problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
12.340. The side of the base of a regular quadrilateral pyramid is equal to $a$, and the plane angle at the vertex of the pyramid is $\alpha$. Find the distance from the center of the base of the pyramid to its lateral edge. | ## Solution.
Let $EO$ be the height of the regular pyramid $EABCD$ (Fig. 12.201), $OF$ be the distance from point $O$ to the lateral edge $EC$, and $\angle ECO = \beta$. Since
$$
\angle CED = \alpha, \text{ then } \angle ECD = 90^\circ - \frac{\alpha}{2}.
$$
$OC$ is the projection of $EC$ onto the plane of the base ... | \frac{}{2}\sqrt{2\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,251 |
12.343. The linear angle of the dihedral angle formed by two adjacent lateral faces of a regular quadrilateral pyramid is twice the plane angle at the vertex of the pyramid. Find the plane angle at the vertex of the pyramid. | ## Solution.
Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.204), $\angle BED$ be the linear angle of the dihedral angle between the adjacent lateral faces $BSC$ and $DSC$. Then $DE \perp SC, EO \perp BD, \angle OED = \frac{1}{2} \angle BED$.
If $F$ is the midpoint of $CD$, then $SF \perp CD, \angle D... | \arccos\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,254 |
12.348. The side of the base of a regular triangular pyramid is equal to $a$, and the plane angle at the vertex of the pyramid is $\alpha$. Find the radius of the sphere inscribed in the pyramid. | Solution.
Let $EO_2$ be the height of the regular tetrahedron $EABC$ (Fig. 12.209), $ED$ the apothem of its lateral face $BEC$. Then $\angle BED = \frac{1}{2} \angle BEC = \frac{\alpha}{2}$. Since the pyramid is regular, the center $O_1$ of the inscribed sphere lies on the height $EO_2$. If $\angle EDO_2 = \gamma$, th... | \frac{}{6}\sqrt{\frac{3\sin(\frac{\pi}{3}-\frac{\alpha}{2})}{\sin(\frac{\pi}{3}+\frac{\alpha}{2})}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,259 |
12.349. The radius of the sphere inscribed in a regular triangular pyramid is four times smaller than the side of the pyramid's base. Find the cosine of the dihedral angle at the pyramid's vertex. | ## Solution.
Let $E O_{2}$ be the height of the regular pyramid $E A B C$ (Fig. 12.210), and $E D$ be the apothem of its lateral face $B E C$. If $\angle B E C=\alpha$, then $\angle D E C=\frac{\alpha}{2}$.
Since the pyramid is regular, the center $O_{1}$ of the inscribed sphere lies on its height $E O_{2}, O_{1} O_{... | \frac{23}{26} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,260 |
12.350. The lateral edges and two sides of the base of a triangular pyramid have the same length $a$, and the angle between the equal sides of the base is $\alpha$. Find the radius of the circumscribed sphere. | ## Solution.
Let $DO$ be the height of the pyramid $DABC$, with $DA = DB = DC = BA = BC = a$ (Fig. 12.211). Then point $O$ is the center of the circle circumscribed around the base of the pyramid, and the points on the line $DO$ are equidistant from the vertices of $\triangle ABC$. The radius of this circle is
$$
\be... | \frac{\cos\frac{\alpha}{2}}{2\sqrt{\sin(\frac{\pi}{3}+\frac{\alpha}{2})\sin(\frac{\pi}{3}-\frac{\alpha}{2})}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,261 |
12.351. A cylinder is inscribed in a cone, the height of which is equal to the diameter of the base of the cone. The total surface area of the cylinder is equal to the area of the base of the cone. Find the angle between the slant height of the cone and the plane of its base. | ## Solution.
Let $D O_{2}$ be the height of the given cone, $O_{2}$ be the common center of the base of the cone and the lower base of the cylinder, $O_{1}$ be the center of the upper base of the cylinder, $D A$ be the generatrix of the cone, $C B$ be the generatrix of the cylinder, $C$ belongs to $D A$ (Fig. 12.212),... | \operatorname{arctg}\frac{2(4+\sqrt{6})}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,262 |
12.354. The base of the pyramid is an acute isosceles triangle, with the base equal to $a$ and the opposite angle equal to $\alpha$. The lateral edge of the pyramid, passing through the vertex of this angle, forms an angle $\beta$ with the plane of the base. Find the volume of the pyramid if the height of the pyramid p... | Solution.
Let $EO$ be the height of the pyramid $EABC$, $AB=BC$, $AC=a$, $\angle ABC=\alpha$, $0^{\circ}<\alpha<90^{\circ}$, $\angle EBO=\beta$, $O$ is the point of intersection of the heights $BD$, $CK$, $AL$ - bases (Fig. 12.215).
Since $\triangle ABC$ is acute-angled, the point $O$ lies inside the triangle, and th... | \frac{^{3}}{12}\operatorname{ctg}\alpha\operatorname{ctg}\frac{\alpha}{2}\operatorname{tg}\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,265 |
12.360. The slant height of the cone is $l$ and it forms an angle $\alpha$ with the base plane. A sphere is inscribed in this cone, and a regular triangular prism, all of whose edges are equal, is inscribed in the sphere. Find the volume of the prism. | Solution.
The center $K$ of the sphere inscribed in the cone lies on its height $SO, SE$ is the generatrix of the cone, $SE=l, \angle SEO=\alpha$, (Fig. 12.220) $\angle KEO=\frac{1}{2} \angle SEO=\frac{\alpha}{2}$.
, with faces $ABE$ and $CBE$ perpendicular to the base plane, making their common edge $EB$ the height of the pyramid.
Then $EB \perp AB, EB \perp BC, \angle ABC$ is the angle between the faces $ABE$ and $CBE$, $\triangle ABC = \beta$.
$BK$ is ... | \frac{2^2\sin\beta\cos^2(\frac{\pi}{4}-\frac{\alpha}{2})}{\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,273 |
12.364. The base of the pyramid $SABC$ is an equilateral triangle $ABC$. The edge $SA$ is perpendicular to the base plane. Find the angle between the lateral face $SBC$ and the base plane, if the lateral surface area of the pyramid is to the base area as $11: 4$. | ## Solution.
Let $E$ be the midpoint of $BC$ (Fig. 12.224). Then $AE \perp BC, SE \perp BC$, and $\angle SEA$ is the angle between the lateral face $SBC$ and the base plane of the pyramid.
Let $\angle SEA = \alpha, AE = 1$.
Then from $\triangle SAE \left(\angle SAE = 90^\circ\right)$:
$AS = AE \operatorname{tg} \an... | \operatorname{arctg}\frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,275 |
12.365. The radius of the base of a cone is $R$, and the angle between the generatrix and the base plane is $\alpha$. A sphere is inscribed in this cone. Through a point $P$ lying on the circle of contact between the sphere and the conical surface, a tangent line to this circle is drawn, and through this line a plane i... | Solution.
Consider the axial section of the given set of bodies passing through point $P$, perpendicular to the tangent line from the problem statement (Fig. 12.225). $P D$ is the diameter of the section of the sphere of the required area.
$$
\angle B A C=\angle B C A=\alpha .
$$
Since $P D \| B A$, then $\angle C P... | \piR^{2}\operatorname{tg}^{2}\frac{\alpha}{2}\sin^{2}2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,276 |
12.366. In a truncated cone, the diagonals of the axial section are mutually perpendicular and the length of each is $a$. The angle between the generatrix and the base plane is $\alpha$. Find the total surface area of the truncated cone. | ## Solution.
Isosceles trapezoid $ABCD$ (Fig. 12.226) is the axial section of the given truncated cone, $AC \perp BD, AC=BD=a, \angle CDA=\alpha, O_{1}$ is the midpoint of $AD$ - the center of the lower base of the truncated cone, $O_{2}$ is the midpoint of $BC$ - the center of the upper base.
Let $O_{2}B=r, O_{1}A=R... | \frac{\pi^{2}\sin(\frac{\alpha}{2}+15)\sin(\frac{\alpha}{2}-15)}{\sin^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,277 |
12.370. The base of the pyramid is a right-angled triangle, one of whose acute angles is $\alpha$. All lateral edges are equally inclined to the plane of the base. Find the dihedral angles at the base if the height of the pyramid is equal to the hypotenuse of the triangle lying in its base. | ## Solution.
Let $S O$ be the height of the pyramid $S A B C, \angle A C B=90^{\circ}, \angle B A C=\alpha$ and, since $\angle S A O=\angle S B O=\angle S C O$, then $O$ is the center of the circle circumscribed around the base of the pyramid, which is the midpoint of the hypotenuse $A B$ (Fig. 12.230), thus, the plan... | 90;\operatorname{arctg}\frac{2}{\sin\alpha};\operatorname{arctg}\frac{2}{\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,281 |
12.371. In the base of the right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ lies an isosceles triangle, where $A B=B C=a$ and $\angle A B C=\alpha$. The height of the prism is $H$. Find the distance from point $A$ to the plane passing through points $B, C$ and $A_{1}$. | Solution.
Draw a perpendicular $AE$ from point $A$ to $BC$ (Fig. 12.231). Then $A_1E \perp BC$ and, consequently, $BC \perp$ plane $AEA_1$. Thus, plane $AEA_1 \perp$ plane $BA_1C$.
In the plane $AEA_1$, drop a perpendicular $AD$ to $A_1E$. Then $AD \perp$ plane $BA_1C$ and the length of the perpendicular $AD$ is the ... | \frac{\sin\alpha}{\sqrt{H^2+^2\sin^2\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,282 |
12.375. The base of the pyramid is a square with side $a$; two lateral faces of the pyramid are perpendicular to the base, and the larger lateral edge is inclined to the base plane at an angle $\beta$. A rectangular parallelepiped is inscribed in the pyramid; one of its bases lies in the plane of the pyramid's base, an... | Solution.
Square $ABCD$ is the base of the pyramid $SABCD$ (Fig. 12.375), the lateral faces $SBA$ and $SBC$ are perpendicular to the base plane, so their common edge $SB$ is the height of the pyramid.
$BA, BC, BD$ are the projections of the edges $SA, SC, SD$ on the base plane, $BA=BC<BD$. Thus, $SD$ is the larger la... | \frac{^3\sqrt{2}\sin\alpha\cos^2\alpha\sin^3\beta}{\sin^3(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,286 |
12.378. A regular truncated quadrilateral pyramid is inscribed in a sphere of radius $R$, with the larger base passing through the center of the sphere, and the lateral edge forming an angle $\beta$ with the plane of the base. Find the volume of the truncated pyramid. | Solution.
The larger base $ABCD$ of the regular truncated pyramid
Fig. 12.238
$ABCD A_{1}B_{1}C_{1}D_{1}$ (Fig. 12.238) passes through the center $O$ of the sphere and is inscribed in the... | \frac{2}{3}R^3\sin2\beta(1+\cos^22\beta-\cos2\beta) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,289 |
12.379. On the segment $A B$, equal to $2 R$, a semicircle is constructed with it as the diameter, and a chord $C D$ parallel to $A B$ is drawn. Find the volume of the solid formed by the rotation of triangle $A C D$ around the diameter $A B$, if the inscribed angle subtending arc $A C$ is $\alpha (A C < A D)$. | ## Solution.
Let $C_{1}$ and $D_{1}$ be the points symmetric to $C$ and $D$ with respect to the diameter $AB$ (Fig. 12.239), $E$ and $F$ be the points of intersection of $CC_{1}$ and $DD_{1}$ with $AB$, $O$ be the center of the semicircle, $V_{1}$ be the volume of the cone with the axial section $\triangle CAC_{1}$, $... | \frac{2}{3}\piR^{3}\sin4\alpha\sin2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,290 |
12.380. The base of a right prism is a right-angled triangle, one of whose acute angles is $\alpha$. The largest lateral face of the prism is a square. Find the angle between the intersecting diagonals of the other two lateral faces. | Solution.
Let $\triangle A B C$ be the base of the right prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.240) $\angle A C B=90^{\circ}, \angle B A C=\alpha$.
Since $A A_{1}=B B_{1}=C C_{1}, A B>A C$ and $A B>B C$, the face $A A_{1} B_{1} B$, which has the largest area, is a square.
Let the desired $\angle A_{1} C B_{1}=\be... | \arccos\frac{2}{\sqrt{8+\sin^{2}2\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,291 |
12.383. The vertex of the cone is located at the center of the sphere, and the base of the cone touches the surface of the sphere. The total surface area of the cone is equal to the surface area of the sphere. Find the angle between the slant height and the height of the cone. | Solution.
Consider the axial section of the given set of bodies (Fig.
Fig. 12.233 12.233). Let the generatrix of the cone $O A=l$, the radius of its base $C A=r$, the radius of the spher... | \operatorname{arctg}\frac{4}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,294 |
12.384. The base of the pyramid is a right-angled triangle, with the hypotenuse equal to $c$, and the smaller of the acute angles equal to $\alpha$. The largest lateral edge forms an angle $\beta$ with the base plane. Find the volume of the pyramid if its height passes through the point of intersection of the medians o... | ## Solution.
Let $EO$ be the height of the pyramid $EABC$ (Fig. 12.234), $O$ be the point of intersection of the medians of triangle $ABC, \angle ACB=90^{\circ}, AB=c, \angle BAC=\alpha, 0^{\circ}<\alpha<90^{\circ}$. Since $OB$ and $OA>OC$ $\Rightarrow EA$ is the largest lateral edge of the pyramid and $\angle EAO=\be... | \frac{^{3}}{36}\sin2\alpha\operatorname{tg}\beta\sqrt{1+3\cos^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,295 |
12.385. The side of a regular triangle is equal to $a$. The triangle rotates around a line lying in the plane of the triangle outside it, passing through the vertex of the triangle and forming an angle $\alpha$ with the side. Find the volume of the solid of revolution and determine at what value of $\alpha$ this volume... | Solution.
Points $B_{1}$ and $A_{1}$ are symmetric to points $B$ and $A$ with respect to the line $l$ passing through the vertex $C$ of the equilateral $\triangle ABC (AB = a)$ and serving as the axis of rotation in the problem's condition (Fig. 12.235), $O$ is the midpoint of $BB_{1}$, $O_{1}$ is the midpoint of $AA_... | \frac{\pi^{3}}{2}\cos(60-\alpha),\alpha=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,296 |
12.386. The lateral face of a regular triangular pyramid $S A B C$ forms an angle $\alpha$ with the base plane. A plane is drawn through the side $B C$ of the base and a point $D$ on the lateral edge $A S$. Find the angle between this plane and the base plane, if $A D: D S=k$. | ## Solution.
Let $SO$ be the height of the regular pyramid $SABC$ (Fig. 12.236), $K$ the midpoint of $BC$, $\triangle BCD$ a section of the pyramid, $\angle SKO = \alpha, AB = a$, and the required $\angle DKA = \gamma$.
Then $\angle SKD = \alpha - \gamma, AK = \frac{a \sqrt{3}}{2}, OK = \frac{a \sqrt{3}}{6}$, and fro... | \arctan(\frac{k}{k+3}\tan\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,297 |
12.389. The larger base of an isosceles trapezoid is equal to $a$, and the acute angle is $\alpha$. The diagonal of the trapezoid is perpendicular to its lateral side. The trapezoid rotates around its larger base. Find the volume of the solid of revolution. | Solution.
Hexagon $A B E C D K$ (Fig. 12.238) is the axial section of a body obtained by rotating an isosceles trapezoid $E C D K$ about its larger base, $E K=a$,
Fig. 12.238
$\angle K E... | \frac{\pi^{3}}{3}\sin^{2}2\alpha\sin(\alpha+\frac{\pi}{6})\sin(\alpha-\frac{\pi}{6}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,299 |
12.390. A sphere is inscribed in a spherical sector of radius $R$ (Fig. 12.239). Find the radius of the circle of contact between the surfaces of the sphere and the sector, if the central angle in the axial section of the spherical sector is $\alpha$. | Solution.
Consider the axial section of the given set of bodies.
$$
\begin{array}{ll}
\text { Fig. } 12.239 & \text { In } \triangle O C A\left(\angle O C A=\frac{\pi}{2}\right):
\end{array}
$$
$O A=\frac{r}{\sin \frac{\alpha}{2}}$, where $r$ is the radius of the given sphere.
$$
R=A B=A O+O B=\frac{\sin \frac{u}{2... | \frac{R\sin\alpha}{4\cos^{2}(\frac{\pi}{4}-\frac{\alpha}{4})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,300 |
13.211. The digits of a certain three-digit number form a geometric progression. If in this number the digits of the hundreds and units are swapped, the new three-digit number will be 594 less than the desired one. If, however, in the desired number the digit of the hundreds is erased and the digits of the resulting tw... | ## Solution.
Let the desired number have the form $100 x+10 y+z$, where $x, y, z$ form a geometric progression, which means $x z=y^{2}$. From the condition, we get:
$$
\begin{aligned}
& 100 z+10 y+x=100 x+10 y+z-594 \Rightarrow \\
& \Rightarrow x-z=6 ; 10 y+z=10 z+y+18 \Rightarrow y-z=2
\end{aligned}
$$
We have the ... | 842 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,301 |
13.212. A sports club allocated $n$ rubles for the purchase of bicycles. Due to a price reduction, the cost of each bicycle decreased by $a$ rubles, which allowed the club to buy $b$ more bicycles than initially planned. How many bicycles were purchased? | ## Solution.
Let $x$ be the number of bicycles bought, and $(x-b)$ be the number intended to be bought. The price of one bicycle before the price reduction: $\frac{n}{x-b}$, after the reduction: $\frac{n}{x-b}-a$.
Then $\left(\frac{n}{x-b}-a\right) x=n$, from which $x=\frac{a b+\sqrt{a^{2} b^{2}+4 n b a}}{2 a}=$
$=\... | \frac{+\sqrt{(b+4n)}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,302 |
13.213. Usually, two mechanisms are involved in performing a certain task simultaneously. The productivity of these mechanisms is not the same, and when working together, they complete the task in 30 hours. However, this time the joint operation of the two mechanisms lasted only 6 hours, after which the first mechanism... | ## Solution.
Let the entire volume of work be equal to one unit, the first mechanism completes the task in $x$ hours, and the second in $y$ hours. Then the productivity of the first mechanism is $\frac{1}{x}$, and the second is $-\frac{1}{y}$, the combined productivity when working together is $\frac{1}{x}+\frac{1}{y}... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,303 |
13.215. A motorcyclist is moving away from point $A$ along a highway at a constant speed of $a$ km/h. After 30 minutes, a second motorcyclist starts from the same point at a constant speed of $1.25 a$ km/h. How many minutes after the start of the first motorcyclist was a third motorcyclist dispatched from the same poin... | Solution.
Let $t$ be the time the third racer drove before catching up with the first. This means he drove $(1.5 a \cdot t)$ km. The first racer drove for $t_{1}$ hours.
So $t_{1} \cdot a=t \cdot 1.5 a$.
The second racer drove $\left(t_{1}-\frac{1}{2}\right)$ hours and covered $1.25 a\left(t_{1}-\frac{1}{2}\right)$ ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,305 |
13.216. Two motorcyclists set off simultaneously towards each other from points $A$ and $B$, the distance between which is 600 km. While the first covers 250 km, the second covers 200 km. Find the speeds of the motorcyclists, assuming their movements are uniform, if the first motorcyclist arrives at $B$ 3 hours earlier... | Solution.
Let $x$ km/h be the speed of the first motorcyclist; $y$ km/h be the speed of the second. The first covers 250 km in $\frac{250}{x}$ h; the second covers 200 km in $\frac{200}{y}$ h. According to the condition, $\frac{250}{x}=\frac{200}{y}$.
The first arrives at $B$ in $\frac{600}{x}$ h; the second arrives ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,306 |
13.217. The road between villages $A$ and $B$ first has an ascent, then a descent. A cyclist, moving downhill at a speed $a$ km/h faster than uphill, spends exactly $t$ hours on the journey from $A$ to $B$, and half of this time on the return journey from $B$ to $A$. Find the cyclist's speed uphill and downhill, if the... | Solution.
Let $x$ km/h be the cyclist's speed on the ascent, and $(x+a)$ km/h on the descent.
Let $y$ km be the length of the ascent, and $(b-y)$ km be the descent. Then:
$\left\{\begin{array}{l}\frac{y}{x}+\frac{b-y}{x+a}=t \\ \frac{b-y}{x}+\frac{y}{x+a}=\frac{t}{2}\end{array}\right.$
Solving the system, we find $... | \frac{43+\sqrt{9^{2}^{2}+16b^{2}}}{6}\text{/;4b>3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,307 |
13.219. The population of the city increases annually by $1 / 50$ of the current number of residents. In how many years will the population triple? | Solution.
Using the compound interest formula $x\left(\frac{1}{50}+1\right)^{n}=3 x$, we find $n=\log _{3} 1.02 \approx 55$.
Answer: $\approx 55$ years. | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,308 |
13.221. Two riders set out simultaneously from points $A$ and $C$ to point $B$, and despite the fact that $C$ was 20 km farther from $B$ than $A$ was from $B$, they arrived at $B$ simultaneously. Find the distance from $C$ to $B$, if the rider who set out from $C$ traveled each kilometer 1 minute 15 seconds faster than... | Solution.
Let $AB = x$ km, $BC = x + 20$ km.
The speed of the rider from $A - \frac{x}{5}$ km/h, the speed of the rider from $C - \frac{x + 20}{5}$ km/h.
According to the condition $\frac{5}{x} - \frac{5}{x + 20} = \frac{1}{48}$, from which $x = 60$ km.
Then $CB = x + 20 = 80$ km.
Answer: 80 km. | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,309 |
13.222. The distance between stations $A$ and $B$ is 103 km. A train left $A$ for $B$ and, after traveling a certain distance, was delayed, and therefore the remaining distance to $B$ was traveled at a speed 4 km/h greater than the initial speed. Find the original speed of the train, given that the remaining distance t... | ## Solution.
Let $x$ km/h be the original speed of the train, $t$ h be the time until its stop.
Then $\left\{\begin{array}{l}x \cdot t=40, \\ (x+4)\left(t+\frac{1}{4}\right)=63,\end{array}\right.$ from which $x=80$ km/h.
Answer: $80 \mathrm{km} /$ h. | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,310 |
13.223. Point $C$ is located 12 km downstream from point $B$. A fisherman set out from point $A$, located upstream from point $B$, to point $C$. After 4 hours, he arrived at $C$, and the return trip took 6 hours. On another occasion, the fisherman used a motorboat, thereby tripling his own speed relative to the water, ... | Solution.
Let $v_{T}$ km/h be the speed of the current; $v_{,}$ km/h be the speed of the boat; $a$ km/h be the distance between $A$ and $B$. Then, according to the problem,
$$
\left\{\begin{array}{l}
\left(v_{T}+v_{T}\right) \cdot 4=12+a, \\
\left(v_{L}-v_{T}\right) \cdot 6=12+a, \\
a=\left(3 v_{T}+v_{T}\right) \cdot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,311 |
13.225. A certain order takes 3.6 hours longer to complete in workshop No.1 compared to workshop No.2, and 10 hours longer compared to workshop No.3. If under the same conditions, workshops No.1 and No.2 combine to complete the order, the completion time will be the same as in workshop No.3 alone. By how many hours mor... | ## Solution.
Let $x$ hours be the time to complete the order in Workshop No.3. Then in the 1st workshop, this time is $x+10$ hours, and in the 2nd workshop, it is $-x+6.4$ hours.
$$
\text { According to the condition } \frac{1}{x+10}+\frac{1}{x+6.4}=\frac{1}{x} \text {, from which } x=8 \text { hours. }
$$
This mean... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,313 |
13.226. A manuscript of 80 pages was given to two typists. If the first typist starts typing the manuscript 3 hours after the second, then each of them will type half of the manuscript. If both typists start working simultaneously, then after 5 hours, 15 pages will remain untyped. How long will it take for each typist ... | Solution.
Let the first typist can retype the manuscript in $x$ hours; the second - in $y$ hours. The first typist prints one page in $\frac{x}{80}$ hours, the second - in $\frac{y}{80}$ hours. The work rate of the first: $\frac{80}{x}$ pages/hour, the second: $\frac{80}{y}$ pages/hour.
According to the condition: $\... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,314 |
13.227. Two workers were assigned a task. The second one started working an hour later than the first. After 3 hours from the time the first one started, they had 9/20 of the entire work left to complete. By the end of the work, it turned out that each had completed half of the entire work. How many hours would each, w... | Solution.
Let the 1st worker can complete the entire job in $x$ hours, the second in $y$ hours. The productivity of the 1st: $\frac{1}{x} ; 2$nd: $\frac{1}{y}$.
$$
\text { According to the condition }\left\{\begin{array}{l}
\frac{1}{x} \cdot 3+\frac{1}{y} \cdot 2=1-\frac{9}{20}, \\
\frac{x}{2}-\frac{y}{2}=1
\end{arra... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,315 |
13.228. Two workers were assigned to manufacture a batch of identical parts. After the first worked for 2 hours and the second for 5 hours, it turned out that they had completed half of the entire work. Working together for another 3 hours, they found that they had 0.05 of the entire work left to complete. In what time... | ## Solution.
Let the 1st worker can complete the entire job in $x$ hours, the second - in $y$ hours.
Productivity of the 1st: $\frac{1}{x}$, the second $-\frac{1}{y}$.
According to the condition $\left\{\begin{array}{l}\frac{1}{x} \cdot 2+\frac{1}{y} \cdot 5=\frac{1}{2}, \\ \frac{1}{2}+\left(\frac{1}{x}+\frac{1}{y}\... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,316 |
13.229. Find four numbers forming a proportion, given that the sum of the extreme terms is 14, the sum of the middle terms is 11, and the sum of the squares of these four numbers is 221. | Solution.
Let the proportion be $\frac{a}{b}=\frac{c}{d}$. Given $a+d=14$, $c+b=11, a^{2}+b^{2}+c^{2}+d^{2}=221$.
Solving the system $\left\{\begin{array}{l}a d=b c, \\ a+d=14, \\ b+c=11, \\ a^{2}+b^{2}+c^{2}+d^{2}=221,\end{array}\right.$
we find $a=12 ; b=8 ; c=3 ; d=2$.
Answer: $12,8,3,2$. | 12,8,3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,317 |
13.230. There are three positive two-digit numbers with the following property: each number is equal to the incomplete square of the sum of its digits. It is required to find two of them, knowing that the second number is 50 units greater than the first. | ## Solution.
Since one number is 50 greater than the other, the number of units in both numbers is the same. Let $10 x_{1}+y, 10 x_{2}+y$ be the desired numbers.
By the condition $10 x_{1}+y-\left(10 x_{2}+y\right)=50$, from which $x_{1}-x_{2}=5$ (1), we also have the system
$$
\left\{\begin{array}{l}
10 x_{1}+y=x_{... | 1363 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,318 |
13.231. Two grades of cast iron with different percentages of chromium were alloyed. If one grade is taken in 5 times the amount of the other, the percentage of chromium in the alloy will be twice the percentage of chromium in the smaller of the alloyed parts. If, however, equal amounts of both grades are taken, the al... | ## Solution.
Let $x \%$ be the chromium content in one alloy, and $y \%$ in the second, i.e., $x$ kg of chromium in 1 kg of cast iron of one type, and $y$ kg in 1 kg of the other type.
According to the problem, we have $\left\{\begin{array}{l}5 x+y=6 \cdot 2 y, \\ x+y=2 \cdot 0.08,\end{array}\right.$
from which $x=0... | 511 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,319 |
13.232. From the railway station to the beach is 4.5 km. A boy and a scheduled bus left the station for the beach simultaneously. After 15 minutes, the boy met the bus returning from the beach, and he managed to walk another $9 / 28$ km from the place of the first meeting with the bus, when he was caught up by the same... | ## Solution.
Let $v_{\text {m }}, v_{\text{a}}$ be the speeds of the boy and the bus, respectively.
According to the problem, we have the system
$\left\{\begin{array}{l}\frac{15}{60} v_{M}+\frac{15-4}{60} v_{\text{a}}=2 \cdot 4.5, \\ \frac{9}{28 v_{M}}=\frac{4}{60}+\frac{2 \frac{15}{60} v_{M}+\frac{9}{28}}{v_{\text{a... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,320 |
13.233. A tourist was returning from vacation on a bicycle. On the first leg of the journey, which was 246 km, he traveled on average 15 km less per day than he did on the last leg of the journey, which was 276 km. He arrived home right on time at the end of his last vacation day. It is also known that it took him one ... | ## Solution.
Let the tourist travel 267 km in $x$ days, and 246 km in $\frac{x}{2}+1$. On the first segment of 246 km, his speed was $v$ km/day, and on the second segment of 276 km, it was $(v+15)$ km/day.
Then $\left\{\begin{array}{l}\left(\frac{x}{2}+1\right) v=246, \\ x(v+15)=276,\end{array}\right.$
from which $x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,321 |
13.234. There were two copper alloys with different percentages of copper in each. The number expressing the percentage of copper in the first alloy is 40 less than the number expressing the percentage of copper in the second alloy. Then, both these alloys were melted together, after which the copper content was $36 \%... | ## Solution.
Let in the first alloy 6 kg of copper constitute $x \%$, in the second 12 kg $-y \%$; $m_{1}$ - the mass of the first alloy, $m_{2}$ - the mass of the second. According to the condition
$$
\left\{\begin{array}{l}
0.01 x \cdot m_{1}=6 \\
0.01 y \cdot m_{2}=12 \\
0.36\left(m_{1}+m_{2}\right)=18 \\
x+40=y
\... | 2060 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,322 |
13.235. Two cars and a motorcycle participated in a race over the same distance. The second car took 1 minute longer to complete the entire distance than the first car. The first car moved 4 times faster than the motorcycle. What part of the distance did the second car cover in one minute, if it covered $1 / 6$ of the ... | Solution.
Let $l$ be the length of the distance, $v_{1}, v_{2}, v_{x}$ be the speeds of the first, second car, and the motorcycle, respectively. According to the problem, we have the system
$\left\{\begin{array}{l}\frac{l}{v_{2}}-\frac{l}{v_{1}}=\frac{1}{60}, \\ v_{1}=4 v_{\mathrm{M}}, \\ \frac{v_{2}}{60}-\frac{v_{\m... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,323 |
13.236. The master is giving a simultaneous chess exhibition on several boards. By the end of the first two hours, he won $10 \%$ of the total number of games played, while 8 opponents drew their games with the master. Over the next two hours, the master won $10 \%$ of the games with the remaining opponents, lost two g... | ## Solution.
Let the game be played on $x$ boards. When in the first 2 hours $(0.1 x+8)$ games were played, in the next 2 hours: $0.1(x-0.1 x-8)+7+2$.
Then
$0.1 x+8+0.1(x-0.1 x-8)+7+2=x$,
from which $x=20$.
Answer: on 20 boards. | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,324 |
13.237. A positive integer was thought of. To its digital representation, a digit was appended on the right. From the resulting new number, the square of the thought number was subtracted. The difference turned out to be 8 times the thought number. What number was thought of and which digit was appended? | ## Solution.
If the number $x$ was thought of, then by appending the digit $y$ to the right, we get the number $10 x + y$.
According to the condition $10 x + y - x^2 = 8 x \Leftrightarrow x^2 - 2 x - y = 0, \quad x = 1 \pm \sqrt{1 + y}$. Then the possible values of $y$ are: $0, 3, 8$. From this, the thought-of number... | 2,3,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,325 |
13.238. The shooter at the shooting range was offered the following conditions: each hit on the target is rewarded with five tokens, but for each miss, three tokens are taken away. The shooter was not very accurate. After the last ($n$-th) shot, he had no tokens left. How many shots were in the series and how many were... | ## Solution.
Let there be $k$ hits, and thus $n-k$ misses.
Then $5 k-3(n-k)=0,8 k=3 n \Rightarrow k=\frac{3 n}{8}$. Since $n$ and $k$ are natural numbers, $n=8 m$. For $10<n<20$, the only suitable value is $n=16$. Therefore, 16 shots were made, 6 of which were successful.
Answer: out of 16 shots, 6 were successful. | outof16shots,6weresuccessful | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,326 |
13.239. To the digital record of a certain two-digit number, the same number was appended to the right, and from the resulting number, the square of the intended number was subtracted. The difference was divided by $4 \%$ of the square of the intended number; in the quotient, half of the intended number was obtained, a... | Solution.
Let $a$ be the number thought of. Then the new number is $100a + a = 101a$.
According to the condition, $101a - a^2 = 0.04a^2 \cdot \frac{1}{2}a + a$, from which $a = 50$.
Answer: 50.
. The area of the annulus is equal to the sum of the areas of all seven disks. Prove that the width of the annulus is equal to the radius of one disk. | ## Solution.
Let $r$ be the radius of the disk; $R_{1}$ be the radius of the outer circle; $R_{2}$ be the radius of the inner circle. The area of the disks is $-7 \cdot \pi r^{2}$; the area of the ring is $-\pi\left(R_{1}^{2}-R_{2}^{2}\right)$.
By the condition, $7 \cdot \pi r^{2}=\pi\left(R_{1}^{2}-R_{2}^{2}\right)(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 49,328 |
13.241. A positive single-digit number was appended to the right of the digital record of a certain thought positive number, and from the new number thus obtained, the square of the thought number was subtracted. This difference turned out to be greater than the thought number by as many times as the complement of the ... | ## Solution.
Let $a$ be the thought number, $x$ be the appended number; $10 a+x$ be the new number.
According to the condition, $10 a+x-a^{2}=(11-x) a$, from which $a=x$.
Which was to be proved. | x | Algebra | proof | Yes | Yes | olympiads | false | 49,329 |
13.242. Two identical pools started filling with water simultaneously. In the first pool, 30 m$^3$ more water flows in per hour than in the second pool. At some point, the total amount of water in both pools was equal to the volume of each of them. After this, the first pool was filled after 2 hours and 40 minutes, and... | Solution.
Let $x+30 \mathrm{~m}^{3} / \mathrm{h}$ be the filling rate of the first pool, $x \mathrm{~m}^{3} / \mathrm{h}$ the second; $V \mathrm{~m}^{3}$ the volume of the pools; $t$ the time when the pools contained $V \mathrm{~m}^{3}$.
According to the problem, $\left\{\begin{array}{l}(x+30) t+x t=V, \\ \left(t+2 \... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,330 |
13.243. One of the three barrels is filled with water, while the others are empty. If the second barrel is filled with water from the first barrel, then $1 / 4$ of the water that was in the first barrel will remain. If the third barrel is then filled from the second, then $2 / 9$ of the amount of water that was in the ... | Solution.
Let $x$ barrels be the capacity of the first barrel, then the capacity of the second is $-\frac{3}{4} x$, and the third is $-\frac{7}{9}\left(\frac{3}{4} x\right)$.
According to the condition $\frac{7}{9} \cdot \frac{3}{4} x + 50 = x$,
from which $x = 120$ barrels,
$$
\frac{3}{4} x = 90 \text{ barrels}
$$... | 120,90,70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,331 |
13.244. Two balls are placed in a cylindrical jar with a diameter of 22 cm (Fig. 13.4). If 5 liters of water are poured into the jar, will both balls, with diameters of 10 and 14 cm, be completely covered by the water? | ## Solution.
Consider the plane $\beta$, which is perpendicular to the base of the cylinder and contains the line on which the centers of the spheres $O_{1}$ and $O_{2}$ lie. $\Delta O_{1} B O_{2}$ is a right triangle, $O_{1} O_{2}=5+7=12 \mathrm{~cm}$,
$ tons of whey with a fat content of $0.5\%$ will remain. Therefore, in 1 ton of milk, there is
$$
\frac{15.5 x}{100}+\frac{0.5(1-x)}{100}=\frac{15 x+0.5}{100} \text { tons of fat. }
$$
According to the condition $\frac{1... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,335 |
13.248. There are two identical pieces of fabric. The cost of the entire first piece is 126 rubles more than the cost of the second. The cost of four meters of fabric from the first piece exceeds the cost of three meters of fabric from the second piece by 135 rubles. A customer bought 3 meters of fabric from the first ... | Solution.
Let $x$ rubles be the cost of the second piece, $x+126$ - the cost of the first, $y$ m - the length of one piece.
Then the cost of 1 m of the first fabric is $\frac{x+126}{y}$ rubles, 1 m of the second is $-\frac{x}{y}$ rubles.
By the condition $\left\{\begin{array}{l}4 \frac{x+126}{y}-3 \frac{x}{y}=135, \... | 5.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,336 |
13.249. It was planned to divide the bonus equally among the most distinguished employees of the enterprise. However, it turned out that there were three more employees deserving the bonus than was initially expected. In this case, each would receive 400 rubles less. The union and administration found a way to increase... | Solution.
Let the number of people who were supposed to receive the award be $x$ and $y$ be the original total amount of the award.
According to the problem,
$$
\left\{\begin{array}{l}
(x+3)\left(\frac{y}{x}-400\right)=y \\
y+9000=2500(x+3)
\end{array}\right.
$$
from which $x=15$ people.
Therefore, the number of p... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,337 |
13.250. A brigade of lumberjacks was supposed to prepare $216 \mathrm{~m}^{3}$ of wood over several days according to the plan. For the first three days, the brigade met the daily planned quota, and then each day they prepared 8 m $^{3}$ more than planned, so by the day before the deadline, they had prepared 232 m $^{3... | Solution.
Let the team be supposed to prepare $x$ m ${ }^{3}$ of wood per day and should have worked $\frac{216}{x}$ days. In the first 3 days, they prepared $3 x \mathrm{~m}^{3}$, and for the remaining days, they prepared $(x+8)\left(\frac{216}{x}-4\right)$.
According to the condition, $3 x+(x+8)\left(\frac{216}{x}-... | 24\mathrm{~}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,338 |
13.251. The hour and minute hands coincide at midnight, and a new day begins. At what hour of this new day will the hour and minute hands coincide again for the first time, assuming that the clock hands move without jumps? | ## Solution.
Let's assume the speed of the minute hand is 1, then the speed of the hour hand is $-\frac{1}{12} \cdot$ The hands will coincide when $\frac{1}{12} \cdot t + 60 = t$, where $t$ is the time of movement of the hands.
Solving the equation, we find $t = \frac{720}{11}$ (min).
Answer: 1 hr $5 \frac{5}{11}$ m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,339 |
13.252. The duty maintenance worker descended on a downward-moving metro escalator. His entire journey from the upper platform to the lower one lasted $24 \mathrm{s}$. Then he climbed up and at the same pace descended again, but this time on a stationary escalator. It is known that the descent lasted 42 s. How many sec... | ## Solution.
Let $l$ be the path along the stationary escalator, $v_{\text {esc. }}$ be the speed of the escalator, and $v_{\text {tech. }}$ be the speed of the technician along the stationary escalator.
Then $\frac{l}{v_{\text {tech. }}}=42, \frac{l}{v_{\text {tech. }}+v_{\text {esc. }}}=24$. We need to find $\frac{... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,340 |
13.253. For hydrodynamic studies, a small model of a channel has been made. Several pipes of the same cross-section, supplying water, and several pipes of another, but also the same cross-section, intended for removing water, are connected to this model. If four inlet and three outlet pipes are opened simultaneously, t... | ## Solution.
Let $x$ m $^{3} /$ h be the flow rate of the inlet pipe, and $y$ m $^{3} /$ h be the flow rate of the outlet pipe. According to the problem,
$\left\{\begin{array}{l}5(4 x-3 y)=1000, \\ 2(2 x-2 y)=180,\end{array}\right.$
from which $x=65 \mathrm{~m}^{3}, y=20 \mathrm{~m}^{3}$.
Answer: 65 and $20 \mathrm{... | 65 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,341 |
13.255. Three swimmers have to swim a lane 50 m long, immediately turn around, and return to the starting point. The first starts first, the second starts 5 s later, and the third starts another 5 s later. At some point in time, before reaching the end of the lane, the swimmers found themselves at the same distance fro... | ## Solution.
Let $t$ be the time it takes for the third swimmer to catch up with the first two; $v_{1}$, $v_{2}, v_{3}$ be the speeds of the swimmers. Then $v_{1}(t+10)=v_{2}(t+5)=v_{3} t$, from which $\frac{v_{2}}{v_{3}-v_{2}}=\frac{2 v_{1}}{v_{3}-v_{1}}(1)$
According to the problem, $\left\{\begin{array}{l}\frac{50... | \frac{22}{15}\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,342 |
13.257. A courier on a moped left city $A$ for city $B$, which are 120 km apart. One hour after this, a second courier on a motorcycle left $A$ and, catching up with the first, immediately handed him an errand and then turned back, returning to $A$ at the moment the first courier reached $B$. What is the speed of the f... | Solution.
Let $v$ km/h be the speed of the first courier; he traveled for $t$ hours to the meeting point. According to the problem, $v t=50(t-1)$. The first courier arrived in $B$ after $\frac{120}{v}$ hours, and the second courier returned to $A$ after $\left(\frac{2 v t}{50}+1\right)$ hours.
$$
\text { According to... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,344 |
13.258. A train is traveling from station $A$ to station $B$. On a certain section of the route near station $B$, repair work was being carried out, and on this section, the train was allowed to travel at a speed that was only $1 / n$ of its original speed, as a result of which the train arrived at station $B$ with a d... | ## Solution.
Let $x$ km/h be the speed of the train, $t$ h be the time the train takes to travel from $A$ to $B$ without stopping, and $S$ km be the distance the train travels after stopping.
According to the problem,
\[
\left\{
\begin{array}{l}
\frac{x t - S}{x} + \frac{S}{\frac{x}{n}} = t + a, \\
\frac{x t - (S - ... | \frac{b(n-1)}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,345 |
13.259. A steamship, 2 hours after departure from pier $A$, stops for 1 hour and then continues its journey at a speed equal to 0.8 of its initial speed, as a result of which it is late to pier $B$ by 3.5 hours. If the stop had occurred 180 km further, then under the same conditions, the steamship would have been late ... | Solution.
Let $AB = x$ km, $v$ km/h be the speed of the steamboat.
According to the problem, $\left\{\begin{array}{l}\frac{x}{v} + 3.5 = 2 + 1 + \frac{x - 2v}{0.8v}, \\ \frac{x}{v} + 1.5 = \frac{2v + 180}{v} + 1 + \frac{x - 2v - 180}{0.8v},\end{array}\right.$ from which $x = 270$ km.
Answer: 270 km. | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,346 |
13.260. Two material particles, being 295 m apart from each other, started moving towards each other simultaneously. The first particle moves uniformly at a speed of $15 \mathrm{~m} / \mathrm{c}$, while the second particle moved 1 m in the first second and 3 m more in each subsequent second than in the previous one. Th... | ## Solution.
Let $t$ seconds have passed before the particles meet. In this time, the first particle has traveled $15 t \text{m}$, and the second particle has traveled $\frac{2+3(t-1)}{2} \cdot t$ (the formula for the sum of the first $n$ terms of an arithmetic progression was used).
According to the condition, $\fra... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,347 |
13.261. At noon, a pedestrian and a cyclist set out from point $A$ to point $B$, and at the same time, a horseback rider set out from $B$ to $A$. All three started their journey simultaneously. After 2 hours, the cyclist and the horseback rider met 3 km from the midpoint of $A B$, and another 48 minutes later, the pede... | Solution.
Let $v_{1}$ km/h be the pedestrian's speed, $2 v_{1}$ km/h be the cyclist's speed, $v_{3}$ km/h be the horse rider's speed, and $x$ km be the distance $A B$.
According to the problem, $\left\{\begin{array}{l}2 v_{1} \cdot 2+2 v_{3}=x, \\ 2.8 v_{1}+2.8 v_{3}=x, \\ 2 v_{3}=\frac{x}{2}-3, \\ 2 v_{1} \cdot 2=\f... | 6,9,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,348 |
13.262. It is known that a freely falling body travels 4.9 m in the first second, and in each subsequent second, it travels 9.8 m more than in the previous one. If two bodies start falling from the same height, one 5 seconds after the other, then after what time will they be 220.5 m apart? | ## Solution.
Let $t \mathrm{c}$ be the required time. In this time, the first body will travel $4.9 t^{2} \mathrm{M}$, and the second body will travel $4.9(t-5)^{2}$ m.
According to the condition, $4.9 t^{2}-4.9(t-5)^{2}=220.5$, from which $t=7 \mathrm{c}$.
Answer: 7 seconds after the first body starts falling.
![]... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,349 |
13.263. The path from $A$ to $B$ is traveled by a passenger train 3 hours and 12 minutes faster than by a freight train. In the time it takes the freight train to travel from $A$ to $B$, the passenger train travels 288 km more. If the speed of each train is increased by $10 \mathrm{~km} / \mathrm{h}$, the passenger tra... | Solution.
In Fig. 13.7, the graphs of train movements before the speed changes are considered.
We have $\operatorname{tg} \angle M C N=v_{\text {pas. }}^{(0)}=288: 3.2=90$ km/h. If $v_{\text {tov. }}^{(0)}=x$ km/h, then $A B=N L=x t \mathrm{km}, C D=90(t-3.2)=x t \Rightarrow t=\frac{288}{90-x}$. After the speed chang... | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,350 |
13.264. To rise in an ordinary elevator to the top floor of an eight-story building (height 33 m) with two 6-second intermediate stops, it takes as much time as it would to rise in an elevator of a high-rise building with one 7-second intermediate stop at the 20th floor (height 81 m). Determine the lifting speed of the... | ## Solution.
Let $v_{1}$ m/s be the speed of a regular elevator ($v_{1} \leq 3.5 \text{ m/s}$), and $v_{2}$ m/s be the speed of the elevator in a high-rise building, $v_{2}=v_{1}+1.5 \text{ m/s}$.
According to the condition,
$$
\frac{33}{v_{1}}+2 \cdot 6=\frac{81}{v_{1}+1.5}+7 \text{, from which } v_{1}=1.5 \text{ m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,351 |
13.265. A material point moves straight-line within the interior of a $60^{\circ}$ angle. After leaving the vertex of this angle, it found itself at a distance $a$ from one side of the angle and at a distance $b$ from the other side after some time. Then it changed its direction of motion and fell onto the side to whic... | ## Solution.
The length of the path traveled by the point is equal to the sum of the lengths of segments $O A$ and $A B=a$ (Fig. 13.8). We obtain
$$
\begin{aligned}
& O A=\frac{a}{\sin \alpha} \text { or } O A=\frac{b}{\sin \left(60^{\circ}-\alpha\right)} \Rightarrow \frac{\sin \left(60^{\circ}-\alpha\right)}{\sin \a... | +2\sqrt{\frac{^{2}++b^{2}}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,352 |
13.266. Two athletes start running simultaneously - the first from $A$ to $B$, the second from $B$ to $A$. They run at different but constant speeds and meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the ... | Solution.
Consider the graphs of the runs of two athletes (Fig. 13.9). Let $AB = x$ m, $C$ and $D$ be the points of the first and second meetings, $v_{1}$ and $v_{2}$ be the speeds of the first and second athlete. Then $\frac{300}{v_{1}}=\frac{x-300}{v_{2}}\left({ }^{*}\right)$. During the time between the meetings $(... | 500\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,353 |
13.267. Two motorcyclists started racing in the same direction from the same starting point simultaneously: one at a speed of 80 km/h, the other at a speed of 60 km/h. Half an hour later, a third racer started from the same starting point in the same direction. Find his speed, given that he caught up with the first rac... | Solution.
Let $x$ km/h be the speed of the third racer. He caught up with the first racer after $t_{1}$ hours, and the second racer after $t_{2}$ hours. According to the problem,
$$
\begin{aligned}
& \left\{\begin{array}{l}
80 t_{1}=x\left(t_{1}-0.5\right), \\
60 t_{2}=x\left(t_{2}-0.5\right) \\
t_{1}-t_{2}=1.25,
\en... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,354 |
13.269. At the pier, two passengers disembarked from a steamship and headed to the same village. One of them walked the first half of the way at a speed of 5 km/h and the second half at a speed of 4 km/h. The other walked the first half of the time at a speed of 5 km/h and the second half at a speed of 4 km/h, arriving... | Solution.
Let $x$ km be the distance between the pier and the village, $t$ h be the time it took the second passenger to walk to the village.
According to the problem, $\left\{\begin{array}{l}\frac{x}{2 \cdot 5}+\frac{x}{2 \cdot 4}=t+\frac{1}{60}, \\ 5 \cdot \frac{t}{2}+4 \cdot \frac{t}{2}=x,\end{array}\right.$
from... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,356 |
13.270. Two cruise ships are due to arrive in Odessa with a 1-hour gap. Both ships are traveling at the same speed, but circumstances have led to the first ship being delayed by \(t_{1}\) minutes, and the second by \(t_{2}\) minutes. Upon receiving a radio instruction to arrive on time, both captains simultaneously inc... | Solution.
Graphical representation of the conditions of the problem in Fig. 13.11.
We obtain $S_{1}=\left(v+v_{1}\right) T, S_{2}=\left(v_{1}+v_{2}\right)(T+1)=v\left(T+1+\frac{t_{2}}{60}\right)$ Simplifying and eliminating $T$, we have $v=\frac{60 v_{1} v_{2}}{v_{1} t_{2}-v_{2} t_{1}}$ km/h.
Answer: $\frac{60 v_{1}... | \frac{60v_{1}v_{2}}{v_{1}t_{2}-v_{2}t_{1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,357 |
13.271. The race track for the bicycle competition is a contour of a right-angled triangle with a difference in the lengths of the legs of 2 km. The hypotenuse runs along a dirt road, while both legs are on the highway. One of the participants covered the segment along the dirt road at a speed of $30 \mathrm{km} / \mat... | ## Solution.
Length of the route
$S=AB+BC+AC=a+a+2+\sqrt{a^{2}+(a+2)^{2}}$ (see Fig. 13.12).
By the condition $\frac{\sqrt{a^{2}+(a+2)^{2}}}{30}=\frac{a+a+2}{42}$, from which $a=8 \text{ km}$.
$S=8+8+2+\sqrt{8^{2}+10^{2}}=24$ km.
Answer: 24 km. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,358 |
13.272. From post office $A$, a car departed in the direction of post office $B$. After 20 minutes, a motorcyclist set off after it at a speed of 60 km/h. Catching up with the car, the motorcyclist handed a package to the driver and immediately turned back. The car arrived at $B$ at the moment when the motorcyclist was... | Solution.
Let $t$ be the time it takes for the motorcyclist to catch up with the car, and $v$ be the speed of the car. According to the problem, we have the system
$\left\{\begin{array}{l}v\left(t+\frac{20}{60}\right)=60 \cdot t, \\ \frac{82.5-60 \cdot t}{v}=\frac{t}{2},\end{array}\right.$
from which $v=45(\mathrm{k... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,359 |
13.273. The ball is rolling perpendicular to the sideline of a football field. Suppose, moving uniformly decelerated, the ball rolled 4 m in the first second, and 0.75 m less in the next second. A football player, initially 10 m away from the ball, started running in the direction of the ball's movement to catch it. Mo... | Solution.
Let $t$ be the desired time. In this time, the ball will roll $\frac{2 \cdot 4-0.75(t-1)}{2} \cdot t \mathrm{M}$, and the footballer will run $\frac{2 \cdot 3.5+0.5(t-1)}{2} \cdot t \mathrm{M}$. (The formula for the sum of $n$ terms of an arithmetic progression was used).
According to the condition,
$$
\fr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,360 |
13.274. According to the schedule, the train covers a 120 km section at a constant speed. Yesterday, the train covered half of the section at this speed and was forced to stop for 5 minutes. To arrive on time at the final point of the section, the driver had to increase the train's speed by 10 km/h on the second half o... | ## Solution.
Let $x$ km/h be the original speed of the train. Considering the stop, it will take $\frac{1}{12}+\frac{60}{x+10}$ to travel the second half of the section, from which $x=80$ km/h.
Let $v$ km/h be the speed of the train today. Then $\frac{60}{80}=0.15+\frac{60}{v}$, from which $v=100$ km/h.
Answer: 100 ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,361 |
13.278. A passenger on a train knows that the speed of this train on this section of the track is 40 km/h. As soon as a passing train started to go by the window, the passenger started a stopwatch and noticed that the passing train took $3 \mathrm{s}$ to pass by the window. Determine the speed of the passing train, giv... | Solution.
Let $x$ be the speed of the oncoming train.
According to the condition, $(40+x) \cdot \frac{3}{3600}=75 \cdot 10^{-3}$, from which $x=50($ km $/ h)$.
Answer: $50 \mathrm{km} /$ h. | 50\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,364 |
13.279. Two checkpoints divide a ski trail into three segments of equal length. It is known that the skier covered the path consisting of the first and second segments together at an average speed of $a$ m/min; the path consisting of the second and third segments together, he covered at an average speed of $b$ m/min. T... | Solution.
Let the length of one section of the track be 1. Let $t_{1}, t_{2}, t_{3}$ be the time it takes for the skier to pass each section of the track. According to the problem,
$$
\left\{\begin{array}{l}
\frac{2}{t_{1}+t_{2}}=a, \\
\frac{2}{t_{2}+t_{3}}=b, \\
\frac{1}{t_{2}}=\frac{2}{t_{1}+t_{3}},
\end{array} \te... | \frac{2}{+b};\frac{2}{3b-};\frac{2}{+b};\frac{2}{3-b}/,where\frac{b}{3}<<3b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,365 |
13.280. To control the skier's movement, the coach divided the track into three equal-length sections. It became known that the average speeds of the skier on these three separate sections were different. On the first and second sections combined, the skier took 40.5 minutes, and on the second and third sections - 37.5... | ## Solution.
Let $x$ be the length of the segment, $t_{1}, t_{2}, t_{3}$ be the time taken to run the first, second, and third segments, respectively. According to the problem, we have the system
$$
\left\{\begin{array}{l}
t_{1}+t_{2}=40.5 \\
t_{2}+t_{3}=37.5, \text { from which } t_{1}+t_{2}+t_{3}=58.5 \text { (min)... | 58.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,366 |
13.282. A communications officer was assigned to arrive at point $B$ from point $A$ by a designated time. The distance between $A$ and $B$ is $s$ km. When the communications officer reached point $C$, located exactly halfway between $A$ and $B$, he calculated that he would be 2 hours late if he continued at the same sp... | Solution.
Let $x$ be the assigned time, $V^{\prime}$ be the speed of the messenger. According to the problem, we have the system
$$
\begin{aligned}
& \left\{\begin{array}{l}
\frac{S}{V^{\prime}}=x+2, \\
\frac{S}{2 V^{\prime}}+1+\frac{S}{2\left(V^{\prime}+v\right)}=x
\end{array}, \text { from which } x=\frac{v+\sqrt{9... | \frac{v+\sqrt{9v^{2}+6vS}}{v} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,368 |
13.283. Two pedestrians set out towards each other simultaneously from two points, the distance between which is 28 km. If the first pedestrian had not stopped for 1 hour at a distance of 9 km from the starting point, the meeting of the pedestrians would have taken place halfway. After the stop, the first pedestrian in... | Solution.
Since pedestrians could meet halfway, their speeds are equal. Let $V$ be the initial speed of one of the pedestrians.
According to the condition $\frac{9}{V}+\frac{4}{V+1}+1=\frac{28-(9+4)}{V}$, from which $V=3$ (km/h).
Answer: initially, both were walking at the same speed of 3 km/h. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,369 |
13.284. Find the speed and length of the train, knowing that it passed a stationary observer at a constant speed for 7 s and spent 25 s to pass a platform of the same speed with a length of 378 m. | Solution.
Let $l$ be the length of the train, $V$ its speed.
According to the problem, we have the system
$$
\left\{\begin{array}{l}
7 V=l, \\
25 V=378+l,
\end{array} \text { from which } l=147(\mathrm{m}), V=21(\mathrm{m} / \mathrm{s})=75.6(\text { km } / \mathrm{h}) .\right.
$$
Answer: 75.6 km $/ \mathbf{h} ; 147... | 75.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,370 |
13.285. On a 10 km stretch of highway, devoid of intersections, the bus stops only for passengers to get on and off. It makes a total of 6 intermediate stops, spending 1 minute at each, and always moves at the same speed. If the bus were to travel without stopping, it would cover the same distance at a speed exceeding ... | Solution.
Let $t$ min be the time the bus is in motion on the segment. Then its average speed without stops is $-\frac{10}{t}$, and with stops, it is $-\frac{10}{t+\frac{1}{60} \cdot 6}$.
According to the condition, $\frac{10}{t+0.1}+5=\frac{10}{t}$, from which $t=24$ (min).
Answer: 24 min. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,371 |
13.286. The schooner is sailing from $A$ to $B$ across the lake, and from $B$ to $C$ upstream, and then it returns. The speed of the schooner relative to still water is maintained at $c$ km/h throughout. The schooner takes $\alpha$ hours to travel from $A$ to $C$, and the return trip takes $\beta$ hours, with the journ... | Solution.
Let $x, y$ be the distances $AB$ and $BC$ respectively, and $V$ be the speed of the river current. According to the problem, we have
$$
\frac{x}{c}+\frac{y}{c-V}=\alpha, \text { (1) } \frac{y}{c+V}+\frac{x}{c}=\beta, \text { (2) } \frac{y}{c+V}=\frac{1}{3} \cdot \frac{x}{c} \text { (3). }
$$
From equations... | AB=\frac{3\beta}{4};BC=\frac{\beta(4\alpha-3\beta)}{4(2\alpha-\beta)}(\alpha>\beta) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,372 |
13.287. Two dairy factories together should process an equal amount of a certain number of liters of milk. The second workshop started working on the task $a$ working days later, but processed $m$ liters of milk more daily than the first. After another $5a/9$ working days from the start of the joint work of these works... | Solution.
Let's take the entire volume of work as 1. Let $x$ be the labor productivity of the first workshop per 1 hour, and $n$ be the number of days required to complete the task. According to the conditions (1) $n x=(n-a)(x+m)$, (2) $a x+\frac{5 a}{9} x+\frac{5 a}{9}(x+m)=1-\frac{1}{3}$ (hours). From (1), $m=\frac{... | 2a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,373 |
13.288. A master and his apprentice were tasked with manufacturing a batch of identical parts. After the master worked for 7 hours and the apprentice for 4 hours, it turned out that they had completed $5 / 9$ of the entire work. After working together for another 4 hours, they found that $1 / 18$ of the entire work rem... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the labor productivity of the master and the apprentice per hour, respectively. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
7 x+4 y=\frac{5}{9} \\
7 x+4 y+4(x+y)=1-\frac{1}{18}
\end{array}\right.
$$
The time required fo... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,374 |
13.289. There are two alloys consisting of zinc, copper, and tin. It is known that the first alloy contains $40 \%$ tin, and the second - $26 \%$ copper. The percentage of zinc in the first and second alloys is the same. By melting 150 kg of the first alloy and 250 kg of the second, a new alloy was obtained, in which t... | ## Solution.
Let $x$ be the percentage of zinc in the first alloy. According to the condition, $\frac{x}{100} \cdot 150+\frac{x}{100} \cdot 250=\frac{30}{100}(150+250)$, from which $x=30 \%$. Then in the second alloy, there is $100 \%-(26 \%+30 \%)=44 \%$ of tin, and the new
$$
\text { alloy contains } \frac{40}{100}... | 170 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,375 |
13.290. If both pipes are opened simultaneously, the pool will be filled in 2 hours and 24 minutes. In reality, however, only the first pipe was opened initially for $1 / 4$ of the time it takes the second pipe to fill the pool on its own. Then, the second pipe was opened for $1 / 4$ of the time it takes the first pipe... | Solution.
Let's take the entire volume of the pool as 1. Let $x, y$ (liters) per hour leak through the first and second pipe, respectively. According to the conditions, we have the system
... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,376 |
13.291. If the execution of an order for a set of several books is entrusted to one of three typesetters, then the first will complete the work 10 hours faster, and the third will complete it 6 hours faster than the second. If one of the ordered books is set by the first typesetter, and another book is set simultaneous... | ## Solution.
Let's assume the entire volume of work is 1. Let $\alpha, \beta, \gamma$ be the work rates (number of pages typed per hour) of the first, second, and third typists, respectively.
According to the problem, we have the system
$$
\left\{\begin{array}{l}
\frac{1}{\beta}-\frac{1}{\alpha}=10, \\
\frac{1}{\bet... | 20,30,24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,377 |
13.292. Two "mechanical moles" of different power, working simultaneously from different ends of the tunnel, could dig it in 5 days. In reality, however, both "moles" were used sequentially from one side of the tunnel, with the first digging $1 / 3$, and the second - the remaining $2 / 3$ of its length. The entire job ... | ## Solution.
Let the length of the tunnel be 1. Let $x, y$ be the powers of the first and second "moles" respectively.
According to the problem, we have the system $\left\{\begin{array}{l}5(x+y)=1, \\ \frac{1}{3 x}+\frac{2}{3 y}=10\end{array}\right.$, from which $x=\frac{1}{15}, y=\frac{2}{15}$.
Then the first "mole... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,378 |
13.293. Two pipes of different cross-sections are connected to a swimming pool. One is a uniform water supply pipe, and the other is a uniform water discharge pipe, with the first pipe filling the pool 2 hours longer than the second pipe empties it. When the pool was filled to $1 / 3$ of its capacity, both pipes were o... | ## Solution.
Let's write down the values of the sought and given quantities in the form of a table:
| Pipe | Time, h | Capacity | Productivity |
| :---: | :---: | :---: | :---: |
| Supply | $x+2$ | 1 | $\frac{1}{x+2}$ |
| Discharge | $x$ | 1 | $\frac{1}{x}$ |
| Both together | 8 | $\frac{1}{3}$ | $\frac{1}{24}$ |
Ac... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,379 |
13.294. Two workers were assigned to manufacture a batch of identical parts; after the first worked for $a$ hours, and the second for $0.6 a$ hours, it turned out that they had completed $5 / n$ of the entire work. After working together for another $0.6 a$ hours, they found that they still had to manufacture $1 / n$ o... | ## Solution.
Let the entire volume of work be 1. Let $x, y$ be the labor productivity (the number of parts produced per hour) of the first and second workers, respectively. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
a x+0.6 a y=\frac{5}{n} \\
a x+0.6 a y+0.6 a(x+y)=1-\frac{1}{n}
\end{a... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,380 |
13.295. A reservoir is supplied by two channels. Water flows out evenly through the first channel, and flows in evenly through the second channel. How many hours will it take for $n$ liters of water to pass through the first channel, if it is known that through the second channel, twice as much water will flow in when ... | Solution.
The values of the sought and given quantities are written in the form of a table:
| Channel | Time, h | Volume of water, l | Productivity |
| :---: | :---: | :---: | :---: |
| First | $x$ | $n$ | $\frac{n}{x}$ |
| Second | $x-a$ | $2 n$ | $\frac{2 n}{x-a}$ |
According to the condition $\frac{2 n}{x-a}-\fra... | \frac{^{2}+n+\sqrt{^{4}+6^{2}n+n^{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,381 |
13.296. Two excavator operators must complete a certain job. After the first one worked for 15 hours, the second one starts and finishes the job in 10 hours. If, working separately, the first one completed $1 / 6$ of the job, and the second one completed $1 / 4$ of the job, it would take an additional 7 hours of their ... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the work efficiency of the first and second excavator operators, respectively. According to the conditions, we have the system $\left\{\begin{array}{l}15 x+10 y=1, \\ 7(x+y)=1-\left(\frac{1}{6}+\frac{1}{4}\right),\end{array}\right.$ from which $x=\frac... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,382 |
13.297. The length of the circular track of the racetrack is $b$ km. Two jockeys, $A$ and $B$, started the race at the same time, and jockey $A$ arrived at the finish line 2 minutes earlier. In another race, jockey $B$ increased his speed by $c$ km/h, while jockey $A$ decreased his speed by $c$ km/h, and as a result, $... | Solution.
Let $x, y$ be the speeds of riders $A$ and $B$ in the first race. According to the problem, we have the system $\left\{\begin{array}{l}\frac{b}{y}-\frac{b}{x}=\frac{2}{60}, \\ \frac{b}{x-c}-\frac{b}{y+c}=\frac{2}{60},\end{array}\right.$ from which $x=\frac{c+\sqrt{c^{2}+120 b c}}{2}$ (km/h), $y=\frac{-c+\sqr... | \frac{+\sqrt{^{2}+120}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,383 |
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