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13.059. On both sides of a 1200 m long street, there are rectangular strips of land allocated for plots, one being 50 m wide, and the other 60 m wide. How many plots is the entire village divided into, if the narrower strip contains 5 more plots than the wider one, given that each plot on the narrower strip is 1200 m$^... | ## Solution.
Let $x$ be the number of plots on the wide strip, $x+5$ plots - on the narrow strip, $y$ m - the width of a plot on the narrow strip, and $y+1200$ m - on the wide strip. According to the problem, $\left\{\begin{array}{l}(x+5) y=1200 \cdot 50, \\ x(y+1200)=1200 \cdot 60 .\end{array}\right.$ Solving the sys... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,560 |
13.060. A load with a mass of 60 kg presses on a support. If the mass of the load is reduced by 10 kg, and the area of the support is reduced by 5 dm², then the mass per square decimeter of the support will increase by 1 kg. Determine the area of the support. | Solution.
Let $x$ dm $^{2}$ be the area of the support. Then $\frac{60}{x}$ kg/dm² is the mass per 1 dm ${ }^{2}$ of the support. After reducing the mass of the load and the area of the support, this mass is $\frac{50}{x-5}$ kg/dm². According to the condition, $\frac{50}{x-5}-1=\frac{60}{x}$, from which $x=15$.
Answe... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,561 |
13.061. To pay for the delivery of four parcels, 4 different postage stamps were needed for a total of 84 kopecks. Determine the cost of the stamps purchased by the sender, if these costs form an arithmetic progression, and the most expensive stamp is 2.5 times more expensive than the cheapest one. | Solution.
Let $x$ kop. - the cost of the cheapest stamp, $x+d, x+2 d$, $x+3 d$ - the costs of the other stamps. By condition
$$
\left\{\begin{array}{l}
x+x+d+x+2 d+x+3 d=84, \\
x+3 d=2.5 x,
\end{array} \text { from which } x=12 ; d=6\right. \text {. Costs of the }
$$
stamps: 12 kop., 18 kop., 24 kop, 30 kop.
Answer... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,562 |
13.062. An apprentice turner is machining pawns for a certain number of chess sets. He wants to learn to produce 2 more pawns per day than he does now; then he would complete the same task 10 days faster. If he could learn to produce 4 more pawns per day than he does now, the time required to complete the same task wou... | ## Solution:
Let the turner provide $x$ sets of chess pieces. He completes the task in $y$ days. According to the problem,
$$
\left\{\begin{array}{l}
\left(\frac{16 x}{y}+2\right)(y-10)=16 x \\
\left(\frac{16 x}{y}+4\right)(y-16)=16 x
\end{array}, \text { from which } x=15\right.
$$
Answer: 15. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,563 |
13.063. In the club's auditorium, there were 320 seats arranged in identical rows. After the number of seats in each row was increased by 4 and one more row was added, the auditorium had 420 seats. How many rows are there in the auditorium now? | Solution:
Let there be $x$ rows in the auditorium; $\frac{320}{x}$ - the number of seats in one row. After the number of seats in a row was increased by 4 and one row was added, the total number of seats in the hall became $\left(\frac{320}{x}+4\right)(x+1)=420$, from which $x=20$. Therefore, there are now 21 rows in ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,564 |
13.064. The hay reserve is such that 96 kg can be issued daily for all the horses. In fact, the daily portion for each horse could be increased by 4 kg because two horses were sold. How many horses were there originally? | Solution.
Let there initially be $x$ horses. The daily portion for each horse was $\frac{96}{x}$ kg. Since two horses were sold to another collective farm, the number of horses became $x-2$, and the daily portion for each horse became $\frac{96}{x}+4$ kg. According to the condition, $\left(\frac{96}{x}+4\right)(x-2)=9... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,565 |
13.065. 108 examinees wrote an essay. They were given 480 sheets of paper, with each girl receiving one sheet more than each boy, and all the girls received as many sheets as all the boys. How many girls and how many boys were there? | Solution.
Let $x$ be the number of girls, and $108-x$ be the number of boys. Boys received $y$ sheets each, and girls received $y+1$ sheets each. According to the problem,
$\left\{\begin{array}{l}x(y+1)+(108-x) y=480, \\ x(y+1)=(108-x) y,\end{array}\right.$ from which $x=48$. There were 48 girls and $108-48=60$ boys.... | 4860 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,566 |
13.066. At a machine-building plant, a new type of parts for generators has been developed. From 875 kg of metal, they now produce three more new-type parts than old-type parts were produced from 900 kg. What is the mass of the new and old type parts, if two new-type parts are 0.1 tons lighter than one old-type part? | ## Solution.
Let $x$ kg be the mass of the new type of part, and $y$ kg be the mass of the old type. According to the problem, $y-2 x=100$. From 875 kg, $\frac{875}{x}$ new type parts are made, and from 900 kg, $\frac{900}{y}$ old type parts were made. According to the problem, $\frac{875}{x}-\frac{900}{y}=3$. Solving... | 175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,567 |
13.067. On the first day of the sports competitions, $1 / 6$ of the boys' team and $1 / 7$ of the girls' team did not meet the qualifying standards and were eliminated from further competition. Over the rest of the competition period, an equal number of athletes dropped out from both teams due to failing to meet the st... | ## Solution:
Let $x$ be the number of boys who met the credit standards and $2x$ be the number of girls. Then the initial number of team members is $x+48$ 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 | 51,568 |
13.068. The working hour of masters $A$ and $B$ is paid differently, but both masters worked the same number of hours. If $A$ had worked one hour less, and $B$ had worked five hours less, then $A$ would have earned 720 rubles, and $B$ 800 rubles. If, on the contrary, $A$ had worked five hours less, and $B$ had worked o... | ## Solution:
Let the masters have worked $x$ hours. If $A$ had worked 1 hour less, he would have earned $\frac{720}{x-1}$ rubles per hour; if $B$ had worked 5 hours less, he would have earned $\frac{800}{x-5}$ rubles per hour. According to the condition, $\frac{800}{x-5}(x-1)-\frac{720}{x-1}(x-5)=360$, from which $x=2... | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,569 |
13.069. In one pool there are 200 m $^{3}$ of water, and in another - 112 m $^{3}$. Taps are opened, through which the pools are filled. After how many hours will the amount of water in the pools be the same, if 22 m $^{3}$ more water is poured into the second pool per hour than into the first? | Solution.
Let the amount of water in the pools become equal after $x$ hours. Let $y$ m$^{3}$ of water be added to the first pool per hour, and $y+22$ m$^{3}$ - to the second. According to the condition, $200+x y=112+x(y+22)$, from which $x=4$.
Answer: in 4 hours. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,570 |
13.072. The city $C$, located between points $A$ and $B$, is supplied with gas from these points, the distance between which is 500 km. From reservoir $A$, 10000 $\mathrm{M}^{3}$ of gas is pumped out every minute, and from reservoir $B$ - 12% more. In this case, the gas leakage in each pipeline is 4 m $^{3}$ per kilome... | ## Solution.
Let $x$ km be the distance between $A$ and $C$, and $500-x$ km be the distance between $C$ and $B$. Considering the gas leakage, $10000-4 x$ m$^3$ of gas is delivered from $A$ to $C$, and $10000 \cdot 1.12 - 4(500-x)$ m$^3$ of gas is delivered from $B$ to $C$. According to the condition, $10000-4 x = 1.12... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,571 |
13.074. A sewing workshop received three stacks of bed linen material, totaling 5000 m. In the first stack, the amount of material was three times less than in the second, and in the third - $22 \%$ of the total amount. From the material of the first stack, 150 sheets and 240 pillowcases were sewn. For the production o... | Solution.
Let $x$ m of material be in the first bale, $3 x$ m - in the second, $5000 \cdot 0.22=1100$ m - in the third. According to the condition, $x+3 x+1100=5000$, from which $x=975$ m - in the first bale. Let one pillowcase be sewn from $y$ m of material, and a sheet - from $3.25+y$ m. A total of $(3.25+y) \cdot 1... | 1.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,573 |
13.075. Two workers together produced 72 parts per shift. After the first worker increased their 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 productivity? | ## Solution.
Initially, the workers produced $x$ and $72-x$ parts per shift, and then $1.15x$ and $1.25(72-x)$ parts. According to the condition, $1.15x + 90 - 1.25x = 86$; $0.1x = 4; x = 40; 1.15 \cdot 40 = 46$.
Answer: 46 and 40 parts. | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,574 |
13.076. The collection of corn from the fields of a livestock farm was 4340 centners. The following year, it is planned to obtain 5520 centners of corn by increasing the area by 14 hectares and increasing the yield by 5 centners per hectare. Determine the area occupied by corn and the yield in centners per hectare (the... | Solution.
Let $x$ ha be the area occupied by corn, $\frac{4340}{x}$ centners/ha the yield. The next year, the area is $x+14$ ha, and the yield is $\frac{4340}{x}+5$ centners/ha. According to the condition, $\left(\frac{4340}{x}+5\right)(x+14)=5520$, from which $x=98$ ha or $x=124$ ha. The yield is $\frac{4340}{98} \ap... | 124 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,575 |
13.077. The older brother on a motorcycle and the younger brother on a bicycle made a two-hour non-stop trip to the forest and back. During this time, the motorcyclist traveled each kilometer 4 minutes faster than the cyclist. How many kilometers did each of the brothers travel in 2 hours, if it is known that the dista... | ## Solution.
Let $x$ km be the distance the younger brother traveled, and $x+40$ km be the distance the older brother traveled. The younger brother took $\frac{120}{x}$ minutes to travel 1 km, and the older brother took $\frac{120}{40+x}$ minutes. According to the condition, $\frac{120}{x}-4=\frac{120}{40+x}$, from wh... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,576 |
13.078. A tourist traveled $5 / 8$ of the total distance by car and the remaining part by boat. The boat's speed is 20 km/h less than the car's speed. The tourist traveled by car for 15 minutes longer than by boat. What are the speeds of the car and the boat if the total distance of the tourist's journey is 160 km? | ## Solution.
The tourist drove 160 * 5/8 = 100 km by car, and 160 - 100 = 60 km by boat. Let \( x \) km/h be the speed of the boat, and \( 20 + x \) km/h the speed of the car. The tourist traveled by car for \( \frac{100}{20+x} \) hours, and by boat for \( \frac{60}{x} \) hours. According to the problem, \( \frac{100}... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,577 |
13.079. 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 \... | ## Solution.
The motion graph is shown in Fig. 13.3. Let $t$ be the time (in hours) it takes for the second tourist to catch up with the first. Since $v_{\text {bike }}=16 \mathrm{km} / \mathbf{h}$, $\mathrm{v}_{\text {motor }}=56 \mathrm{km} /$ h, then $S_{\text {bike }}=(2.5+t) 16, S_{\text {motor }}=56 t$. Therefor... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,578 |
13.081. 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 after an hour of travel. After this stop, he continued his journey to $A$, increasing his speed by 6 km/h. What was the initial speed of the motorcycli... | ## Solution.
By the condition $A C=C D$ (Fig. 13.4). We have $A C=\frac{120}{x}$, where $x-$ is the initial speed; $C D=1+\frac{1}{6}+\frac{120-x}{x+6} \Rightarrow \frac{120}{x}=\frac{7}{6}+\frac{120-x}{x+6}$, from which $x=48($ km $/ h)$.
Answer: 48 km $/$ h | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,580 |
13.082. Two groups of tourists must walk towards each other from tourist bases $A$ and $B$, the distance between which is 30 km. If the first group leaves 2 hours earlier than the second, they will meet 2.5 hours after the second group leaves. If the second group leaves 2 hours earlier than the first, the meeting will ... | Solution.
Let $x$ km/h be the speed of the first group, $y$ km/h be the speed of the second. If the first group starts earlier than the second, then $2.5 y + (2.5 + 2) x = 30$. If the second group starts earlier than the first, then $3 x + (3 + 2) y = 30$. Solving the system $\left\{\begin{array}{l}2.5 y + 4.5 x = 30,... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,581 |
13.083. A freight train was delayed on the way for 12 minutes, and then at a distance of 60 km, it made up for the lost time by increasing its speed by 15 km/h. Find the original speed of the train. | ## Solution.
Let the speed of the train before the delay be $x$ km/h, and after the delay $(x+15)$ km/h. Then (Fig. 13.5) $A B=\frac{x}{5}, C E=60, C D=60-\frac{x}{5}$, $B D=\frac{60-\frac{x}{5}}{x}, A E=\frac{60}{x+15}$. Since $B D=A E$, then $\frac{60-\frac{x}{5}}{x}=\frac{60}{x+15}$, from which $x=60($ km $/ \mathb... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,582 |
13.085. Two brothers took their bicycles and set off simultaneously with the intention of riding $42 \mathrm{km}$. The older brother maintained the same speed throughout the journey; while the younger brother fell behind by 4 km every hour. However, since the older brother rested for a whole hour during the trip, and t... | ## Solution.
Let $x$ km/h be the speed of the older brother, then $(x-4)$ km/h is the speed of the younger brother. The older brother was on the road for $t_{\text {old }}=\frac{42}{x}+1$, and the younger brother for $t_{\text {young }}=\frac{42}{x-4}+\frac{1}{3}$. According to the problem, $t_{\text {old }}=t_{\text ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,584 |
13.086. A positive integer is thought of. To its notation, 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 reduced by $75\%$ of this remainder, and the thought number is subtracted again. In the final result, zero is obtained. Wha... | Solution.
Let the number be $x$. Consider the numbers $10 x+7, 10 x+7-x^{2}$, and the remainder $\frac{25}{100}\left(10 x+7-x^{2}\right)$. Then $\frac{1}{4}\left(10 x+7-x^{2}\right)-x=0, x^{2}-6 x-7=0 \Rightarrow$ $\Rightarrow x=7$
Answer: 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,585 |
13.087. A positive integer is thought of. The digit 5 is appended to its right, and from the resulting new number, the square of the thought number is subtracted. The difference is divided by the thought number, and then the thought number is subtracted, and the result is one. What number is thought of? | ## Solution.
Let $x$ be the number thought of. Then the new number can be represented as $10 x + 5$. According to the condition, $\frac{10 x + 5 - x^2}{x} - x = 1$, from which $x = 5$.
Answer: 5.
 | Cost (rubles) | Transportation (rubles) | Total cost (rubles) |
| :---: | :---: | :---: | :---: | :---: |
| № 1 | $x$ | $37.5 x$ | $3 x$ | $40.5 x$ |
| № 2 | $300-x$ | $37.5(300-x)$ | $4.5(300-x)$ | $42(300-x)$ |
Acc... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,590 |
13.092. A monetary prize was distributed among three inventors: the first received half of the entire prize minus $3 / 22$ of what the other two received together. The second received $1 / 4$ of the entire prize and $1 / 56$ of the money received by the other two together. The third received 30000 rubles. How large was... | ## Solution.
Let $x$ rubles be the prize for the first inventor, and $y$ rubles be the prize for the second. According to the condition, the total prize is $x+y+30000$ rubles. Then
$$
\left\{\begin{array}{l}
x=\frac{1}{2}(x+y+30000)-\frac{3}{22}(y+30000) \\
y=\frac{1}{4}(x+y+30000)+\frac{1}{56}(x+30000)
\end{array} \... | 95000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,591 |
13.093. An alloy of copper and silver contains 1845 g more silver than copper. If a certain amount of pure silver, equal to $1 / 3$ of the mass of pure silver originally contained in the alloy, were added to it, then a new alloy would be obtained, containing $83.5\%$ silver. What is the mass of the alloy and what was t... | ## Solution.
Let $x$ g be the mass of silver, $x-1845$ g be the mass of copper. Then $2x - 1845$ g is the mass of the entire alloy. After adding silver, its mass will become $x + \frac{1}{3}x$ g, and the mass of the alloy will be $2x - 1845 + \frac{1}{3}x$ g; $x + \frac{1}{3}x$ g make up $83.5\%$. Therefore, the mass ... | 3165;\approx79.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,592 |
13.094. In 500 kg of ore, there is a certain amount of iron. After removing 200 kg of impurities from the ore, which on average contain $12.5\%$ iron, the iron content in the remaining ore increased by $20\%$. How much iron is left in the ore? | Solution.
In 200 kg of impurities, there will be $x=\frac{200 \cdot 12.5}{100}=25$ kg of iron. Initially, $z$ kg of iron constituted $y \%$ of 500 kg of ore, i.e.,
$z=\frac{500 \cdot y}{100}=5 y$ kg. In the remaining ore, $(5 y-25)$ kg of iron constitutes $(y+20) \%$. Thus, $\frac{5 y-25}{300}=\frac{y+20}{100} \Righta... | 187.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,593 |
13.095. On a flat horizontal platform, two masts stand 5 m apart from each other. At a height of $3.6 \mathrm{M}$ above the platform, one end of a wire segment $13 \mathrm{M}$ long is attached to each mast. The wire is stretched in the plane of the masts and is attached to the platform, as shown in Fig. 13.7. At what d... | Solution.
Let the required distance be $x$. From $\triangle A B C$ we get:
$A B^{2}+B C^{2}=A C^{2} ; A C=y,(x+5)^{2}+3.6^{2}=y^{2}$. From $\triangle A E D$ we get:
$A E^{2}+E D^{2}=A D^{2} ; A D=z \cdot x^{2}+3.6^{2}=z^{2}$. According to the condition: $z+y=13$. Solving the system: $\left\{\begin{array}{l}(x+5)^{2}... | 2.7\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,594 |
13.097. The distance from $A$ to $B$ by railway is 88 km, while by river it is 108 km. The train from $A$ departs 1 hour later than the riverboat and arrives in $B$ 15 minutes earlier. Find the average speed of the train, given that it is 40 km/h greater than the average speed of the riverboat. | ## Solution.
Let $x$ km/h be the speed of the train, and $x-40$ km/h be the speed of the steamer. The train was on the way for $\frac{88}{x}$ hours, and the steamer for $\frac{108}{x-40}$ hours. According to the condition, $\frac{108}{x-40}-\frac{88}{x}=1+\frac{1}{4}$, from which $x=88$.
Answer: 88 km/h. | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,596 |
13.098. A pedestrian and a cyclist set off simultaneously towards each other from cities $A$ and $B$, the distance between which is 40 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 speeds of th... | Solution.
We will 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 $/ \mathbf{4}$ | time, h | distance, | speed, km $/ \mat... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,597 |
13.099. The distance between villages $A$ and $B$ is $s$ km. Two motorists left $A$ for $B$ simultaneously along the same road and were supposed to arrive in $B$ at the same time. In reality, the first motorist arrived in $B$ $n$ hours earlier than scheduled, while the second was $3n$ hours late, as the latter traveled... | Solution.
Let $t$-time, $v_{1}$ and $v_{2}$-speeds; then $\frac{s}{v_{1}}=t-n$, (1) $\frac{s}{v_{2}}=t+3 n$,
(2) $v_{1}-v_{2}=r$. Subtracting (1) from (2), we get $s\left(\frac{1}{v_{2}}-\frac{1}{v_{1}}\right)=4 n \Rightarrow$ $\Rightarrow\left(v_{1}-v_{2}\right) s=4 n v_{1} v_{2} \Rightarrow v_{1} v_{2}=\frac{s r}{4... | \frac{nr+\sqrt{nr(nr+)}}{2n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,598 |
13.100. Determine the positive integer based on the following data: if it is written in digits and the digit 4 is appended to the right, the resulting number is divisible without a remainder by a number that is 4 greater than the sought number, and the quotient is a number that is 27 less than the divisor. | Solution.
Let $x$ be the required number. According to the condition $\frac{10 x+4}{x+4}=x+4-27$, from which $x=32$.
Answer: 32. | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,599 |
13.102. Given two two-digit numbers, the second of which is denoted by the same digits as the first but written in reverse order. The quotient of dividing the first number by the second is 1.75. The product of the first number and its tens digit is 3.5 times the second number. Find these numbers. | Solution.
Let's represent the first number as $10 x+y$, then the second number is $-10 y+x$.
According to the condition: $\left\{\begin{array}{l}\frac{10 x+y}{10 y+x}=1.75, \\ (10 x+y) x=3.5(10 y+x) .\end{array}\right.$ Solving the system, we find $x=2$, $y=1$. The required numbers are 21 and 12.
Answer: 21 and 12. | 2112 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,600 |
13.103. From the railway station to the tourist base, one can walk along the highway or a path, with the path being 5 km shorter. Two friends agreed that one would walk along the highway at a constant speed of $\mathrm{v}$ km/h, while the other would take the path at a speed of 3 km/h. The second one arrived at the tou... | Solution.
Let $x$ km be the distance from the station to the tourist base by road, and $x-5$ km by trail. According to the problem, $\frac{x}{v}-\frac{x-5}{3}=1$, from which $x=\frac{2 v}{v-3}$. The expression makes sense for $v>3$. By trial, we find: at $v=4$ km/h, $x=8$ km. Other solutions to this equation do not sa... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,601 |
13.104. The length of the bus route is 16 km. During peak hours, the bus switches to express mode, i.e., it significantly reduces the number of stops, as a result of which the travel time from the beginning to the end of the route is reduced by 4 minutes, and the average speed of the bus increases by 8 km/h. At what sp... | Solution.
Let $x$ km/h be the speed of the bus in express mode. Then the travel time in this mode will be $\left(\frac{16}{x-8}-\frac{1}{15}\right)$ hours. According to the condition, $\left(\frac{16}{x-8}-\frac{1}{15}\right) x=16$, from which $x=48$ km/h.
Answer: 48 km/h. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,602 |
13.105. On one of the tram lines, trams of a new design have started to operate. A trip of 20 km now takes 12 minutes less, as the average speed of the new design tram is 5 km/h higher than the average speed of the outdated design tram. How much time does the new design tram take for the trip and what is its average sp... | ## Solution.
Let $x$ km/h be the average speed of the new tram model.
Then the time it takes for the trip is $-\frac{20}{x}$ hours. According to the problem, $\frac{20}{x}=\frac{20}{x-5}-\frac{1}{5}$, from which $x=25$ km/h, and the time $-\frac{20}{25}=\frac{4}{5}$ hours $=48$ minutes.
Answer: 48 min, 25 km/h. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,603 |
13.106. The plane must fly 2900 km. Having flown 1700 km, it made an emergency landing for 1 hour 30 minutes, after which it flew at a speed 50 km/h less than before. Find the original speed of the plane, given that it arrived at its destination 5 hours after takeoff. | ## Solution.
Let $x$ km/h be the original speed of the airplane. Before landing, it was in the air for $\frac{1700}{x}$ hours, and after landing, it was in the air for $\frac{1200}{x-50}$ hours. According to the condition, $\frac{1700}{x} + 1.5 + \frac{1200}{x-50} = 5$, from which we find $x = 850$ km/h.
Answer: 850 ... | 850 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,604 |
13.107. Two brigades, working together, were supposed to repair a given section of a highway in 18 days. In reality, however, only the first brigade worked at first, and the second brigade, which has a higher labor productivity than the first, finished the repair of the road section. As a result, the repair of the give... | Solution.
Let $x$ be the work rate of the first team, and $y$ be the work rate of the second team. Working together, the teams will repair the section in $\frac{1}{x+y}$ days. According to the problem, $\frac{1}{x+y}=18$. The productivity of the first team is $\frac{1}{x}$, and the second team's productivity is $\frac... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,605 |
13.108. On the plots allocated by the agrolaboratory for experiments, $c$ from two plots, 14.7 tons of grain were collected. The next year, after the application of new agricultural techniques, the yield on the first plot increased by $80 \%$, and on the second - by $24 \%$, as a result of which from these same plots, ... | ## Solution.
Let $x$ tons of grain be collected from the first plot after the application of new methods, and $y$ tons - from the second plot. According to the condition, $x+y=21.42$. Initially, $\frac{x}{1.8}$ tons were collected from the first plot, and $-\frac{y}{1.24}$ tons from the second plot. According to the c... | 10.26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,606 |
13.109. Two cyclists set off simultaneously towards each other from two places, the distance between which is 270 km. The second cyclist travels 1.5 km less per hour than the first, and meets him after as many hours as the first cyclist travels in kilometers per hour. Determine the speed of each cyclist. | ## Solution.
Let $x$ km/h be the speed of the first cyclist, and $x-1.5$ km/h be the speed of the second cyclist. Before they meet, the first cyclist traveled $x \cdot x$ km, and the second cyclist traveled $x(x-1.5)$ km. According to the problem, $x^{2}+x(x-1.5)=270$, from which $x=12$ km/h; $12-1.5=10.5$ km/h.
Answ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,607 |
13.111. The train was delayed by $t$ hours. By increasing the speed by $m$ km/h, the driver eliminated the delay on a section of $s$ km. Determine what speed the train should have had on this section if there had been no delay. | Solution.
Let $m_{1}$ km/h be the speed of the train if it were to travel without delay; $t_{1}$ h be the travel time of the train in this case. By condition $m_{1} \cdot t_{1}=s$. Traveling faster at a speed of $m_{1}+m$ km/h, the train was on the road for $\left(t_{1}-t\right)$ h. Then $s=\left(m_{1}+m\right)\left(t... | \frac{\sqrt{(4+)}-}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,608 |
13.112. Two bodies move towards each other from two places, the distance between which is 390 km. The first body traveled 6 m in the first second, and in each subsequent second, it traveled 6 m more than in the previous one. The second body moved uniformly at a speed of $12 \mathrm{~m} / \mathrm{c}$ and started moving ... | ## Solution.
Let $t$ s be the time of movement of the first body until the meeting; $t-5$ s of the second. The first traveled until the meeting $\frac{a t^{2}}{2}=\frac{6 t^{2}}{2}=3 t^{2}$ km; the second $12(t-5)$ km. According to the condition $3 t^{2}+12(t-5)=360$, from which $t=10$ s.
Answer: in $10 \mathrm{s}$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,609 |
13.113. A material particle entered the pipe through an opening, and 6.8 minutes later, a second particle entered the same opening. 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... | ## Solution.
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 with $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 distanc... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,610 |
13.114. The distance between two cities is $a$ km. Two motorists, setting out from these cities towards each other, will meet halfway if the first one leaves $t$ hours earlier than the second. If, however, they set out towards each other simultaneously, the meeting will occur after $2t$ hours. Determine the speed of ea... | ## Solution.
Let $x$ km/h be the speed of the first car, and $y$ km/h be the speed of the second car. The first car will travel half the distance in $\frac{a}{2 x}$ hours; the second car will take $\frac{a}{2 y}$ hours. According to the condition, $\frac{a}{2 x}+t=\frac{a}{2 y}$. In $2 t$ hours, they will travel $2 t(... | \frac{(3-\sqrt{5})}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,611 |
13.115. Tourist $A$ set off from city $M$ to city $N$ at a constant speed of 12 km/h. Tourist $B$, who was in city $N$, upon receiving a signal that $A$ had already traveled 7 km, immediately set off towards him and traveled 0.05 of the total distance between $M$ and $N$ each hour. From the moment $B$ set off until his... | ## Solution.
Let $x$ km be the distance between $M$ and $N$. Then the speed of tourist $B$ is $B-0.05 x$ km/h. Before the meeting, $A$ traveled $7+0.05 x \cdot 12$ km; $B$ traveled $(0.05 x)^{2}$ km. According to the condition, $7+0.05 x \cdot 12+(0.05 x)^{2}=x$, from which $x_{1}=20$ km; $x_{2}=140$ km. Considering t... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,612 |
13.117. A cyclist traveled 60 km from point $A$ to point $B$. On the return trip, he rode the first hour at the same speed, then stopped for 20 minutes. Resuming his journey, he increased his speed by 4 km/h and therefore spent as much time traveling from $B$ to $A$ as he did traveling from $A$ to $B$. Determine the cy... | ## Solution.
Let $x$ km/h be the speed of the cyclist on the way from $A$ to $B$. The time spent on the way from $A$ to $B$ is $\frac{60}{x}$ hours; the time spent on the return trip is $-1+\frac{1}{3}+\frac{60-x}{x+4}$. According to the condition, $\frac{60}{x}=1+\frac{1}{3}+\frac{60-x}{x+4}$, from which $x=20$ km/h.... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,613 |
13.118. Two buses left the factory simultaneously and headed to the recreation area, to the lake. The distance between the factory and the lake is 48 km. The first bus arrived at the lake 10 minutes earlier than the second, and the average speed of the second bus is 4 km/h less than the average speed of the first. Calc... | ## Solution.
Let $x$ km/h be the speed of the first bus, and $x-4$ km/h be the speed of the second bus. The first bus arrived at the lake in $\frac{48}{x}$ hours, and the second bus in $\frac{48}{x-4}$ hours. According to the condition, $\frac{48}{x-4}-\frac{48}{x}=\frac{1}{6}$, from which $x=36$ km/h, $36-4=32$ km/h.... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,614 |
13.122. Two brothers had tickets to a stadium located 20 km from their home. To get to the stadium, they decided to use their bicycle and agreed that they would leave simultaneously, one on the bicycle and the other on foot; after covering part of the distance, the first would leave the bicycle, and the second, upon re... | Solution.
Let $x$ km be the distance the second brother walks. He will spend $\frac{x}{4}$ hours on this path. The remaining $20-x$ km he will travel in $\frac{20-x}{4 \cdot 5}$ hours. The total time the second brother spends traveling is $\frac{x}{4}+\frac{20-x}{20}$ hours. The first brother will travel $x$ km in $\f... | at\the\midpoint\of\the\journey;\3\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,615 |
13.123. The motorcyclist was delayed at the barrier for 24 minutes. After increasing his speed by $10 \mathrm{km} / \mathrm{h}$, he made up for the delay on an 80 km stretch. Determine the motorcyclist's speed before the delay. | Solution.
Let $x$ km/h be the motorcyclist's speed before the delay, and $x+10$ km/h on the segment. Without the delay, the motorcyclist would have traveled 80 km in $t$ hours,
but he traveled 80 km at a speed of $x+10$ km/h in $t-0.4$ hours. Then $\left\{\begin{array}{l}(x+10)(t-0.4)=80, \\ x \cdot t=80,\end{array}\r... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,616 |
13.124. Two steamers left the port simultaneously, with one heading south and the other heading east. After 2 hours, the distance between them was 174 km. Find the average speed of each steamer, given that one of them traveled 3 km more on average per hour,
$ km, and the second $2x$ km. From the right triangle (Fig. 13.8), by the Pythagorean theorem, $(2x)^2 + (2(x+3))^2 = 174^2$, from which $x = 60$ km/h.
Answer: 60 a... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,617 |
13.126. Two points rotate uniformly along two circles. One of them completes a full revolution 5 seconds faster than the other, and therefore manages to make two more revolutions in 1 minute. How many revolutions per minute does each point make? | ## Solution.
Let the first point make a revolution in $x$ s; the second - in $x+5$ s. In a minute, the first point makes $\frac{60}{x}$ revolutions, and the second $-\frac{60}{x+5}$. According to the condition, $\frac{60}{x}=\frac{60}{x+5}+2$, from which $x=10$. The first point makes $\frac{60}{10}=6$ revolutions per ... | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,618 |
13.127. At the trainer's signal, two ponies simultaneously started running uniformly along the outer circumference of the circus arena in opposite directions. The first pony ran slightly faster than the second, and by the time they met, the first pony had run 5 m more than the second. Continuing their run, the first po... | ## Solution.
Let $x$ m be the distance run by the second pony until the meeting, and $x+5$ m - the first pony. Then the length of the entire arena $\pi d=2 x+5$. The speed of the first pony is $\frac{x}{9}$ m/s, and the second is $-\frac{x+5}{16}$ m/s. According to the condition $9 \frac{x+5}{x}=16 \frac{x}{x+5}$, fro... | 11\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,619 |
13.128. Above point $A$, the helicopter was at 8:30 AM. After flying a straight distance of $s$ km, the helicopter arrived at point $B$. Hovering over point $B$ for 5 minutes, the helicopter then returned along the same route. It returned to point $A$ at 10:35 AM. From $A$ to $B$, it flew with the wind, and on the retu... | Solution.
Let $x$ hours be the time taken to travel from $A$ to $B$, and $y$ hours from $B$ to $A$. According to the condition, $x+y=2$. The speed of the helicopter from $A$ to $B$ was $v+v_{\mathrm{B}}=\frac{s}{x}$, and on the return trip - $v-v_{\mathrm{B}}=\frac{s}{y}$. Solving the system $\left\{\begin{array}{l}x+... | \sqrt{v(v-)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,620 |
13.129. At 9 AM, a self-propelled barge left $A$ upstream and arrived at point $B$; 2 hours after arriving at $B$, the barge set off on the return journey and arrived back at $A$ at 7:20 PM on the same day. Assuming the average speed of the river current is 3 km/h and the barge's own speed is constant throughout, deter... | Solution.
Let $v$ km/h be the own speed of the barge. From $A$ to $B$, the barge sailed $\frac{60}{v-3}$ hours, and on the way back $-\frac{60}{v+3}$ hours. According to the condition, $\frac{60}{v-3}+2+\frac{60}{v+3}=10 \frac{1}{3}$, from which $v=15$ km/h. The barge arrived in $B$ at $9+\frac{60}{15-3}=14$ hours.
A... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,621 |
13.130. Two friends in one boat traveled along the riverbank and returned along the same route 5 hours after departure. The entire trip was 10 km. According to their calculations, on average, it took them as much time to travel 2 km against the current as it did to travel 3 km with the current. Find the speed of the cu... | Solution.
Let $v$ km/h be the own speed of the boat, $v_{\text {r. }}$ km/h be the speed of the current. According to the problem, $\left\{\begin{array}{l}\frac{2}{v-v_{\mathrm{T}}}=\frac{3}{v+v_{\mathrm{T}}}, \\ \frac{5}{v-v_{\mathrm{T}}}+\frac{5}{v+v_{\mathrm{T}}}=5,\end{array}\right.$ from which $v_{\mathrm{T}}=\fr... | \frac{5}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,622 |
13.131. The beacon keeper, inspecting his river section in an ordinary rowboat, traveled 12.5 km upstream and then returned to the starting point along the same route. In this voyage, he overcame every 3 km against the current and every 5 km with the current in the same average time intervals, and was on the way for ex... | ## Solution.
Let $v$ km/h be the own speed of the boat, $v_{\mathrm{T}}$ km/h be the speed of the current. According to the condition, $\left\{\begin{array}{l}\frac{3}{v-v_{T}}=\frac{5}{v+v_{\mathrm{T}}} \\ \frac{12.5}{v-v_{T}}+\frac{12.5}{v+v_{T}}=8,\end{array}\right.$ from which $v_{\mathrm{T}}=\frac{5}{6}$ km/h, $v... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,623 |
13.132. In a laboratory setup, a certain liquid flows into a vessel through three inlet valves. If all valves are opened simultaneously, the vessel will be filled in 6 minutes. If the vessel is filled only through the second valve, it will take 0.75 of the time it takes to fill the vessel only through the first valve. ... | ## Solution.
Let the first tap be opened for $x$ min, the second for $y$ min, and the third for $z$ min. If all taps are opened simultaneously, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{6}$. According to the conditions, $y=0.75 x ; z=y+10$. Solving the system
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y... | \frac{56}{3},14,24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,624 |
13.133. A swimming pool has three pipes of different cross-sections for draining water using a uniformly pumping pump. The first and second pipes together, with the third pipe closed, empty a full pool in $a$ minutes; the first and third pipes together, with the second pipe closed, empty a full pool in $b$ minutes; and... | ## Solution.
Let the first pipe empty the pool in $x$ min; the second pipe in $y$ min; the third pipe in $z$ min. According to the problem:
$$
\begin{aligned}
& z=\frac{2 a b c}{a b+a c-... | \frac{2}{+-},\frac{2}{+-},\frac{2}{+-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,625 |
13.134. According to the program, two machines on the assembly line should process the same number of parts in $a$ hours. The first machine completed the task. The second machine turned out to be not quite serviceable, worked with interruptions, as a result of which it processed $n$ fewer parts than the first in the sa... | Solution.
Let $x$ be the number of parts processed by the first machine, and $x-n$ be the number of parts processed by the second machine. The time spent on processing one part on the first machine is $\frac{a}{x}$ hours, and on the second machine it is $-\frac{a}{x-n}$ hours. According to the condition, $\frac{a}{x-n... | \frac{+\sqrt{b^{2}n^{2}+240n}}{2b}\text{}\frac{-+\sqrt{b^{2}n^{2}+240n}}{2b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,626 |
13.135. A team of mechanics can complete a certain task of processing parts 15 hours faster than a team of apprentices. If the team of apprentices works for 18 hours on this task, and then the team of mechanics continues working on the task for 6 hours, only 0.6 of the entire task will be completed. How much time does ... | ## Solution.
Let $x$ hours be necessary for the students to complete the task; $x-15$ hours for the team of mechanics. In 18 hours, the students will complete $\frac{1}{x} \cdot 18$ of the entire task, and in 6 hours, the mechanics will complete $\frac{6}{x-15}$ of the entire task. According to the condition, $\frac{1... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,627 |
13.137. Three machines from different systems perform a certain counting task. If the entire task is assigned to only the second or the first machine, then the second machine alone will take 2 minutes longer to complete the entire task than the first machine. The third machine can complete the entire task in a time twi... | Solution.
Let the first machine complete the work in $x$ min, the second in $y$ min, and the third in $z$ min. According to the conditions, $\left\{\begin{array}{l}y=x+2, \\ z=2 x, \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{8},\end{array}\right.$ from which $x=6$ min, $y=8$ min, $z=12$ min.
Answer: in 6, 8, and ... | 6,8,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,628 |
13.138. Two workers, the second of whom started working 1.5 days later than the first, independently wallpapered several rooms in 7 days, counting from the moment the first worker started. If this work had been assigned to each separately, the first would have needed 3 days more than the second to complete it. How many... | Solution.
Let the first worker complete the entire job in $x$ days, the second in $x-3$ days. In 7 days, the first worker completed $\frac{7}{x}$ of the entire job, and in $7-1.5=5.5$ days, the second worker completed $\frac{5.5}{x-3}$ of the job. Therefore, $\frac{7}{x}+\frac{5.5}{x-3}=1$, from which $x=14$ days.
An... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,629 |
13.139. Find a two-digit number, the quotient of which when divided by the product of its digits is $8 / 3$, and the difference between the number and the number written with the same digits but in reverse order is 18. | ## Solution.
Let's represent the desired number as $10 x + y$. Then, according to the condition,
$$
\left\{\begin{array}{l}
\frac{10 x + y}{x y} = \frac{8}{3}, \\
10 x + y = 10 y + x + 18,
\end{array} \text { from which } x = 6, y = 4. \text { The desired number is } 64. \right.
$$
Answer: 64. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,630 |
13.140. On one of two machines, a batch of parts is processed 3 days longer than on the other. How many days would it take for each machine to process this batch of parts separately, if it is known that when working together, these machines processed a batch of parts three times larger in 20 days? | Solution.
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 $V=W t$. Let $V=1$ and fill in the following table:
| Machine | Time, days | Volume of work | Productivity |
| :---: | :---: | :---: | :---: |
| First | $x$ | 1 | $\frac{1}{x... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,631 |
13.141. An integer was given. It was required to increase it by 200000 and then triple the resulting number. Instead, the digit 2 was appended to the right of the digital representation of the given number, and the correct result was obtained: What number was given? | ## Solution.
Let $x$ be the desired number. By appending the digit 2 to this number, we get the number $10 x+2$. According to the condition, $10 x+2=3(x+200000)$, from which $x=85714$.
Answer: 85714. | 85714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,632 |
13.143. A completes a certain work in a time that is $a$ days longer than $B$, and $b$ days longer than $C$. Working together, $A$ and $B$ complete this work in the same number of days as $C$. Determine the time it takes for each to complete the work individually. Under what relationship between the given quantities do... | ## Solution.
Let $A$ complete the work in $x$ days, $B$ in $x-a$ days, and $C$ in $x-b$ days. According to the condition, $\frac{1}{x}+\frac{1}{x-a}=\frac{1}{x-b}$, from which $x=b+\sqrt{b(b-a)}$ days; for $b>a$, $x-a=b-a+\sqrt{b(b-a)}$ days; $x-b=\sqrt{b(b-a)}$ days.
Answer: $\quad b+\sqrt{b(b-a)} ; b-a+\sqrt{b(b-a)... | b+\sqrt{b(b-)};b-+\sqrt{b(b-)};\sqrt{b(b-)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,634 |
13.144. The sum of all even two-digit numbers is divisible by one of them without a remainder. The quotient obtained differs from the divisor only in the order of the digits, and the sum of its digits is 9. What two-digit number was the divisor? | Solution.
Let's represent the divisor in the form $10a + b$. Then, according to the condition, we have $\left\{\begin{array}{l}\frac{2430}{10a + b} = 10b + a, \\ a + b = 9,\end{array}\right.$ from which $a = 4, b = 5$. The required number is 54.
Answer: 54. | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,635 |
13.145. First, the motorboat traveled 10 km downstream, and then twice that distance - across the lake into which the river flows. The entire trip lasted 1 hour. Find the boat's own speed, if the river current speed is 7 km/h. | ## Solution.
Let $v$ km/h be the own speed of the boat. Downstream, the boat traveled $\frac{10}{v+7}$ hours, and on the lake $-\frac{20}{v}$ hours. According to the condition, $\frac{10}{v+7}+\frac{20}{v}=1$, from which $v^{2}-23 v-140=0$, i.e., $v=28$ (km/h).
Answer: 28 km/h. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,636 |
13.146. Find three numbers, the first of which is as many times greater than the second as the second is greater than the third. If the first number is decreased by the sum of the other two, the result is 2, and if the first number is increased by half the difference of the second and the third, the result is 9. | Solution.
Let $x, y, z$ be the required numbers. According to the conditions $\left\{\begin{array}{l}\frac{x}{y}=\frac{y}{z}, \\ x-(y+z)=2, \\ x+\frac{y-z}{2}=9,\end{array}\right.$ $x=8, y=4, z=2$ or $x=-6.4 ; y=11.2 ; z=-19.6$.
Answer: $8 ; 4 ; 2$ or $-6.4 ; 11.2 ;-19.6$. | 8;4;2or-6.4;11.2;-19.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,637 |
13.147. A sheet of metal in the form of a rectangle, with the ratio of length to width being $2: 1$. From this sheet, an open-top box is made by cutting out a square with a side of 3 cm from each corner and folding up the resulting flaps. Determine the dimensions of the metal sheet if the volume of the box turned out t... | Solution.
Let $2a$ cm be the length of the sheet, and $a$ cm be its width. The length of the box is $(2a-6)$ cm, its width is $(a-6)$ cm, and its height is 3 cm. The volume of the box is $3(a-6)(2a-6)=168$, from which we find $a=10$ cm is the width of the sheet, and 20 cm is its length.
Answer: $10 \times 20$ cm. | 10\times20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,638 |
13.148. A photograph measuring $12 \times 18$ cm is inserted into a frame of constant width. Determine the width of the frame if its area is equal to the area of the photograph itself. | ## Solution.
Let $x$ cm be the width of the frame (Fig. 13.9). Then its area is $2 \cdot 12 x + 2(18 + 2 x) x$ cm. According to the condition, $2 \cdot 12 x + 2(18 + 2 x) x = 12 \cdot 18$, from
$. Then $\left\{\begin{array}{l}x+y=44, \\ \frac{x-y}{y} \cdot 100=y,\end{array}\right.$ from which $x=264, y=-220$.
Answer: -220 and 264. | -220264 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,640 |
13.150. In the manuscript of an arithmetic problem book, an example was written in which a given number had to be multiplied by 3 and 4 subtracted from the result. In the printing house, a misprint was made: instead of the multiplication sign, a division sign was placed, and instead of a minus sign, a plus sign. Nevert... | ## Solution.
Let $x$ be the given number. Then the assumed example is $-3 x-4$, and the printed one is $-\frac{x}{3}+4$. According to the condition, $3 x-4=\frac{x}{3}+4$, from which $x=3$.
Answer: $3 \cdot 3-4$. | 3\cdot3-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,641 |
13.153. A plot in the shape of a rectangle with a diagonal of 185 m was allocated for a sports field. During construction, the length of each side was reduced by 4 m. The rectangular shape was maintained, but the area was reduced by \(1012 \mathrm{~m}^{2}\). What are the actual dimensions of the sports field?
(y-4)=x y-1012$, from which $x y-4(x+y)=$ $=x y-1028 ; x+y=257 ; y=257-x$. Solving the quadratic equation $x^{2}+(257-x)^{2}-185^{2}=0$, we find $x_{1}=104, x_{2}=153$.
Answer: $100... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,642 |
13.155. In the first week of their vacation trip, the friends spent 60 rubles less than $2 / 5$ of the amount of money they brought with them; in the second week, $1 / 3$ of the remainder and another 12 rubles on theater tickets; in the third week, $3 / 5$ of the new remainder and another 31 rubles 20 kopecks on boat r... | Solution.
Let the initial amount be $x$ rubles.
| Week | Spent | Balance |
| :--- | :---: | :---: |
| First | $\frac{2}{5} x-60$ (rubles) | $\frac{3}{5} x+60$ (rubles) |
| Second | $\frac{1}{3}\left(\frac{3}{5} x+60\right)+12=\frac{1}{5} x+32$ (rubles) | $\frac{2}{5} x+28$ (rubles) |
| Third | $\frac{3}{5}\left(\frac... | 2330 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,643 |
13.156. A motorboat with a speed of 20 km/h traveled the distance between two points along the river and back without stopping in 6 hours and 15 minutes. The distance between the points is 60 km. Determine the speed of the river current. | Solution.
Let $v_{\text {r }}$ km/h be the speed of the river current; the boat traveled 60 km downstream in $\frac{60}{20+v_{\mathrm{r}}}$ hours, and upstream in $\frac{60}{20-v_{\mathrm{r}}}$ hours. According to the problem, $\frac{60}{20+v_{\mathrm{r}}}+\frac{60}{20-v_{\mathrm{r}}}=6.25$, from which $v_{\mathrm{r}}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,644 |
13.157. Find a two-digit number such that if it is divided by the product of the digits it is composed of, the quotient is $16 / 3$, and if 9 is subtracted from it, the difference will also be a two-digit number that differs from the desired number only in the order of the digits. | ## Solution.
Let $10 x+y$ be the desired two-digit number. Then
$$
\left\{\begin{array}{l}
\frac{10 x+y}{x y}=\frac{16}{3}, \\
10 x+y-9=10 y+x,
\end{array} \text { from which } x=3, y=2 . \text { The desired number is } 32 .\right.
$$
Answer: 32. | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,645 |
13.158. Apples of the 1st grade for a total of 228 rubles and apples of the 2nd 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 1st... | ## Solution.
Let $x$ kg of apples of the first grade and $x+5$ kg of apples of the second grade be delivered to the store. In total, $2x+5$ kg of apples were delivered. A kilogram of apples of the 1st grade costs $\frac{228}{x}$ rubles. The new price of apples $-\left(\frac{228}{x}-0.9\right)$ rubles. The amount recei... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,646 |
13.159. Three departments of the institute have submitted applications for the purchase of additional equipment for laboratories. The cost of the equipment in the application of the first department is $45\%$ of the application of the second department, and the cost of the equipment in the application of the second dep... | ## Solution.
Let $x, y, z$ thousand rubles be the cost of equipment in the applications of the first,
second, and third departments, respectively. According to the conditions $\left\{\begin{array}{l}x=0.45 y, \\ y=0.8 z, \\ z=x+640,\end{array}\right.$ from
which $x=360$ thousand rubles, $y=800$ thousand rubles, $z=1... | 2160 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,647 |
13.160. If a two-digit number is divided by the sum of its digits, the quotient is 4 and the remainder is 3. If this number is divided by the product of its digits, the quotient is 3 and the remainder is 5. Find this number. | Solution.
Let the desired number be $10 x+y$. Then, according to the condition, we have the system $\left\{\begin{array}{l}10 x+y-3=4(x+y), \\ 10 x+y-5=3 x y\end{array} \Rightarrow x=2, y=3\right.$.
Answer: 23. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,648 |
13.161. The cost of transporting a ton of cargo from point $M$ to point $N$ by rail is $b$ rubles more expensive than by water. How many tons of cargo can be transported from $M$ to $N$ by rail for $a$ rubles, if by water the same amount can transport $k$ tons more than by rail? | ## Solution.
Let $x$ tons of cargo can be transported by railway, $(x+k)$ tons by waterway. The cost of transporting one ton of cargo by railway is $\frac{a}{x}$ rubles, by waterway $-\frac{a}{x+k}$ rubles. According to the condition, $\frac{a}{x}-\frac{a}{x+k}=b$, from which
$$
\begin{aligned}
x= & \frac{-b k+\sqrt{... | \frac{-+\sqrt{b^{2}k^{2}+4k}}{2b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,649 |
13.162. A certain product was purchased in the fall for 825 rubles. A kilogram of this product in the fall was 1 ruble cheaper than in the spring, and therefore, for the same amount in the spring, 220 kg less was purchased. How much does 1 kg of the product cost in the spring and how much of it was purchased in the fal... | ## Solution.
Let $x$ kg of the product be purchased in the fall. In the spring, $x-220$ kg of the product was purchased for the same amount of money. The cost of 1 kg in the fall is $\frac{825}{x}$ rubles, and in the spring it is $\frac{825}{x-220}$ rubles. According to the condition, $\frac{825}{x-220}-\frac{825}{x}=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,650 |
13.163. When harvesting, 210 centners of wheat were collected from each of two plots. The area of the first plot is 0.5 hectares less than the area of the second plot. How many centners of wheat were collected per hectare on each plot if the wheat yield on the first plot was 1 centner per hectare more than on the secon... | Solution.
Let $x$ ha be the area of the first plot, $(x+0.5)$ ha be the area of the second. From 1 ha on the first plot, $\frac{210}{x}$ centners of wheat were collected, and from the second plot, $\frac{210}{x+0.5}$ centners. According to the condition, $\frac{210}{x}-\frac{210}{x+0.5}=1$, from which $x=10$ ha. From ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,651 |
13.164. The cost of 60 copies of the first volume and 75 copies of the second volume is 2700 rubles. In reality, only 2370 rubles were paid for all these books, as a discount was applied: 15% on the first volume and 10% on the second. Find the original price of these books. | Solution.
Let $x$ and $y$ rub. - the original price of the first and second volumes, respectively. According to the condition, $60 x + 75 y = 2700$. After the discount, the price of the first volume became $0.85 x$ rub., the second - $0.9 y$ rub. According to the condition, $60 \cdot 0.85 x + 75 \cdot 0.9 y = 2370$. S... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,652 |
13.165. There are 140 cans of two capacities. The volume of the larger can is 2.5 liters more than the volume of the smaller can. The total volume of the larger cans is equal to the total volume of the smaller cans and is 60 liters. Determine the number of large and small cans. | ## Solution.
Let $x$ be the number of large jars, $(140-x)$ be the number of small jars, $y$ liters be the volume of the smaller jar, and $(y+2.5)$ liters be the volume of the larger jar. According to the problem, $\left\{\begin{array}{l}x(y+2.5)=60, \\ (140-x) y=60,\end{array}\right.$ from which $x=20$ is the number ... | 20120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,653 |
13.167. A motorboat and a sailboat, being 30 km apart on a lake, move towards each other and meet after 1 hour. If the motorboat were 20 km behind the sailboat and was catching up to it, it would take 3 hours and 20 minutes. Determine the speeds of the boat and the sailboat, assuming they are constant and unchanged in ... | ## Solution.
Let $x$ km/h be the speed of the boat, and $y$ km/h be the speed of the sailboat. According to the problem, $x \cdot 1 + y \cdot 1 = 30$. In the second case, in 3 hours and 20 minutes, the boat would have traveled
$3 \frac{1}{3} y + 20$ km. Then $\frac{3 \frac{1}{3} y + 20}{x} = 3 \frac{1}{3}$. Solving th... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,655 |
13.168. A one-digit number was increased by 10 units. If the resulting number is increased by the same percentage as the first time, the result is 72. Find the original number. | Solution.
Let $x$ be the number we are looking for, which was increased by $y \cdot 100\%$. Then $x+10=y x$, from which $y=\frac{x+10}{x}$. According to the condition $(x+10) \cdot \frac{x+10}{x}=72$, from which $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,656 |
13.169. A crystal, while in the formation stage, uniformly increases its mass. Observing the formation of two crystals, it was noted that the first one over 3 months gave the same mass increase as the second one over 7 months. However, after a year, it turned out that the first crystal increased its initial mass by $4 ... | ## Solution.
Let the annual increase in mass $x$ be $a$; then the annual increase in mass $y$ is $\frac{3 a}{7}$. We have $a=0.04 x, \frac{3 a}{7}=0.05 y \Rightarrow$ $\Rightarrow \frac{4}{5} \cdot \frac{x}{y}=\frac{7}{3} \Rightarrow x: y=35: 12$.
Answer: $35: 12$. | 35:12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,657 |
13.170. One tractor brigade plowed 240 ha, and the other plowed $35 \%$ more than the first. The first brigade processed 3 ha less daily than the second, but finished the work 2 days earlier than the second. How many hectares did each brigade process per working day, given that the planned daily norm of 20 ha was excee... | Solution.
The second brigade plowed $240 \cdot 1.35=324$ hectares. Let $x$ hectares be the amount the first brigade processed daily, and $(x+3)$ hectares - the second brigade; $y$ days the first brigade worked, and $(y+2)$ days - the second brigade. According to the problem,
$$
\left\{\begin{array}{l}
x \cdot y=240, ... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,658 |
13.171. In a family, there is a father, a mother, and three daughters; together they are 90 years old. The age difference between the girls is 2 years. The mother's age is 10 years more than the sum of the daughters' ages. The difference in age between the father and the mother is equal to the age of the middle daughte... | Solution.
Let $x$ years be the age of the youngest daughter. Then the age of the middle daughter is $-x+2$ years, the oldest is $-x+4$ years, the mother is $x+x+2+x+4+10=3x+16$ years old, and the father is $3x+16+x+2=4x+18$ years old. According to the condition, $x+x+2+x+4+3x+16+4x+18=90$, from which $x=5$ years is th... | 38,31,5,7,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,659 |
13.172. Two vessels with salt solutions were set for evaporation. The daily evaporated portions of salt are constant for each vessel. From the first vessel, 48 kg of salt was obtained, and from the second, which stood for 6 days less, 27 kg. If the first vessel had stood as many days as the second, and the second as ma... | ## Solution.
Let $x$ be the number of days the first solution stood, and $y$ be the number of days the second solution stood. The daily evaporated portion for the first container is $\frac{48}{x}$ kg of salt, and for the second container, it is $\frac{27}{y}$ kg. According to the problem, we have the system of equatio... | 1824 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,660 |
13.173. If an unknown two-digit number is divided by the number represented by the same digits but in reverse order, the quotient is 4 and the remainder is 3. If the sought number is divided by the sum of its digits, the quotient is 8 and the remainder is 7. Find this number. | Solution.
Let $10 x+y$ be the desired number. Then $\left\{\begin{array}{l}10 x+y=(10 y+x) \cdot 4+3, \\ 10 x+y=(y+x) \cdot 8+7,\end{array}\right.$ from which $x=7, y=1$. The desired number is 71.
Answer: 71. | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,661 |
13.174. In four boxes, there is tea. When 9 kg were taken out of each box, the total that remained in all of them together was as much as there was in each one. How much tea was in each box? | ## Solution.
Let there be $x$ kg of tea in each box, leaving $(x-9)$ kg. According to the condition, $4(x-9)=x$, from which $x=12$ kg.
Answer: $12 \mathrm{kg}$. | 12\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,662 |
13.175. A motorboat left the pier simultaneously with a raft and traveled downstream $40 / 3$ km. Without stopping, it turned around and went upstream. After traveling $28 / 3$ km, it met the raft. If the speed of the river current is 4 km/h, what is the motorboat's own speed | Solution.
The boat and the raft were on the way $\frac{\frac{40}{3}-\frac{28}{3}}{4}=1$ hour. Let $v$ km $/ h$ be the boat's own speed. Downstream, the boat traveled $\frac{\frac{40}{3}}{4+v}$ hours, and upstream $\frac{28}{3(v-4)}$ hours. According to the condition, $\frac{40}{3(v+4)}+\frac{28}{3(v-4)}=1$, from which... | \frac{68}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,663 |
13.176. The total capacity of three tanks is 1620 liters. Two of them are filled with kerosene, while the third one is empty. To fill it, you need to use either all the contents of the first tank plus $1 / 5$ of the contents of the second, or all the contents of the second plus $1 / 3$ of the contents of the first. Fin... | Solution.
Let $x$ be the capacity of the first tank in liters, $y_{\text {l }}$ be the capacity of the second tank in liters, and $z \pi-$ be the capacity of the third tank in liters. According to the conditions, $\left\{\begin{array}{l}x+y+z=1620, \\ z=x+\frac{1}{5} y, \\ z=y+\frac{1}{3} x,\end{array}\right.$ from wh... | 540,450,630 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,664 |
13.178. 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 the plots 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 wer... | Solution.
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 $S: n=Q$. We will fill in the table with the given and required values in the sequence indicated by the numbers (1), (2),
| Park | Initially | | | With new division | | |
| :---: | :---: | :---: ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,665 |
13.181. A team of workers was supposed to manufacture 7200 parts per shift, with each worker making the same number of parts. However, three workers fell ill, and therefore, to meet the entire quota, each of the remaining workers had to make 400 more parts. How many workers were in the team? | Solution.
Let there be $x$ workers in the team. Each worker made $\frac{7200}{x}$ parts. After the team size was reduced to $x-3$ workers, each worker started making $\frac{7200}{x}+400$ parts. According to the condition, $\left(\frac{7200}{x}+400\right)(x-3)=7200$, from which $x=9$.
## Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,666 |
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