problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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12.046. The base of the pyramid is an equilateral triangle. One lateral edge is perpendicular to the base plane and equals $l$, while the other two form an angle $\alpha$ with the base plane. A right prism is inscribed in the pyramid; three of its vertices lie on the lateral edges of the pyramid, and the other three li... | ## Solution.
Given $A B=B C=A C, S A \perp A B, S A \perp A C, \angle S B A=\alpha=\angle S C A$, $\angle O A E=\beta, S A=l$ (Fig. 12.48). From $\triangle S A B\left(\angle S A B=90^{\circ}\right): A B=\dot{B C}=A C=$ $=l \cdot \operatorname{ctg} \alpha, S B=\frac{l}{\sin \alpha}$. Let $A E=F E=A F=x$, then from $\tr... | \frac{\cos\alpha\sin\beta}{\sin(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,429 |
12.049. The plane angle at the vertex of a regular hexagonal pyramid is equal to the angle between a lateral edge and the plane of the base. Find this angle. | ## Solution.
Let OABCDEF be a regular hexagonal pyramid, $O O_{1} \perp (A B C)$, $\angle F O E = \angle O F O_{1}$ (Fig. 12.51). Let $O F = l$ and $\angle F O E = \alpha$; then $F E = \sqrt{2 l^{2} - 2 l^{2} \cos \alpha} = 2 l \sin \frac{\alpha}{2}$. Considering that $O_{1} F = F E$, in $\triangle O O_{1} F$ we have ... | 2\arcsin\frac{\sqrt{3}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,432 |
12.050. A plane is drawn through vertex $C$ of the base of a regular triangular pyramid $S A B C$ perpendicular to the lateral edge $S A$. This plane forms an angle with the base plane, the cosine of which is $2 / 3$. Find the cosine of the angle between two lateral faces. | ## Solution.
Let $C K B$ be the given section (Fig. 12.52). This means that $K C \perp A S$, $K B \perp A S$, $L$ is the midpoint of $B C$, and $\angle A L K=\alpha$. We need to find $\angle C K B=\beta$. Let $A B=B C=A C=a \cdot \triangle B K C$ be isosceles, with $K B=K C$ (since $C K \perp A S$, $K B \perp A S$, $A... | \frac{1}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,433 |
12.051. At the base of a right triangular prism lies an isosceles triangle $ABC$; where $AB = BC = a$ and $\angle BAC = \alpha$. A plane is drawn through the side $AC$ at an angle $\varphi (\varphi < \pi / 2)$ to the base. Find the area of the section, given that a triangle is obtained in the section. | ## Solution.
Let $A L C$ be the given section (Fig. 12.53), $L K$ be the perpendicular to $A C$, so $K B \perp A C$ (by the theorem of the $3$ perpendiculars), $K$ is the midpoint of $A C$. From $\triangle A B K \quad A K=A B \cos \alpha=a \cos \alpha . A C=2 A K=2 a \cos \alpha$. The area of $\triangle A B C \quad S_... | \frac{^{2}\sin2\alpha}{2\cos\varphi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,434 |
12.052. Triangle $ABC$ rotates around a line lying in the plane of this triangle, passing outside it through vertex $A$ and equally inclined to sides $AB$ and $AC$. Find the volume of the solid of revolution if $AB=a$, $AC=b$, and $\angle BAC=\alpha$. | ## Solution.
The volume of the given body can be obtained by subtracting the volumes of the cones with vertex at $A$ and diameters $B B_{1}$ and $C C_{1}$ in the bases from the volume of the frustum of a cone with diameters $B B_{1}$ and $C C_{1}$ in the bases (Fig. 12.54). Drop perpendiculars $B K$ and $C L$ to the l... | \frac{\pi}{3}(+b)\sin\alpha\cos\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,435 |
12.053. The lateral surface area of a regular triangular pyramid is 5 times the area of its base. Find the plane angle at the vertex of the pyramid. | ## Solution.
Let $SABC$ be a regular triangular pyramid, $S_{\text {side}}=5 S_{\triangle ABC}$ (Fig. 12.55); we need to find $\angle ASB$. Let $SA=l$ and $\angle ASB=\alpha$; then $S_{\text {side}}=\frac{3}{2} l^{2} \sin \alpha$. From $\triangle ASB$ using the cosine theorem, we find $AB^{2}=2 l^{2}-2 l^{2} \cos \alp... | 2\operatorname{arctg}\frac{\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,436 |
12.057. The base of a right prism is a right-angled triangle with an acute angle $\alpha$. The diagonal of the larger lateral face is $d$ and forms an angle $\beta$ with the lateral edge. Find the volume of the prism. | Solution.
Let $\triangle ABC$ be a right triangle, $\angle C=90^{\circ}, \angle BAC=\alpha$ (Fig. 12.59). By the condition, $BA_1=d, \angle BA_1A=\beta$. Then, from $\triangle BA_1A$, the height of the prism $AA_1=d \cos \beta$, and $AB=d \sin \beta$, since $AB$ is the hypotenuse of $\triangle ABC$, then $AC=AB \cos \... | \frac{^3\sin\beta\sin2\beta\sin2\alpha}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,440 |
12.059. Find the acute angle of the rhombus, knowing that the volumes of the bodies obtained by rotating the rhombus around its larger diagonal and around its side are in the ratio of $1: 2 \sqrt{5}$. | ## Solution.
Let the side of the rhombus be $a$, and the acute angle be $\alpha$ (Fig. 12.61a). This means that $\angle DAB = \alpha$, and $AC$ is the larger diagonal. When the rhombus rotates around the larger diagonal, a body consisting of 2 equal cones is formed (Fig. 12.61b). The volume of each cone $V' = \frac{1}... | \arccos\frac{1}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,442 |
12.060. The base of the pyramid is an isosceles triangle, with the lateral side equal to ' $a$, and the angle at the vertex equal to $\alpha$. All lateral edges are inclined to the plane of the base at an angle $\beta$. Find the volume of the pyramid. | Solution.
Given that $SABC$ is a triangular pyramid, $AB=BC=a$, $\angle ABC=\alpha, SO \perp (ABC), \angle SAO=\angle SBO=\angle SCO=\beta$ (Fig. 12.62). Since $\triangle SOA = \triangle SOB = \triangle SOC$ (by the leg and adjacent angle), then $OA=OB=OC$, i.e., $O$ is the center of the circle circumscribed around $\... | \frac{1}{6}^3\sin\frac{\alpha}{2}\tan\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,443 |
12.063. Each lateral edge of a quadrilateral pyramid forms an angle $\alpha$ with the height. The base of the pyramid is a rectangle with an angle $\beta$ between the diagonals. Find the volume of the pyramid if its height is $h$. | ## Solution.
Since each of the lateral edges of the quadrilateral pyramid forms an angle $\alpha$ with the height, the vertex of the pyramid projects onto the point of intersection of the diagonals of the rectangle lying at the base (Fig. 12.65). Thus, if $d$ is the length of the diagonal, then $\frac{d}{2}=h \operato... | \frac{2}{3}^{3}\mathrm{tg}^{2}\alpha\sin\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,446 |
12.064. A square with side length $a$ is inscribed in the base of a cone. A plane passing through one of the sides of this square and the vertex of the cone intersects the surface of the cone to form an isosceles triangle, with the angle at the vertex being $\alpha$. Find the volume of the cone. | Solution.
Let $ABCD$ be the given square with side $a$ (Fig. 12.66) and let the cutting plane pass through the side $AD$. Then, by the condition, $\angle ASD = \alpha$. Let $L$ be the midpoint of
$, $\angle S A O=\alpha$ (Fig. 12.67). From $\triangle S A O$ we find $S O=l \sin \alpha$, $A O=l \cos \alpha$, hence $A B=A O \sqrt{3}=l \sqrt{3} \cos \alpha$. Therefore,
\[
\begin{aligned}
& S_{\triangle A B C}=\frac{A B^{2} \sqrt{3}}{4}=\... | \frac{^{3}\sqrt{3}\sin2\alpha\cos\alpha}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,448 |
12.067. The base of a right prism is an isosceles triangle with an angle $\alpha$ at the vertex. The diagonal of the face opposite this angle is $l$ and forms an angle $\beta$ with the base plane. Find the volume of the prism. | Solution.
Let $A B C A_{1} B_{1} C_{1}$ be a right prism, $A C=C B, \angle A C B=\alpha$, $A_{1} B=l, \angle A_{1} B A=\beta$ (Fig. 12.69), we need to find $V_{\text {pr }}=S_{\triangle A B C} \cdot A_{1} A$. From $\triangle A_{1} A B$ we find $A A_{1}=l \sin \beta, A B=l \cos \beta$, and from $\triangle A D C$ we hav... | \frac{1}{8}^{3}\sin2\beta\cos\beta\operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,450 |
12.068. The lateral edge of a regular triangular pyramid forms an angle $\alpha$ with the side of the base. Find the angle between the lateral edge and the height of the pyramid and the permissible values of $\alpha$. | Solution.
Let the side of the base be denoted by \(a\) (Fig. 12.70), and the lateral edge by \(l\). Then, from the isosceles triangle, which is a lateral face of the pyramid, we find \(l = \frac{a}{2 \cos \alpha}\). Since the pyramid is regular, the base of the height is the center of the circle circumscribed around t... | \arcsin\frac{2\cos\alpha}{\sqrt{3}};\frac{\pi}{6}<\alpha<\frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,451 |
12.070. The lateral edges of a regular triangular pyramid are pairwise perpendicular. Find the angle between a lateral face and the plane of the base. | ## Solution.
By the condition $\angle A S C=\angle C S B=\angle A S B=90^{\circ}$ (Fig. 12.72). Moreover, $A S=S B=S C, \triangle A B C$ is equilateral; $S K \perp B C, H K \perp B C$, where $H$ is the foot of the altitude. Since $\triangle S C B$ is a right and isosceles triangle, then $\angle S C B=45^{\circ}$. Let ... | \arccos\frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,453 |
12.071. A cone is inscribed in a hemisphere; the vertex of the cone coincides with the center of the circle, which is the base of the hemisphere; the planes of the bases of the cone and the hemisphere are parallel. A line passing through the center of the base of the cone and an arbitrary point on the circumference of ... | Solution.
Let the radius of the circle at the base of the hemisphere be denoted by $R$, i.e., $O K=R$, and the radius at the base of the cone by $r$, i.e., $O_{1} L=r$, $\angle O_{1} K O=\alpha$ (Fig. 12.73). Then the height $O_{1} O$ of the cone will be equal to $h=K O \cdot \operatorname{tg} \alpha=R \operatorname{t... | \frac{2\cos^{2}\alpha}{\cos2\alpha\operatorname{tg}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,454 |
12.072. The base of the pyramid is a right-angled triangle with an acute angle $\alpha$. The height of the pyramid is $H$. All lateral edges form the same angle with the base plane, which is equal to $\beta$. Find the volume of the pyramid. | ## Solution.
Since all the lateral edges are inclined at the same angle to the plane of the base, the base of the height of the pyramid will be the center of the circle circumscribed around the base. Since the base is a right triangle, the center of the circle circumscribed around it will be the midpoint of the hypote... | \frac{H^{3}\sin2\alpha}{3\operatorname{tg}^{2}\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,455 |
12.074. A sphere is inscribed in a cone. The ratio of the radius of the circle of contact between the spherical and conical surfaces to the radius of the base of the cone is $k$. Find the cosine of the angle between the slant height of the cone and the plane of the base. | Solution.
From the problem statement, the radius of the circle of tangency between the spherical and conical surfaces will be $K O_{1}$, and the radius of the base of the cone will be $L O_{2}$ (Fig. 12.76). $\angle S L O_{2}=\alpha$. From the diagram, it is clear that $\triangle S K O \sim \Delta S O_{2} L$, and $\tr... | 1-k | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,457 |
12.076. At the base of a right prism lies a rhombus with an acute angle $\alpha$. The ratio of the height of the prism to the side of the base is $k$. A plane is drawn through a side of the base and the midpoint of the opposite lateral edge. Find the angle between this plane and the plane of the base. | Solution.
Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a right prism, $A B C D$ be a rhombus, $\angle B A D=\alpha$ ( $\alpha<90^{\circ}$ ), $A A_{1}: A B=k, C_{1} E=E C, A D E F$ be a section of the prism (Fig. 12.78); we need to find $\angle(A B C) ;(F A D)$. Let $A B=a$; then $A A_{1}=k a$. Draw $B K \perp A D$; then $... | \operatorname{arctg}\frac{k}{2\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,459 |
12.077. The sides of the base of a right parallelepiped are in the ratio $1: 2$, and the acute angle in the base is $\alpha$. Find the angle between the smaller diagonal of the parallelepiped and the base plane, if the height of the parallelepiped is equal to the larger diagonal of the base. | Solution.
Let the sides of the base of the parallelepiped be denoted by $a$ and $b$. Then $a: b=1: 2$. The larger diagonal of the base
$$
d_{2}=\sqrt{a^{2}+b^{2}-2 a b \cos \left(180^{\circ}-\alpha\right)}=\sqrt{a^{2}+b^{2}+2 a b \cos \alpha} \text {. Considering that }
$$
. Then $h=a \operatorname{tg} \alpha=b \operatorname{tg} \beta$, from which $\frac{b}{a}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}$. The diagonal of the base
. Draw $S O \perp(A B C)$ and connect $O$ with points $A, B, M$, and
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,465 |
12.087. Find the angle between the slant height and the height of a cone, the lateral surface of which is the mean proportional between the area of the base and the total surface area. | ## Solution.
According to the problem, $S_{\text {bok }}=\sqrt{S_{\text {osn }} \cdot S}=\sqrt{S_{\text {osn }}\left(S_{\text {bok }}+S_{\text {osn }}\right)}$. Let the slant height of the cone be $l$, and the radius of the base circle be $R$. Then $S_{\text {bok }}=\pi R l, S_{\text {osn }}=\pi R^{2}$, and $\sin \alp... | \arcsin\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,469 |
12.090. A plane passing through a generatrix of a cylinder forms an acute angle $\alpha$ with the plane of the axial section containing the same generatrix. The diagonal of the rectangle obtained by the intersection of the cylinder with this plane is equal to $l$ and forms an angle $\beta$ with the plane of the base. F... | ## Solution.
The volume of the cylinder is found as follows: $V=\pi R^{2} H$. The height of the cylinder $H=C A=l \sin \beta$ (from $\triangle C A B$) (Fig. 12.91). From $\triangle C A B: A B=l \cos \beta$. Consider $\triangle O F B\left(\angle O F B=90^{\circ}\right): F B=\frac{1}{2} A B=\frac{l}{2} \cos \beta$. The ... | \frac{\pi^{3}\sin2\beta\cos\beta}{8\cos^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,472 |
12.091. The side of the rhombus is $a$, and its acute angle is $\alpha$. The rhombus rotates around a line passing through its vertex and parallel to the longer diagonal. Find the volume of the solid of revolution. | ## Solution.
The volume of the solid of revolution is the sum of the volumes of two identical figures $V_{\text {s.r }}=2 V$ (Fig. 12.92). The volume of such a figure is: $V=V_{\text {tr.c }}-V_{\text {c }}$. The volume of the frustum of a cone is: $V_{\text {tr.c }}=\frac{\pi H}{3}\left(R^{2}+r^{2}+R r\right)$. The r... | 2\pi^{3}\sin\frac{\alpha}{2}\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,473 |
12.095. The lateral face of a regular truncated triangular pyramid forms an acute angle $\alpha$ with the base plane. Find the angle between the height and the lateral edge of the pyramid. | Solution.
From $\triangle S O D\left(\angle S O D=90^{\circ}\right): S O=D O \operatorname{tg} \alpha$, and $D O=\frac{x \sqrt{3}}{6}$ (radius of the inscribed circle) (Fig. 12.96). Therefore, $S O=\frac{x \sqrt{3}}{6} \operatorname{tg} \alpha$. From $\triangle S O A$ $\left(\angle S O A=90^{\circ}\right) \operatornam... | \operatorname{arctg}(2\operatorname{ctg}\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,477 |
12.098. A cone is inscribed in a sphere. The area of the axial section of the cone is $S$, and the angle between the height and the slant height is $\alpha$. Find the volume of the sphere. | Solution.
The volume of the sphere is: \( V = \frac{4}{3} \pi R^{3} \). The area of the axial section of the cone is: \( S = \frac{1}{2} l^{2} \sin 2 \alpha \). Hence, \( l = \sqrt{\frac{2 S}{\sin 2 \alpha}} \). The following relationship holds between the elements of the sphere and the inscribed cone: \( l = 2 R \sin... | \frac{1}{3}\piS\frac{\sqrt{2S\sin2\alpha}}{\sin^{2}2\alpha\cos^{3}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,478 |
12.099. The base of a quadrilateral pyramid is a rhombus with side $a$ and acute angle $\alpha$. All lateral faces are inclined to the base plane at the same angle $\beta$. Find the total surface area of the pyramid. | Solution.
Let $S A B C D$ be a pyramid, $A B C D$ be a rhombus, $\angle B A D=\alpha\left(\alpha<90^{\circ}\right)$, $A B=a,((S A B)(A B C))=((S B C) ;(A B C))=((S C D) ;(A B C))=((S D A) ;(A B C))=\beta$, $S O \perp(A B C)$ (Fig. 12.99). Draw the apothems of the pyramid $S E, S F, S K$, $S L$; then $O E \perp A B, O ... | \frac{2^{2}\sin\alpha\cos^{2}\frac{\beta}{2}}{\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,479 |
12.101. The side of the base of a regular quadrilateral prism is equal to $a$. The angle between the intersecting diagonals of two adjacent lateral faces is $\boldsymbol{\alpha}$. Find the volume of the prism. | Solution.
Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a regular quadrilateral prism, $A B=a, \angle A_{1} D C_{1}=\alpha$ (Fig. 12.101). We have $A_{1} C_{1}=a \sqrt{2}$. Let $A_{1} D=x ;$
Fi... | \frac{^{3}\sqrt{2\cos\alpha}}{2\sin\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,481 |
12.102. The volume of a cone is $V$. A pyramid is inscribed in the cone, with an isosceles triangle as its base, where the angle between the lateral sides is $\alpha$. Find the volume of the pyramid. | Solution.
The volume of the pyramid is: $V_{\text {p }}=\frac{1}{3} S H$ (Fig. 12.102). The area of the base of the pyramid $S=\frac{A B^{2} \cdot C B}{4 R}$. From $\triangle A B C \quad C B=2 A B \sin \frac{\alpha}{2}$. Then
$S=\frac{2 A B^{3} \sin \frac{\alpha}{2}}{4 R}=\frac{2 x^{3} \sin \frac{\alpha}{2}}{4 R}$. He... | \frac{2V}{\pi}\sin\alpha\cos^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,482 |
12.103. A plane is drawn through two generators of a cone, the angle between which is $\alpha$. Find the ratio of the area of the section to the total surface area of the cone, if the generator of the cone makes an angle $\beta$ with the base plane. | ## Solution.
The area of the section is $S_{\mathrm{ce4}}=\frac{1}{2} S A^{2} \sin \alpha$ (Fig. 12.103), the total surface area of the cone is $S_{\text {pol }}=S_{\text {bok }}+S_{\text {osn }}=\pi R l+\pi R^{2}=\pi A O \cdot S A+\pi A O^{2}$. From $\triangle S O A\left(\angle S O A=90^{\circ}\right) A O=S A \cos \b... | \frac{\sin\alpha}{4\pi\cos\beta\cos^{2}\frac{\beta}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,483 |
12.104. The ratio of the lateral surface area of a regular triangular pyramid to the area of its base is $k$. Find the angle between a lateral edge and the height of the pyramid. | Solution.
By the condition, $SABC$ is a regular triangular pyramid, $S_{\text {side }}: S_{\triangle ABC}=k, SO \perp(ABC)$; we need to find $\angle ASO$ (Fig. 12.104). Let $SO=h$ and $\angle ASO=\alpha$; then in $\triangle SOA$ we have $AO=h \operatorname{tg} \alpha$, from which $AB=AO \sqrt{3}=h \operatorname{tg} \a... | \operatorname{arcctg}\frac{\sqrt{k^{2}-1}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,484 |
12.107. A truncated cone is described around a sphere, with the area of one base being four times the area of the other base. Find the angle between the slant height of the cone and the plane of its base. | Solution.
According to the condition $S_{2}=4 S_{1}$ (Fig. 12.107). From this, $\pi R_{2}^{2}=4 \pi R_{1}^{2} \Rightarrow R_{2}=2 R_{1}$ $\left(A O=2 B O_{1}\right)$. The required angle can be found as follows: $\cos \angle C D O=\frac{F D}{C D}=\frac{B O_{1}}{C D}$. It is known that $C F^{2}=B C \cdot A D=2 B O_{1} \... | \arccos(\frac{1}{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,487 |
12.108. A plane is drawn through the side of the lower base of a cube, dividing the volume of the cube in the ratio $m: n$, counting from the lower base. Find the angle between this plane and the plane of the base, if $m \leq n$. | Solution.
By the condition $\frac{V_{1}}{V_{2}}=\frac{m}{n}$. Volume
Fig. 12.108
$V_{1}=S \cdot H=\frac{1}{2} K P \cdot K O \cdot A D$ (Fig. 12.108). From $\triangle O K P\left(\angle O K... | \operatorname{arctg}(\frac{2}{+n}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,488 |
12.109. The height of a regular triangular prism is $H$. A plane passing through the midline of the lower base and the parallel side of the upper base forms an acute dihedral angle $\alpha$ with the plane of the lower base. Find the area of the section formed by this plane. | ## Solution.
Let $A B C A_{1} B_{1} C_{1}$ be a regular triangular prism, $A A_{1}=H$, $A N=N B, B M=M C,\left(\left(A_{1} C_{1} M\right) ;(A B C)\right)=\alpha$ (Fig. 12.109). Draw $B E \perp \dot{A} C$ and $E F \| A A_{1} ; \angle F E B=\alpha$ as the linear angle of the dihedral angle. Let $B E \cap M N=O$; then $F... | \frac{H^{2}\sqrt{3}\operatorname{ctg}\alpha}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,489 |
12.112. The distance from the center of the base of a cone to its generatrix is $d$. The angle between the generatrix and the height is $\alpha$. Find the total surface area of the cone. | ## Solution.
The total surface area of the cone $S_{\text {pol }}=S_{\text {bok }}+S_{\text {osn }}=\pi R l+\pi R^{2}$ (Fig. 12.112). From $\triangle O B A\left(\angle O B A=90^{\circ} ; \angle B O A=\angle A S O=\alpha\right) \quad R=O A=\frac{d}{\cos \alpha}$. From $\triangle S O A \quad\left(\angle S O A=90^{\circ}... | \frac{\pi^{2}}{2\sin\alpha\sin^{2}(\frac{\pi}{4}-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,491 |
12.113. The base of the pyramid $ABCD$ is a right triangle $ABC\left(\angle C=90^{\circ}\right)$. The lateral edge $AD$ is perpendicular to the base. Find the acute angles of triangle $ABC$, if $\angle DBA=\alpha$ and $\angle DBC=\beta(\alpha<\beta)$. | ## Solution.
Let $DABC$ be a pyramid, $\angle ACB=90^{\circ}, AD \perp(ABC), \angle DBA=\alpha$, $\angle DBC=\beta(\alpha<\beta)$ (Fig. 12.113). Let $AD=h$; then from $\triangle ADB$ we find $AB=h \operatorname{ctg} \alpha, DB=\frac{h}{\sin \alpha}$. Since $BC \perp AC$, then $DC \perp BC$ and from $\triangle DCB$ we ... | \arcsin\frac{\cos\beta}{\cos\alpha};\arccos\frac{\cos\beta}{\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,492 |
12.114. In a regular hexagonal prism, a plane passing through a side of the base and the midpoint of the segment connecting the centers of the bases forms an acute angle $\alpha$ with the base plane. Find the area of the section formed by this plane if the side of the base of the prism is equal to $a$. | Solution.
The area of the section $S_{\mathrm{sec}}=2 S_{B A F E}=2\left(\frac{B E+A F}{2} \cdot P K\right)$ (Fig. 12.114). The radius of the inscribed circle in the base of the prism: $O K=r=\frac{a \sqrt{3}}{2}$. From $\triangle P O K\left(\angle P O K=90^{\circ}\right): P K=\frac{O K}{\cos \alpha}=\frac{a \sqrt{3}}... | \frac{3^{2}\sqrt{3}}{2\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,493 |
12.115. A sphere is inscribed in a cone, the surface of which is equal to the area of the base of the cone. Find the cosine of the angle at the vertex in the axial section of the cone. | ## Solution.
By the condition $S_{\mathrm{w}}=S_{\text {osn }}$ (Fig. 12.115). Hence, $4 \pi R_{\mathrm{u}}^{2}=\pi R_{\mathrm{k}}^{2} \Rightarrow R_{\mathrm{k}}=2 R_{\mathrm{w}}$. Consider $\triangle S C A$. By the Law of Cosines, $(2 A O)^{2}=2 S A^{2}-2 S A^{2} \cos \alpha=2 S A^{2}(1-\cos \alpha)$. Therefore, $\co... | \frac{7}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,494 |
12.116. The base of a right prism is an isosceles triangle, with the lateral side equal to $a$, and the angle between the lateral sides equal to $\alpha$. Find the volume of the prism if its lateral surface area is $S$. | Solution.
The volume of the prism \( V = S_{\text{base}} \cdot H \) (Fig. 12.116). The area of the base is: \( S_{\text{base}} = \frac{1}{2} a^2 \sin \alpha \). The lateral surface area \( S = P H \), hence \( H = \frac{S}{P} \). Let \( AC = x \). By the cosine theorem:
\[
x^2 = 2a^2 - 2a^2 \cos \alpha = 2a^2 (1 - \c... | \frac{}{2}\sin\frac{\alpha}{2}\tan\frac{\pi-\alpha}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,495 |
12.117. The base of a pyramid inscribed in a cone is a quadrilateral with adjacent sides equal in pairs, and the angle between one pair of adjacent sides is $\alpha$. Find the ratio of the volume of the pyramid to the volume of the cone. | Solution.
The volume of the pyramid $V_{\mathrm{n}}=\frac{1}{3} S_{\text {osn }} \cdot H$ (Fig. 12.117). The area of the base of the pyramid $S_{\text {osn }}=2 S_{\triangle A B C}=2 \cdot \frac{1}{2} A B \cdot B C\left(\angle A B C=90^{\circ}\right)$. From $\triangle A B C$ $A B=2 O C \cdot \sin \frac{\alpha}{2} ;$ a... | \frac{2\sin\alpha}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,496 |
12.118. Find the cosine of the angle between adjacent lateral faces of a regular quadrilateral pyramid, where the lateral edge is equal to the side of the base. | Solution.
From $\triangle A O C$ by the cosine theorem $A C^{2}=2 A O^{2}-2 A O^{2} \cos \alpha$ (Fig. 12.118). From this, $\cos \alpha=\frac{2 A O^{2}-A C^{2}}{2 A O^{2}}$. From $\triangle A B C\left(\angle A B C=90^{\circ}\right)$: $A C=\sqrt{2 a^{2}}=a \sqrt{2}$. From $\triangle A O B \quad\left(\angle A O B=90^{\c... | -\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,497 |
12.119. A sphere is inscribed in a cone. The radius of the circle where the cone and the sphere touch is $r$. Find the volume of the cone if the angle between the height and the slant height of the cone is $\alpha$. | ## Solution.
The volume of the cone is: $V=\frac{1}{3} \pi R_{\mathrm{k}}^{2} \cdot H$ (Fig. 12.119). From $\triangle A O_{1} O$ $\left(\angle A O_{1} O=90^{\circ}\right) O A=R_{\text {m }}=\frac{r}{\cos \alpha} ; O O_{1}=r \operatorname{tg} \alpha$. From $\triangle S O_{1} A$ $\left(\angle S O_{1} A=90^{\circ}\right)... | \frac{\pir^{3}\operatorname{ctg}^{3}(\frac{\pi}{4}-\frac{\alpha}{2})}{3\cos^{2}\alpha\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,498 |
12.120. The lateral edge of a regular quadrilateral pyramid is equal to $m$ and is inclined to the base plane at an angle $\alpha$. Find the volume of the pyramid. | Solution.
The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} H$ (Fig. 12.120). From $\triangle S O A$ we find: $S O=H=m \sin \alpha ; A O=m \cos \alpha=\frac{A C}{2}$. Hence, $A C=2 m \cos \alpha$. The area of the base $S_{\text {base }}=\frac{A C^{2}}{2}=\frac{4 m^{2} \cos ^{2} \alpha}{2}=2 m^{2} \cos ^{2} \a... | \frac{1}{3}^{3}\cos\alpha\sin2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,499 |
12.121. The base of the pyramid is an equilateral triangle with side $a$. Two lateral faces of the pyramid are perpendicular to the base plane, and the equal lateral edges form an angle $\alpha$. Find the height of a right triangular prism equal in volume to the given pyramid and having a common base. | ## Solution.
Let $DABC$ be a pyramid, $ABC$ be an equilateral triangle, $(ABD) \perp (ABC), (BCD) \perp (ABC), \angle ADC = \alpha, AB = a$ (Fig. 12.121). We need to find the height of the prism with the base equal to $\triangle ABC$, such that $V_{\text{pr}} = V_{\text{pir}}$. The height of the pyramid coincides with... | \frac{}{3\sin\frac{\alpha}{2}}\cdot\sqrt{\sin(\frac{\pi}{6}+\frac{\alpha}{2})\sin(\frac{\pi}{6}-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,500 |
12.122. Find the volume of a cone if a chord of length $a$ in its base subtends an arc of $\alpha$ radians, and the height of the cone makes an angle $\beta$ with the slant height. | Solution.
Let $SAK$ be a cone, $SO$ its height, $BK$ a chord, $BK=a$, $\cup B n K=\alpha, \angle O S K=\beta$ (Fig. 12.122). Since $\cup B n K=\alpha$, then $\angle B O K=\alpha$. Let $R$ be the radius of the base of the cone; then in $\triangle O B K$ we have $a^{2}=2 R^{2}-2 R^{2} \cos \alpha=4 R^{2} \sin ^{2} \frac... | \frac{\pi^{3}\operatorname{ctg}\beta}{24\sin^{3}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,501 |
12.123. The angle at the vertex of the axial section of a cone is $2 \alpha$, and the sum of the lengths of its height and slant height is $a$. Find the volume of the cone. | ## Solution.
By the condition $S A+S O=a \Rightarrow S O=a-S A$ (Fig. 12.123). The volume of the cone $V=\frac{1}{3} \pi R^{2} H$. From $\triangle S O A \quad S O=S A \cos \alpha$. Then $S A \cos \alpha=a-S A$. $S A=\frac{a}{1+\cos \alpha}$ and $H=S O=\frac{a \cos \alpha}{1+\cos \alpha}$, and $A O=R=S A \sin \alpha=\f... | \frac{^{3}\pi\cos\alpha\cdot\sin^{2}\frac{\alpha}{2}}{6\cos^{4}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,502 |
12.126. Find the angle in the axial section of a cone if a sphere with its center at the vertex of the cone, touching its base, divides the volume of the cone in half. | ## Solution.
The volume of the sector $V_{\text {sec }}=\frac{2}{3} \pi R_{w}^{2} \cdot h$ (Fig. 12.126). The volume of the cone $V_{\mathrm{k}}=\frac{1}{3} \pi R^{2} \cdot H$. According to the condition $V_{\mathrm{sec}}=\frac{V_{\mathrm{k}}}{2} ; \frac{2}{3} \pi R_{\mathrm{m}}^{2} h=\frac{1}{6} \pi R^{2} H$. The hei... | 2\arccos\frac{1+\sqrt{17}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,505 |
12.127. The development of the lateral surface of a cylinder is a rectangle, the diagonals of which intersect at an angle $\alpha$. The length of the diagonal is $d$. Find the lateral surface area of the cylinder. | ## Solution.
Let $A A_{1} B_{1} B$ be the development of the lateral surface of the cylinder, $A A_{1}$ and $B B_{1}$ be its generators, $A B$ and $A_{1} B_{1}$ be the lengths of the circumferences of its bases, $A B_{1}=d, \angle A O A_{1}=\alpha$ (Fig. 12.127). Since $\angle A O A_{1}$ is the external angle of the i... | \frac{1}{2}^{2}\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,506 |
12.128. A regular hexagonal prism with equal edges is inscribed in a cone with a generatrix of length $l$. Find the lateral surface area of the prism if the angle between the generatrix and the height of the cone is $\boldsymbol{\alpha}$. | ## Solution.
The lateral surface area of the prism $S_{\text {bok }}=P \cdot h$ (Fig. 12.128). Let the edge of the prism be $x$. Then $h=x$ and $P=6 x$. From $\triangle S O A \quad S O=l \cos \alpha$. From $\Delta S O_{1} B \quad S O_{1}=\frac{x}{\operatorname{tg} \alpha}$. We find $S O=O O_{1}+O_{1} S=x+\frac{x}{\ope... | \frac{3^{2}\sin^{2}2\alpha}{4\sin^{2}(\frac{\pi}{4}+\alpha)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,507 |
12.129. Find the volume of a regular quadrilateral pyramid if the side of its base is equal to $a$, and the dihedral angle at the base is $\alpha$. | Solution.
The volume of the pyramid $V=\frac{1}{3} S H$ (Fig. 12.129). The area of the base: $S=a^{2}$. The height of the pyramid from $\triangle S O E \quad S O=H=\frac{a}{2} \operatorname{tg} \alpha$. Therefore, $V=\frac{1}{3} \frac{a^{3}}{2} \operatorname{tg} \alpha=\frac{a^{3}}{6} \operatorname{tg} \alpha$.
Answe... | \frac{^{3}}{6}\operatorname{tg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,508 |
12.130. The development of the lateral surface of a cylinder is a rectangle, in which the diagonal is equal to $a$ and forms an angle $\alpha$ with the base. Find the volume of the cylinder. | ## Solution.
The volume of the cylinder $V=\pi R^{2} H$ (Fig. 12.130). The height of the cylinder from $\triangle A B C$ is $H=C B=a \sin \alpha$. The length of the circumference of the base $c=A B=a \cos \alpha$. On the other hand, $c=2 \pi R$. Then $R=\frac{a \cos \alpha}{2 \pi}$. Substituting, we get: $V=\pi \frac{... | \frac{^{3}\cos^{2}\alpha\sin\alpha}{4\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,509 |
13.001. Of the four given numbers, the first three are in the ratio $1 / 5: 1 / 3: 1 / 20$, and the fourth number is $15 \%$ of the second number. Find these numbers, given that the second number is 8 more than the sum of the others. | Solution.
The required numbers are: $k / 5 ; k / 3 ; k / 20 ; 0.15 \cdot k / 3=k / 20$. According to the condition $\frac{k}{3}-8=\frac{k}{5}+\frac{k}{20}+\frac{k}{20}$, from which $k=240$.
Answer: $48 ; 80 ; 12 ; 12$. | 48;80;12;12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,510 |
13.002. How many kilograms of water need to be evaporated from 0.5 t of cellulose mass containing $85 \%$ water to obtain a mass with $75 \%$ water content? | Solution.
In the cellulose mass, there is $0.85 \cdot 500=425$ kg of water. Let $x$ kg of water be evaporated; then we get $425-x=0.75(500-x)$, from which $x=200(\mathrm{kg})$.
Answer: $200 \mathrm{kg}$. | 200\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,511 |
13.003. In two barrels, there are 70 liters of milk. If 12.5% of the milk from the first barrel is poured into the second barrel, then both barrels will have the same amount. How many liters of milk are in each barrel? | ## Solution.
Let the initial amount of milk in the first bucket be $x$, and in the second bucket $70-x$ liters. After transferring, the first bucket has $x-0.125x=0.875x$ liters left, and the second bucket has $70-x+0.125x$ liters. According to the condition, $70-x+0.125x=0.875x$, from which $x=40$. There were 40 lite... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,512 |
13.005. The sum of the digits of a two-digit number is 12. If 36 is added to the desired number, the result is a number written with the same digits but in reverse order. Find the number. | ## Solution.
Let the desired number be of the form $10x + y$. Then, according to the condition, $x + y = 12$ (1) and $10x + y + 36 = 10y + x$, i.e., $x - y + 4 = 0$ (2). Adding (1) and (2), we get $2x = 8$, from which $x = 4$, and $y = 8$.
Answer: 48. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,513 |
13.006. A tractor driver plowed three plots of land. The area of the first is $2 / 5$ of the area of all three plots, and the area of the second is to the area of the third as $3 / 2: 4 / 3$. How many hectares were there in all three plots if the third plot was 16 hectares less than the first? | Solution.
Let the areas of the plots be $x, y, x-16$ (ha). According to the condition, $x=\frac{2}{5}(x+y-16+x)$ and $y:(x-16)=3 / 2: 4 / 3$, from which $y=\frac{9}{8}(x-16)$. We have
$x=\frac{2}{5}\left(2 x-16+\frac{9}{8} x-18\right) \Leftrightarrow x=\frac{2}{5}\left(\frac{25 x}{8}-34\right), x=\frac{272}{5}$. Next,... | 136 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,514 |
13.007. The price of the product was first reduced by $20 \%$, then the new price was reduced by another $15 \%$, and finally, after recalculation, a further reduction of $10 \%$ was made. By what total percentage was the original price of the product reduced? | ## Solution.
Let $x$ be the original price of the item. After the first reduction, the price became $x-0.2x=x(1-0.2)$; after the second, $-x(1-0.2)(1-0.15)$; after the third, $-x(1-0.2)(1-0.15)(1-0.1)=0.612x$. Therefore, the total reduction from the original price is $1-0.612=0.388=38.8\%$.
Answer: $38.8\%$. | 38.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,515 |
13.008. Seawater contains $5 \%$ salt by mass. How much fresh water needs to be added to 30 kg of seawater to make the salt concentration $1.5 \%$? | ## Solution.
Let $x$ kg of salt in 30 kg of seawater. $x$ kg is $5 \%$. Hence, $x=1.5$ kg, $30 \text{kg}-100 \%$. In the diluted seawater: 1.5 kg of salt is $1.5 \%$, $a$ kg is $100 \%$, from which $a=100$ kg of diluted seawater. Therefore, we need to add $100-30=70$ kg of fresh water.
Answer: $70 \text{kg}$. | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,516 |
13.009. In the library, there are books in English, French, and German. English books make up $36 \%$ of all foreign language books, French books make up $75 \%$ of the English books, and the remaining 185 books are German. How many foreign language books are there in the library? | ## Solution.
Let $x$ be the total number of books in foreign languages. The number of English books is $0.36x$; the number of French books is $0.75 \cdot 0.36x$. According to the problem, $0.36x + 0.75 \cdot 0.36x + 185 = x$, from which we find $x = 500$.
Answer: 500 | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,517 |
13.010. A pump can extract $2 / 3$ of the water from a pool in 7.5 min. After working for 0.15 h, the pump stopped. Find the capacity of the pool, if after the pump stopped, there was still 25 m $^{3}$ of water left in the pool. | Solution.
Let $x$ m $^{3}$ be the capacity of the pool. In 7.5 minutes, the pump can extract $\frac{2}{3} x$ water. In $0.15 \text{ h} = 9$ minutes, the pump extracted $\frac{9 \cdot \frac{2}{3} x}{7.5}=0.8 x$ water. In the pool, $0.2 x$ water remained. According to the problem, $0.2 x=25$, from which $x=125$.
Answer... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,518 |
13.011. Due to the reconstruction of equipment, the labor productivity of a worker increased twice during the year by the same percentage. By what percentage did the labor productivity increase each time, if during the same time a worker used to produce goods worth 2500 rubles, and now produces goods worth 2809 rubles? | Solution.
Let a worker produce a parts in 8 hours of work. Then the rate is $\frac{2500}{a}$ rubles per part, and the labor productivity is $\frac{a}{8}$ parts per hour. After the first increase in productivity by $x \%$, the worker started producing $\frac{a}{8}+\frac{x \cdot a}{100 \cdot 8}$ parts per hour; after th... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,519 |
13.012. The working day has been reduced from 8 to 7 hours. By what percentage does labor productivity need to increase to ensure that, with the same rates, the wage increases by $5 \%$? | Solution.
Let the master produce $a$ parts and earn $b$ rubles for 8 hours of work. Then the rate is $b / a$ rubles per part, and the labor productivity is $a / 8$ parts per hour. After increasing productivity by $x \%$, the master started producing $\frac{a}{8}+\frac{x a}{8 \cdot 100}$ parts per hour. Therefore, in 7... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,520 |
13.013. In January, the plant fulfilled $105\%$ of the monthly production plan, and in February, it produced $4\%$ more than in January. By what percentage did the plant exceed the two-month production plan? | Solution.
Let $x / 2$ be the monthly plan of the plant; $x$ be the two-month plan. In January, the plant exceeded the plan by $0.05 \cdot x / 2$; in February by $0.04 \cdot 0.05 \cdot x / 2$. Over two months, the plant exceeded the plan by $0.05 \frac{x}{2} + 0.04 \cdot 0.05 \frac{x}{2} = 0.071 x$. Therefore, the plan... | 7.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,521 |
13.014. Find three numbers if the first is $80 \%$ of the second, the second is related to the third as $0.5: 9 / 20$, and the sum of the first and third is 70 more than the second number. | Solution.
Let $x$ be the first number, $y$ be the second number, and $z$ be the third number.
Then, according to the condition, $\left\{\begin{array}{l}x=0.8 y, \\ \frac{y}{z}=\frac{0.5}{9 / 20}, \\ x+z=y+70 .\end{array}\right.$ Solving the system, we get $x=80 ; y=100 ; z=90$.
Answer: $80,100,90$. | 80,100,90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,522 |
13.015. A tourist traveled the distance between two cities in 3 days. On the first day, he traveled $1 / 5$ of the total distance and an additional 60 km, on the second day $1 / 4$ of the total distance and an additional 20 km, and on the third day $23 / 80$ of the total distance and the remaining 25 km. Find the dista... | Solution.
Let $x$ km be the distance between the cities. We can form the following table:
| Day | Distance traveled in a day |
| :---: | :---: |
| First | $\frac{1}{5} x+60$ (km) |
| Second | $\frac{1}{4} x+20$ (km) |
| Third | $\frac{23}{80} x+25$ (km) |
According to the condition, $\frac{1}{5} x+60+\frac{1}{4} x+2... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,523 |
13.016. The numerators of three fractions are proportional to the numbers 1, 2, and 3, while the reciprocals of the corresponding denominators are proportional to the numbers $1, 1 / 3$, and 0.2. Find these fractions, given that their arithmetic mean is $136 / 315$. | Solution.
Let's compile the following table:
| Fraction | Numerator | Denominator |
| :--- | :---: | :---: |
| First | $x$ | $1 / y$ |
| Second | $2 x$ | $3 / y$ |
| Third | $3 x$ | $1 / 0.2 y$ |
First fraction $-x y$; second $-\frac{2}{3} x y$; third $-0.6 x y$. The arithmetic mean of the fractions $\frac{x y+\frac... | \frac{4}{7};\frac{8}{21};\frac{12}{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,524 |
13.017. Find the sum of three numbers, knowing that the third is related to the first as $18.48: 15.4$ and constitutes $40 \%$ of the second, and the sum of the first and second is 400. | Solution.
Let $x, y, z$ be the given numbers. According to the condition, we have: $\frac{z}{x}=\frac{1848}{1540}=\frac{6}{5}$, $z=0.4 y, x+y=400$. Since $y=\frac{5}{2} z$ and $z=\frac{6}{5} x$, then $y=3 x$ and $x+3 x=400 \Rightarrow$ $\Rightarrow x=100, z=\frac{6}{5} x=120$. Then
$S=x+y+z=400+120=520$.
Answer: 520... | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,525 |
13.018. The depositor withdrew from his savings account at the savings bank first $1 / 4$ of his money, then $4 / 9$ of the remaining money and another 640 rubles. After this, he had $3 / 20$ of all his money left on the savings book. How large was the deposit? | Solution.
Let the deposit be $x$ rubles. Then the first remainder is $\frac{3 x}{4}$; the second remainder $-\frac{3 x}{4}-\frac{4}{9} \cdot \frac{3 x}{4}-640=\frac{3 x}{20}$. We have $\frac{3 x}{4}-\frac{x}{3}-\frac{3 x}{20}=640$, from which $x=2400$ (rubles).
Answer: 2400 rubles. | 2400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,526 |
13.019. Two snow-clearing machines are working on snow removal. The first can clear the entire street in 1 hour, while the second can do it in $75\%$ of this time. Starting the cleaning simultaneously, both machines worked together for 20 minutes, after which the first machine stopped. How much more time is needed for ... | ## Solution.
Let's consider the entire volume of work as 1. The productivity of the first machine is 1 (per hour), and the second machine's productivity is $1: \frac{3}{4}=\frac{4}{3}$ (per hour). Working together for $\frac{1}{3}$ of an hour, they will complete $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{4}{3} = \... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,527 |
13.020. The sum of the first three terms of the proportion is 58. The third term is $2 / 3$, and the second term is $3 / 4$ of the first term. Find the fourth term of the proportion and write it down. | Solution.
Let the proportion be $\frac{a}{b}=\frac{c}{d}$. Given that $a+b+c=58$; $c=\frac{2}{3} a$; $b=\frac{3}{4} a$. Therefore, $a+\frac{3}{4} a+\frac{2}{3} a=58$, from which $a=24$; $c=16$; $b=18$. Thus, $d=\frac{b c}{a}=12$.
Answer: $12 ; \frac{24}{18}=\frac{16}{12}$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,528 |
13.021. One brigade can harvest the entire field in 12 days. The second brigade needs $75\%$ of this time to complete the same work. After the first brigade worked alone for 5 days, the second brigade joined, and together they finished the work. How many days did the brigades work together? | ## Solution.
Let the teams work together for $x$ days. The productivity of the first team is $-\frac{1}{12}$; the productivity of the second team is $-\frac{1}{12 \cdot 0.75}=\frac{1}{9}$. According to the condition, $\frac{1}{12} \cdot 5+\left(\frac{1}{12}+\frac{1}{9}\right) \cdot x=1$, from which $x=3$ days.
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,529 |
13.022. On the mathematics entrance exam, $15 \%$ of the applicants did not solve a single problem, 144 people solved problems with errors, and the number of those who solved all problems correctly is to the number of those who did not solve any at all as 5:3. How many people took the mathematics exam that day | Solution.
Let $x$ be the total number of people who took the exam. $0.15x$ people did not solve any problems; $0.25x$ people solved all the problems, which is $\frac{0.15x \cdot 5}{3}=0.25x$ people. According to the problem, $0.15x + 144 + 0.25x = x$, from which we get $x = 240$.
Answer: 240. | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,530 |
13.023. Identical parts are processed on two machines. The productivity of the first machine is $40 \%$ higher than that of the second. How many parts were processed during the shift on each machine, if the first machine worked for 6 hours during this shift, and the second - 8 hours, and both machines together processe... | Solution.
Let the productivity of the second machine be $x$; then the productivity of the first machine is $1.4x$ parts per hour. The first machine processed $1.4x \cdot 6$ parts in the shift; the second machine processed $x \cdot 8$ parts. According to the condition, $1.4x \cdot 6 + x \cdot 8 = 820$, from which $x = ... | 420 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,531 |
13.024. A tractor team can plow 5/6 of a plot of land in 4 hours and 15 minutes. Before the lunch break, the team worked for 4.5 hours, after which 8 hectares remained unplowed. How large was the plot? | ## Solution.
Let $x$ ha be the entire plot. Before lunch, the team plowed $\frac{4.5 \cdot \frac{5}{6} x}{4.25}=\frac{15}{17} x$ ha. According to the condition, $\frac{15}{17} x + 8 = x$, from which $x = 68$ ha.
Answer: 68 ha. | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,532 |
13.026. A tourist traveled 90 km by boat along the river and walked 10 km on foot. For the walking route, 4 hours less were spent compared to the river journey. If the tourist had walked for as long as he had rowed, and rowed for as long as he had walked, these distances would have been equal. How long did he walk and ... | ## Solution.
Let $x$ be the number of hours the tourist walked. Then he spent $(x+4)$ hours on the boat. The tourist's walking speed is $\frac{10}{x}$ km/h; the speed on the boat is $-\frac{90}{x+4}$ km/h. According to the problem, $\frac{10}{x}(x+4)=\frac{90}{x+4} \cdot x$, from which $x=2$ hours.
Answer: 2 and 6 ho... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,533 |
13.028. The numerators of three fractions are proportional to the numbers $1,2,5$, and the denominators are proportional to the numbers $1,3,7$ respectively. The arithmetic mean of these fractions is 200/441. Find these fractions. | Solution.
The required fractions have the form: $\frac{x}{y} ; \frac{2 x}{3 y} ; \frac{5 x}{7 y}$. According to the condition
$\frac{\frac{x}{y}+\frac{2 x}{3 y}+\frac{5 x}{7 y}}{3}=\frac{200}{441}$, from which $\frac{x}{y}=\frac{4}{7} ; \frac{2 x}{3 y}=\frac{8}{21} ; \frac{5 x}{7 y}=\frac{20}{49}$.
Answer: $\frac{4}... | \frac{4}{7};\frac{8}{21};\frac{20}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,534 |
13.030. Three brigades of workers built an embankment. The entire work is valued at 325500 rubles. What salary will each brigade receive if the first one consisted of 15 people and worked for 21 days, the second one - of 14 people and worked for 25 days, and the number of workers in the third brigade, which worked for ... | ## Solution.
Let $x$ rubles be received by one person for one day of work. Then the first team will receive $x \cdot 15 \cdot 21$ rubles; the second - $x \cdot 14 \cdot 25$ rubles; the third $x \cdot 1.4 \cdot 15 \cdot 20$ rubles. According to the condition $15 \cdot 21 x + 14 \cdot 25 x + 1.4 \cdot 15 \cdot 20 x = 32... | 126000,105000,94500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,536 |
13.031. A group of students went on a hike through the Moscow region during their vacation. The first 30 km they walked, $20 \%$ of the remaining part of the route they traveled by raft along the river, and then they walked again, covering a distance 1.5 times greater than the distance they traveled by raft. The remain... | ## Solution.
Let $x$ km be the total length of the route. Then the students floated on a raft for $(x-30) \cdot 0.2$ km, and then walked another $1.5(x-30) \cdot 0.2$ km. According to the problem, $30+(x-30) \cdot 0.2+1.5(x-30) \cdot 0.2+40 \cdot 1.5=x$, from which $x=150$ km.
Answer: 150 km. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,537 |
13.032. In 3.5 hours of operation, one stamping press can produce $42\%$ of all ordered parts. The second press can produce $60\%$ of all parts in 9 hours, and the work speeds of the third and second presses are in the ratio of $6:5$. How long will it take to complete the entire order if all three presses work simultan... | Solution.
Let $x$ parts be the entire order. Then the working speed of the first press is $-\frac{0.42 x}{3.5}$ parts/hour; the second press is $\frac{0.6 x}{9}$ parts/hour; the third press is $\frac{6}{5} \cdot \frac{0.6 x}{9}$ parts/hour. Therefore, all three presses working simultaneously will complete the entire o... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,538 |
13.033. Each of the two typists retyped a manuscript of 72 pages. The first typist retyped 6 pages in the same time it took the second to retype 5 pages. How many pages did each typist retype per hour, if the first finished the work 1.5 hours faster than the second? | Solution.
Let $x$ pages per hour be the typing speed of the first typist, then the second typist's speed is $\frac{5}{6} x$. The first typist worked $\frac{72}{x}$ hours; the second - $\frac{72}{5}$ hours. According to the problem, $\frac{72}{x}=\frac{72}{\frac{5}{6} x}-1.5$, from which $x=9.6($ pages/hour); the secon... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,539 |
13.034. In a store, textbooks on physics and mathematics were received for sale. When $50\%$ of the mathematics textbooks and $20\%$ of the physics textbooks were sold, which amounted to a total of 390 books, the remaining mathematics textbooks were three times as many as the remaining physics textbooks. How many textb... | Solution.
Let $x$ be the number of mathematics textbooks and $y$ be the number of physics textbooks available for sale. The number of textbooks sold is $0.5 x + 0.2 y = 390$. The remaining textbooks are $0.5 x$ mathematics textbooks and $0.8 y$ physics textbooks. According to the condition, $\frac{0.5 x}{0.8 y} = 3$. ... | 720150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,540 |
13.035. A shoe factory completed $20 \%$ of the monthly plan in the first week, produced $120 \%$ of the amount of products made in the first week in the second week, and produced $60 \%$ of the products made in the first two weeks combined in the third week. What is the monthly production plan for shoes, if it is know... | ## Solution.
Let $x$ be the monthly production plan for shoes. In the first week, the factory produced $0,2 x$ pairs of shoes, in the second week - $1,2 \cdot 0,2 x$; in the third week - $0,6(0,2 x+1,2 \cdot 0,2 x)$ pairs. According to the condition,
$0,2 x+1,2 \cdot 0,2 x+0,6(0,2 x+1,2 \cdot 0,2 x)+1480=x$, from whi... | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,541 |
13.036. Fresh mushrooms contain $90 \%$ water by mass, and dried mushrooms contain $12 \%$. How many kilograms of dried mushrooms can be obtained from 22 kg of fresh mushrooms? | ## Solution.
Let $x$ kg of dried mushrooms be obtained from 22 kg of fresh mushrooms. Then the water in the dried mushrooms is $0.12 x$. According to the condition, $x - 0.12 x = 22(1 - 0.9); 0.88 x = 2.2$. Hence, $x = 2.5$ kg.
Answer: 2.5 kg. | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,542 |
13.037. One mill can grind 19 centners of wheat in 3 hours, another - 32 centners in 5 hours, and a third - 10 centners in 2 hours. How should 133 tons of wheat be distributed among these mills so that, starting work simultaneously, they also finish it simultaneously? | ## Solution.
Let $x$ hours be the time the mills work. The first mill will grind $\frac{19}{3} x$ tons of wheat in this time; the second $-\frac{32}{5} x$ tons; the third $-\frac{10}{2} x$ tons. According to the condition, $\frac{19}{3} x+\frac{32}{5} x+\frac{10}{2} x=1330$, from which $x=75$ hours. Therefore, the fir... | 475,480,375 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,543 |
13.038. In three sections of a sports school, there were 96 athletes. The number of members in the speed skating section was 0.8 of the number of members in the skiing section, and the number of members in the hockey section was $33 \frac{1}{3} \%$ of the total number of members of the first two sections. How many athl... | Solution.
Let $x$ people be the number of members in the skiing section; $0.8 x$ people be the number of members in the skating section; $\frac{33 \frac{1}{3}}{100}(x+0.8 x)$ people be the number of members in the hockey section. According to the condition, $x+0.8 x+\frac{33 \frac{1}{3}}{100}(x+0.8 x)=96$, from which ... | 40,32,24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,544 |
13.039. For the first quarter, the car factory completed $25 \%$ of the annual production plan for cars. The number of cars produced in the second, third, and fourth quarters turned out to be proportional to the numbers $11,25,12$ and 13.5. Determine the overfulfillment of the annual plan in percentages, if in the seco... | Solution.
Let $x$ be the annual plan of the car factory. In the first quarter, the factory completed $0.25 x$ of the plan. In the second quarter, the factory produced $11.25 y$ cars, which amounted to $1.08 \cdot 0.25 x = 0.27 x$ of the plan: $11.25 y = 1.08 \cdot 0.25 x = 0.27 x$, hence $y = 0.024 x$. In the third qu... | 13.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,545 |
13.040. Three employees received a bonus of 2970 rubles, with the second receiving $1 / 3$ of what the first received, plus 180 rubles, and the third receiving $1 / 3$ of the second's amount, plus 130 rubles. How much did each receive? | ## Solution.
Let $x$ rubles be the bonus of the first employee. Then $\frac{1}{3} x+180$ rubles for the second; $\frac{1}{3}\left(\frac{1}{3} x+180\right)+130$ rubles for the third. According to the condition, $x+\frac{1}{3} x+180+\frac{1}{3}\left(\frac{1}{3} x+180\right)+130=2970$, from which $x=1800$ rubles - the bo... | 1800,780,390 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,546 |
13.041. A 30% hydrochloric acid solution was mixed with a 10% solution to obtain 600 g of a 15% solution. How many grams of each solution were taken? | ## Solution.
Let $x$ g of a $30\%$ hydrochloric acid solution and $(600-x)$ g of a $10\%$ solution be taken. The mass of hydrochloric acid in the $30\%$ solution is $0.3x$ g; in the $10\%$ solution, it is $0.1(600-x)$ g. In the resulting $15\%$ solution, the mass of hydrochloric acid is $0.15 \cdot 600 = 90$ g. Accord... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,547 |
13.043. The distance between Moscow and Smolensk by railway is 415 km. On this route, there are the cities of Mozhaisk and Vyazma. The distance between Moscow and Mozhaisk is to the distance between Mozhaisk and Vyazma as $7: 9$, and the distance between Mozhaisk and Vyazma is $27 / 35$ of the distance between Vyazma a... | Solution.
Let $x$ km be the distance between Mozhaisk and Vyazma. Then $\frac{7}{9} x$ km is the distance between Moscow and Mozhaisk; $\frac{35}{27} x$ km between Vyazma and Smolensk. According to the condition, $\frac{7}{9} x + x + \frac{35}{27} x = 415$, from which $x = 135$ km. Between Moscow and Mozhaisk $-\frac{... | 105,135,175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,548 |
13.044. Sugar and granulated sugar were delivered to the store in 63 bags, totaling 4.8 tons, with the number of bags of granulated sugar being $25 \%$ more than those of sugar. The mass of each bag of sugar was $3 / 4$ of the mass of a bag of granulated sugar. How much sugar and how much granulated sugar were delivere... | Solution.
Let $x$ be the number of bags of sugar; $1.25 x$ be the number of bags of sand. According to the condition, $x + 1.25 x = 63$, from which $x = 28$ bags of sugar; $1.25 \cdot 28 = 35$ bags of sand. Now let the total be $y$ tons of sand; $4.8 - y$ tons of sugar. Then $\frac{y}{35}$ tons is the weight of one ba... | 1.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,549 |
13.045. A piece of copper and zinc alloy weighing 36 kg contains $45 \%$ copper. What mass of copper needs to be added to this piece so that the resulting new alloy contains $60 \%$ copper? | Solution.
Initially, the alloy contains $36 \cdot 0.45 = 16.2$ kg of copper. Let $x$ kg of copper be added. Then the mass of the new alloy is $36 + x$ kg, and the mass of copper in it is $16.2 + x$ kg. According to the condition, $16.2 + x = (36 + x) \cdot 0.6$, from which $x = 13.5$ kg.
Answer: 13.5 kg. | 13.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,550 |
13.047. A musical theater announced a competition for admission to the orchestra. Initially, it was planned that the number of places for violinists, cellists, and trumpeters would be distributed in the ratio $1.6: 1: 0.4$. However, it was then decided to increase the intake, and as a result, 25% more violinists and 20... | ## Solution.
Let it be planned to recruit $1.6 x$ violinists, $x$ - cellists, $0.4 x$ trumpeters. Then, they recruited $1.25 \cdot 1.6 x$ violinists, $(1-0.2) x$ cellists, $0.4 x$ trumpeters. According to the condition, $1.25 \cdot 1.6 x + (1-0.2) x + 0.4 x = 32$, from which $x=10$. Therefore, they recruited $1.25 \cd... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,551 |
13.049. The first of the unknown numbers is $140\%$ of the second, and the ratio of the first to the third is 14/11. Find these numbers if the difference between the third and the second is 40 units less than the number that is $12.5\%$ of the sum of the first and second numbers. | Solution.
Let $x$ be the first number. Then $\frac{1}{1.4} x$ is the second number, and $\frac{11}{14} x$ is the third number. According to the condition, $\frac{11}{14} x - \frac{1}{1.4} x + 40 = 0.125 \left(x + \frac{1}{1.4} x\right)$, from which $x = 280$. The second number is $-\frac{280}{1.4} = 200$; the third is... | 280,200,220 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,553 |
13.050. The worker's wages for October and November were in the ratio of $3 / 2: 4 / 3$, and for November and December, they were in the ratio of $2: 8 / 3$. In December, he received 450 rubles more than in October, and for exceeding the quarterly plan, the worker was awarded a bonus of $20 \%$ of his three-month earni... | Solution.
Let $x$ rubles be the worker's salary for November. Then for October, it is $\left(\frac{3}{2} : \frac{4}{3}\right) x = \frac{9}{8} x$ rubles; for December, it is $\left(\frac{8}{3} : 2\right) x = \frac{4}{3} x$ rubles. According to the condition, $\frac{4}{3} x = \frac{9}{8} x + 450$, from which $x = 2160$ ... | 1494 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,554 |
13.051. Two cylinders roll down an inclined board 6 m long, one of which has a circumference of 3 dm, and the other 2 dm. Can the circumferences of both cylinders be increased by the same amount so that on the same path one of them makes 3 more revolutions than the other? | Solution.
Assume that the lengths of the circumferences can be increased by $x$ dm. Then the first cylinder will make $\frac{60}{3+x}$ revolutions, and the second will make $-\frac{60}{2+x}$. According to the condition, $\frac{60}{3+x}+3=\frac{60}{2+x}$, from which we get $x^{2}+5 x-14=0$. Solving this equation, we fi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 51,555 |
13.052. An artificial pond has the shape of a rectangle with a difference in sides of 1 km. Two fishermen, located at one vertex of this rectangle, simultaneously set off for a point located at the opposite vertex. One fisherman swam directly along the diagonal, while the other walked along the shore. Determine the dim... | ## Solution.
Let $x$ km be the length of the shorter side of the pond, and $x+1$ km be the length of the second side (Fig. 13.1). Then the first fisherman swam $\sqrt{x^{2}+(x+1)^{2}}$ km, and the second walked $x+x+1$ km. According to the condition, $(x+x+1) / 4 - 0.5 = \sqrt{x^{2}+(x+1)^{2}} / 4$, from which $x=3$ k... | 3\times4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,556 |
13.053. A crystal, while in the formation stage, uniformly increases its mass. Observing the formation of two crystals, it was noted that over a year, the first crystal increased its initial mass by $4 \%$, while the second increased by $5 \%$, at the same time the mass increase of the first crystal over 3 months was e... | Solution.
Let $a$ be the initial mass of the first crystal; $b$ be the second; and let $x$ be the mass increase of the first crystal over 3 months. Over a year, the mass increase of the first crystal is $4x$; of the second crystal - $3x$. According to the problem, $a+4x=1.04a, b+3x=1.05b$. According to the problem, $\... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,557 |
13.054. One farmer obtained an average buckwheat yield of 21 tons per hectare, while another, who had 12 hectares less under buckwheat, achieved an average yield of 25 tons per hectare. As a result, the second farmer collected 300 tons more buckwheat than the first. How many tons of buckwheat were collected by each far... | Solution.
Let's compile the following table:
| Farmer | Area, ha | Yield, c/ha | Mass, c |
| :---: | :---: | :---: | :---: |
| First | $x$ | 21 | $21 x$ |
| Second | $x-12$ | 25 | $25(x-12)$ |
According to the condition $25(x-12)-21 x=300$, from which $x=150$. Then $21 x=3150$ (c); $25(x-12)=3450$ c.
Answer: 3150 a... | 3150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,558 |
13.058. In the laboratory, the speed at which sound propagates along rods made of different materials is measured. In the first experiment, it turned out that the entire path, consisting of three sequentially connected rods, is traveled by sound in $a$ s, and the path consisting of the second and third rods is traveled... | Solution.
Let $t_{1}, t_{2}, t_{3}, t_{4}$ be the times it takes for sound to travel along the 1st, 2nd, 3rd, and new rods, respectively. The fourth rod is the new one that we use to replace the second rod in the second experiment. According to the problem, in the first experiment:
$$
\left\{\begin{array}{l}
t_{1}+t_... | \frac{3}{2(b-)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,559 |
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