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5. In Ali-Baba's cave, there is a lot of gold and diamonds. A full bag of gold weighs 200 kg, a full bag of diamonds 40 kg.
75 What is better?
Ali-Baba can carry 100 kg at a time. One kilogram of gold is worth 20 dinars, one kilogram of diamonds is worth 60 dinars. How much money can he get for the gold and diamonds ... | 5. First, note that 5 kg of gold has the same volume as 1 kg of diamonds, but is more expensive. We will prove that: 1. Ali-Baba can get 3000 dinars for the treasures. Indeed, the bag can hold 40 kg of diamonds. If we replace 15 kg of diamonds with 75 kg of gold, the volume of the bag will remain the same, and its valu... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Satisfied Pikes. 40 pikes were released into the pond. A pike is satisfied if it has eaten three other pikes (satisfied or hungry). What is the maximum number of pikes that can be satisfied? | 9. We will feed the pikes in three stages. Initially, we will feed 9 pikes, after which there will be 9 full and 4 hungry pikes left. Then, with the nine full pikes, we will feed three hungry pikes, and with three "newly" full pikes, we will feed the last hungry pike. In total, 13 pikes have been fed. Note that as each... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. From Alcuin's problems. How many jumps will it take for the hound to catch up with the hare if initially they are separated by a distance of 150 feet, the hare moves away from the dog by 7 feet with each jump, and the dog runs faster than the hare and gets 9 feet closer with each jump? | 1. For each jump, the greyhound gets 2 feet closer to the hare. Therefore, she will catch up to it in 75 jumps. | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is average speed? A car travels from point $A$ to point $B$ at a speed of 40 km/h, and returns at a speed of 60 km/h. What is the average speed of the car? | 2. If the distance from $A$ to $B$ is $n$, then the average speed on the path $2 A B$ is $V_{\text {avg }}=2 n:\left(\frac{n}{40}+\frac{n}{60}\right)=\frac{240 n}{5 n}=48 \mathrm{km} /$ h. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. From Nizhny Novgorod to Astrakhan, a steamboat takes 5 days, and on the return trip - 7 days. How long will it take for rafts to float from Nizhny Novgorod to Astrakhan? | 3. When the steamboat goes from Nizhny Novgorod to Astrakhan (downstream), it covers $\frac{1}{5}$ of the distance in a day, and when it goes back - $\frac{1}{7}$ of the distance. Therefore, $\frac{1}{5}-\frac{1}{7}=\frac{2}{35}$ - two current speeds. From this, $\frac{1}{35}$ of the distance per day is the speed of th... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Two pedestrians are walking towards each other along a straight road at a speed of 5 km/h. The initial distance between them is 10 kilometers. A fly, which flies at a speed of $14 \mathrm{km} / \mathrm{h}$, takes off from the first pedestrian, flies straight to the second, lands on him, and without losing a second, ... | 6. If the pedestrians walk at a speed of $5 \mathrm{km} / \mathrm{h}$, they will meet in one hour. In this time, the fly will fly $14 \times 1=14$ km. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Train and Pole. A train 18 m long passes by a pole in 9 seconds. How much time will it need to pass a bridge 36 m long? | 7. Draw a picture! The lead car will pass the bridge in 18 seconds, and the last car will still be traveling on the bridge for another 9 seconds. Answer: 27 seconds. | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Task by L. Carroll. Couriers from places $A$ and $B$ move towards each other uniformly, but at different speeds. After meeting, one needed another 16 hours, and the other needed another 9 hours to reach their destination. How much time does it take for each to travel the entire distance from $A$ to $B$? | 8. Let $t$ be the time it took for the couriers to reach the meeting point, and $v_1$ and $v_2$ be the speeds of the couriers. Then $\left(v_{1}+v_{2}\right) t=v_{1}(16+t)=v_{2}(9+t) ; 16 v_{1}=v_{2} t$ and $v_{1} t=9 v_{2} ; \frac{16 v_{1}}{v_{1} t}=\frac{v_{2} t}{9 v_{2}}$ or $\frac{16}{t}=\frac{t}{9}$ and $t^{2}=16 ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Draw a diagram! Two ferries leave simultaneously from opposite banks of a river and cross it perpendicular to the banks. The speeds of the ferries are constant. The ferries meet each other 720 meters from the nearest bank. Upon reaching the bank, they immediately head back. On the return trip, they meet 400 meters ... | 10. The total distance they traveled by the time of their first meeting (720 m from one of the banks) is equal to the width of the river. When they meet for the second time, the total distance is three times the width of the river, which required three times the time. By the time of the first meeting, one of the ferrie... | 1760 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. A Trip for Milk. Anton went to a dairy store. He had no money, but he had empty bottles - six one-liter bottles (worth 20 kopecks each) and six half-liter bottles (worth 15 kopecks each). In the store, there was milk sold by the liter for 22 kopecks. What is the maximum amount of milk he could bring home? He had no... | 11. By returning six half-liter bottles and one liter bottle, Anton will receive 1 ruble 10 kopecks, which will be the cost of 5 liters of milk. The 5 liters of milk he buys can be carried home in the remaining liter bottles. Let's ensure that he won't be able to carry more than 5 liters. If he returns not one liter bo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12. How many parts should a worker produce to exceed the plan by 40 parts by no less than $47 \%$? | 12. Let's pay attention to the fact that $1 \%$ of the plan for 40 parts is 0.4 parts. Next, $47 \% - 18.8$ parts. But you can't make 0.8 of a part, so the worker must make no less than 59 parts. | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Balls. In a box, there are 100 black and 100 white balls. What is the smallest number of balls that need to be taken out, without looking into the box, to ensure that among them there are 2 balls of the same color? To ensure there were 2 white balls? | 2. Out of three balls, there will definitely be 2 of the same color. Therefore, three balls are sufficient. Note that two balls are not enough, as they can be of different colors. It may happen that we initially draw 100 black balls, and only then - 2 white ones. In total: 102 balls. | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. A three-digit number is thought of, which with any of the numbers 543, 142, and 562, matches one digit in the same place, while the other two do not match. What is the number thought of? | 11. If the first digit of the desired number is 5, then either the second digit is 4 or the third is 2 (since a match with the second number is required). Both cases lead to a contradiction: a match with either the first or the third number will be in two digits, so the first digit cannot be 5. Reasoning similarly, we ... | 163 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12. Given a 1998-digit number, every two adjacent digits of which form a two-digit number divisible by 17 or 23. The last digit of the number is 1. What is the first? | 12. Let's start recording this number from the end. At some point, we will notice that the number has the form ...92346...9234692346851. Further, we see that the digits 92346 are repeating, so we subtract from 1998 the number of digits that do not belong to this cycle - 3. We divide the resulting number by 5 (the numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. The warehouse has nails in boxes weighing $24, 23, 17$ and 16 kg. Can the warehouse keeper issue 100 kg of nails from the warehouse without opening the boxes? | 13. For example: 4 boxes - at 17 kg each and 2 boxes - at 16 kg each. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. Find the sum: $1+2+3+\ldots+1999$. | 15. Let $S=1+2+3+\ldots+1999$. Then: $S=1999+1998+\ldots+3+2+1$ and $2 S=(1+1999)+(2+1998)+\ldots+(1999+1)=2000 \cdot 1999$. Therefore: $S=1000 \cdot 1999$. Answer: 1999000. | 1999000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. Find the sum: $1+3+5+\ldots+1999$. | 16. Let $S=1+3+5+\ldots+1999$. Rewrite it as: $S=1999+1997+\ldots+1$. Then $2 S=2000 \cdot 1000 . S=1000000$. | 1000000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25. Find the number that when divided by 2 gives a remainder of 1, when divided by 3 gives a remainder of 2, when divided by 4 gives a remainder of 3, and when divided by 5 gives a remainder of 4. | 25. A number greater than the desired one by 1 will be divisible by $2,3,4,5$, that is, by 60. The smallest suitable number is 59. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
26. For a three-digit number $x$, it is known that if you subtract 7 from it, the result will be divisible by 7, if you subtract 8, the result will be divisible by 8, and if you subtract 9, the result will be divisible by 9. What is this number? | 26. The desired number is divisible by 7, 8, and 9. Therefore, this number is 504. There are no other such three-digit numbers. | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
30. What digit does the number $3^{100}$ end with? | 30. The number $3^{100}=81^{25}$, and therefore, ends in 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
31. What is the remainder when the number $2^{99}$ is divided by 7?
95 More problems! | 31. Note that $2^{99}=8^{33}=(7+1)^{33}=(7+1) \ldots(7+1)$. Expand the brackets. The resulting terms will be divisible by 7, except for 1. Thus, $2^{99}$ can be written in the form: $7 x$ +1 . Answer: 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
35. How many diagonals does a thirty-four-sided polygon have? | 35. Each vertex of the polygon is connected by diagonals to all other vertices, except for two, which it is connected to by a side. Thus, each vertex of the thirty-four-sided polygon is connected by diagonals to 31 vertices. Note that in this counting system, each diagonal is counted twice. In total: 527 diagonals. | 527 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
37. In two boxes, there are 70 coins. It is known that in the first box, $\frac{5}{9}$ of the total number of coins are gold, and the rest are silver, in the second box, $\frac{7}{17}$ of the number of coins are silver, and the rest are gold. How many coins are in each box | 37.Let $9 x$ be the number of coins in the first box, and $17 y$ be the number in the second, then $9 x + 17 y = 70$, where $x$ and $y$ are natural numbers, and $1 \leq y \leq 4$. By enumeration, we find that $y=2$, and $x=4$. Answer: in the first -36 coins, in the second -34. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
39. Calculate: $2379 \cdot 23782378-2378 \cdot 23792379$. | 39. Let $a=2378$, then the desired expression is: $(a+1)(10000 a+a)-a(10000(a+1)+(a+1))=$ $=(a+1) \cdot 10001 a-a \cdot 10001 \cdot(a+1)=0$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
41.On the farmyard, geese and piglets were wandering around. A boy counted the number of heads, there were 30, then he counted the total number of legs, there were 84. Can you find out how many geese and how many piglets were on the farmyard? | 41.If only geese were wandering around the farmyard, there would be 60 legs in total, the "extra" legs, which number 24, belong to the piglets - two for each. Therefore, there were 12 piglets, and 18 geese. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
42.A brick weighs 2 kg and half a brick. How much does the brick weigh? | 42.It follows from the condition that half a brick weighs 2 kg. Therefore, a whole brick weighs 4 kg. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
43. A cup and a saucer cost 2500 rubles, while 4 cups and 3 saucers cost 8870 rubles. Find the price of a cup and the price of a saucer. | 43.4 cups and 4 saucers cost 10000 rubles, while 4 cups and 3 saucers cost 8870 rubles, therefore, the price of one saucer: $10000-8870=1130$ rubles, the price of one cup: $2500-1130=1370$ rubles. | 1370 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
47. 6 carps are heavier than 10 perches, but lighter than 5 pikes; 10 carps are heavier than 8 pikes. What is heavier: 2 carps or 3 perches? | 47. Since 6 carp are heavier than 10 perch, it is clear that 6 carp are even heavier than 9 perch. Therefore, 2 carp are heavier than 3 perch. This means that two of the three conditions in the problem are redundant. | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
50.What is the 1997th digit in the decimal expansion of the fraction $\frac{1}{7}=0.142857 \ldots ?$ | 50.If we divide 1 by 7 using long division, we get that $\frac{1}{7}=0.(142857)$. The remainder of 1997 divided by 6 is 5, Therefore, the digit at the 1997th place is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
51. Little One eats a jar of jam in six minutes, while Karlson is twice as fast. How long will it take them to eat the jam together | 51. The question of the problem can also be formulated as follows: "How long would it take for three Little Ones to eat the jam?" (According to the condition of the problem, Carlsson can be equated to two Little Ones). It is clear that three Little Ones would finish the jam three times faster than one. Answer: in 2 min... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
56. 180 g of gold with a fineness of 920 was alloyed with 100 g of gold with a fineness of 752. What is the fineness of the resulting alloy? | 56. Assay of an alloy: $p=\frac{a}{b} \cdot 1000$, where $a$ is the weight of gold in the alloy, and $b$ is the weight of the alloy. That is: $p=\frac{100 \cdot 0.752 + 180 \cdot 0.920}{280} \cdot 1000=\frac{100 \cdot 752 + 180 \cdot 920}{280}=860$. | 860 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
60.Yura left the house for school 5 minutes later than Lena, but walked at twice her speed. How long after leaving will Yura catch up to Lena? | 60.Let Lena walk $s$ km in 5 minutes. Then in the next 5 minutes, Yura will walk $2 s$ km, and Lena will walk another $s$ km, that is, a total of $2 s$ km. Therefore, in 5 minutes, Yura will catch up with Lena. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
61.If a cyclist rides at a speed of 10 km/h, he will be 1 hour late. If he rides at a speed of 15 km/h, he will arrive 1 hour early. At what speed should he ride to arrive on time? | 61.If there were two cyclists, with the first one's speed being $10 \mathrm{km} / \mathrm{h}$ and the second one's speed being $15 \mathrm{km} / \mathrm{h}$. Then, according to the problem, if the first cyclist started 2 hours earlier than the second, they would arrive at the destination simultaneously. In this case, i... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
64.A motorboat travels 90 km downstream in the same time it takes to travel 70 km upstream. What distance can a raft drift in the same time? | 64.Let in $t$ hours the motorboat covers 90 km downstream and 70 km upstream. Then the speed of the motorboat downstream is $\frac{90}{t}$ km/hour, and the speed upstream is $\frac{70}{t}$ km/hour. From this, the doubled speed of the current will be $\frac{90}{t}-\frac{70}{t}=\frac{20}{t}$ km/hour. The speed of the cur... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
69. Someone has 12 pints of wine and wants to pour out half, but he does not have a 6-pint container. However, he has two containers with capacities of 5 and 8 pints. How can he measure exactly 6 pints of wine? | 69. The solution is visible from the table:
| steps: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 12 l | 12 | 4 | 4 | 9 | 9 | 1 | 1 | 6 |
| 8 l | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| 5 l | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 | | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
71. A biologist discovered an amazing variety of amoebas. Each of them divided into two every minute. The biologist puts an amoeba in a test tube, and exactly an hour later, the test tube is filled with amoebas. How much time would it take for the entire test tube to be filled with amoebas if, instead of one, two amoeb... | 71. At the beginning of the experiment, there is one amoeba in the test tube, after one minute there are already two, so 2 amoebas will fill the test tube in 59 minutes. | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
75. When Vanya was asked how old he was, he thought for a moment and said: “I am three times younger than Dad, but three times older than Seryozha.” At this point, little Seryozha ran up and reported that Dad is 40 years older than him. How old is Vanya? | 75. Dad is 3 times older than Vanya, who, in turn, is 3 times older than Seryozha, so Dad is 9 times older than Seryozha. Therefore, Dad is 8 times older than Seryozha, which is 40 years. Hence, Seryozha's age is 5 years. Vanya is three times older than Seryozha, so he is 15 years old.
109 Such problems! | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
76. In the box, there are 100 white, 100 red, 100 blue, and 100 black balls. What is the smallest number of balls that need to be pulled out, without looking into the box, to ensure that among them there are at least 3 balls of the same color? | 76. Answer: 9 balls. If we randomly draw 8 balls, there might not be three balls of the same color (2 white + 2 red + 2 blue + 2 black). If we add one more ball, then there will definitely be 3 balls of the same color. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
78. How many circles of radius 1 are needed to cover a square with a side length of $2$? | 78. A circle of radius 1 can only cover one vertex of a $2 \times 2$ square, so at least four circles are required. However, a circle of unit radius can completely cover a $1 \times 1$ square, so four circles are sufficient. Answer: 4 circles. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
81. In the box, there are pencils: 7 red and 5 blue. In the dark, pencils are taken. How many pencils need to be taken to ensure that there are at least two red and at least three blue among them? | 81. In order to definitely take no less than two red pencils, you need to take no less than 7 pencils, and to definitely take no less than 3 blue ones, you need to take no less than 10 pencils. Therefore, to definitely take no less than two red and no less than 3 blue, you need to take no less than 10 pencils. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
86. The distance between points $A B=30, B C=80, C D=236$, $D E=86, E A=40$. What is the distance $E C ?$ | 86. Since $D E+E A+A B+B C=D C$, the points $E, A$ and $B$ lie on the segment $D C$ in the given order. Therefore, $E C=150$. | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11. Find the maximum value of the function $y=3 x+4 \sqrt{1-x^{2}}$
$$ | \text { S o l u t i o n. }
$$
First, let's find the domain of the given function: $D(f)=$ $=\left\{x \mid 1-x^{2} \geqslant 0\right\}=\{x \mid-1 \leqslant x \leqslant 1\} ; \quad E(f)=\{y \mid \quad$ the equation $y=3 x+4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $y-3 x=4 \sqrt{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 16. Points $\boldsymbol{A}$ and $\boldsymbol{B}$ are located on a straight highway running from west to east. Point B is 9 km east of A. A car departs from point A heading east at a speed of 40 km/h. Simultaneously, a motorcycle departs from B in the same direction with a constant acceleration of 32 km/h².
$ km.
20
The distance between them is the absolute value of the difference between $16 t^{2}+9$ and $40 t$. Let's denote this distance by $y$:
$$
y=\left|16 t^{2}+9-4... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 26. A company has 2 workshops and 3 warehouses. It is necessary to determine the most profitable organization of transportation.
| Warehouses | No. 1 | No. 2 | No. 3 | Produced |
| :---: | :---: | :---: | :---: | :---: |
| I | 3 | 3 | 2 | 10000 |
| II | 6 | 5 | 1 | 5000 |
| Required to deliver | 4000 | 8000 | ... | ## S o l u t i o n.
Let $x$ units be transported from workshop I to warehouse 1, and $y$ units to warehouse 2. Then, from workshop I to warehouse 3, $(10000-x-y)$ units will be transported, and from workshop II to warehouses 1, 2, and 3, $4000-x$, $8000-y$, and $x+y-7000$ units will be transported, respectively (Table... | 43000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 35. For the fattening of animals, two types of feed I and II are used. Each kilogram of feed I contains 5 units of nutrient $\boldsymbol{A}$ and 2.5 units of nutrient B, while each kilogram of feed II contains 3 units of nutrient A and 3 units of nutrient B. Experimental data has shown that the fattening of ani... | ## S o l u t i o n.
1. Let's construct a mathematical model of this problem. Let $x$ and $y$ be the number of kilograms of feed of types I and II, respectively, consumed daily. Then the system of constraints is:
$$
\left\{\begin{array}{l}
5 x+3 y \geqslant 30 \quad(a) \\
2.5 x+3 y \geqslant 22.5 \\
x \geqslant 0, y \... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 37. On the set of solutions of the system of constraints
$$
\left\{\begin{array}{l}
2-2 x_{1}-x_{2} \geqslant 0 \\
2-x_{1}+x_{2} \geqslant 0 \\
5-x_{1}-x_{2} \geqslant 0 \\
x_{1} \geqslant 0, \quad x_{2} \geqslant 0
\end{array}\right.
$$
find the minimum value of the function $F=x_{2}-x_{1}$. | S o l u t i o n.
The set of feasible plans is the polygon $A B C D E$ (Fig. 35). The line $2-$ $-x_{1}-x_{2}=0$ is parallel to the level lines of the function $F=x_{2}$ $-x_{1}$. Therefore, all points on the segment $C D$ give the same minimum value of the objective function $F=x_{2}-x_{1}$ on the set of points of the... | -2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 39. Minimize $\boldsymbol{F}=\boldsymbol{x}_{2}-\boldsymbol{x}_{1}$ for non-negative $x_{1}$ and $x_{2}$, satisfying the system of constraints:
$$
\left\{\begin{aligned}
-2 x_{1}+x_{2}+x_{3} & =2 \\
x_{1}-2 x_{2}+x_{4} & =2 \\
x_{1}+x_{2}+x_{5} & =5
\end{aligned}\right.
$$
56 | ## S o l u t i o n.
These constraints can be considered as derived from inequalities, since each of the variables $x_{3}, x_{4}, x_{5}$ appears only in one equation.
1. Write the constraints as equations expressing the basic variables in terms of the non-basic variables:
$$
\left\{\begin{array}{l}
x_{3}=2+2 x_{1}-x_... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
To find the maximum value of $F=4 x_{1}+6 x_{2}$
subject to
$$
\left\{\begin{array}{c}
x_{1}+x_{2} \leqslant 18 \\
0.5 x_{1}+x_{2} \leqslant 12 \\
0 \leqslant x_{1} \leqslant 12 \\
0 \leqslant x_{2} \leqslant 9
\end{array}\right.
$$ | S o l u t i o n.
This problem has already been solved graphically (see problem 32).
When solving by the simplex method, the system of constraints, consisting of inequalities, must be transformed into a system of linear equations by introducing additional variables:
$$
\begin{gathered}
x_{1}+x_{2}+x_{3}=18 \\
0.5 x_{... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5 (1246). By what percentage will the area of a rectangle increase if its length is increased by $20 \%$ and its width by $10 \%$? | Solution. Let the length of the rectangle be $x$, and the width be $y$, then its area is $x y$. After increasing the length and width of the rectangle by $20 \%$ and $10 \%$ respectively, its area became $1.2 x \cdot 1.1 y = 1.32 x y$, i.e., the area of the rectangle increased by $0.32 x y$, which is $32 \%$ of $x y$.
... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6 (1247). Three boxes are filled with nuts. In the second box, there are $10 \%$ more nuts than in the first, and $30 \%$ more than in the third. How many nuts are in each box if there are 80 more nuts in the first box than in the third? | Solution. Let there be $x$ nuts in the first box and $y$ in the third. Then in the second box, there were $x+0.1 x=1.1 x$ or $y+0.3 y=1.3 y$. Considering that there were 80 more nuts in the first box than in the third, we form the system of equations:
$$
\left\{\begin{array}{l}
1.1 x=1.3 y \\
x-y=80
\end{array}\right.... | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13(1254). To a certain two-digit number, a one was added to the left and right. As a result, a number was obtained that is 23 times greater than the original. Find this two-digit number. | Solution. Let the desired two-digit number be $\bar{a} \bar{b}$. According to the problem, we form the equation: $\overline{\mid a b 1}=23 \cdot \overline{a b}$. We have $1000+$ $+10 \overline{a b}+1=23 \cdot \overline{a b}$, from which $13 \cdot \overline{a b}=1001, \overline{a b}=77$.
Answer: 77.
Remark 1. The unkn... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. If a zero is appended to the right of the thought number and the result is subtracted from 143, the result will be three times the thought number. What number was thought of? | Solution. Let the number be $x$. We have the equation $143-10 x=3 x$, from which $x=11$. A n s w e r: 11 . | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If to a given number we append the digit 9 on the right and add to the resulting number twice the given number, the sum will be equal to 633. Find the given number. | Solution. The problem boils down to solving the equation $10 x+$ $+9+2 x=633$, where $x$ is the given number; $x=52$. A n s w e r: 52 . | 52 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19(1260). Find the smallest natural number that, when multiplied by 2, becomes a square, and when multiplied by 3, becomes a cube of a natural number. | Solution. Let $x$ be the smallest natural number such that $2 x=b^{2}, 3 x=c^{3}$, where $b$ and $c$ are natural numbers. From the equation $2 x=b^{2}$, it follows that $x$ is divisible by 2. Since $3 x=c^{3}$, $x$ is divisible by $2^{3}=8$ and by $3^{2}=9$, i.e., $x=2^{3} \cdot 3^{2} a^{6}=72 a^{6}$, where $a$ is any ... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
41(1283). The path from $A$ to $B$ goes 3 km uphill, 6 km downhill, and 12 km on flat ground. The motorcyclist covered this path in 1 hour and 7 minutes, and the return path in 1 hour and 16 minutes. Find the motorcyclist's speed uphill and downhill, if his speed on flat ground was 18 km/h. (Note that the motorcyclist ... | Solution. I method. On a flat surface in one direction, the motorcyclist traveled $\frac{2}{3}$ hours $\left(12: 18=\frac{2}{3}\right)$, or 40 minutes. Then, 3 km uphill and 6 km downhill, the motorcyclist traveled 27 minutes, and 6 km uphill and 3 km downhill, the motorcyclist traveled 36 minutes. If we denote the mot... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
42 (1284). How old is the brother and how old is the sister if 2 years ago the brother was twice as old as the sister, and 8 years ago - five times as old? | Solution. Method I. Let the current age of the sister be $x$ years, and the brother's age be $y$ years. According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
y-2=2(x-2) \\
y-8=5(x-8)
\end{array}\right.
$$
from which we find $x=10, y=18$.
Method II. Let 2 years ago the sister was $x$... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
43 (1285). From two cities $A$ and $B$, the distance between which is 180 km, a bus and a car set out towards each other at 6:20 AM. Their meeting occurred at 7:50 AM. If the bus had left 1 hour and 15 minutes earlier, and the car 15 minutes later, they would have met at 7:35 AM. What are the speeds of the bus and the ... | Solution. Let the speed of the bus be $v_{1}$ km/h, and the speed of the car be $v_{2}$ km/h. Since they met after 1.5 hours, we have the equation:
$$
1.5 v_{1} + 1.5 v_{2} = 180
$$
If the bus had left 1 hour and 15 minutes earlier, it would have been on the road for 2 hours and 30 minutes (7:35 - 5:05 = 2 hours 30 m... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
45 (1287). A rider and a pedestrian simultaneously set off from point $A$ to point $B$. The rider, arriving in $B$ 50 minutes earlier than the pedestrian, returned back to $A$. On the return trip, he met the pedestrian 2 kilometers from $B$. The rider spent 1 hour and 40 minutes on the entire journey. Find the distance... | S o l u t i o n. Since the rider spent 1 hour 40 minutes on the entire journey, he spent 50 minutes on the journey from $A$ to $B$ (1 hour 40 minutes: $2=50$ minutes). Then the pedestrian spent 1 hour 40 minutes on the journey from $A$ to $B$, as he arrived in $B$ 50 minutes later than the rider. Therefore, the pedestr... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
46 (1288). Freshly mined coal contains $2 \%$ water, and after two weeks of exposure to air, it contains $12 \%$ water. By how many kilograms has the mass of a ton of coal increased after the coal has been lying in the air for two weeks? | Solution. Since the freshly mined coal contains $2 \%$ water, in one ton of such coal, there are 20 kg of water and 980 kg of dry matter.
After two weeks of exposure to the air, the 980 kg of dry matter constitutes $88 \% (100 \% - 12 \% = 88 \%)$, meaning the total weight of the coal will be approximately 1114 kg $\l... | 114 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
47 (1289). Two brothers walk together from school to home at the same speed. One day, 15 minutes after leaving school, the first brother ran back to school and, upon reaching it, immediately set off to catch up with the second. Left alone, the second continued walking home at half his usual speed. When the first brothe... | Solution. Let the initial speed of the brothers be denoted by $v$ (m/min), and the segment of the path that the second brother traveled at a speed of $\frac{v}{2}$ be denoted by $s$. For this segment, he spent $\frac{2 s}{v}$ minutes $\left(s: \frac{v}{2}\right)$, which is 6 minutes more than the usual time spent (at s... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
52 (1138). In a box, there are multicolored balls: 5 white, 12 red, and 20 black. What is the smallest number of balls that need to be taken out of the box without looking inside, to ensure that among them there will be: a) at least one ball of each of the specified colors; b) 10 balls of one color? | Solution. a) In the worst case, 20 black and 12 red balls can be drawn, then the next ball drawn (white) will ensure the presence of balls of all the specified colors. Therefore, it is sufficient to draw 33 balls $(20+12+1=33)$ to ensure that there is at least one ball of each of the specified colors among them.
b) In... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
53(1139). I have thought of an integer not exceeding 1000. How can you find out what number I have thought of by asking no more than 10 questions, to which I will only answer “yes” or “no”? | Solution. When solving the problem, it is convenient to use a table of powers with base 2, as the number not exceeding 1000 is in the numerical interval $\left[2^{0}, 2^{10}\right)$. Questions should be asked in such a way that after each answer, the interval is reduced by half. (The found intervals will be marked on F... | 300 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
64(1168). Can: a) the sum of five consecutive natural numbers be a prime number; b) the sum of the squares of five consecutive natural numbers be a prime number? | Solution. a) Let's check on specific examples whether the sum of five consecutive natural numbers is a prime or composite number: $1+2+3+4+5=15 ; 2+3+4+5+6=$ $=20 ; 3+4+5+6+7=25 ; \ldots ; 10+11+12+13+14=60$.
The numbers $15,20,25,60$ are composite numbers. We get the hypothesis: the sum of five consecutive natural nu... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
67 (1183). One vessel contains 49 liters of water, and the other contains 56 liters. If the first vessel is filled to the top with water from the second vessel, the second vessel will be only half full. If the second vessel is filled to the top with water from the first, the first vessel will be only one-third full. Wh... | Solution. Method I. Let the capacity of the first vessel be $(49+x)$ liters, and the capacity of the second vessel be $(56+y)$ liters. If $x$ liters are poured from the second vessel into the first, then $(56-x)$ liters will remain in the second vessel, or $\frac{1}{2}(56+y)$. We have the equation:
$$
56-x=\frac{1}{2}... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
69 (1185). From $A$ to $B$, the distance between which is 37 km, two buses with the same speed left at 7:18 AM and 7:48 AM, respectively. A cyclist, who left $B$ for $A$ at 7:28 AM, met the first bus at 7:58 AM and the second bus at 8:19 AM. Find the speeds of the cyclist and the buses. | Solution. Let the speed of the buses be $v_{1}$ km/min, and the speed of the cyclist be $v_{2}$ km/min. Since the first bus traveled for 40 minutes (7:58 AM - 7:18 AM = 40 minutes) before meeting the cyclist, the second bus for 31 minutes (8:19 AM - 7:48 AM = 31 minutes), and the cyclist spent 30 minutes (7:58 AM - 7:2... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
71 (1187). A tourist left from $A$ to $B$. After 1 hour and 20 minutes, a cyclist left from $A$ in the same direction and overtook the tourist after 30 minutes. Upon arriving at $B$, the cyclist, without stopping, turned back and met the tourist 1.5 hours after the first meeting. Find the speeds of the tourist and the ... | Solution. Let the speed of the cyclist be $v_{1}$ km/h, and the speed of the tourist be $v_{2}$ km/h. Since the cyclist overtook the tourist 30 minutes $\left(\frac{1}{2}\right.$ hours) after departure and traveled $\frac{1}{2} v_{1}$ km, while the tourist walked 1 hour 20 minutes longer than the cyclist $\left(1 \frac... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
85 (954). Prove that the value of the numerical expression $11^{11}+$ $+12^{12}+13^{13}$ is divisible by 10. | Solution. It is sufficient to prove that the value of the numerical expression ends in zero. Indeed, the number $11^{11}$ ends in the digit 1. To determine which digit the number $12^{12}$ ends in, it is advisable to represent $12^{12}$ as $\left(12^{4}\right)^{3}$ and note that $12^{4}$ ends in the digit 6. Therefore,... | 10989 | Number Theory | proof | Yes | Yes | olympiads | false |
91 (960). Prove the validity of the equality:
1) $(n-2)(n-1) n(n+1)+1=\left(n^{2}-n-1\right)^{2}$;
2) $n(n+1)(n+2)(n+3)+1=\left(n^{2}+3 n+1\right)^{2}$. | Solution.
1) $(n-2)(n-1) n(n+1)+1=(n-2)(n+1)(n-1) n+1=$ $\left(n^{2}-n-2\right)\left(n^{2}-n\right)+1=\left(n^{2}-n\right)^{2}-2\left(n^{2}-n\right)+1=$ $\left(n^{2}-n-1\right)^{2}$
2) $n(n+1)(n+2)(n+3)+1=n(n+3)(n+1)(n+2)+1=$ $=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 ... | 3926341 | Algebra | proof | Yes | Yes | olympiads | false |
105(974). Determine the value of $b$, if the graph of the function defined by the formula passes through the point with coordinates $(3 ; 10)$:
1) $y=x+b$;
2) $y=3 x+b$
3) $y=-\frac{1}{3} x+b$;
4) $y=-\frac{1}{2} x+b$.
A n s w e r: 1) $b=7$; 2) $b=1$; 3) $b=11$; 4) $b=11.5$.
106(975). Define the function by a formul... | Solution. Let $x$ be the number of turns in the spring. Considering that the number of gaps in the spring is one less than the number of turns, according to the problem, we set up the equation $5x + 8(x-1) = 122$, from which $x = 10$.
Answer: 10 turns. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
108(977). A person, living in a cottage near a railway station, usually manages to walk from the cottage to the station in time for the train departure in 18 minutes. Once, this person was delayed at home before leaving by several minutes. Although after that, he walked 1.2 times faster than usual, he still missed the ... | Solution. If the usual speed of the gardener is $v$ m/min, then the distance from the garden to the station is $18 v$ meters. When the gardener walked at a speed of $1.2 v$ m/min, it took 15 minutes to travel from home to the garden $(18 v: 1.2 v=15)$. Since the gardener was 2 minutes late, the train left 13 minutes af... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
113(982). A cyclist arrived from point $A$ to point $B$ at the appointed time, moving at a certain speed. If he had increased this speed by 3 km/h, he would have arrived at the destination one hour earlier, and if he had traveled 2 km less per hour than he actually did, he would have been an hour late. Determine the di... | Solution. Let the first pipe fill the pool in $x$ hours, then the second pipe - in $(x+6)$ hours; $\frac{2}{3}$ of the pool the second pipe will fill in $\frac{2}{3}(x+6)$ hours.
According to the problem, we have the equation
$$
x=\frac{1}{2} \cdot \frac{2}{3}(x+6)
$$
from which $x=3, x+6=9$.
Answer: the first pipe... | 1040 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
119(988). In a flask, there is a solution of table salt. From the flask, $\frac{1}{5}$ of the solution is poured into a test tube and evaporated until the percentage of salt in the test tube doubles. After this, the evaporated solution is poured back into the flask. As a result, the salt content in the flask increases ... | Solution. Before solving the problem, it is reasonable to clarify how the expression "percentage content of salt" should be understood (this is the ratio of the mass of salt to the mass of the solution, expressed as a percentage).
Let the initial mass of the solution in the flask be $m$ grams, the mass of salt in it b... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
120 (989). The road from point $A$ to point $B$ is 11.5 km long and goes first uphill, then on flat ground, and finally downhill. The pedestrian spent 2 hours and 54 minutes on the journey from $A$ to $B$, and 3 hours and 6 minutes on the return journey. The pedestrian's walking speed uphill was 3 km/h, on flat ground ... | Solution. Let the length of the road uphill be $x$ kilometers, on the plain be $y$ kilometers, and downhill be $z$ kilometers. According to the problem, we form a system of three linear equations with three unknowns:
88
$$
\left\{\begin{array}{l}
\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=2.9 \\
\frac{z}{3}+\frac{y}{4}+\fra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
125. A steamship from Kyiv to Kherson takes three days, and from Kherson to Kyiv - four days (without stops). How long will it take for rafts to float from Kyiv to Kherson? | Solution. To answer the question of the problem, the distance from Kyiv to Kherson needs to be divided by the speed of the Dnieper current. If $s$ is the distance from Kyiv to Kherson, then the speed of the steamboat downstream is $-\frac{s}{3}$ (km/day), and the speed of the steamboat upstream is $-\frac{s}{4}$ (km/da... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
127. We collected 100 kg of mushrooms. It turned out that their moisture content was $99 \%$. When the mushrooms were dried, the moisture content decreased to $98 \%$. What was the mass of these mushrooms after drying? | Solution. According to the condition, in 100 kg of mushrooms, there is 1 kg of dry matter $(100-0.99 \cdot 100=1)$. Since the mass of dry matter in the total mass of mushrooms is constant (1 kg) and after drying it became $2 \%(100-98=2)$, the mass of mushrooms after drying became 50 kg (if $2 \%-1$ kg, then $100 \%-50... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
130. How many four-digit numbers can be formed from the digits $0,1,2,3,4,5$, if the numbers must be odd and digits may be repeated? | S o l u t i o n. For the first position, we can choose any of the five digits (excluding zero). For the second and third positions, we can choose any of the six digits, as digits can be repeated. For the last position, we can choose any of the three odd digits. Therefore, the total number of four-digit numbers is $5 \c... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
139. On a meadow, grass grows. If 9 cows were let onto the meadow, they would empty it in 4 days. If 8 cows were let onto the meadow, they would eat all the grass in 6 days. How many cows can graze on the meadow all the time while the grass grows? | Solution. Let $x$ kilograms of grass grow on the meadow per day, and one cow eats $y$ kilograms of grass per day. We have the system of equations:
$$
\left\{\begin{array}{l}
a+4 x-9 \cdot 4 y=0 \\
a+6 x-8 \cdot 6 y=0
\end{array}\right.
$$
where $a$ is the amount of grass that grew on the meadow before the cows were l... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
158. It is known that the number $a$ is $n$ times greater than the number $b$, and the sum of the numbers $a$ and $b$ is $m$ times greater than their difference. Find the sum of the numbers $m$ and $n$, if $m$ and $n$ are natural numbers.
94 | Given the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
\frac{a}{b}=n \\
\frac{a+b}{a-b}=m
\end{array}\right.
$$
From the second equation, express $\frac{a}{b}$. We have:
$$
\begin{gathered}
a+b=m a-m b, a(1-m)=-b(m+1) \\
\frac{a}{b}=\frac{m+1}{m-1}=\frac{m-1+2}{m-1}=1+\frac{2}{m-1} .
\end{g... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
173. In the Olympiad, 55 schoolchildren participated. All of them submitted their work. During the checking of each problem, one of three grades was given: «+» - the problem was solved, «-» - the problem was attempted, «0» - the problem was not attempted. After checking all the works, it turned out that in no two works... | Instruction. If the number of tasks was $a$, then
$$
\frac{(a+1)(a+2)}{2}=55 \text { when } a=9 .
$$
A n s w e r: 9 tasks. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
176. Among 150 schoolchildren, only boys collect stamps. 67 people collect stamps of the USSR, 48 people - of Africa, and 32 people - of America, 11 people only collect stamps of the USSR, 7 people - only of Africa, 4 people - only of America, and only Ivanov collected stamps of the USSR, Africa, and America. Find the ... | Solution. Let's represent the condition of the problem using Euler circles (Fig. 17).
We have a system of three equations:
\[
\begin{cases}
x + y + 1 = 56 & (56 = 67 - 11), \\
x + z + 1 = 41 & (41 = 48 - 7), \\
y + z + 1 = 28 & (28 = 32 - 4)
\end{cases}
\]
. Prove that the value of the expression $\sqrt{11+6} \sqrt{ } 2+$ $+\sqrt{11-6 \sqrt{2}}$ is a natural number. | Proof. I method. Let
$$
\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=A
$$
Then
$$
A^{2}=11+6 \sqrt{2}+11-6 \sqrt{2}+2 \sqrt{121-72}=36
$$
from which
$$
\sqrt{A^{2}}=|A|=6, \text { i.e. }|\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}|=6
$$
Since under the modulus sign is the sum of two radicals, each of which is a positi... | 6 | Algebra | proof | Yes | Yes | olympiads | false |
23(1099). A bus left city $M$ for city $N$ at a speed of 40 km/h. A quarter of an hour later, it met a car traveling from city $N$. This car reached city $M$, 15 minutes later it set off back to city $N$ and overtook the bus 20 km from city $N$. Find the distance between cities $M$ and $N$, if the speed of the car is 5... | Solution. Since the bus traveled 10 km in a quarter of an hour $(40: 4=10)$, the bus met the car 10 km from city $M$. If the distance between the cities is $s$ kilometers, then by the time the car overtook the bus, the bus had traveled $(s-30)$ kilometers from the moment of the meeting, spending $\frac{s-30}{40}$ hours... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
25(1101). The distance between the ports $A$ and $B$ is covered by the steamboat downstream in 5 hours, and upstream in 6 hours. How many hours will it take for a raft to drift downstream this distance? | Solution. I method. To answer the question of the problem, the distance $AB$ needs to be divided by the speed of the current. Let $AB = s$ kilometers. Then the speed of the steamer downstream is $\frac{s}{5}$ kilometers per hour, the speed upstream is $-\frac{s}{6}$ kilometers per hour, the speed of the current is $-\l... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33(1109). Find the members of the proportion $x_{1}: x_{2}=x_{3}: x_{4}$, where the first term is 6 more than the second, and the third is 5 more than the fourth. The sum of the squares of all terms is 793. | Solution. By the condition $x_{1}=x_{2}+6, x_{3}=x_{4}+5,\left(x_{2}+6\right): x_{2}=$ $=\left(x_{4}+5\right): x_{4}$.
We have a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
\left(x_{2}+6\right)^{2}+x_{2}^{2}+\left(x_{4}+5\right)^{2}+x_{4}^{2}=793 \\
\left(x_{2}+6\right) \cdot x_{4}=x_{2} \cd... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
46(1122). Several points are marked on a plane, no three of which lie on the same line. A line is drawn through every two points. How many points are marked on the plane if it is known that a total of 45 lines have been drawn? | Solution. The problem reduces to solving the equation $\frac{x(x-1)}{2}=45$ in natural numbers, where $x$ is the number of points marked on the plane. We have $x^{2}-x-90=0$, from which $x_{1}=10, x_{2}=-9$. Only the root $x=10$ satisfies the condition of the problem.
Answer: 10 points.
$\triangle 47(969)$ '. Prove t... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
51 (974). What is the last digit of the difference $43^{43}-17^{17}$? | Solution. Let's determine the last digits of various powers of the numbers 43 and 17. The number 43 ends with the digit $3, 43^2$ ends with the digit $9, 43^3$ ends with the digit $7, 43^4$ ends with the digit 1. Therefore, the last digits in subsequent powers of the number 43 will repeat: $3, 9, 7, 1$, and so on. Sinc... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
55 (978). There are 4 balls of different masses. Using a balance scale without weights, how many weighings are needed to arrange these balls in order of decreasing mass? | Solution. Let $m_{A}, m_{B}, m_{C}, m_{D}$ be the masses of four different balls. We will use a graph to show how the weighings are performed (the arrow points from the heavier ball to the lighter one).
Suppose, for example, after two weighings, we have $m_{A}>m_{B}$ and $m_{C}>m_{D}$ (Fig. $29, a$ ). Next, we will co... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
59(982). In the cabinet, there are books that need to be packed. If they are bundled in groups of 4, 5, or 6, there will always be one extra book left each time, but if they are bundled in groups of 7, there will be no extra books left. What is the number of books in the cabinet, given that there are no more than 400 b... | Solution. The problem reduces to finding a natural number that is a multiple of seven, not exceeding 400, which gives a remainder of 1 when divided by 4, 5, and 6. This number, obviously, has the form $60 n+1$, where 60 is the least common multiple of the numbers $4,5,6, n \in \boldsymbol{N}$, $n<7$. By trial, we find ... | 301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
60(983). There is a sheet of paper. It is cut into 4 pieces, then some (or all) of the resulting pieces are again cut into 4 pieces, and so on. Prove that it is impossible to obtain 50 pieces of the sheet in this way. | Solution. It is easy to understand that after each cut, the number of pieces of the sheet increases by 3. Thus, after $m$ cuts, the number of pieces will be $1+3 m$, where $m \in \boldsymbol{N}$. Since $1+3 m \neq 50$, it is impossible to obtain 50 pieces of the sheet in the manner specified in the problem.
$61(984)$.... | 7875 | Combinatorics | proof | Yes | Yes | olympiads | false |
62 (985). In a chess tournament, more than 9 but fewer than 25 grandmasters and masters participated. At the end of the tournament, it turned out that each participant scored half of their points against grandmasters. How many people participated in the tournament? How many of them were masters? | Solution. Let $n$ be the number of chess players in the tournament, of which $m$ are grandmasters, and thus $n-m$ are masters. Therefore, all participants played $\frac{n(n-1)}{2}$ games and scored $\frac{n(n-1)}{2}$ points (both games and points). Among them, the grandmasters played $\frac{m(m-1)}{2}$ games (and score... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
65 (998). An alloy consists of zinc and copper, entering into it in a ratio of $1: 2$, and another alloy contains the same metals in a ratio of $2: 3$. From how many parts of both alloys can a third alloy be obtained, containing the same metals in a ratio of 17:27? | S o l u t i o n. Let the third alloy contain $x$ kg of the first alloy and $y$ kg of the second alloy. We need to find the ratio $\frac{x}{y}$. Since zinc and copper are in the ratio $1:2$ in the first alloy and in the ratio $2:3$ in the second alloy, $x$ kg of the first alloy contains $\frac{1}{3} x$ (kg) of zinc and ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
69(1003). A cyclist and a motorcyclist are moving towards an intersection along two mutually perpendicular roads. At a certain moment, the cyclist is 8 km away, and the motorcyclist is 15 km away from the intersection. After how many minutes will the distance between them be 5 km, if the cyclist's speed is $\frac{1}{3}... | Solution. Let $OA=8$ km, $OB=15$ km (Fig. 30). If the distance of 5 km between the motorcyclist and the cyclist will be in $t$ minutes $\left(A_{1} B_{1}=5\right.$ km), then $A A_{1}=\frac{1}{3} t$ (km), $B B_{1}=t$ (km). Then $O A_{1}=O A-\frac{1}{3} t=$
. A master and his apprentice were supposed to complete a job by a certain deadline. However, when half of the work was done, the apprentice fell ill, and the master, left alone, finished the job 2 days late. How many days would it take for each of them to complete the entire job working alone, if the master wo... | Solution. Let the master be able to complete the entire job in $x$ days, then the apprentice - in $(x+5)$ days. In one day, the master completed $\frac{1}{x}$ of the work, the apprentice completed $\frac{1}{x+5}$ of the work; when working together, the master and the apprentice completed $\frac{2 x+5}{x(x+5)}$ of the w... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
74 (1008). A tourist who set out from $A$ to $B$ walked the first half of the distance at a speed of $v_{1}$, and the second half at a speed of $v_{2}\left(v_{1} \neq v_{2}\right)$. On the return trip, he walked half of the total time at a speed of $v_{1}$, and the other half at a speed of $v_{2}$. In which case did th... | Solution. I method. Let the time spent by the tourist on the journey from $B$ to $A$ be $2 t$. Then half of the time ($t$) he walked at a speed of $v_{1}$, and the second half of the time at a speed of $v_{2}$. The entire distance, therefore, is $v_{1} t + v_{2} t$, and half the distance is $\frac{v_{1} t + v_{2} t}{2}... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
76(1010). A collective farm team must harvest a potato crop by a certain deadline. After 60% of the potatoes were harvested, a combine was sent to assist the team, which reduced the harvesting time by 5 days. How many days would it have taken to harvest the potatoes without the combine's help, if it is known that the c... | Let's solve it. Let the brigade need $t$ days to harvest potatoes without the help of a combine, then the combine needs $(t-8)$ days to complete the entire work.
In one day, the brigade can complete $\frac{1}{t}$ of the entire work, and the combine can complete $\frac{1}{t-8}$ of the entire work in one day. With the h... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
79. Calculate the sum
$$
1^{2}-2^{2}+3^{2}-4^{2}+\ldots+99^{2}-100^{2}+101^{2}
$$ | $$
\begin{gathered}
1^{2}-2^{2}+3^{2}-4^{2}+\ldots+101^{2}= \\
=1+(101^{2}-100^{2})+(99^{2}-98^{2})+\ldots+(3^{2}-2^{2})= \\
=201+197+\ldots+9+5+1= \\
=\frac{1}{2}(201+1) \cdot 51=101 \cdot 51=5151
\end{gathered}
$$
Answer: 5151. | 5151 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
84. Nine identical books cost 11 rubles and some kopecks, while 13 such books cost 15 rubles and some kopecks. Determine the exact cost of one book | S o l u t i o n. It is clear that each book costs 1 r. $m$ k., where $m$ is an integer, and $m<100$. From the first condition of the problem, it follows that $200<9 m<300$, i.e., $23 \leqslant m \leqslant 33$. From the second condition, $200<13 m<300$, i.e., $16 \leqslant m \leqslant 23$. Therefore, $m=23$.
A n s w e ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
85. The percentage of VII grade students attending gymnastics classes is between 2.9 and $3.1 \%$. Determine the smallest possible number of students in this class. | Solution. If there are $x$ students in the class and $y$ of them do gymnastics, then by the condition $2.9 < 100 \cdot \frac{y}{x} < 3.1$, or $29x < 1000y < 31x$. If $x \leqslant 32$, then $31x \leqslant 992 \leqslant 1000y$, so $x \geqslant 33$. Since $29 \cdot 33 < 1000 \cdot y < 31 \cdot 33$, the number of students ... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
89. Calculate the value of the expression $\frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}$, given that $10 a^{2}-3 b^{2}+5 a b=0$ and $9 a^{2}-b^{2} \neq 0$. | Solution. Since $5 a b=3 b^{2}-10 a^{2}$, then
$$
\begin{aligned}
& \quad \frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}=\frac{3 a^{2}+15 a b-6 b^{2}}{9 a^{2}-b^{2}}=\frac{3 a^{2}+3\left(3 b^{2}-10 a^{2}\right)-6 b^{2}}{9 a^{2}-b^{2}}= \\
& \text { Answer: }-3 . \\
& =\frac{-3\left(9 a^{2}-b^{2}\right)}{9 a^{2}-b^{2}}=-3 .
\... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
90. If the first digit of a four-digit number, which is a perfect square, is decreased by 3, and the last digit is increased by 3, the result is also a perfect square. Find this number. | S o l u t i o n. Let the given number be $a^{2}$, and the modified number $b^{2}$, then $a^{2}-3000+3=b^{2}$, or $a^{2}-b^{2}=2997$, from which $(a-b) \times$ $\times(a+b)=2997=3 \cdot 3 \cdot 3 \cdot 3 \cdot 37$. Clearly, $a+b$ is greater than 37, and from the problem statement, it follows that $a+b$ is less than 200 ... | 4761 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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