=\frac{y}{6} \\
x_{2}+x_{4}+x_{5}=\frac{y}{4}
\end{array}\right.
$$
Multiplying the... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,229 |
13.385 At model aircraft competitions with motors, two models turned out to be the best. The first model stayed in the air for $\boldsymbol{m}$ minutes less than the second, but flew $\boldsymbol{h}$ meters farther. The wind speed is $с$ m/min, but the wind does not affect the duration of the flight; the wind only affe... | Solution. Let $z$ and $z+m$ be the flight durations of the first and second models (which are independent of the wind), and $x$ and $y$ be their own flight speeds. Then the first model flew $(x-c) z$ meters, and the second model flew $(y-c)(z+m)$ meters. Therefore,
$(x-c) z=(y-c)(z+m)+h' \Rightarrow x z-c z=y(z+m)-c z... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,230 | |
13.386 On the sides $AB, BC, AC$ of an equilateral triangle $ABC$, points $A_1, B_1, C_1$ are located such that $AA_1 = BB_1 = CC_1 = x$. The side of the triangle is $a$. Find such $x$ for which the ratio of the areas of triangles $A_1B_1C_1$ and $ABC$ is equal to $m$. Within what limits can the value of $m$ vary? | Solution. By the cosine theorem for $\Delta A_{1} B B_{1}$, we have:
$$
\begin{aligned}
& A_{1} B_{1}^{2}=x^{2}+(a-x)^{2}-2 x(a-x) \cos 60^{\circ} \Rightarrow m=\frac{A_{1} B_{1}^{2}}{A B^{2}}=\frac{x^{2}+(a-x)^{2}-x(a-x)}{a^{2}} \Rightarrow \\
& \Rightarrow 3 x^{2}-3 a x+a^{2}-m a^{2}=0 \Rightarrow x=\frac{a}{6}(3 \p... | \frac{}{6}(3\\sqrt{12-3})for\frac{1}{4}\leq<1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,231 |
13.387 A motorboat departed from point A upstream, and simultaneously a raft set off downstream from point B. They met after a hours and continued moving without stopping. Upon reaching point B, the boat turned back and caught up with the raft at point A. The boat's own speed remained constant throughout. How long were... | Solution. Let $x$ be the own speed of the boat, $y$ be the speed of the river current. Then, up to the first meeting, the boat has traveled a distance of $a(x-y)$, and the raft has traveled $ay$. Therefore, the distance between A and B is $ax$. The time it takes for the raft to travel from B to A is $a \frac{x}{y}$. Th... | (1+\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,232 |
13.388 Three swimmers must swim a lane 50 m long in a pool, immediately turn around, and return to the starting point. The first starts first, the second starts after $a$ s, and the third starts after another $a$ s. At some point in time, before reaching the end of the lane, the swimmers found themselves at the same di... | Solution. Let $x, y, z$ be the speeds (m/s) of the swimmers. Then $\frac{50+r}{z}$ and $\frac{50-r}{x}$ are the times taken by the third and first swimmers to reach the moment of their second meeting, and $\frac{50+s}{z}$ and $\frac{50-s}{y}$ are the times taken by the third and second swimmers to reach the moment of t... | \frac{100-50r-r}{(3-r)},\frac{100-50r-r}{(r-)},<r<\frac{100}{50+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,233 |
13.389 From two pieces of alloy of the same mass but with different percentage content of copper, pieces of equal mass were cut off. Each of the cut pieces was melted with the remainder of the other piece, after which the percentage content of copper in both pieces became the same. How many times smaller is the cut pie... | Solution. Let $y$ be the mass of each alloy piece, $x$ be the mass of the cut-off piece, $p$ and $q$ be the concentration of copper in the first and second pieces of the alloy. Then $p x+q(y-x)$ and $q x+p(y-x)$ are the amounts of copper in the new alloys, respectively. According to the problem,
$$
\frac{p x+q(y-x)}{y... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,234 |
13.390 A column of cars moving uniformly at the same speed has a length of $5 \mathrm{km}$. The head of the column is in the last car, and a motorcyclist is next to him. On the order of the head, the motorcyclist increased his speed, caught up with the lead car, delivered a package, instantly turned around, and returne... | Solution. Let $x$ be the speed of the motorcyclist, $y$ be the speed of the column. Then $\frac{5}{x-y}$ is the time it takes for the motorcyclist to travel from the end of the column to its beginning, $\frac{5}{x+y}$ is the time it takes for the motorcyclist to travel from the beginning of the column to its end, and $... | 5(1+\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,235 |
13.391 Two cars start simultaneously from points A and B and meet at 12 o'clock noon. If the speed of the first car is doubled while the speed of the second car remains the same, the meeting will occur 56 minutes earlier. If, however, the speed of the second car is doubled while the speed of the first car remains the s... | Solution. Let $x$ and $y$ be the initial speeds (km/min) of the cars, and $2z$ be the time (in minutes) after which they met after leaving from A and B. Then $2z(x+y)$ is the distance between A and B, hence
\[
\left\{
\begin{array}{l}
2z(x+y) = (2z-56)(2x+y) \\
2z(x+y) = (2z-65)(x+2y)
\end{array}
\right.
\]
\[
\left\{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,236 |
13.392 A car left the airport for the city center at the same time a bus left the city center for the airport. When the first had traveled half the distance, the second had 19.2 km left to reach the end of the route, and when the second had traveled half the distance, the first had 12 km left to reach the end of the ro... | Solution. Let the speed of the car be $x$ times the speed of the bus, and $z$ be the distance between the airport and the city center. Then, according to the condition, we have:
$\left\{\begin{array}{l}\frac{z / 2}{z-19.2}=x, \\ \frac{z-12}{z / 2}=x\end{array} \Leftrightarrow\left\{\begin{array}{l}z=32, \\ x=\frac{4}{... | 6.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,237 |
13.393 The distance between two points is $d$. Under the influence of certain forces, both points start moving uniformly towards each other. To meet at the midpoint of the path, the first point needs to start moving $t$ units of time earlier than the second. If the points start moving towards each other simultaneously,... | Solution. Let $x$ and $y$ be the speeds of the points. According to the first condition of the problem, $\frac{d}{2 x}-t=\frac{d}{2 y}$. According to the second condition, $T(x+y)=\frac{(k-1)}{k} d$. From these two equations, we have: $y^{2}-y(a-b)-\frac{a b}{2}=0$, where $a=\frac{(k-1) d}{k T}, b=\frac{d}{t} \Rightarr... | \frac{k-1}{2kT}\\frac{}{2}(1-\sqrt{1+\frac{(k-1)^{2}}{k^{2}T^{2}}^{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,238 |
13.394 Two brothers had tickets to a stadium located 10 km from their home. At first, they planned to walk to the stadium, but they changed their mind and decided to use a bicycle, agreeing that one would go by bicycle while the other would walk at the same time. After covering part of the distance, the first brother w... | Solution. Since the brothers arrived at the stadium simultaneously (i.e., spent the same amount of time on the journey), they walked (and cycled) the same distance. But in total, they walked 10 km, so each of them walked 5 km. Therefore, the time gain will be $5 \cdot 12=60$ minutes.
Answer: 1 hour. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,239 |
13.395 A sportsman, walking along a highway, noticed that every 6 minutes a trolleybus catches up with him and every 3 minutes a trolleybus passes him in the opposite direction. Find the intervals at which trolleybuses depart from the terminal points and how many times slower the sportsman was walking compared to the t... | Solution. Let $x$ and $y$ be the speeds of the athlete and the trolleybuses, and $z$ be the interval of the trolleybuses' movement. Then $y z$ is the distance between the trolleybuses. According to the problem, $\left\{\begin{array}{l}y z=3(x+y), \\ y z=6(y-x),\end{array} \Rightarrow x=\frac{y}{3} ; z=4\right.$.
Answe... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,240 |
13.396 According to the schedule, a communications officer should first depart from point A, and after 6 hours - a second communications officer at such a speed to catch up with the first one 180 km from point A. However, at the moment of departure, the first communications officer received an order to travel at a spee... | Solution. Let $x$ and $y$ be the initial speeds of the first and second messengers, and $z$ be the time the first messenger was on the road. According to the initial plan, $\frac{180}{x}-\frac{180}{y}=6$. In reality, the speed became $x+a$, and the second messenger was on the road for $z-3$ hours.
$$
\begin{aligned}
&... | \frac{-3+3\sqrt{^{2}+240}}{2};\frac{-9+3\sqrt{^{2}+240}}{2},where<30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,241 |
13.397 Two trains depart simultaneously from A and B towards each other and meet at a distance of $p$ km from B. After $t$ hours from the meeting, the second train, having passed point A, was $q$ km away from it, while the first train, having passed point B, was at a distance from the second train that was twice the di... | Solution. Let $x$ and $y$ be the speeds of the trains, and $z$ be the distance between A and B. Then $\frac{z}{x+y}=\frac{p}{y}$ is the time the trains travel until their first meeting;
$t y=q+(z-p)-$ is the distance traveled by the second train after the meeting, $t(x+y)-$ is the total distance traveled by both train... | 3p-q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,242 |
13.398 Two friends decided to go hunting. One of them lives 46 km from the hunting base, the other, who has a car, lives 30 km from the base between the base and his friend's house. They set off at the same time, with the car owner driving towards his friend who was walking. Upon meeting, they drove together to the bas... | Solution. Let $x$ and $y$ be the speeds (km/h) of the pedestrian and the car. Then $y \cdot 1$ is the total distance the car has traveled, from which $\frac{y-30}{2}$ is the distance the car traveled before meeting the pedestrian, $16-\frac{y-30}{2}=\frac{62-y}{2}$ is the distance the pedestrian traveled before the mee... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,243 |
13.399 The train was delayed at the departure station for 1 hour 42 minutes. Upon receiving the departure signal, the driver followed this schedule: on the section constituting 0.9 of the entire route from the departure station to the destination station, he maintained a speed 20% higher than the usual speed, and on 0.... | Solution. Let $z$ be the distance between stations, $x$ be the usual speed of the train.
$$
\begin{aligned}
& \text { Then } \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}-\text { is the travel time of the train, hence } \\
& \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}=\frac{z}{x}-1 \frac{42}{60} \Rightarrow \frac{z}{x}=10 .
\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,244 |
13.400 On the highway, points D, A, C, and B are arranged sequentially. A motorcyclist and a cyclist set off from A and B simultaneously, heading to C and D, respectively. Meeting at E, they exchanged vehicles and each continued their journey. As a result, the first one spent 6 hours on the trip from A to C, while the ... | Solution. Let $x=DA, y=AE, z=EC, u=CB$. Then $\frac{y}{60}+\frac{z}{25}$ is the total travel time of the first, $\frac{z+u}{25}+\frac{x+y}{60}$ is the total travel time of the second, $\frac{y}{60}=\frac{z+u}{25}$ is the time of their travel until they meet. In the end, we have:
$\frac{y}{60}+\frac{z}{25}=6$
$\left\{... | 340 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,245 |
13.401 Two speed skaters started simultaneously on a running track for a distance of $s$ m. When the winner reached the finish line, the other had one more lap to go. Determine the length of the running track, given that the winner, running each lap $a$ s faster than the loser, completed the distance in $t$ min. | Solution. Let $x$ - be the length of the running track (or circle). According to the condition, $\frac{s}{60 t}$ - is the speed (m/sec) of the first athlete. Hence, $\frac{x}{s / 60 t}+a$ - is the time it takes for the second athlete to run one lap. Since the number of laps to the finish is $\frac{s}{x}$, then $\left(\... | \frac{}{120}(\sqrt{^{2}+240}-) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,246 |
13.402 The capacities of three vessels $A$, B, C, each of which has the shape of a cube, are in the ratio $1: 8: 27$, and the volumes of water poured into them are in the ratio 1:2:3. After transferring part of the water from vessel A to vessel B and from vessel B to vessel C, the water level in all three vessels becam... | Solution. Let $x$ be the initial volume of water in vessel $\mathrm{A}$, and $a$ be the linear size of vessel A. According to the problem, $2x$ and $3x$ are the volumes of water in B and C, and $2a$ and $3a$ are the linear sizes of B and C. Since after the first transfer, the water level in all vessels became the same,... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,247 |
13.403 Three logging brigades are competing. The first and third brigades processed wood twice as much as the second, while the second and third brigades processed wood three times as much as the first. Which brigade won this competition? | Solution. Let $x, y, z$ be the volumes of work of the teams. Then, according to the conditions, $\left\{\begin{array}{l}x+z=2 y, \\ y+z=3 x,\end{array} \Rightarrow 4 x=3 y \Rightarrow y>x\right.$.
But from the first equation $z=(y-x)+y$, hence $z>y$.
## Answer: the third team. | thethirdteam | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,248 |
13.405 Two pedestrians set out simultaneously from A to B and from B to A. When the first had walked half the distance, the second had 24 km left to walk to the end, and when the second had walked half the distance, the first had 15 km left to walk to the end. How many kilometers will the second pedestrian have left to... | Solution. Let $x$ be the distance between A and B, $y$ be the ratio of the speeds of the first and second pedestrians. Then
$\left(\frac{x}{2}\right):(x-24)=y,(x-15):\left(\frac{x}{2}\right)=y$, from which $\frac{x}{2(x-24)}=\frac{2(x-15)}{x} \Rightarrow x=40$ (the value $x=12$ is not suitable, as $x>24) \Rightarrow y=... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,249 |
13.406 Three motorcyclists travel the same section AB of the road at constant but different speeds. First, the first motorcyclist passed point A, and 5 s later, in the same direction, the second and third motorcyclists. After some time, the third motorcyclist overtook the first, and 10 s later, the second overtook him.... | Solution. Let $z$ be the distance between A and B, $x$ be the time (sec) it takes for the first motorcyclist to cover this distance. Then $\frac{z}{x}, \frac{z}{60}, \frac{z}{40}$ are the speeds of the motorcyclists, $\frac{5 z}{x}$ is the difference in distances between them at the moment the second and third pass poi... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,250 |
13.407 Three people, A, B, and C, approached the shore of a reservoir; A set off to the opposite shore swimming at a speed of v km/h; simultaneously, B and C set off in a motorboat at a speed of $10 v$ km/h. After some time, C decided to cover the remaining distance by swimming at the same speed as A. At the same momen... | Solution. Let $T_{1}$ be the point on the shore where the crossing began, $T_{4}$ - the point where it ended, $T_{2}$ - the point where A met the boat, $T_{3}$ - the point where B disembarked from the boat. Since A swam the segment $T_{1} T_{2}$, and C swam the segment $T_{3} T_{4}$, and they both arrived at $T_{4}$ si... | \frac{31b}{130v} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,251 |
13.408 A material particle $m_{1}$ oscillates between points $A$ and $B$, which are 3.01 m apart. The particle's speed is constant, and it does not stop at the endpoints. After 11 s from the departure of particle $m_{1}$ from point $A$, another particle $m_{2}$ starts moving from point $B$ with a constant but lower spe... | Solution. Let $x$ and $y$ be the speeds (cm/s) of particles $m_{1}$ and $m_{2}$. By the time particle $m_{2}$ exits, the distance between $m_{1}$ and $m_{2}$ is $301-11 x$ cm, hence $301-11 x=10(x+y)$. Let $C$ be the point of the first meeting of the particles. Then $B C=10 y$, and therefore, by the time of the second ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,252 |
13.409 A self-propelled roller can compact a strip 0.85 m wide, with each subsequent strip overlapping the previous one by $1 / 4$ of its width. At what speed should this roller move to ensure that a highway section 750 m long and 6.5 m wide can be compacted twice within a time not exceeding 6 hours and not less than 5... | Solution. According to the condition, one pass of the roller over the highway section adds a strip of width $\frac{3}{4} \cdot 0.85=0.6375$ m, i.e., 10 passes will be sufficient to compact the entire strip. In the end, the roller will need to pass 20 times the entire section, i.e., 15 km.
From this, if $x$ is its spee... | 2.5\leqx\leq3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,253 |
13.410 Along the sides of a right angle, towards the vertex, two spheres with radii of 2 and 3 cm are moving, with the centers of these spheres moving along the sides of the angle at unequal but constant speeds. At a certain moment, the center of the smaller sphere is 6 cm from the vertex, and the center of the larger ... | Solution. Let $x$ and $y$ be the speeds of the balls. After 1 second, the distances from the centers of the balls to the corner are $6-x$ and $16-y$, and after 3 seconds, they are $6-3x$ and $16-3y$. By the Pythagorean theorem, we have: $\left\{\begin{array}{l}(6-x)^{2}+(16-y)^{2}=13^{2}, \\ (6-3 x)^{2}+(16-3 y)^{2}=5^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,254 |
13.411 Two points $\boldsymbol{A}$ and $\boldsymbol{B}$, initially at a distance of $a$ from each other, simultaneously started moving towards the vertex of a right angle along different sides with the same constant speed $v$. Point $B$ reaches the vertex $t$ units of time earlier than point $A$ (all measurements are i... | Solution. Let $x$ be the time of movement of point $A$. Then point $A$ travels a distance $\boldsymbol{x} \cdot v$ to the vertex, and point $B-(x-t) v$.
By the Pythagorean theorem, $x^{2} v^{2}+(x-t)^{2} v^{2}=a^{2} \Rightarrow x=\frac{v t \pm \sqrt{2 a^{2}-v^{2} t^{2}}}{2 v}$.
Since $x>t$, only one value of $x$ rema... | \frac{v+\sqrt{2^{2}-v^{2}^{2}}}{2v} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,255 |
13.412 Three state farms are not located on the same straight line. The distance from the first to the third via the second is four times longer than the direct route between them; the distance from the first to the second via the third is $a$ km longer than the direct route; the distance from the second to the third v... | Solution. Let $x$ be the distance between the first and second collective farms, $y$ be the distance between the second and third, and $z$ be the distance between the first and third. Then, according to the condition,
$$
\left\{\begin{array} { l }
{ x + z = 85, } \\
{ x + y = 4 z, } \\
{ z + y = x + a }
\end{array} \... | \frac{425-}{7},\frac{255+5}{7},\frac{170+}{7}for0<<68 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,256 |
13.413 An alloy consists of tin, copper, and zinc. If 20 g is separated from this alloy and melted with 2 g of tin, then in the newly obtained alloy, the mass of copper will be equal to the mass of tin. If, however, 30 g is separated from the original alloy and 9 g of zinc is added, then in this new alloy, the mass of ... | Solution. Let $x, y, z$ be the concentrations of copper, tin, and zinc in the alloy. According to the first condition of the problem, $20 x = 20 y + 2$. According to the second condition, $30 y = 30 z + 9$. In the end, we have:
$$
\left\{\begin{array} { l }
{ 1 0 x = 1 0 y + 1 , } \\
{ 1 0 y = 1 0 z + 3 , } \\
{ x + ... | 50,40,10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,257 |
13.414 Two inspectors set off simultaneously from points $\boldsymbol{A}$ and $B$ to the scene of an incident, point $C$. The first inspector arrived at point $C$ after $a$ minutes. If the second inspector aims to reach point $C$ from point $B$ simultaneously with the first, he would have to spend $c$ minutes less on e... | Solution. Let $x$ be the distance between $A$ and $C$. Then $\dot{x}+b$ is the distance between $B$ and $C$, $\frac{a}{x}$ and $\frac{a}{x+b}$ are the times taken by the first and second inspectors respectively to travel 1 km.
According to the condition, $\frac{a}{x}=\frac{a}{x+b}+c \Rightarrow c x^{2}+b c x-a b=0$.
... | \frac{-+\sqrt{b^{2}^{2}+4}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,258 |
13.415 Two cyclists set out simultaneously from points $A$ and $B$ towards each other. Four hours after their meeting, the cyclist who had set out from $A$ arrived at $B$, and nine hours after their meeting, the cyclist who had set out from $B$ arrived at $A$. How many hours was each cyclist on the road? | Solution. Let $x$ and $y$ be the travel times of the first and second cyclists, respectively. Then $x-4=y-9$ is the time each of them traveled until they met. Since each of them traveled a distance after the meeting equal to the distance the other had traveled before the meeting, $\frac{4}{y-9}=\frac{x-4}{9}$.
$$
\tex... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,259 |
13.416 There is a certain number of barrels of two samples (sizes) with a total capacity of 7000 liters. If all the barrels were of the first sample, the total capacity would increase by 1000 liters. If all the barrels were of the second sample, the total capacity would decrease by 4000 liters. Calculate the capacity o... | Solution. Let $x$ and $y$ be the total capacities of the barrels of the first and second types, respectively, and $n$ and $k$ be their quantities. Then $\frac{x}{n}$ and $\frac{y}{k}$ are the capacities of one barrel of each type. According to the conditions of the problem,
$$
\left\{\begin{array} { l }
{ x + y = 7 0... | 6400,600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,260 |
13.417 A bumblebee flew to an apple tree at a speed of $v_{1}$ m/min. At the same time, a bee flew to another apple tree at a speed of $v_{2}$ m/min. The bumblebee had to cover a distance of $2 a$ m, while the bee had to cover a distance of $2 b$ m. Suppose the trajectories of their flights are mutually perpendicular s... | Solution. Let $O$ be the point of intersection of the trajectories of the bumblebee and the bee. Then, after time $x$, the distances of the bumblebee and the bee from point $O$ will be respectively $\pm\left(x v_{1}-a\right)$ and $\pm\left(x v_{2}-b\right)$. From here, by the Pythagorean theorem, the distance between t... | \frac{v_{1}+v_{2}b}{v_{1}^{2}+v_{2}^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,261 |
13.418 Two cyclists leave point $A$ simultaneously at different speeds and head for point $B$. Upon reaching it, they immediately turn back. The first cyclist, who was faster than the second, meets the second one on the return trip at a distance of $a$ km from $B$; then, after reaching $A$, he heads again towards $B$ a... | Solution. Let $z$ be the distance between $A$ and $B$, $x$ and $y$ be the speeds of the cyclists. By the time of the first meeting, the first cyclist has traveled a distance of $z+a$, and the second cyclist has traveled $z-a$, hence $\frac{z+a}{z-a}=\frac{x}{y}$. By the time of the second meeting, the first cyclist has... | 2ak | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,262 |
13.419 Two trains, 490 m and 210 m long, are moving towards each other at a constant speed on parallel tracks. The driver of one of them noticed the oncoming train at a distance of $700 \mathrm{~m}$; after this, the trains met after 28 s. Determine the speed of each train, given that the first train takes 35 s longer t... | Solution. Let $x$ and $y$ be the speeds (m/s) of the trains. According to the first condition, $700=28(x+y)$. According to the second condition of the problem, $\frac{490}{x}-\frac{210}{y}=35$.
In the end, we have: $\left\{\begin{array}{l}x+y=25, \\ \frac{14}{x}-\frac{6}{y}=1\end{array} \Rightarrow\left\{\begin{array}... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,263 |
13.420 A motorcade of cars with cosmonauts is moving uniformly along the avenue at a speed of $v$ km/h. The length of the motorcade is constantly maintained at $\boldsymbol{m}$ m. A bouquet of flowers, thrown from a window of a house, hit the motorcycle sidecar of a motorcyclist who was riding behind the motorcade. The... | Solution. Let $x$ be the speed (km/h) of the motorcyclist. Then he will catch up with the beginning of the convoy in $\frac{m}{1000(x-v)}$ hours, and then he will be at the end of the convoy in $\frac{m}{1000(x+v)}$ hours. According to the condition, $\frac{m}{1000(x-v)}+\frac{m}{1000(x+v)}=\frac{t}{60} \Rightarrow 25 ... | \frac{3+\sqrt{9^{2}+2500^{2}v^{2}}}{50} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,264 |
13.421 If a two-digit number is divided by a certain integer, the quotient is 3 and the remainder is 8. If the digits in the dividend are swapped while the divisor remains the same, the quotient becomes 2 and the remainder is 5. Find the original value of the dividend. | Solution. Let $\overline{x y}$ be the desired number, $z$ be the divisor. Then, according to the condition,
\[
\left\{\begin{array}{l}
10 x+y=3 z+8 \\
10 y+x=2 z+5
\end{array}\right.
\]
By eliminating $z$ from these equations, we have: $17 x-28 y=1 \Leftrightarrow 28(x-y)=1+11 x$.
But $12 \leq 1+11 x \leq 100$. In t... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,265 |
13.422 Watermelons delivered to the base are intended for two stores. The first store immediately began transporting the watermelons and transported them daily in portions of equal mass. The second store began transporting the watermelons $a$ days later and also transported them daily in portions of equal mass, but dif... | Solution. Let $z$ be the number of days it took to transport all the watermelons from the base, $x$ and $y$ be the daily portions of the transported products by the first and second stores, respectively. Then $x z$ is the mass of watermelons transported by the first store, and $y(z-a)$ by the second store. According to... | b+\sqrt{b(b-)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,266 |
13.423 In a brigade of earthworkers, each works the same number of hours daily. It is known that the labor productivity is the same for all workers in the brigade, and the brigade can dig a trench for laying a cable in 6 days. However, before the work began, it was found that the working day was reduced by 1 hour, and ... | Solution. Let $x$ be the original number of workers, $y+2$ be the originally planned duration of their working day. Then the volume of work, according to the first condition of the problem, is $6 x(y+2)$, according to the second condition - $9(x-5)(y+1)$; according to the third condition - $12(x-7) y$. Since the volume... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,267 |
13.424 Three machines perform a certain job. If the first one does the job alone, it will finish $a$ days later than if all machines work together. If the second one does the job alone, it will finish $b$ days later than all together, and if the third one does the job alone, it will take $c$ times longer than all machi... | Solution. Let $x, y, z$ be the number of days it takes for the first, second, and third machines, respectively, to complete the entire work $V$ individually. Then $\frac{V}{x}, \frac{V}{y}, \frac{V}{z}$ are the efficiencies of the machines, $\frac{z}{c}$ is the time it takes for them to complete the entire work togethe... | +\frac{-(+b)+\sqrt{(-b)^{2}+4^{2}}}{2(+1)};b+\frac{-(+b)+\sqrt{(-b)^{2}+4^{2}}}{2(+1)};\frac{-(+b)+\sqrt{(-b)^{2}+4^{2}}}{2(+1)},\text | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,268 |
13.425 There are $n$ measuring cylinders with liquid. From the first cylinder, $1 / n$ of the liquid in it was poured into the second cylinder, then $1 / n$ of the liquid that ended up in the second cylinder after the transfer from the first cylinder was poured into the third cylinder, and so on. Finally, $1 / n$ of th... | Solution. Let $x_{k}$ be the initial volume of liquid in the $k$-th beaker. After taking $\frac{1}{n}$ part from the $k$-th beaker (for $k \geq 2$), there remained, according to the problem, $a$ cm$^{3}$, which is $\frac{n-1}{n}$ of the volume that was there before the removal. Therefore, after the transfer from the $(... | x_{1}=\frac{(n-2)}{(n-1)^{2}},x_{2}=\frac{(n^{2}-2n+2)}{(n-1)^{2}},x_{3}=\ldots=x_{n}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,269 |
13.426 Two pumps were installed to fill a swimming pool with water. The first pump can fill the pool 8 hours faster than the second one. Initially, only the second pump was turned on for a time equal to twice the amount of time it would take to fill the pool if both pumps were working simultaneously. Then, the first pu... | Solution. Let $x$ and $y$ be the pump efficiencies, and $z$ be the volume of work. Then $\frac{z}{x+y}$ is the time it takes to fill the pool when both pumps are working simultaneously. According to the conditions of the problem,
$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=8, \\ \frac{2 z}{x+y} \cdot y+1.5(x+y)=z\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,270 |
13.427 After passing through the filter, the liquid is evenly poured into a 40-bucket barrel and can be drained through a tap at the bottom of the barrel. If this tap is open, the inflow and outflow of the liquid are such that every 4 minutes, one bucket is reduced in the barrel. How long will it take for the filtered ... | Solution. Let $x$ and $y$ be the rates (bucket/min) of liquid passing through the filter and the tap, respectively. Then, according to the conditions of the problem,
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 4 ( y - x ) = 1 , } \\
{ \frac { 6 6 } { y } - \frac { 4 0 } { x } = 3 }
\end{array} \Rightarrow \left... | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,271 |
13.428 A batch of identical parts was processed on three machines of different designs in the following sequence: first, only the first machine worked for as many hours as it would take for the second and third machines to complete the entire job together; then, only the second machine worked for as many hours as it wo... | Solution. Let $x, y, z$ be the efficiencies of the machines, and $V$ be the volume of work. According to the problem, $\frac{V}{y+z}$ is the time taken by the first machine, $\frac{V}{x+z}$ is the time taken by the second machine, and $\frac{V}{x+y}$ is the time taken by the third machine.
From this, $\frac{V x}{y+z}+... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,272 |
13.429 First, the motorboat traveled $a$ km across the lake, and then half of this distance up the river flowing into the lake. The entire trip lasted 1 hour. Find the own speed of the motorboat if the speed of the river current is $c$ km/h. | Solution. If $x$ is the speed of the motorboat (km/h), then $\frac{a}{x}$ is the time it takes to travel on the lake, and $\frac{a}{2(x-c)}$ is the time it takes to travel on the river.
From this,
$\frac{a}{x}+\frac{a}{2(x-c)}=1 \Rightarrow 2 x^{2}-x(2 c+3 a)+2 a c=0 \Rightarrow x=\frac{3 a+2 c \pm \sqrt{9 a^{2}-4 a ... | \frac{3+2+\sqrt{9^{2}-4+4^{2}}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,273 |
13.430 A passenger can travel from Moscow to city $N$ by train. In this case, he will be on the way for 20 hours. If, however, he waits for the departure of the plane (and he will have to wait more than 5 hours after the train departs), the passenger will reach city $N$ in 10 hours, including the waiting time. How many... | Solution. Let $x$ and $y$ be the speeds (km/h) of the train and the airplane, respectively, and $z$ be the waiting time for the airplane to depart. Then $\frac{8}{9} y$ is the distance the airplane will fly until it meets the train. According to the problem, $\frac{8}{9} y = x \left(z + \frac{8}{9}\right)$. Additionall... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,274 |
13.431 It is known that the difference between the variable quantities $y$ and $z$ is proportional to the quantity $x$, and the difference between the quantities $z$ and $x$ is proportional to the quantity $\boldsymbol{y}$. The coefficients of these proportions are $k_{1}$ and $k_{2}$, respectively, and $k_{1} \neq k_{... | Solution. According to the condition,
$$
\left\{\begin{array} { l }
{ y - z = k _ { 1 } x , } \\
{ z - x = k _ { 2 } y , } \\
{ z = 3 ( x - y ) }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ y - 3 ( x - y ) = k _ { 1 } x , } \\
{ 3 ( x - y ) - x = k _ { 2 } y }
\end{array} \Rightarrow \left\{\begin{array}{l}... | (k_{1}+3)(k_{2}+3) | Algebra | proof | Yes | Yes | olympiads | false | 52,275 |
13.432 Two athletes are running on a closed track of a stadium. The speed of each is constant, but the first one takes $a$ seconds less to run the entire track than the second one. If they start running from a common starting point in the same direction, they meet every $b$ seconds. How long will it take for them to me... | Solution. Let $x, y$ be the speeds of the athletes, $z$ be the length of the running track. According to the conditions of the problem,
$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=a, \\ \frac{z}{x-y}=b\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=a \\ \frac{x}{z}-\frac{y}{z}=\frac{1}{b}... | \frac{}{\sqrt{^{2}+4}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,276 |
13.433 Enterprise $A$, which consumes ice, purchases it at point $B$ at a price of $a$ rubles per ton. Sometimes this enterprise has to buy ice from another point $C$ at a price of 1.5 $a$ rubles per ton. Both manufacturers deliver the purchased ice to consumer $A$ themselves, charging $p$ rubles per ton-kilometer for ... | Solution. Let $x$ and $y$ be the distances between $B$ and $A$ and between $C$ and $A$ respectively, and $z$ be the actual (real) price of 1 ton of ice. Then, in $A$, $\left(1 \frac{x n}{1000}\right)$ fraction of each ton of ice sent from $B$ and $\left(1 \frac{y n}{1000}\right)$ fraction of each ton of ice sent from $... | \frac{(2.5+p)1000}{2000-n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,277 |
13.434 Prove that the cube of the largest of three consecutive natural numbers cannot be equal to the sum of the cubes of the other two numbers. | Solution. Let $n-1, n, n+1$ be arbitrary consecutive natural numbers. Suppose that $(n+1)^{3}=n^{3}+(n-1)^{3} \Rightarrow n^{3}=2\left(3 n^{2}+1\right) \Rightarrow n=2 k$, i.e., $4 k^{3}=12 k^{2}+1$. A contradiction is obtained, since $12 k^{2}+1$ is an odd number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,278 |
13.435 The desired number is greater than 400 and less than 500. Find it, if the sum of its digits is 9 and it is equal to 47/36 of the number represented by the same digits but written in reverse order. | Solution. The condition that the desired number is within the range from 400 to 500 makes the problem trivial. Therefore, let's solve this problem by replacing this redundant condition with the condition that the desired number is a three-digit number.
Let $\overline{x y z}$ be the desired number. According to the con... | 423 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,279 |
13.436 On the river section from $A$ to $B$, the current is so weak that it can be neglected; on the section from $B$ to $C$, the current affects the boat's movement. The boat covers the distance downstream from $A$ to $C$ in 6 hours, and upstream from $C$ to $\boldsymbol{A}$ in 7 hours. If the current on the section f... | Solution. Let $x$ be the speed of the boat relative to the shore downstream, and $y$ be the speed of the boat relative to the shore upstream. Then $\frac{x+y}{2}$ is the boat's own speed. Let $z$ be the time it takes for the boat to travel the segment from $A$ to $B$ (or from $B$ to $A$), where there is no current. The... | 7.7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,280 |
13.437 What whole positive number should 180 be divided by, so that the remainder is $25\%$ of the quotient? | Solution. Let $n$ be the required number, $m$ be the remainder of the division of 180 by $n$, and $k$ be the quotient.
Then $\left\{\begin{array}{l}180=n \cdot k+m, \\ 4 m=k, \\ m4 m n>4 m^{2}$, i.e., $m^{2}<45 \Rightarrow m=4 \Rightarrow 4 n+1=45 \Rightarrow n=11$.
Answer: 11. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,281 |
13.438 By mixing $2 \mathrm{~cm}^{3}$ of three substances, 16 g of the mixture was obtained. It is known that $4 \mathrm{r}$ of the second substance occupies a volume that is $0.5 \mathrm{~cm}^{3}$ larger than $4 \mathrm{r}$ of the third substance. Find the density of the third substance, given that the mass of the sec... | Solution. Let $x, y, z$ be the densities of three substances. Then, according to the condition $\left\{\begin{array}{l}2 x+2 y+2 z=16, \\ \frac{4}{y}-\frac{4}{z}=0.5, \quad \Rightarrow z=4 . \\ y=2 x\end{array}\right.$
Answer: 4 g $/ \mathrm{cm}^{3}$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,282 |
13.439 If the desired two-digit number is increased by 46, the resulting number will have a product of its digits equal to 6. Find this number given that the sum of its digits is 14. | Solution. Let $\overline{x y}$ be the desired number $z$ - the number obtained from $\overline{x y}$ by adding 46. Then $y=14-x, 10 x+y+46=z \Rightarrow 9 x+60=z$. Consider two cases.
1) $z=\overline{a b}$. Then $9 x+60=10 a+b=9 a+(a+b)$, i.e., $a+b$ is divisible by 3. But by the condition $a \cdot b=6$. Hence, one of... | 77or86 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,283 |
13.440 Given two mutually perpendicular axes $O x$ and $O y$, and a point $A(a, a)$, where $a>0$. It is required to find the coordinates of a point $M$ on the axis $O x$ and a point $P$ on the axis $O y$, such that the triangle $A M P$ is equilateral. | Solution. Let $(o, y)$ and ( $x, o$ ) be the coordinates of points $P$ and $M$ respectively. Since $\triangle A M P$ is equilateral, then
$$
x^{2}+y^{2}=(a-x)^{2}+a^{2}=(a-y)^{2}+a^{2} \Rightarrow(a-x)^{2}=(a-y)^{2} \Leftrightarrow\left[\begin{array}{l}
x=y \\
2 a=x+y
\end{array}\right.
$$
If $x=y$, then $2 x^{2}=(a-... | M((\sqrt{3}-1);0),P(0;(\sqrt{3}-1)) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,284 |
13.441 At a distance of $l$ m from bridge $A$ downstream along the river is located bridge $B$. When the athlete was passing by bridge $A$, heading towards bridge $B$, two balls were thrown to him. He caught the first ball and let the second one float downstream. After swimming a certain section of the river with the b... | Solution. Let $D$ be the point where the swimmer left the first ball and turned back, $C$ be the point where he picked up the second ball, $x$ be the speed (m/min) of the river current, and $y$ be the distance between $C$ and $D$.
Since the first ball freely drifted the distance $D B$, and the second ball drifted the ... | \cdot\frac{3k+1}{k+3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,285 |
13.442 In the store, goods of the first and second grades arrived for a total amount of 450 rubles. An inspection determined that all the goods could only be sold at the price of the second grade, as a result of which the company would incur a loss of 50 rubles. The company eliminated the defects in the first-grade goo... | Solution. Let $x$ and $y$ be the cost of the first and second grade goods, respectively; $a$ and $b$ be the quantity of the first and second grade goods. Then $\frac{x}{a}, \frac{y}{b}$ are the cost per unit of the first and second grade goods, respectively.
According to the conditions of the problem, we have:
$$
\le... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,286 |
13.443 In a flask, there is a salt solution. From the flask, $1 / n$ part of the solution is poured into a test tube, and the solution remaining in the flask is evaporated until the percentage of salt content doubles. After that, the solution from the test tube is poured back into the flask. As a result, the salt conte... | Solution. Let $x$ be the original percentage of salt, and $y$ be the volume of the solution. After removing $\frac{1}{n}$ of the solution, $\frac{n-1}{n} \cdot y$ of the solution remains in the flask. Since the percentage of salt in the flask increased by two times after the water evaporated, the volume of the solution... | \frac{n+1}{n-1}\cdotp;\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,287 |
13.444 The values of the lengths of the sides of a triangle and its area are four consecutive integers, respectively. Find the lengths of the sides of this triangle. | Solution. According to the condition, the sides of the triangle are $n-1, n, n+1$, and the area is $n+2$, where $n$ is a natural number. By Heron's formula,
$n+2=\sqrt{\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n\right) \cdot\left(\frac{3 n}{2}-n+1\right) \cdot\left(\frac{3 n}{2}-n-1\right)} \Rightarrow 16(n+2)=3 n^{2}(n-2... | 3,4,5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,288 |
13.445 On the table stands a cylindrical can with water. The radius of the can's base is $R$. If a ball of radius $r$ is placed into the can, it will lie on the bottom of the can, and the water surface will rise so much that it will be tangent to the ball. Prove that the same will happen if, instead of the given ball, ... | Solution. Let $x$ be the radius of the new ball. The volume $V$ of water in the jar is equal to the volume of a cylinder with a base radius of $R$ and a height equal to the diameter of the ball, minus the volume of the ball, i.e., $V=\pi R^{2} \cdot 2 x-\frac{4}{3} \pi x^{3}$.
Since the amount of water remains constan... | \frac{-r+\sqrt{6R^{2}-3r^{2}}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 52,289 |
13.446 From the same point, two pedestrians set out simultaneously in the same direction along a straight section of the highway at constant but different speeds. After 2 hours, the distance between them was $s$ km. After that, the pedestrians started walking faster and spent 10 minutes less on each kilometer of the wa... | Solution. Let $x$ be the distance (km) traveled by the first pedestrian in the first 2 hours. Then $\frac{120}{x}$ is the number of minutes spent by him to travel 1 km.
Similarly, $\frac{120}{x+s}$ is the number of minutes spent by the second pedestrian to travel 1 km. After increasing their speed, the first and secon... | \frac{24--\sqrt{^{2}+288}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,290 |
13.447 Comparing two bars having the shape of a rectangular parallelepiped, establish that the length, width, and height of the second bar are respectively 1 cm larger than those of the first bar, and the volume and total surface area of the second bar are respectively $18 \mathrm{~cm}^{3}$ and $30 \mathrm{~cm}^{2}$ la... | Solution. Let $x, y, z$ be the linear dimensions of the first block. Then $x+1$, $y+1, z+1$ are the linear dimensions of the second. By the condition
$$
\begin{aligned}
& \left\{\begin{array}{l}
x y z+18=(x+1)(y+1)(z+1) \\
2(x y+y z+x z)+30=2[(x+1)(y+1)+(y+1)(z+1)+(x+1)(z+1)]
\end{array} \Rightarrow\right. \\
& \Right... | 22\mathrm{~}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,291 |
13.448 Two trains departed from station A with an interval of 12 min and almost immediately developed the same speed of 50 km/h. They are traveling in the same direction without stopping, maintaining the indicated speed unchanged. At what speed was the oncoming train traveling if it met these trains 5 min apart? | Solution. Let $x$ be the speed of the oncoming train. The distance between the other two trains is $50 \cdot \frac{1}{5}=10$ km. Therefore, the oncoming train met the second of these two trains after $\frac{10}{x+50}$ hours from meeting the first one.
$$
\text { According to the condition } \frac{10}{x+50}=\frac{1}{12... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,292 |
13.449 The desired three-digit number starts with the digit 1. If it is erased and then written as the last digit of the number, the new three-digit number obtained will be greater than the desired number by $9 a^{1 / \lg a}$. Find this number. | Solution. Let $\overline{1 x y}$ be the desired number. According to the condition
$$
\begin{aligned}
& \overline{1 x y}=\overline{x y 1}-9 \cdot a^{\log _{a} 10}=\overline{x y 1}-90 \text { or } 100+10 x+y=100 x+10 y+1-90 \Leftrightarrow \\
& \Leftrightarrow 21=10 x+y \Rightarrow x=2, y=1
\end{aligned}
$$
Answer: 12... | 121 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,293 |
13.450 If at the beginning of the time measurement there were $m_{0}$ g of substance $A$ and $2 m_{0}$ g of substance $B$, then after any number $t$ years, as a result of the radioactive decay of these substances, there will remain respectively $\boldsymbol{m}=\boldsymbol{m}_{0} \cdot 2^{-\lambda_{1} t}$ and $M=2 m_{0}... | Solution. Let $x$ and $2x$ be the half-lives of the second and first substances, respectively. According to the conditions,
$$
\left\{\begin{array}{l}
m_{0} \cdot 2^{-\lambda_{1} \cdot 2x} = \frac{m_{0}}{2}, \\
2m_{0} \cdot 2^{-\lambda_{2} \cdot x} = m_{0}, \\
m_{0} \cdot 2^{-\lambda_{1} \cdot 20} + 2m_{0} \cdot 2^{-\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,294 |
5.075 Prove that $\frac{C_{n}^{1}+2 C_{n}^{2}+3 C_{n}^{3}+\ldots+n C_{n}^{n}}{n}=2^{n-1}$. | Solution. Since $(1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+C_{n}^{2} x^{2}+\ldots+C_{n}^{n} x^{n}$, then, differentiating both sides of the equation with respect to $x$, we will have
$$
n(1+x)^{n-1}=C_{n}^{1}+2 C_{n}^{2} x+3 C_{n}^{3} x^{2}+\ldots+n C_{n}^{n} x^{n-1}
$$
Substituting $x=1$ into this equation, we obtain the requ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,295 |
5.076 Prove the identity $C_{n}^{k}+C_{n}^{k-1}=C_{n+1}^{k}$. Using this identity, show that $C_{n}^{m}+C_{n-1}^{m}+\ldots+C_{n-10}^{m}=C_{n+1}^{m+1}-C_{n-10}^{m+1}$. | ## Solution.
$$
\begin{gathered}
C_{n}^{k}+C_{n}^{k-1}=\frac{n!}{(n-k)!k!}+\frac{n!}{(n-k+1)!(k-1)!}=\frac{n!}{(n-k)!(k-1)!}\left(\frac{1}{k}+\frac{1}{n-k+1}\right)= \\
=\frac{n!(n+1)}{(n-k)!(k-1)!k(n-k+1)}=\frac{(n+1)!}{(n-k+1)!k!}=C_{n+1}^{k} .
\end{gathered}
$$
Using the proven relation, we get
$$
\begin{aligned}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,296 |
### 5.078 Prove that $C_{2 n+x}^{n} \cdot C_{2 n-x}^{n} \leq\left(C_{2 n}^{n}\right)^{2}$. | ## Solution.
$$
\begin{aligned}
& (n!)^{2} \cdot C_{2 n+x}^{n} \cdot C_{2 n-x}^{n}=(n!)^{2} \cdot \frac{(2 n+x)!}{n!(n+x)!} \cdot \frac{(2 n-x)!}{n!(n-x)!}=(n+x+1) \cdot \ldots \cdot(n+x+n) \times \\
& \times(n-x+1) \ldots \cdot(n-x+n)=((n+1)+x)((n+1)-x)((n+2)+x)((n+2)-x) \ldots \\
& \ldots((n+n)+x)((n+n)-x)=\left((n+... | 2^{36} | Combinatorics | proof | Yes | Yes | olympiads | false | 52,298 |
### 5.080 Find the largest term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$. | Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$.
Then $\frac{T_{k+1}}{T_{k}}=\frac{C_{20}^{k} \cdot(\sqrt{5})^{20-k} \cdot(\sqrt{2})^{k}}{C_{20}^{k-1} \cdot(\sqrt{5})^{21-k} \cdot(\sqrt{2})^{k-1}}=\frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{21-k}{k}$.
The ratio $\frac{T_{k+1}}{T_{k}}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,299 |
5.081 For what values of $x$ is the fourth term the largest in the expansion of $(5+3 x)^{10}$? | Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(5+3 x)^{10}$.
Then $T_{4}=C_{10}^{3} \cdot 5^{7} \cdot(3 x)^{3}$ and the maximum of $T_{4}$ is reached when $x>0$.
We calculate $\frac{T_{k}}{T_{k-1}}=\frac{C_{10}^{k-1} \cdot 5^{11-k} \cdot(3 x)^{k-1}}{C_{10}^{k-2} \cdot 5^{12-k} \cdot(3 x)^{k-2}}=\frac{... | x\in(\frac{5}{8},\frac{20}{21}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,300 |
5.082 If we expand all the brackets in the expression $(1+x)^{9}+(1+x)^{10}+\ldots$ $\ldots+(1+x)^{14}$ and combine like terms, we will get some polynomial. Determine the coefficient of $x^{9}$ in this polynomial without expanding the brackets. | Solution. The coefficient of $x^{9}$ is $A=C_{9}^{9}+C_{10}^{9}+C_{11}^{9}+C_{12}^{9}+C_{13}^{9}+C_{14}^{9}$. According to the condition of problem 5.076,
\[
\begin{aligned}
& A+C_{10}^{10}=1+C_{10}^{9}+C_{11}^{9}+\ldots+C_{14}^{9}+C_{10}^{10}=1+\left(C_{10}^{9}+C_{10}^{10}\right)+C_{11}^{9}+\ldots+C_{14}^{9}= \\
& =1... | 3003 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,301 |
5.083 Twelve students are given two versions of a test (a total of six identical tickets of each version). In how many ways can the students be seated in two rows so that those sitting next to each other do not have the same ticket, and those sitting one behind the other have the same version? | Solution. Six students with version №1 can be arranged in one row in 6! ways (the number of permutations of six elements). The same is true for six students with version №2. The number of different ways (without considering which row will get version №1) is (6!) ${ }^{2}$. The final result is $(6!)^{2}+(6!)^{2}$ ways, ... | (6!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,302 |
5.084 Each of the ten radio operators at point A is trying to establish contact with each of the twenty radio operators at point B. How many different variants of such contact are possible? | Solution. The number of pairs containing one radio operator from each point is $10 \times 20=200$. Since for each pair there are only 2 communication options (either communication exists or it does not), there are a total of $2^{200}$ such options.
Answer: $2^{200}$ | 2^{200} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,303 |
5.085 Six boxes of different materials are delivered to eight floors of a construction site. In how many ways can the materials be distributed across the floors? In how many of these ways will at least two materials be delivered to the eighth floor? | Solution. For each box, there are 8 ways to distribute it across the floors. Therefore, for six boxes, there are a total of $8^{6}$ such ways.
If the 8th floor does not have any boxes, then for each box, there are only 7 ways to distribute it across the seven floors, i.e., a total of $7^{6}$ ways for 6 boxes. If there... | 8^{6};8^{6}-13\cdot7^{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,304 |
5.086 In how many ways can the players of two football teams be arranged in a single row so that no two players from the same team stand next to each other? | Solution. All players of the first team must occupy only even (or only odd) positions in the line. The number of ways to distribute the players of one team to even (or odd) positions is 11! (the number of permutations of 11 elements). Therefore, the total number of ways to form a line where all players of the first tea... | 2\cdot(11!)^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,305 |
5.087 On a bookshelf, there are mathematics and logic books - a total of 20 books. Show that the maximum number of combinations containing 5 mathematics books and 5 logic books is possible when the number of books on the shelf for each subject is 10. | Solution. Suppose the number of mathematics and logic books is $10-k$ and $10+k$, respectively, where $0 \leq k \leq 10, k$ - an integer. Then there are $C_{10+k}^{5}$ ways to choose 5 logic books for the set. Therefore, the total number of such sets is $C_{10-k}^{5} \cdot C_{10+k}^{5}$. By problem 5.078, this number d... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,306 |
### 5.088 A lift with 9 passengers can stop at ten floors. Passengers exit in groups of two, three, and four people. In how many ways can this happen? | Solution. The number of ways in which 3 fixed groups of passengers can be distributed across 10 floors is $A_{10}^{3}$ (the number of arrangements of 10 elements taken 3 at a time). Let's calculate the number of ways to form three groups of 2, 3, and 4 people from 9 people. There are $C_{9}^{2}$ ways to form a group of... | \frac{10!}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,307 |
5.089 "Early in the morning, smiling Igor was dashing barefoot to go fishing." How many different meaningful sentences can be formed using part of the words from this sentence without changing their order of succession?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Solution. Let's break down the original sentence into phrases and number them with Roman numerals, and the words in these phrases with Arabic numerals:
"early in the morning" - I, "to go fishing" - II, "smiling Igor" - III, "ran barefoot" - IV.
In each meaningful sentence composed of phrases I-IV, either III or III. ... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,308 |
5.090 In a chess match between two teams of 8 people, the participants of the games and the color of the pieces for each participant are determined by a draw. What is the number of different outcomes of the draw? | Solution. For each pair of players, there are 2 ways to distribute the color of the pieces. There are a total of 8 pairs. Therefore, the total number of ways to distribute the color of the pieces is $2^{8}$.
Number the players of the first team as $1, 2, \ldots, 8$. Then player №1 can be paired with one of 8 players f... | 2^{8}\cdot8! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,309 |
Robako A.v.
On the board, $2 \cdot n$ consecutive integers are written. In one move, you can split the written numbers into pairs in any way and replace each pair of numbers with their sum and difference (it is not necessary to subtract the smaller number from the larger one, all replacements occur simultaneously). Pr... | By the well-known formula $\$ 1 \wedge 2+2 \wedge 2+\ldots+(2 n) \wedge 2 \$=1 / 3 \$ n(n+1)(4 n+1) \$$, let $\$ m=\backslash f r a c\{n\}\{$ GCD(n,3) $\} \$$, then $\$ 1 \wedge 2+2 \wedge 2+\ldots+(2 n)^{\wedge} 2 \$$ is divisible by $\$ m \$$, but not by $2 \$ m \$$$. This is true not only for the sum of squares of n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,312 |
Frankin B. Given a square grid $n \times n$ where $n > 1$. A crossword is any non-empty set of its cells, and a word is any horizontal or vertical strip (a cell grid rectangle one cell wide) that is entirely composed of cells of the crossword and is not contained in any larger strip of cells of the crossword (neither h... | Example. For a rectangle \$nıtimes $2 \$$ we get $\$ \mathrm{x}=\mathrm{n}+2 \$, \$ \mathrm{y}=2 \$$.
Estimate. Method 1. Let there be $\$ z \$$ cells in the crossword. Choose some of its covering with the smallest number of words. Words from this covering are called correct, and the rest are called incorrect. Each ce... | 1+\frac{n}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,314 |
a) A rectangular table with $m$ rows and $n$ columns is filled with numbers. If we rearrange the numbers in each row in ascending order, and then rearrange the numbers in each column in ascending order, the numbers in each row will still be in ascending order. Prove this.
b) What will happen if we act in a different o... | a) Consider a table where the numbers in each row are in ascending order.
We will prove that the $k$-th largest number $b_{k}$ in the $r$-th column is not less than the $k$-th number $a_{k}$ in the $l$-th column. Indeed, in the $l$-th column, there are $k$ numbers not less than $a_{k}$. In the same rows in the $r$-th ... | notfound | Combinatorics | proof | Yes | Yes | olympiads | false | 52,317 |
Avor: Koniazin S.B.
In a table of size $m \times n$, numbers are arranged - one number in each cell. In each column, the $k$ largest numbers are underlined ($k \leq m$), and in each row, the $l$ largest numbers are underlined ($l \leq n$). Prove that at least $k l$ numbers are underlined twice. | We will prove a more general statement by induction on $m+n$: we will assume that in each column, at least $k$ of the largest numbers are underlined, and in each row, at least $l$ of the largest numbers are underlined.
Base case. When $m=n=k=l=1$, the statement is obvious.
Inductive step. If in an $m \times n$ table ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,318 |
|
Avo: : Alexeev B.E.
Several numbers are written in a circle. If for some four consecutive numbers $a, b, c, d$ the product of the numbers $a-d$ and $b-c$ is negative, then the numbers $b$ and $c$ can be swapped. Prove that such operations can be performed only a finite number of times.
# | Consider the sum $P$ of pairwise products of adjacent numbers. Note that with our operations, $P$ increases. Indeed, with a valid permutation, the terms $a b$ and $c d$ are replaced by $a c$ and $b d$, while the other terms remain the same ($b c$ becomes $c b$). Expanding the brackets in the inequality given in the con... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,319 |
Looshim A.L.
The sequence $x_{0}, x_{1}, x_{2}, \ldots$ is defined by the following conditions: $x_{0}=1, x_{1}=\lambda$, for any $n>1$ the equality holds
$$
(\alpha+\beta)^{n} x_{n}=\alpha^{n} x_{n} x_{0}+\alpha^{n-1} \beta x_{n-1} x_{1}+\alpha^{n-2} \beta^{2} x_{n-2} x_{2}+\ldots+\beta^{n} x_{0} x_{n} .
$$
Here $\... | We will prove by induction that $x_{n}=\frac{\lambda^{n}}{n!}$. The base case $(n=0,1)$ is given in the condition.
Induction step. Suppose $x_{k}=\frac{\lambda^{k}}{k!}$ for all $k \leq n-1$. Then from the equalities
$$
\left((\alpha+\beta)^{n}-\alpha^{n}-\beta^{n}\right) x_{n}=\sum_{k=1}^{n-1} \frac{\alpha^{n-k} \be... | x_{n}=\frac{\lambda^{n}}{n!};atn=[\lambda] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,320 |
Limakov L.g.
In a $4 \times 4$ square table, the numbers $1, 2, 3, \ldots, 16$ are arranged such that the sum of four numbers in each row, each column, and each of the two diagonals is the same, and the numbers 1 and 16 are placed in opposite corners of the table. Prove that in this "magic square," the sum of any two ... | The total sum of all numbers in the square is $17 \cdot 16: 2=34 \cdot 4$; the sums of the numbers in the rows are equal to each other, so the sum of the numbers in each row, column, or diagonal is 34.
Therefore, we need to prove that in the squares satisfying our condition, the sum of the numbers symmetric with respe... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,321 |
In a rectangular table with $m$ rows and $n$ columns, $m n$ positive numbers are written. We find the product of the numbers in each column and sum all $n$ such products. Prove that if the n... | b) Induction on the number of columns in the table. The base case (one column) is obvious.
Induction step. Let $P_{i}$ denote the product of the numbers in the $i$-th column. Now, rearrange the columns (this does not change the sum of the products) so that the product $P_{1}$ becomes minimal. Let $x_{ij}$ be the numbe... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,322 |
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