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13.259. A steamship, 2 hours after departure from pier $A$, stops for 1 hour and then continues its journey at a speed equal to 0.8 of its initial speed, as a result of which it is late to pier $B$ by 3.5 hours. If the stop had occurred 180 km further, then under the same conditions, the steamship would have been late ... | Solution.
Let $AB = x$ km, $v$ km/h be the speed of the steamboat.
According to the problem, $\left\{\begin{array}{l}\frac{x}{v} + 3.5 = 2 + 1 + \frac{x - 2v}{0.8v}, \\ \frac{x}{v} + 1.5 = \frac{2v + 180}{v} + 1 + \frac{x - 2v - 180}{0.8v},\end{array}\right.$ from which $x = 270$ km.
Answer: 270 km. | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.260. Two material particles, being 295 m apart from each other, started moving towards each other simultaneously. The first particle moves uniformly at a speed of $15 \mathrm{~m} / \mathrm{c}$, while the second particle moved 1 m in the first second and 3 m more in each subsequent second than in the previous one. Th... | ## Solution.
Let $t$ seconds have passed before the particles meet. In this time, the first particle has traveled $15 t \text{m}$, and the second particle has traveled $\frac{2+3(t-1)}{2} \cdot t$ (the formula for the sum of the first $n$ terms of an arithmetic progression was used).
According to the condition, $\fra... | 60 | 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 |
13.263. The path from $A$ to $B$ is traveled by a passenger train 3 hours and 12 minutes faster than by a freight train. In the time it takes the freight train to travel from $A$ to $B$, the passenger train travels 288 km more. If the speed of each train is increased by $10 \mathrm{~km} / \mathrm{h}$, the passenger tra... | Solution.
In Fig. 13.7, the graphs of train movements before the speed changes are considered.
We have $\operatorname{tg} \angle M C N=v_{\text {pas. }}^{(0)}=288: 3.2=90$ km/h. If $v_{\text {tov. }}^{(0)}=x$ km/h, then $A B=N L=x t \mathrm{km}, C D=90(t-3.2)=x t \Rightarrow t=\frac{288}{90-x}$. After the speed chang... | 360 | 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,
$$
\frac{33}{v_{1}}+2 \cdot 6=\frac{81}{v_{1}+1.5}+7 \text{, from which } v_{1}=1.5 \text{ m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.267. Two motorcyclists started racing in the same direction from the same starting point simultaneously: one at a speed of 80 km/h, the other at a speed of 60 km/h. Half an hour later, a third racer started from the same starting point in the same direction. Find his speed, given that he caught up with the first rac... | Solution.
Let $x$ km/h be the speed of the third racer. He caught up with the first racer after $t_{1}$ hours, and the second racer after $t_{2}$ hours. According to the problem,
$$
\begin{aligned}
& \left\{\begin{array}{l}
80 t_{1}=x\left(t_{1}-0.5\right), \\
60 t_{2}=x\left(t_{2}-0.5\right) \\
t_{1}-t_{2}=1.25,
\en... | 100 | 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 |
13.271. The race track for the bicycle competition is a contour of a right-angled triangle with a difference in the lengths of the legs of 2 km. The hypotenuse runs along a dirt road, while both legs are on the highway. One of the participants covered the segment along the dirt road at a speed of $30 \mathrm{km} / \mat... | ## Solution.
Length of the route
$S=AB+BC+AC=a+a+2+\sqrt{a^{2}+(a+2)^{2}}$ (see Fig. 13.12).
By the condition $\frac{\sqrt{a^{2}+(a+2)^{2}}}{30}=\frac{a+a+2}{42}$, from which $a=8 \text{ km}$.
$S=8+8+2+\sqrt{8^{2}+10^{2}}=24$ km.
Answer: 24 km. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.272. From post office $A$, a car departed in the direction of post office $B$. After 20 minutes, a motorcyclist set off after it at a speed of 60 km/h. Catching up with the car, the motorcyclist handed a package to the driver and immediately turned back. The car arrived at $B$ at the moment when the motorcyclist was... | Solution.
Let $t$ be the time it takes for the motorcyclist to catch up with the car, and $v$ be the speed of the car. According to the problem, we have the system
$\left\{\begin{array}{l}v\left(t+\frac{20}{60}\right)=60 \cdot t, \\ \frac{82.5-60 \cdot t}{v}=\frac{t}{2},\end{array}\right.$
from which $v=45(\mathrm{k... | 45 | 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 |
13.274. According to the schedule, the train covers a 120 km section at a constant speed. Yesterday, the train covered half of the section at this speed and was forced to stop for 5 minutes. To arrive on time at the final point of the section, the driver had to increase the train's speed by 10 km/h on the second half o... | ## Solution.
Let $x$ km/h be the original speed of the train. Considering the stop, it will take $\frac{1}{12}+\frac{60}{x+10}$ to travel the second half of the section, from which $x=80$ km/h.
Let $v$ km/h be the speed of the train today. Then $\frac{60}{80}=0.15+\frac{60}{v}$, from which $v=100$ km/h.
Answer: 100 ... | 100 | 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 |
13.285. On a 10 km stretch of highway, devoid of intersections, the bus stops only for passengers to get on and off. It makes a total of 6 intermediate stops, spending 1 minute at each, and always moves at the same speed. If the bus were to travel without stopping, it would cover the same distance at a speed exceeding ... | Solution.
Let $t$ min be the time the bus is in motion on the segment. Then its average speed without stops is $-\frac{10}{t}$, and with stops, it is $-\frac{10}{t+\frac{1}{60} \cdot 6}$.
According to the condition, $\frac{10}{t+0.1}+5=\frac{10}{t}$, from which $t=24$ (min).
Answer: 24 min. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.288. A master and his apprentice were tasked with manufacturing a batch of identical parts. After the master worked for 7 hours and the apprentice for 4 hours, it turned out that they had completed $5 / 9$ of the entire work. After working together for another 4 hours, they found that $1 / 18$ of the entire work rem... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the labor productivity of the master and the apprentice per hour, respectively. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
7 x+4 y=\frac{5}{9} \\
7 x+4 y+4(x+y)=1-\frac{1}{18}
\end{array}\right.
$$
The time required fo... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.289. There are two alloys consisting of zinc, copper, and tin. It is known that the first alloy contains $40 \%$ tin, and the second - $26 \%$ copper. The percentage of zinc in the first and second alloys is the same. By melting 150 kg of the first alloy and 250 kg of the second, a new alloy was obtained, in which t... | ## Solution.
Let $x$ be the percentage of zinc in the first alloy. According to the condition, $\frac{x}{100} \cdot 150+\frac{x}{100} \cdot 250=\frac{30}{100}(150+250)$, from which $x=30 \%$. Then in the second alloy, there is $100 \%-(26 \%+30 \%)=44 \%$ of tin, and the new
$$
\text { alloy contains } \frac{40}{100}... | 170 | 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 |
13.292. Two "mechanical moles" of different power, working simultaneously from different ends of the tunnel, could dig it in 5 days. In reality, however, both "moles" were used sequentially from one side of the tunnel, with the first digging $1 / 3$, and the second - the remaining $2 / 3$ of its length. The entire job ... | ## Solution.
Let the length of the tunnel be 1. Let $x, y$ be the powers of the first and second "moles" respectively.
According to the problem, we have the system $\left\{\begin{array}{l}5(x+y)=1, \\ \frac{1}{3 x}+\frac{2}{3 y}=10\end{array}\right.$, from which $x=\frac{1}{15}, y=\frac{2}{15}$.
Then the first "mole... | 15 | 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 |
13.296. Two excavator operators must complete a certain job. After the first one worked for 15 hours, the second one starts and finishes the job in 10 hours. If, working separately, the first one completed $1 / 6$ of the job, and the second one completed $1 / 4$ of the job, it would take an additional 7 hours of their ... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the work efficiency of the first and second excavator operators, respectively. According to the conditions, we have the system $\left\{\begin{array}{l}15 x+10 y=1, \\ 7(x+y)=1-\left(\frac{1}{6}+\frac{1}{4}\right),\end{array}\right.$ from which $x=\frac... | 20 | 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.
$$
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.306. Two hours after departure, the train stopped for 30 minutes. On the remaining part of the route to the station, repair work was being carried out, and the train was allowed a speed that was $1 / 3$ of its initial speed, as a result of which the train arrived at the station 1 hour and 10 minutes late. The next d... | Solution.
Let $x$ be the distance between the stations, $V$ be the speed of the train. According to the problem, we have the system
$$
\left\{\begin{array}{l}
2+\frac{30}{60}+\frac{x-2 V}{\frac{1}{3} V}=\frac{x}{V}+1+\frac{10}{60} \\
\frac{2 V+14}{V}+\frac{30}{60}+\frac{x-(2 V+14)}{\frac{1}{4} V}=\frac{x}{V}+\frac{50... | 196 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.307. Find a three-digit number whose digits form a geometric progression, given that after reducing it by 495, the resulting number is written with the same digits as the desired number but in reverse order; if the digits of the number obtained after subtraction are decreased (from left to right) by 1, 1, and 2, res... | Solution.
Let $100 x+10 y+z$ be the desired number. According to the condition, we have the system
$$
\left\{\begin{array}{l}
x z=y^{2}, \\
100 x+10 y+z-495=100 z+10 y+x, \text { from which } x=9, y=6, z=4 \\
(z-1)+(x-2)=2(y-1)
\end{array}\right.
$$
Answer: 964. | 964 | Number Theory | 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 |
13.310. There is scrap steel of two grades with nickel content of 5 and $40 \%$. How much of each of these grades of metal is needed to obtain 140 tons of steel with a $30 \%$ nickel content? | Solution.
Let $x, y$ be the mass of each type of metal taken. According to the condition
Answer: 40 and 100 tons. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.311. Two trains, a passenger train and an express train, set off towards each other simultaneously from two points that are 2400 km apart. Each train travels at a constant speed, and at some point in time, they meet. If both trains had traveled at the speed of the express train, they would have met 3 hours earlier t... | Solution.
Let $V_{1}, V_{2}$ be the speeds of the passenger and express trains, respectively, and $t$ be the time of travel until the trains meet. According to the problem, we have the system $\left\{\begin{array}{l}t\left(V_{1}+V_{2}\right)=2400, \\ 2 V_{2}(t-3)=2400, \text { from which } V_{1}=60(\text { km } / \tex... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.315. The initial cost price of a unit of product was 50 rubles. During the first year of production, it increased by a certain percentage, and during the second year, it decreased (relative to the increased cost price) by the same percentage, as a result of which it became 48 rubles. Determine the percentages of the... | ## Solution.
Let the cost of a unit of production increase by x percent.
After the increase, the cost became $50+\frac{x}{100} \cdot 50=50+\frac{x}{2}$, and after the decrease - $\left(50+\frac{x}{2}\right)-\frac{x}{100} \cdot\left(50+\frac{x}{2}\right)=\left(50+\frac{x}{2}\right)\left(1-\frac{1}{100}\right) \cdot$ A... | 20 | 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 |
13.323. For the production of wheat bread, as many kilograms of flour were taken as the percentage that the leaven constitutes for this flour. For the production of rye bread, 10 kg more flour was taken, i.e., as many kilograms as the percentage that the leaven constitutes for rye flour. How many kilograms of each type... | Solution.
Let $x$ be the mass of flour for wheat bread, and $(x+10)$ be for rye bread.
According to the condition, $x+x \cdot \frac{x}{100}+(x+10)+\frac{x+10}{100}(x+10)=112.5$ from which $x=35$.
Answer: 35 and 45 kg. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.324. In the first week of his vacation, the engineer spent a few rubles less than $3 / 5$ of the amount of money he had taken with him; in the second week, $1 / 4$ of the remainder plus 30 rubles; in the third week, $2 / 5$ of the new remainder plus 12 rubles; after which $6 / 35$ of the amount of money taken remain... | Solution.
Let $S$ be the total amount of money taken. We fill in the table:
| Period | Spent, rubles | Balance, rubles |
| :---: | :---: | :---: |
| first week | $x$ | $S-x$ |
| second week | $\frac{S-x}{4}+30$ | $\frac{3(S-x)}{4}-30$ |
| third week | $\frac{2}{5}\left(\frac{3(S-x)}{4}-30\right)+12$ | $\frac{6}{35} S... | 1160 | 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.
$$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.326. Three cars are dispatched from $A$ to $B$ at equal time intervals. They arrive in $B$ simultaneously, then proceed to point $C$, which is 120 km away from $B$. The first car arrives there one hour after the second. The third car, upon arriving at $C$, immediately turns back and meets the first car 40 km from $C... | Solution.
Let $V_{1}, V_{2}, V_{3}$ be the speeds of the cars. Compare the time intervals (see Fig. 13.16), expressing them as the ratio of distance to speed:
$A P=\frac{x}{V_{1}}, M P=\frac{x}{V_{2}}, N P=\frac{x}{V_{3}}$.
By the condition:
$$
\frac{x}{V_{1}}-\frac{x}{V_{2}}=\frac{x}{V_{2}}-\frac{x}{V_{3}}, \text ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.328. A fishing brigade planned to catch 1800 tons of fish within a certain period. For $1 / 3$ of this period, there was a storm, as a result of which the planned daily target was underachieved by 20 tons. However, in the remaining days, the brigade managed to catch 20 tons more than the daily norm, and the planned ... | ## Solution.
Let $x$ be the planned period, and $y$ be the number of centners per day according to the plan. According to the conditions, we have the system $\left\{\begin{array}{l}x y=1800, \\ \frac{x}{3}(y-20)+\left(\frac{2 x}{3}-1\right)(y+20)=1800,\end{array}\right.$ from which $y=100$.
Answer: 100 centners. | 100 | 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
$$
\left\{\begin{array}{l}
\frac{1}{x+y}=6, \\
\frac{3}{5 x}+\frac{2}{5 y}=12,
\end{array} \text { from which } x=\frac{1}{10}, y=\frac{1}{15}... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.331. Several balls of equal mass for bearings and several piston rings, also of equal mass, were made from a certain grade of metal. If the number expressing the mass of each ball in grams were 2 less than the number of rings made, and the number expressing the mass of each ring in grams were 2 more than the number ... | Solution.
Let $x$ be the number of balls, $m_{x}$ be the mass of one ball, $y$ be the number of rings, $m_{y}$ be the mass of one ring. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
m_{x}+2=y \\
m_{y}=x+2 \\
x \cdot m_{x}+y \cdot m_{y}-800=2(y-x), \text{ from which } x=25, y=16 \text{ or ... | 25 | 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 |
13.335. If a two-digit number is divided by the product of its digits, the quotient is 3 and the remainder is 8. If the number, formed by the same digits but in reverse order, is divided by the product of the digits, the quotient is 2 and the remainder is 5. Find this number. | ## Solution.
Let $10 x+y$ be the desired number. According to the condition, we have the system
$$
\left\{\begin{array}{l}
10 x+y=3 x y+8 \\
10 y+x=2 x y+5
\end{array}\right.
$$
Answer: 53. | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.338. The recording of a six-digit number starts with the digit 2. If this digit is moved from the first position to the last, keeping the order of the other five digits, the newly obtained number will be three times the original number. Find the original number. | Solution.
The original six-digit number has the form $2 \cdot 10^{5}+x$. After moving the digit 2 to the last place, we get the number $10 x+2$. According to the condition, $10 x+2=3\left(2 \cdot 10^{5}+x\right)$, from which $x=85714$.
Answer: 285714. | 285714 | Number Theory | 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 |
13.352. A girl, an acquaintance of a young man sitting by the window of a tram, was walking towards the moving tram. 8 seconds after she passed by the window, the young man got off the tram and followed her. How much time passed from that moment until he caught up with the girl? The young man's speed is twice the girl'... | Solution.
Let $x, 2x, 10x$ be the speeds of the girl, the boy, and the tram, respectively. According to the problem, $8 \cdot x + 8 \cdot 10x + t \cdot x = t \cdot 2x$, where $t$ is the desired time. Solving the equation, we find $t=88$ (s).
Answer: 88 s. | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.355. A car, having traveled a distance from $A$ to $B$, equal to 300 km, turned back and after 1 hour 12 minutes from leaving $B$, increased its speed by 16 km/h. As a result, it spent 48 minutes less on the return trip than on the trip from $A$ to $B$. Find the original speed of the car. | Solution.
Let $V$ be the original speed of the car. According to the condition, we have
$$
\frac{300}{V}-\left(1 \frac{12}{60}+\frac{300-1 \frac{12}{60} V}{V+16}\right)=\frac{48}{60}, \quad \text { from which } V=60 \text { (km/h) }
$$
Answer: 60 km/h. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.356. The distance between points $A$ and $B$ is 308 m. A point moves from point $A$ towards $B$, covering 15 m in the first second and 1 m less in each subsequent second. Another point moves from point $B$ in the opposite direction, covering 20 m in the first second and 3 m more in each subsequent second. At what di... | Solution.
Let's find the laws of motion of the points.
For the point from $A:\left\{\begin{array}{l}15=1 \cdot V_{1}-\frac{a_{1} \cdot 1^{2}}{2}, \\ 15+14=2 V_{1}-\frac{a_{1} \cdot 2^{2}}{2},\end{array}\right.$ from which $V_{1}=15.5(\mathrm{~m} / \mathrm{c})$, $a_{1}=1\left(\mathrm{m} / \mathrm{c}^{2}\right)$, and t... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.357. A cyclist traveled 96 km 2 hours faster than he had planned. During this time, for each hour, he traveled 1 km more than he had planned to travel in 1 hour and 15 minutes. At what speed did he travel? | Solution.
Let $V$ be the speed at which the cyclist was traveling, $\frac{1 \cdot V-1}{1 \frac{15}{60}}$ - the speed at which the cyclist intended to travel. According to the condition $\frac{96}{V}+2=\frac{96}{\frac{V-1}{1 \frac{1}{4}}}$, from which $V=16$ (km/h).
Answer: 16 km/h. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.358. Find a six-digit number starting with the digit 1 such that if this digit is moved to the end, the resulting number is three times the original number. | Solution.
The original six-digit number has the form $1 \cdot 10^{5}+x$. After moving the digit 1 to the last place, we get the number $10 x+1$. According to the condition, $10 x+1=3\left(1 \cdot 10^{5}+x\right)$, from which $x=42857$.
Answer: 142857. | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.361. A red pencil costs 27 kopecks, a blue one - 23 kopecks. No more than 9 rubles 40 kopecks can be spent on purchasing pencils. It is necessary to purchase the maximum possible total number of red and blue pencils. At the same time, the number of red pencils should be as few as possible, but the number of blue pen... | ## Solution.
Let $x$ red and $y$ blue pencils be bought. According to the condition, $27 x + 23 y \leq 940$ and $y - x \leq 10$. Let's construct the lines $27 x + 23 y = 940$ (1) and $y - x = 10$ (2). From Fig. 13.18, it is clear that these lines intersect at point $A$, the coordinates of which satisfy equations (1) a... | 14 | 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 |
13.367. Several students decided to buy a tape recorder priced between 170 and 195 dollars. However, at the last moment, two of them refused to participate in the purchase, so each of the remaining had to contribute 1 dollar more. What was the price of the tape recorder? | Solution.
Let $x$ be the amount each student was originally supposed to contribute; $y$ be the number of students. According to the problem,
\[
\left\{
\begin{array}{l}
2 x = y - 2, \\
x y \geq 170, \\
x y \leq 195,
\end{array}
\right.
\left\{
\begin{array}{l}
x = \frac{y}{2} - 1, \\
y\left(\frac{y}{2} - 1\right) \geq... | 180 | 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:
$$
\begin{aligned}
& \frac{a b c+4}{a}+4 \sqrt{\frac{b c}{a}}=b c+\frac{4}{a}+4 \sqrt{\frac{b c}{a}}= \\
& =(\sqrt{b c})^{2}+4 \sqrt{\frac{b c}{a}}+\left(\frac{2}{\sqrt{a}}\right)^{2}=\left(\sqrt{b c}+\frac{2}{\sqrt{a}}\right)^{2} \\
& \frac{\sqrt{\left(\sqrt{b c}+\frac{2}{\sqrt{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.26 Given the expression
$$
\begin{aligned}
& \frac{x^{3}-a^{-\frac{2}{3}} \cdot b^{-1}\left(a^{2}+b^{2}\right) x+b^{\frac{1}{2}}}{b^{\frac{3}{2}} x^{2}} \\
& \text { Substitute } x=a^{\frac{2}{3} b}-\frac{1}{2} \text { and simplify the result }
\end{aligned}
$$
Eliminate the irrationality in the denominator of the ... | 1.26 After performing the specified substitution, we find
$$
\begin{aligned}
& \frac{\left(a^{\frac{2}{3}} b^{-\frac{1}{2}}\right)^{3}-a^{-\frac{2}{3}+\frac{2}{3}} b^{-1-\frac{1}{2}}\left(a^{2}+b^{2}\right)+b^{\frac{1}{2}}}{b^{\frac{3}{2}}\left(a^{\frac{2}{3}} b^{\left.-\frac{1}{2}\right)^{2}}\right.}= \\
& =\frac{a^{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.29 Calculate the sum of the cubes of two numbers if their sum and product are 11 and 21, respectively. | 1.29 Let $a$ and $b$ be the required numbers; then $a+b=11$ and $ab=21$. According to formula (1.11), we have $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$. Therefore, $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$, i.e., $a^{3}+b^{3}=$ $=11^{3}-3 \cdot 21 \cdot 11=638$.
Answer: 638. | 638 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.53 \frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}-\sqrt{3}=2$. | 1.53 Consider the equality
$$
\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}=2+\sqrt{3}
$$
Obviously, if this equality is true, then the given equality is also true. Let
$$
a=\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}, b=2+\sqrt{3}
$$
It is easy to establish that $a>0$ and ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.54 \sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=4$. | ### 1.54 Let
$\sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=x$.
Cubing both sides of this equation, and using formula (1.11), we get
$38+\sqrt{1445}+38-\sqrt{1445}+3 \sqrt[3]{(38+\sqrt{1445})(38-\sqrt{1445})} x=x^{3}$,
or $x^{3}+3 x-76=0$. By substitution, we verify that $x=4$ is one of the roots of the resulti... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.1 \quad \frac{2}{3-x}+\frac{1}{2}=\frac{6}{x(3-x)}$. | 2.1 Let's move all terms of the equation to the left side and transform the obtained equation to the form
$$
\frac{x^{2}-7 x+12}{2 x(3-x)}=0
$$
From the equation $x^{2}-7 x+12=0$, we find $x_{1}=3, x_{2}=4$. When $x=3$, the denominator becomes zero; therefore, 3 is not a root. Thus, $x=4$.
Answer: $x=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.13 \sqrt{x-2}=x-4$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.13 \sqrt{x-2}=x-4$. | 2.13 Let's square both sides of the equation:
$$
\begin{aligned}
& (\sqrt{x-2})^{2}=(x-4)^{2} ; x-2=x^{2}-8 x+16 \\
& x^{2}-9 x+18=0 ; x_{1}=3 ; x_{2}=6
\end{aligned}
$$
Let's check the found roots by substituting them into the original equation. If $x=3$, we get $1=-1$ - an incorrect equality; if $x=6$, we get $2=2$... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.14 \sqrt{15-x}+\sqrt{3-x}=6$. | 2.14 We have $\sqrt{15-x}=6-\sqrt{3-x}$. Squaring both sides of the equation, we get $15-x=36-12 \sqrt{3-x}+3-x$; $\sqrt{3-x}=2 ; 3-x=4 ; x=-1$. Checking shows that this value is a root of the equation.
Omвem: $x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.18 \frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$. | 2.18 Reducing the fractions in the left part of the equation, we get
$\sqrt[3]{x^{2}}+1-\sqrt[3]{x}+1=4$, or $\sqrt[3]{x^{2}}-\sqrt[3]{x}-2=0$.
From this, $\sqrt[3]{x}=-1$ or $\sqrt[3]{x}=2$. Since $\sqrt[3]{x}$ cannot be equal to -1 (as this would make the denominators of the original fractions zero), then $x=8$.
O... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.19 \frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$. | 2.19 To solve the equation, we will use the following statement: if the equality $\frac{a}{b}=\frac{c}{d}$ is true, then the equality (check it!) $\frac{a+b}{a-b}=\frac{c+d}{c-d}$ is also valid. We have
$$
\begin{aligned}
& \frac{(\sqrt{x}+\sqrt[3]{x})+(\sqrt{x}-\sqrt[3]{x})}{(\sqrt{x}+\sqrt[3]{x})-(\sqrt{x}-\sqrt[3]{... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.20 \sqrt[7]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | 2.20 Instruction. Use the substitution $\sqrt[7]{\frac{5-x}{x+3}}=z$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$. | 2.49 We have
$$
\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)=-2(x+1)(x-2)
$$
Performing the transformations, we get $2 x^{3}+x^{2}+x-4=0$. Since the sum of all coefficients is zero, then $x=1$. Let's write the last equation in the form
$2 x^{3}-2 x^{2}+3 x^{2}-3 x+4 x-4=0$,
or $(x-1)\left(2 x^{2}+3 x+4\right... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$. | 2.55 By squaring both sides of the equation, we get
$$
\begin{aligned}
& x+1+4 x+13+2 \sqrt{(x+1)(4 x+13)}=3 x+12 \\
& \sqrt{(x+1)(4 x+13)}=-(x+1)
\end{aligned}
$$
Another squaring would eliminate the irrationality, but there is no need for this transformation here. We notice that the derived equation can have a solu... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$. | ### 2.58 We have
$\sqrt{x}=2-\frac{2 x+1}{x+2}$, or $\sqrt{x}=\frac{3}{x+2}$.
After squaring and performing transformations, the equation will take the form $x^{3}+4 x^{2}+4 x-9=0$. Notice that the sum of the coefficients of the last equation is zero, which means it has a root $x=1$. Therefore,
$$
x^{3}-x^{2}+5 x^{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$. | 2.59 We have $\sqrt{x+2}=\sqrt[3]{3 x+2}$. After raising the equation to the sixth power, we get $(x+2)^{3}=(3 x+2)^{2}$, or $x^{3}-3 x^{2}+4=0$, or $(x+1)\left(x^{2}-4 x+4\right)=0$.
From this, $x_{1}=-1$ (extraneous root), $x_{2}=2$.
Answer: $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.78 For what values of a do the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$ have a common root? | 2.78 Subtracting the second equation from the first, we get
$$
a x - x + 1 - a = 0, \text{ or } (a-1)(x-1) = 0
$$
$$
x^{2} + x + 1 = 0
$$
such an equation has no real roots. If $a \neq 1$, then $x = 1$ and, therefore, $1 + a + 1 = 2$, i.e., $a = -2$.
Answer: when $a = -2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1 The tractor driver plowed three plots of land. The area of the first is $\frac{2}{5}$ of the total area of all three plots, and the area of the second is to the area of the third as $\frac{3}{2}: \frac{4}{3}$. How many hectares were there in all three plots if the third plot was 16 hectares smaller than the first? | 3.1 The areas of the plots are equal to $x, y, x-16$ (ha). According to the condition, $x=\frac{2}{5}(x+y+x-16)$ and $y:(x-16)=\frac{3}{2}: \frac{4}{3}$, from which $y=\frac{9}{8}(x-16)$. Further, we have $x=\frac{2}{5}\left(2 x-16+\frac{9}{8} x-18\right)$, or $x=\frac{2}{5}\left(\frac{25 x}{8}-34\right)$, from which $... | 136 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.2 Two workers together produced 72 parts per shift. After the first worker increased their labor productivity by $15 \%$, and the second by $25 \%$, together they started producing 86 parts per shift. How many parts does each worker produce per shift after the increase in labor productivity? | 3.2 Initially, workers produced $x$ and $72-x$ parts per shift, and then $1.15 x$ and $1.25(72-x)$ parts. According to the condition, $1.15 x + 90 - 1.25 x = 86 ; 0.1 x = 4 ; x = 40 ; 1.15 \cdot 40 = 46$.
Answer: 46 and 40 parts. | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.3 The areas of three plots of land are in the ratio $2 \frac{3}{4}: 1 \frac{5}{6}: 1 \frac{3}{8}$. It is known that 72 centners of grain more were harvested from the first plot compared to the second. Find the area of all three plots, if the average yield is 18 centners per hectare. | 3.3 Let $x, y, z$ be the areas of the plots. Then $\frac{x}{y}=\frac{6}{4}, \frac{y}{z}=\frac{4}{3}$, from which $x=6 k, y=4 k, z=3 k$. According to the condition, $(6 k-4 k) \cdot 18=72$, from which $k=2$. Therefore, the total area of all plots is $x+y+z=2(6+4+3)=26$ (ha).
Omeem: 26 ha. | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.4 The working day has been reduced from 8 to 7 hours. By what percentage does labor productivity need to increase to ensure that, with the same rates, the wage increases by $5 \%$? | 3.4 Let a master produce $a$ parts and earn $b$ rubles for 8 hours of work. Then the rate is $\frac{b}{a}$ rubles per part, and the labor productivity is $\frac{a}{8}$ parts per hour. After increasing productivity by $x \%$, the master started producing $\frac{a}{8}+\frac{x a}{8 \cdot 100}$ parts per hour. Therefore, i... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.5 On the first day of the sports competition, $\frac{1}{6}$ of the boys' team and $\frac{1}{7}$ of the girls' team did not meet the qualifying standards and were eliminated from further competition. Over the remaining period of the competition, an equal number of athletes from both teams were eliminated due to non-co... | 3.5 Let $x$ be the number of boys who met the credit standards and $2x$ be the number of girls. Thus, the initial number of team members consists of $x+48$ boys and $2x+50$ girls. By the end of the first day of the competition, $\frac{1}{6}(x+48)$ boys and $\frac{1}{7}(2x+50)$ girls dropped out. Later, an equal number ... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.7 Two parks with a total area of 110 hectares are divided into an equal number of plots. The plots of each park are equal in area to each other, but differ from those of the other park. If the first park were divided into plots of the same area as the second, it would have 75 plots, and if the second park were divide... | 3.7 Let $S$ be the area of the park, $n$ be the number of equal-sized plots, and $Q$ be the area of a plot. Then $\frac{S}{n}=Q$. We will fill in the table with the given and required values in the sequence indicated by the numbers (2) and (12).
| Park | Initially | | | With new arrangement | | |
| :---: | :---: |... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.8 The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $\frac{1}{4}$ of its length, and the circumference of the rear wheel is increased by $\frac{1}{5}$ of its length, then over the same distance, the front wh... | 3.8 The circumference of a wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $C n=s$. We will fill in the table of values of these quantities in the order indicated by the numbers (1), (2), .., (12).
| Wheel | Before change | | | After change | | |
| :---: | :---: | :---: | ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.10 Two skins with a total cost of 2250 rubles were sold at an auction with a profit of $40 \%$. What is the cost of each skin if a profit of $25 \%$ was made on the first one, and a loss of $-50 \%$ on the second one? | 3.10 Let $x$ (rubles) be the cost of the first pelt, then $2250-x$ (rubles) is the cost of the second pelt. After selling, the first pelt was sold for $1.25 x$ (rubles), and the second for $1.5(2250-x)$ (rubles). According to the condition, we set up the equation
$1.25 x + 1.5(2250 - x) = 1.4 \cdot 2250$,
from which ... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.11 Apples of the first grade for a total of 228 rubles and apples of the second grade for a total of 180 rubles were delivered to the store. During unloading, the delivered apples were accidentally mixed. It was found that if all the apples are now sold at one price - 90 kopecks lower than the price per kilogram of f... | 3.11 Let's construct the following table:
| Apple Variety | Cost, rub | Quantity, tons | Price, rub/kg |
| :---: | :---: | :---: | :---: |
| First | 228 | $x$ | $\frac{228}{x}$ |
| Second | 180 | $x+5$ | $\frac{180}{x+5}$ |
According to the problem, we have the equation
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.12. The sum of the digits of a two-digit number is 12. If 36 is added to this number, the result is a number written with the same digits but in reverse order. Find the original number. | 3.12 Let the desired number be of the form $10 x+y$. Then, by the condition,
$x+y=12$
and $10 x+y+36=10 y+x$, i.e.,
$x-y+4=0$.
Adding (1) and (2), we get $2 x=8$, hence $x=4$, and $y=8$.
Answer: 48. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.13 A three-digit number ends with the digit 2. If it is moved to the beginning of the number, the resulting number will be 18 more than the original. Find the original number. | 3.13 Let the desired three-digit number be of the form $100x + 10y + 2$; then after moving the digit 2, it will take the form
$200 + 10x + y$.
By the condition,
$200 + 10x + y - (100x + 10y + 2) = 18$
from which $10x + y = 20$. Substituting this expression into (1), we get $200 + 20 = 220$. Thus, the original three... | 202 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.14 A positive integer is thought of. To its representation, the digit 7 is appended on the right, and from the resulting new number, the square of the thought number is subtracted. The remainder is then reduced by $75\%$ of this remainder, and the thought number is subtracted again. In the final result, zero is obtai... | 3.14 Let a number $x$ be thought of. Then, following the text of the condition, we get the numbers
$10 x+7, 10 x+7-x^{2}$ and the remainder $\frac{25}{100}\left(10 x+7-x^{2}\right)$.
Then $\frac{1}{4}\left(10 x+7-x^{2}\right)-x=0$, or $x^{2}-6 x-7=0$. Only the value $x=7$ works.
Answer: 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.16 A material particle entered the hole of a pipe, and 6.8 minutes later, a second particle entered the same hole. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 m/min, while the second particle covered 3 m in the first minute a... | 3.16 Let $t$ be the time (in minutes) it takes for the second particle to catch up with the first. The distance traveled by the second particle is equal to the sum of $t$ terms of an arithmetic progression, where $a_{1}=3, d=0.5$; therefore,
$s=\frac{2 a_{1}+d(t-1)}{2} t=\frac{6+0.5(t-1)}{2} t$.
The same distance tra... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.18 The distance between points $A$ and $B$ is 270 m. A body moves uniformly from $A$ to $B$; upon reaching $B$, it immediately returns with the same speed. A second body, which leaves $B$ for $A$ 11 s after the first body leaves $A$, moves uniformly but more slowly. On its way from $\boldsymbol{B}$ to $\boldsymbol{A}... | 3.18 A convenient model of the problem is a graph of uniform motion in the coordinate system "path" ($s$ - in meters), "time" ($t$ in seconds). Let $AC$ (Fig. 3.3) be the graph of the motion from $A$ to $B$ of the first body with speed $v_{1}=\operatorname{tg} \alpha$ (time axis $At$); $CD$ be the graph of the motion f... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.19 The first tourist, after riding a bicycle for 1.5 hours at a speed of $16 \mathrm{km} / \mathrm{h}$, makes a stop for 1.5 hours, and then continues the journey at the initial speed. After 4 hours from the departure of the first tourist, the second tourist sets off in pursuit on a motorcycle at a speed of 56 km/h. ... | 3.19 An approximate graph of the movement is shown in Fig. 3.4.
Fig. 3.4
Let $t$ be the time (in hours) it takes for the second tourist to catch up with the first. Since
$$
\begin{aligne... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.21 A motorcyclist set off from point $A$ to point $B$, which is 120 km away from $A$. On the way back, he set off at the same speed, but had to stop for 10 minutes one hour after departure. After this stop, he continued his journey to $A$, increasing his speed by 6 km/h. What was the initial speed of the motorcyclist... | 3.21 According to the condition, $A C=C D$ (Fig. 3.6).
Fig. 3.6
We have $A C=\frac{120}{x}$, where $x$ is the initial speed;
$C D=1+\frac{1}{6}+\frac{120-x}{x+6}$. Therefore,
$$
\frac{1... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.22 A pedestrian and a cyclist set off simultaneously towards each other from cities $A$ and $B$, the distance between which is $40 \mathrm{km}$, and meet 2 hours after departure. Then they continue their journey, with the cyclist arriving in $A$ 7 hours and 30 minutes earlier than the pedestrian in $B$. Find the spee... | 3.22 Fill in the table of speed, distance, and time values in the order indicated by the numbers (1), (2), .., (12):
| Tourist | Before Meeting | | | After Meeting | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | speed, km/h | time, h | distance, km | speed, km/h | time, h | distance, km |
| Pe... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.23 A team of workers completed a certain task. If the team is reduced by 20 people, then the same task will be completed 5 days later than with the original composition, and if the team is increased by 15 people, then the task will be completed 2 days earlier. How many workers were originally in the team and how many... | 3.23 Let $x$ workers completed the task in $y$ days; then according to the condition $x y=(x-20)(y+5)$ and $x y=(x+15)(y-2)$. We can write both equations as proportions:
$\frac{x-20}{x}=\frac{y}{y+5}$ and $\quad \frac{x+15}{x}=\frac{y}{y-2}$.
Each proportion of the form $\frac{a}{b}=\frac{c}{d}$ can be replaced by an... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.24 One of the two machines processes a batch of parts for 3 days longer than the other. How long would the processing of this batch of parts take on each machine separately, if it is known that when working together, these machines processed a batch of parts three times larger in 20 days? | 3.24 Time ( $t$ ), the amount of work done per unit of time, i.e., productivity ( $W$ ), and the total volume of work ( $V$ ) are related by the equation $V=W t$. Let $V=1$ and fill in the following table:
| Machine | Time, days | Volume of work | Productivity |
| :---: | :---: | :---: | :---: |
| First | $x$ | 1 | $\... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.25 Two snow-clearing machines are working on snow removal. The first can clear a street in 1 hour, while the second can do it in $75\%$ of this time. Starting the cleaning simultaneously, both machines worked together for 20 minutes, after which the first machine stopped. How much more time is needed for the second m... | 3.25 Let's accept the entire volume of work as a unit. The productivity of the first machine is 1 (per hour), and the second is $1: \frac{3}{4}=\frac{4}{3}$ (per hour). Working together for $\frac{1}{3}$ hour, they will complete $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{4}{3} = \frac{7}{9}$ of the entire work. The... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.29 The volume of substance $A$ is half the sum of the volumes of substances $B$ and $C$, and the volume of substance $B$ is $\frac{1}{5}$ 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 $B$. | 3.29 According to the condition, $2 V_{A}=V_{B}+V_{C}$ 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,
$\frac{V_{C}}{V_{A}+V_{B}}=\frac{1}{x+y}=1$
Answer: 1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.30 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 equal to a positive integer $k$. A certain value of the q... | 3.30 I n s t r u c t i o n. Write the condition as a system of three equations. From the first two equations, express $\frac{x}{y}$ and $\frac{y}{z}$ in terms of $k$, and then substitute the obtained expressions for $\frac{x}{y}$ and $\frac{y}{z}$ into the third equation and calculate the desired value of $k$.
A n s w... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.33 From milk with a fat content of $5 \%$, cottage cheese with a fat content of $15.5 \%$ is produced, leaving behind whey with a fat content of $0.5 \%$. How much cottage cheese is obtained from 1 ton of milk?
## Geometric and Physical Problems | 3.33 Let $x$ tons of cottage cheese with a fat content of $15.5\%$ be obtained; then $1-x$ tons of whey with a fat content of $0.5\%$ will remain. Therefore, in 1 ton of milk, there is
$\frac{15.5 x}{100}+\frac{0.5(1-x)}{100}=\frac{15.5 x+0.5}{100}$ tons of fat.
According to the condition, $\frac{15.5 x+0.5}{100}=\fr... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.38 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 s after the other, then after what time will they be 220.5 m apart from each other? | 3.38 Instruction. Use the well-known physics formula $s=4.9 t^{2}$.
Answer: 7 seconds after the start of the fall of the first body. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.40 A red pencil costs 27 kopecks, a blue one - 23 kopecks. No more than 9 rubles 40 kopecks can be spent on purchasing pencils. It is necessary to purchase the maximum possible total number of red and blue pencils. At the same time, the number of red pencils should be as few as possible, but the number of blue pencil... | 3.40 Let $x$ red and $y$ blue pencils be bought. According to the conditions, $27 x+23 y \leqslant 940$ and $y-x \leqslant 10$. Let's construct the lines
\[
\begin{aligned}
& 27 x+23 y=940 ... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.41 Several students decided to buy an imported tape recorder priced between 170 and 195 dollars. However, at the last moment, two of them refused to participate in the purchase, so each of the remaining had to contribute 1 dollar more. What was the price of the tape recorder?
## Ratios between natural numbers | 3.41 Let $x$ be the original number of students, and $p$ be the cost of the tape recorder. Then $\frac{p}{x}$ is the amount each student was originally supposed to contribute, and $\frac{p}{x-2}$ is the amount each of the tape recorder buyers contributed. Using one of the conditions of the problem, we get the equation
... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.42 The digits of a certain three-digit number form a geometric progression. If in this number the digits of the hundreds and units are swapped, the new three-digit number will be 594 less than the desired one. If, however, in the desired number the digit of the hundreds is erased and the digits of the resulting two-d... | 3.42 The desired number has the form $100 x+10 y+z$, where $x, y, z$ form a geometric progression, i.e.
$x z=y^{2}$.
According to the condition, we have:
$100 z+10 y+x=100 x+10 y+z-594$, from which
$x-z=6$
$10 y+z=10 z+y+18$, from which
$y-z=2$.
The system of equations (1), (2), (3) is satisfied by $x=8$, $\bold... | 842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.45 The recording of a six-digit number starts with the digit 2. If this digit is moved from the first position to the last, keeping the order of the other five digits, the newly obtained number will be three times the original number. Find the original number.
## Distance: path, speed, time | 3.45 The original six-digit number has the form $2 \cdot 10^{5}+x$. After moving the digit 2 to the last place, we get the number $10 x+2$. According to the condition, $10 x+2=3\left(2 \cdot 10^{5}+x\right)$, from which $x=85714$. Therefore, the original number is 285714.
Answer: 285714. | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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