SCP-116K: A High-Quality Problem-Solution Dataset and a Generalized Pipeline for Automated Extraction in the Higher Education Science Domain
Paper • 2501.15587 • Published
Datasets:
id stringlengths 38 38 | images images listlengths 1 14 | problem stringlengths 9 6.9k | matched_solution stringlengths 1 9.42k |
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scpvl_2de65fac8fd243649851ba06945f0b9b | 1.1 Damped Harmonic Oscillator\n\nConsider a damped harmonic oscillator such that \(\omega_0^2 < \eta/2m\) (consider also the case \(\omega_0^2 > \eta/2m\)), where \(\omega_0\) is the eigenfrequency of the oscillator, \(m\) is the mass and \(\eta\) is the friction constant of the oscillator moving in a viscous medium (... | Solution:\n\n(a) The mass displacement with respect to its equilibrium position will be called \(x\). The general solution for \(\omega_0^2 > \eta/2m\) is (equation 1.24 from [1]):\n\n\[ x(t) = A e^{-\gamma t} \sin(\Omega_0 t + \delta) \]\n\nbeing: \( \Omega_0 = \sqrt{\omega_0^2 - \gamma^2}; \quad \omega_0^2 = \frac{k}... | |
scpvl_e9246fad9b394f1180edde52398e8e0e | Green’s Function of an Oscillator with Friction\n\nFind Green's function of an oscillator in a viscous medium in the limit of very low friction.\n\nFigure 1.2: Illustration of an oscillating function labeled G(t), showing a graph with t on the horizontal axis and a delta function δ(t–t₀) at a point t₀. The curve illust... | Solution:\nWe can write the following equation of motion u(t) (normalized by the mass m) of a harmonic oscillator with a spring constant k in a viscous medium with friction coefficient η, and under the action of an external force F(t):\n\nd²u/dt² + 2β du/dt + ω₀² u = f(t)\n\nwith β = η/(2m), ω₀² = k/m, f(t) = F/m. If ... | |
scpvl_bf94413b320b45b8955b6226b3726ad7 | ### 1.11 Forced Harmonic Oscillator, Solved with Green’s Functions\n\nAn oscillator with mass m = 1 kg and an eigenfrequency ω₀ = 1 Rad/s is initially at rest.\n\nStarting at t = 0, a force starts acting on the oscillator with a value that exponentially decays with time e^(−t). Find the particular solution for this cas... | For this oscillator with natural frequency ω₀ = 1, Green’s function is sin(t − x), which allows to write the particular solution as:\n\n\[\nu_{\text{part}}(t) = \int_{-\infty}^{t} f(x)\sin(t-x)\,dx = \int_{0}^{t} e^{-x}\sin(t-x)\,dx \tag{1.102}\]\n\nFrom integral tables we have:\n\n\[\n\int_{0}^{t} e^{-x}\sin(t-x)\,dx ... | |
scpvl_3b5b24b1e7ae4f33bbae1a3bc7780f31 | String with a Point Mass Hanging from One of Its Ends\n\nConsider a string of length L, tension T and linear density of mass \(\rho\). The string has the left end fixed and the right end can move freely in the transversal direction. The string is placed in the Earth’s gravitational field (\(g\)). Determine the stationa... | Mathematical formulation\n\nWe assume that the mass is at a distance \(\varepsilon \to 0\) from the right end at \(x = L\). First we will discuss how to find the density of forces. The total force applied to the string is directed in the negative direction:\n\n\[\nf(x) = \frac{F}{L} = -\left[\rho g + mg \delta(x - L + ... | |
scpvl_c4660cd18fda4765af1a1fa855ee77fc | ## 2.7 Static Form of a String with a Mass\n\nIn the middle point of a tense string a mass m is placed. The string has a fixed end and the other can move transversally. Determine the shape of the string when, under the action of gravity, is in mechanical equilibrium.\n\nIllustration of a string with mechanical equilibr... | Mathematical formulation:\n\n\[\nT \frac{d^2 u(x)}{dx^2} = -f(x) = -[-mg\,\delta(x-x_0) - g\rho]\n\]\n(2.73)\n\nIn the indicated ranges the equation is non-homogeneous (uniform density approximation):\n\n\[\nT \frac{d^2u(x)}{dx^2} = g\rho\n\]\n(2.75)\n\n\[\n\rightarrow u(x) = \n\begin{cases}\n\frac{g\rho}{2T} x^2 + A_1... | |
scpvl_b4aeee98d2af4750bdecd12190716977 | 2.34 Oscillations in a String Interrupted by a Spring\n\nA homogeneous string of length L and tension T is connected to a spring with constant β on its mid-point (L/2). Its ends can move freely in the direction transversal to the string. From t = −∞ a local, periodic force acts on the string at x = x₀. Find the station... | Sturm–Liouville problem\nWe will seek the general solution as a function with a forced temporal variation and eigenfunctions of the system:\n\nu(x,t) = ∑ₙ Aₙ(t) Xₙ(x) = ∑ₙ Qₙ sin(ω t) Xₙ(x) (2.540)\n\nReplacing this into the wave equation we get:\n\n∑ₙ Qₙ [ρ(−ω²) Xₙ(x) − T d²Xₙ(x)/dx² + δ(x−L/2) β Xₙ(x)] = δ(x−x₀) ... | |
scpvl_9bf8092a225a401a8c3607d143d15985 | 3.2 Oscillations of a Membrane Fixed at Two Boundaries\n\nA square membrane, whose sides are of length π have two opposite boundaries free to move (at y = 0 and y = π) and the other two (x = 0, x = π), fixed. Starting at t = 0 the membrane is subject to a periodic force with areal density of the form sin(t) sin(x) cos(... | Mathematical formulation\n\n ∂^2u/∂t^2 = c^2 ( ∂^2u/∂x^2 + ∂^2u/∂y^2 ) + sin(t) ⋅ sin(x) ⋅ cos(y)\n Boundary conditions:\n ∂u/∂y |_{y=0} = 0\n ∂u/∂y |_{y=π} = 0\n u(x=0) = 0\n u(x=π) = 0\n Initial conditions:\n u(x,y,0) = 0\n ∂u/∂t |_{t=0} = 0\n\nSturm–Liouville proble... | |
scpvl_0f0798c43a0545e897e6e6f9b7e3efd6 | Find the electrostatic potential inside a semi-infinite region, limited by conductor plates at (y = 0, y = b, x = 0) if the plate at x = 0 is connected to a V0 potential (see figure). The plates at y = 0, y = b are grounded and there are no charges inside the region. | Mathematical formulation\n\n\[\n\begin{cases}\n\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \\\n u(0, y) = V_0 \\\n u(x, 0) = u(x, b) = 0 \\\n u(x \to \infty, y) = 0\n\end{cases}\n\]\n\n(3.24)\n\nSturm–Liouville problem\n\nSeparating variables and taking advantage of the homogeneous boundar... | |
scpvl_f903d036276445ff9ae407a38a6ca355 | 3.13 Temperature Distribution inside a Box Heated by Two Transistors\n\nFind the stationary distribution of temperature T(x, y) inside a box with thermal conductivity k. The box is infinite in the z direction and semi-infinite in the y > 0 direction. The face at x = a is in contact with a thermal reservoir at T = 0. Th... | Solution\n\nSturm–Liouville problem\nWe first seek the general solution of the problem.\n\nSeparation of variables:\n\nu(x, y) = X(x) Y(y) (3.205)\n\nWe arrive at two second-order differential equations. In the x-direction we have the Sturm–Liouville problem:\n\nd^2X/dx^2 + λ X = 0;\n∂X/∂x|_{x=0} = 0; X(a) = 0 (3.20... | |
scpvl_7dda703a217d41f7bdcc4aacb0074597 | Determine the distribution of carriers inside the bar as a function of time if at t = 0 the laser is turned off and never used again.\n\nFigure 4.18: The figure depicts a rectangular prism representing the metallic bar. The dimensions of the bar are denoted as a, b, and c along the x, y, and z axes, respectively. The o... | Solution:\n\nWe first seek the stationary distribution of carriers up until t = 0.\n\nIf we consider that the sources of photoelectrons are at the interface, we need to solve Laplace’s equation in a prism in which five out of the six faces have homogeneous boundary conditions. The central part of the face at z = 0 inje... | |
scpvl_5135e1723d4e4d94b9b33961b72a94b2 | Find the distribution of electric potential inside a circular sector without charges, which spans an angle \( (0 < \varphi < \alpha) \), if the electric potential at the boundaries is as specified in the figure.\n\nDescription of Figure 5.2:\n\nThe figure illustrates a circular sector with boundaries marked by \( u = 0... | Sturm–Liouville problem\n\nWe separate variables to get to the eigenfunctions of the problem:\n\n\[\nu(\rho, \varphi) = Q(\rho) \Phi(\varphi) \tag{5.25}\]\n\nThe Sturm–Liouville problem for \(\Phi(\varphi)\) is:\n\n\[\n\begin{cases}\n\frac{d^2 \Phi}{d \varphi^2} + \lambda \Phi(\varphi) = 0 \\\n\Phi(0) = \Phi(\alpha) = ... | |
scpvl_636c40b892094e388cf1b25da6d42945 | Oscillations of a Quarter of a Membrane\n\nFind the amplitude of the main tone (the lowest frequency) of a membrane with fixed boundaries and the shape of 1/4 of a circle, with tension T and surface mass density ρ. A point‐like hit hits the membrane at rest at t = 0 in the indicated location, with a total transfer of ... | Sturm–Liouville problem\n\n\[\n\begin{cases}\n\Delta V(r,\varphi)+\lambda V(r,\varphi)=0,\\\nV(R,\varphi)=0,\\\nV(r,0)=0,\\\nV\Bigl(r,\tfrac{\pi}{2}\Bigr)=0.\n\end{cases}\n\tag{5.157}\n\]\n\nSeparate variables: let \(V(r,\varphi)=\mathcal{R}(r)\Phi(\varphi)\). One finds the angular eigenfunctions and values\n\n\[\n\P... | |
scpvl_28e379beff3b4fd88e9bdd037a419680 | Find the stationary distribution of temperature u(ρ, φ) in an infinite cylindrical tube with radii ρ₁ = 1 and ρ₂ = 2 if the temperature on the inner surface is u(1) = sin²(φ), whereas the temperature on the outer surface is u(2) = 0.\n\nFigure 5.19: A cylindrical tube is depicted with two concentric circles representin... | Mathematical formulation:\n\nDue to the symmetry we will solve the problem in cylindrical coordinates. Since the cylinder is infinite there is no dependence on the z coordinate. The mathematical formulation is that of Laplace’s problem in a disk with the specified boundary conditions.\n\nGeneral solution:\n\nIn these c... | |
scpvl_b6a8185d128b406d81664f8931e628d1 | General Solution of the Heat Equation in a Finite Cylinder with a Hole\n\nFind the general solution for the temporal variation of temperature in a finite cylinder of height h with inner and outer radii R1 and R2. Both curved surfaces are thermally insulated. The flat surfaces are in contact with a thermal reservoir at ... | Sturm–Liouville problem\n\nWe separate variables:\n\n\[\nu u = W(\rho, \varphi, z)\cdot T(t) \quad (6.2)\]\n\nAuxiliary problem:\n\n\[\n\begin{cases}\n\Delta W + \lambda W = 0 \\\n\left. \frac{\partial W}{\partial \rho} \right|_{\rho=R_1} = 0 \\\n\left. \frac{\partial W}{\partial \rho} \right|_{\rho=R_2} = 0 \\\nW(\rho... | |
scpvl_96de09f4cab347fd95cf7eb69345d6a8 | Half Cylinder with Two Inward Fluxes\n\nFind the stationary distribution of temperature inside a half cylinder of radius R, height L and thermal conductivity coefficient k. The flat back surface is thermally insulated and the curved surface at the front has a temperature T0. Each of the remaining surfaces (upper and lo... | Subtracting T0 from the solution we achieve the curved boundary with homogeneous boundary conditions, and the conditions in the rest of the boundaries don't change.\n\nThe solution can be expanded in a sum of orthogonal functions in the radial and angular variable. We will apply the method of separation of variables. I... | |
scpvl_f54a6af3fefc4b088dc698985c921019 | 6.12 Heat Flux through Half a Cylinder\n\nFind the stationary distribution of temperature in a half cylinder of length L, radius ρ_0, and thermal conductivity k if a heat flux of value J = J_0(Lz - z^2) enters through the curved surface and exits through the bases, being homogeneously distributed. Consider that the fla... | Solution:\n\nTo find all boundary conditions we must equate the flux that enters through the curved surface to the flux that exits through the lateral faces. The total flux that enters through the curved surface is\n\n- \int^π_0 \int^L_0 J_0 [z^2 - Lz] R dφ dz = - R J_0 π \left[\int^L_0 z^2 dz - L \int^L_0 z dz \right]... | |
scpvl_db4236e1ac0c4c89803a05e897322c5c | 7.2 Distribution of Temperature inside a Sphere\n\nA sphere has radius \( R \). Its surface is kept at a temperature that depends on the angles as: \( T_0 \sin(3\theta) \cos(\varphi) \). Find the distribution of temperature inside this object.\n\nA diagram of a sphere labeled with the equation for the surface temperatu... | Mathematical formulation\n\n\[\n\begin{cases} \n\Delta u(r, \theta, \varphi) = 0 & (0 < r < R) \\\n\frac{1}{r^2} \frac{\partial}{\partial r} \left[ r^2 \left( \frac{\partial u}{\partial r} \right) \right] + \frac{1}{r^2} \Delta_{\theta, \varphi} u = 0 \\\nu(r = R) = f(\theta, \varphi) = T_0 \sin(3\theta) \cos(\varphi) ... | |
scpvl_87fc314b9ec847fa9d6f0c9a477d6ef7 | ## 7.5 Electric Potential of a Metallic Sphere inside a Homogeneous Electric Field\n\nA uniform electric field \(\vec{E}\) occupies all space, directed in the \(z\) direction, so that the corresponding electric potential is \(u = -Ez\). A metallic sphere of radius \(R\) is placed in this field. Calculate the equilibriu... | Mathematical formulation\n\n\[\n\begin{cases} \n\Delta u(r, \theta, \varphi) = 0 & (R < r < \infty) \\\n u(r = \infty, \theta, \varphi) = -Ez = -Er \cos(\theta) \\\n u(r = R, \theta, \varphi) = 0 \n\end{cases}\n\]\n\nThe general solution of the Laplace problem in spheric coordinates (problem 7.1) has the form:\n\n\[\nu... | |
scpvl_174f48b878a041d898fa6eaed1674041 | 8.10 Autoconvolution of a Rectangular Pulse\n\nFind the convolution of the following rectangular pulse with itself. Use Fourier transforms from tables to facilitate calculations. Present the result graphically.\n\nIllustration of a rectangular pulse function, f(t), which equals 1 for -1 ≤ t ≤ 1 and 0 otherwise.\n\nMath... | We know that F(ω) = \sqrt{\tfrac{2}{π}} \frac{sin(ω)}{ω}. Then:\n\n\[\n\frac{1}{\sqrt{2π}} \int_{-∞}^{+∞} f(t') f(t−t') dt' = 𝓕^{−1}\bigl([F(ω)]^2\bigr) = \frac{1}{\sqrt{2π}} \int_{-∞}^{+∞} \Bigl[ \frac{2}{π} \frac{sin(ω)}{ω} \Bigr]^2 e^{iωt} dω =\n\]\n\n(8.55)\n\nFrom integral tables:\n\n\[\n\begin{cases}\n\sqrt{\tfr... | |
scpvl_989f4ea058994991a485fdf85bb20091 | 8.11 Fourier Transform of a Bipolar Triangular Pulse\n\nFind the Fourier transform of the pulse shown in the figure. How is the Fourier transform of the convolution of the signal with itself? | We describe mathematically the function:\n\nf(x) = \n\begin{cases} \n1 - x & (0 < x \le 1) \\\n-1 - x & (-1 \le x < 0) \\\n0 & (|x| > 1) \n\end{cases}\n\nAlternatively:\n\nf(x) = \n\begin{cases} \n\mathrm{sign}(x) - x & (|x| < 1) \\\n0 & (|x| > 1) \n\end{cases} \n(8.63)\n\n\mathcal{F}[f(x)] = \frac{1}{\sqrt{2\pi}} \int... | |
scpvl_03fe7c1f096840a4951d5d135f7825ce | 8.13 Fourier Transform of the Convolution of a Triangular Pulse with Itself\n\nFind the Fourier transform of the convolution of the triangular pulse shown in the figure below with itself.\n\nf(x) = \n\begin{cases} \n1 - |x| & (-1 \le x \le 1) \\\n0 & (x < -1; x > 1) \n\end{cases}\n\nWe know that, by definition, the con... | Applying the Fourier transform to this relation and supposing f = g, we get:\n\n\mathcal{F}[f \ast f] = [F(k)]^2\n(8.82)\n\nThe transform of the triangular pulse is\n\nF(k) = \sqrt{\frac{2}{\pi}} \frac{1 - \cos(k)}{k^2}\n(8.83)\n\nUsing 1 - \cos(k) = 2 \sin^2\bigl(\tfrac{k}{2}\bigr)\n\nF(k) = \sqrt{\frac{8}{\pi} \frac{... | |
scpvl_fe3d8aef13644401877284bc35138f2e | The diagram shows a British 50 pence coin. The seven arcs AB, BC, …, FG, GA are of equal length and each arc is formed from the circle of radius a having its centre at the vertex diametrically opposite the mid-point of the arc. Show that the area of the face of the coin is\n\na^2/2 (π − 7 tan(π/14)). | In the figure, the point O is equidistant from each of three vertices A, B and E. The plan is to find the area of the sector AOB by calculating the area of AEB and subtracting the areas of the two congruent isosceles triangles OBE and OAE. The required area is 7 times this.\n\nFirst, we need ∠AEB. We know ∠AOB = 2π/7 a... | |
scpvl_0a99b2b1e42f4052b4e1a225370fbb52 | Inclusion of a sojourn probability: Here is a slightly more generalized version of the biased random walk discussed above. Suppose the walker jumps from any site j to neighboring sites j + 1 and j - 1 with respective probabilities α and β, as before, and stays at the site j with a probability γ, where α + β + γ = 1. Se... | (a) The recursion relation is obviously\n\n P(ja, nτ) = α P(ja − a, nτ − τ) + β P(ja + a, nτ − τ) + γ P(ja, nτ − τ).\n\n(b) The recursion relation can again be re-written as in Eq. (30.21), on using the fact that α + β + γ = 1. The limits to be taken are once again a → 0, τ → 0 and α − β → 0. The same Smoluchowski e... | |
scpvl_b9762cbe0ea242b6bbdde86295bdd150 | ★ 11. In its region of analyticity, an analytic function f(z) = u + iv may also be regarded as a map from (a region of) the complex plane to (a region of) the complex plane (Fig. 22.4).\n\n(a) Show that the Jacobian determinant of the transformation (x, y) → (u, v) is just |f'(z)|^2, where f'(z) denotes the derivative ... | (a) Let us use the convenient notation ∂u/∂x = u_x, ∂u/∂y = u_y, etc. Then the determinant of the Jacobian of the transformation is\n\n|∂(u, v)/∂(x, y)| =\n | u_x u_y |\n | v_x v_y |\n= u_x v_y − u_y v_x = u_x² + v_x²,\n\non using the Cauchy–Riemann conditions. But\n\nf'(z) = df/dz = ∂f/∂x · ∂x/∂z + ∂f/∂y · ∂y/∂z =... | |
scpvl_b397d2279d2b4450aeecaa40273f5d43 | **Q. 11** Which of the following is not the graph of a quadratic polynomial?\n\n(a) \n(b) \n(c) \n(d) | **Sol. (d)** A quadratic polynomial \(y=ax^2+bx+c\) has graph a parabola opening upwards (if \(a>0\)) or downwards (if \(a<0\)), and can cross the \(x\)-axis at at most two points. Option (d) crosses three times, so it cannot be a quadratic. | |
scpvl_faae43b2be304864a90fc8aebc82e1f9 | Q. 8 Find the values of x and y in the following rectangle\n\nD x + 3y C\n ← 3x + y → \nA → B Height = 7 | Solution. By property of a rectangle, lengths are equal, i.e.,\n\nCD = AB\nx + 3y = 13 ...(i)\n\nBreadth are equal, i.e.,\n\nAD = BC\n3x + y = 7 ...(ii)\n\nOn multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get\n\n9x + 3y = 21\nx + 3y = 13\n—————\n8x = 8\nx = 1\n\nOn putting x = 1 in Eq. (i), we get\n\... | |
scpvl_2ef857f25ab4423e869e28dd53c75ef6 | In the figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, \( \angle APB = 50^\circ \) and \( \angle CDP = 30^\circ \). Then, \( \angle PBA \) is equal to\n\n(a) \( 50^\circ \)\n(b) \( 30^\circ \)\n(c) \( 60^\circ \)\n(d) \( 100^\circ \) | Sol. (d): In \( \triangle APB \) and \( \triangle CPD \,\),\n\n\[\angle APB = \angle CPD = 50^\circ \quad[\text{vertically opposite angles}]\]\n\n\[\frac{AP}{PD} = \frac{6}{5} \tag{i}\]\n\[\frac{BP}{CP} = \frac{3}{2.5} = \frac{6}{5} \tag{ii}\]\n\nFrom (i) and (ii),\n\n\[\frac{AP}{PD} = \frac{BP}{CP} \]\n\nHence by SAS ... | |
scpvl_9e63663fba6c4addb679cbc92aeb0d8f | In figure, BD and CE intersect each other at the point P. Is \(\triangle PBC \sim \triangle PDE\)? Why? | True\n\nIn \(\triangle PBC\) and \(\triangle PDE\),\n\[ \angle BPC = \angle EPD \quad [\text{vertically opposite angles}]\]\nNow,\n\[ \frac{PB}{PD} = \frac{5}{10} = \frac{1}{2} \quad (i)\]\nand \n\[ \frac{PC}{PE} = \frac{6}{12} = \frac{1}{2} \quad (ii)\]\nFrom Eqs. (i) and (ii),\n\[ \frac{PB}{PD} = \frac{PC}{PE} \]\nSi... | |
scpvl_9517f4603aa04b399d12a0bc37578fa1 | In \triangle PQR and \triangle MST, \angle P = 55^\circ, \angle Q = 25^\circ, \angle M = 100^\circ and \angle S = 25^\circ. Is \triangle PQR \sim \triangle TSM? Why?\n\nTriangle Diagram | *False*\n\nWe know that, the sum of three angles of a triangle is 180^\circ.\n\nIn \triangle PQR,\n\n\n\angle P + \angle Q + \angle R = 180^\circ\n\n\Rightarrow 55^\circ + 25^\circ + \angle R = 180^\circ\n\n\Rightarrow \angle R = 180^\circ - (55^\circ + 25^\circ) = 180^\circ - 80^\circ = 100^\circ\n\nIn \triangle TSM,\... | |
scpvl_42581700ac954402ad80c8efca609de9 | In figure, if \angle 1 = \angle 2 and \triangle NSQ \equiv \triangle MTR, then prove that \triangle PTS \sim \triangle PRQ. | Thinking Process: Firstly, show that ST \parallel QR with the help of given information, then use AAA similarity criterion.\n\nSolution:\n\nSince, \triangle NSQ \equiv \triangle MTR\n\nSo, SQ = TR ...(i)\n\nAlso, \angle 1 = \angle 2 \Rightarrow PT = PS ...(ii)\n\nFrom (i) and (ii):\n\nPS/SQ = PT/TR \Rightarrow ST \p... | |
scpvl_c9c4c9a1409443e889bbdac9df6de162 | Diagonals of a trapezium \( PQRS \) intersect each other at the point \( O \). \( PQ \parallel RS \) and \( PQ = 3 \, RS \). Find the ratio of the areas of \( \triangle POQ \) and \( \triangle ROS \).\n\nTrapezium Diagram | Thinking Process\n\nFirstly, show that \( \triangle POQ \) and \( \triangle ROS \) are similar by AAA similarity, then use the property of area of similar triangles to get required ratio.\n\nSolution. Given \( PQRS \) is a trapezium in which \( PQ \parallel RS \) and \( PQ = 3 \, RS \)\n\n\[\n\Rightarrow \frac{PQ}{RS} ... | |
scpvl_7d2dd91744924eb5937cb04d3650d035 | In figure, if \( AB \parallel DC \) and \( AC, PQ \) intersect each other at the point \( O \). Prove that \( OA \cdot CQ = OC \cdot AP \).\n\nIllustration of triangles and lines intersecting \n*Illustration shows a quadrilateral formed by two intersecting triangles \( \triangle AOP \) and \( \triangle COQ \), with a ... | Solution: Given \( AC \) and \( PQ \) intersect each other at the point \( O \) and \( AB \parallel DC \).\n\nTo prove \( OA \cdot CQ = OC \cdot AP \).\n\nProof: In \( \triangle AOP \) and \( \triangle COQ \),\n\n\[\n\begin{align*}\n\angle AOP &= \angle COQ &\quad& \text{[vertically opposite angles]} \\\n\angle APO &= ... | |
scpvl_bc6f89e5d6dc43f08c74fbd010fef971 | In figure, if \( DE \parallel BC \), then find the ratio of \( \text{ar} \, (\triangle ADE) \) and \( \text{ar} \, (DECB) \). | Sol. Given, \( DE \parallel BC \), \( DE = 6 \, \text{cm} \) and \( BC = 12 \, \text{cm} \).\n\nIn \( \triangle ABC \) and \( \triangle ADE \),\n\n\[\n\angle ABC = \angle ADE \quad [\text{corresponding angle}]\n\]\n\[\n\angle ACB = \angle AED \quad [\text{corresponding angle}]\n\]\n\nand\n\[\n\angle A = \angle A \quad ... | |
scpvl_fc7ad3cf1f004e4f87641f8c337f55ac | Q. 14 In the figure, PA, QB, RC and SD are all perpendiculars to a line l. AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS. | Given, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm\nAlso, PA, QB, RC and SD are all perpendiculars to line l.\n\nPA ∥ QB ∥ RC ∥ SD\nBy the basic proportionality theorem,\n PQ : QR : RS = AB : BC : CD = 6 : 9 : 12\nLet\n PQ = 6x, QR = 9x and RS = 12x\nSince, the length of PS = 36 cm,\n PQ + QR + RS = 36\n ... | |
scpvl_46fdc5f48a104d09adad42f660e9390f | Q. 16 In figure, line segment DF intersects the side AC of a \Delta ABC at the point E such that E is the mid-point of CA and \angle AEF = \angle AFE. Prove that \frac{BD}{CD} = \frac{BF}{CE}. | Given \Delta ABC, E is the mid-point of CA and \angle AEF = \angle AFE\nTo prove \frac{BD}{CD} = \frac{BF}{CE}\nConstruction Take a point G on AB such that CG ∥ EF.\n\nProof Since, E is the mid-point of CA,\n CE = AE ...(i)\nIn \triangle ACG, CG ∥ EF and E is mid-point of CA,\n CE = GF ...(ii)\n [by mid-... | |
scpvl_f9291431ccdb414f8097fbae463efb66 | Q. 15 The coordinates of the point which is equidistant from the three vertices of the Δ A0B as shown in the figure is\n\n- (a) (x, y)\n- (b) (y, x)\n- (c) (\frac{x}{2}, \frac{y}{2})\n- (d) (\frac{y}{2}, \frac{x}{2}) | Thinking Process\n\n(i) Firstly consider the new point be P(h, k) \n(ii) Secondly, determine the distance PO, PA, and PB by using the formula, \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} and equating them, i.e., PO = PA = PB \n(iii) Further, solving two-two terms at a time and solving them to get required point.\n\nSolution... | |
scpvl_25f655545eec4d6c9aaf5cde099a6c6c | **Q. 2** In figure, if ∠AOB = 125°, then ∠COD is equal to (a) 62.5° (b) 45° (c) 35° (d) 55°\n\n![Illustration: Diagram showing quadrilateral ABCD inscribed in a circle. The angle ∠AOB is marked as 125°.] | **Sol. (d)** We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.\n\ni.e., \n∠AOB + ∠COD = 180° \n⇒ ∠COD = 180° – ∠AOB \n= 180° – 125° = 55° | |
scpvl_d763dc0cb01147078a117058f039d0bd | **Q. 3** In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to (a) 45° (b) 60° (c) 50° (d) 55°\n\n*Illustration Description:* A circle with center O is shown. The diameter AOC is marked, and a chord AB is drawn. ... | **Sol. (c)** In figure, AOC is a diameter of the circle. We know that, diameter subtends an angle 90° at the circle. \nSo, ∠ABC = 90°\n\nIn ΔACB, \n∠A + ∠B + ∠C = 180°\n⇒ ∠A + 90° + 50° = 180°\n⇒ ∠A = 40° \n⇒ ∠OAB = 40°\n\nNow, AT is the tangent at A, so OA ⟂ AT. \nTherefore, ∠OAT = 90°\n\nIn right triangle OAT, \... | |
scpvl_2b3046d15b6b47779b4f3854499b4e1b | In the figure, AT is a tangent to the circle with center O such that OT = 4 cm and angle OTA = 30°. Then, AT is equal to\n\n(a) 4 cm\n(b) 2 cm\n(c) 2√3 cm\n(d) 4√3 cm | Join OA.\n\nWe know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.\n\nTherefore, angle OAT = 90°.\n\nIn triangle OAT,\n\ncos 30° = AT/OT\n\n⇒ √3/2 = AT/4\n\n⇒ AT = 2√3 cm | |
scpvl_1deef1a8123a428f932bdbd8d5735a6e | In the figure, if O is the center of a circle, PQ is a chord, and the tangent PR at P makes an angle of 50° with PQ, then angle POQ is equal to\n\n(a) 100°\n(b) 80°\n(c) 90°\n(d) 75° | Given, angle QPR = 50°\n\nWe know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.\n\nTherefore, angle OPR = 90°\n\n⇒ angle OPQ + angle QPR = 90° [from figure]\n\n⇒ angle OPQ = 90° - 50° = 40° [because angle QPR = 50°]\n\nNow,\n\nOP = OQ = Radius of circle\n\nThe... | |
scpvl_1dc28b5009d54255a9004435e2df5093 | In figure, if PA and PB are tangents to the circle with centre O such that angle APB = 50°, then angle OAB is equal to\n\n(a) 25°\n(b) 30°\n(c) 40°\n(d) 50° | Given, PA and PB are tangent lines.\n\nTherefore PA = PB [since, the length of tangents drawn from an external point to a circle is equal]\n\n⇒ angle PBA = angle PAB = θ [say]\n\nIn triangle PAB,\n\nangle P + angle A + angle B = 180°\n\n⇒ 50° + θ + θ = 180°\n\n⇒ 2θ = 180° - 50° = 130°\n\n⇒ θ = 65°\n\nAlso,\n\nOA ⟂ PA... | |
scpvl_f7c3af49a5e6415fba1b61ec979244d9 | In figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and \(\angle BQR = 70^\circ\), then \(\angle AQB\) is equal to\n\n1. (a) 20°\n2. (b) 40°\n3. (c) 35°\n4. (d) 45° | (b) Given, \(AB \parallel PR\)\n\n*Diagram Description:* The figure shows a circle with center O. A tangent PQR touches the circle at Q, and chord AB is parallel to PR. Angle BQR is marked as \(70^\circ\).\n\nTherefore, \(\angle ABQ = \angle BQR = 70^\circ\) [alternate angles]\n\nAlso, \(QD\) is perpendicular to \(AB\)... | |
scpvl_f01484c0238c4c599883d6ed4858acc4 | In figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD. | Sol. Given AB and CD are common tangent to two circles of unequal radius\n\nTo prove AB = CD\n\nDescription of Extended Figure: The circles are shown with extended lines meeting at a point P.\n\nConstruction Produce AB and CD, to intersect at P.\n\nProof\nPA = PC\n[the length of tangents drawn from an internal point to... | |
scpvl_56204667fd4d4d2abe7b44b48a07b8cf | In the figure, AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. | Solution\n\nGiven AB and CD are tangents to two circles of equal radii.\n\nTo prove AB = CD\n\nConstruction Join OA, OC, O′B, and O′D.\n\nProof\n\nNow, ∠OAB = 90°\n[tangent at any point of a circle is perpendicular to radius through the point of contact]\nThus, AC is a straight line.\n\nAlso,\n∠OAB + ∠OCD = 180°\nAB ∥ ... | |
scpvl_685c1f9b152942af990736ccf0f97768 | In the figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD. | Sol. Given common tangents AB and CD to two circles intersecting at E.\n\nTo prove AB = CD\n\nProof\nEA = EC ...(i)\nEB = ED ...(ii)\n[the lengths of tangents drawn from an internal point to a circle are equal]\n\nOn adding Eqs. (i) and (ii), we get\n\nEA + EB = EC + ED\n⇒ AB = CD\n\nHence proved. | |
scpvl_69bd080e58024e908e10575ebe06eb33 | If \( AB \) is a chord of a circle with centre \( O \), \( AOC \) is a diameter and \( AT \) is the tangent at \( A \) as shown in figure. Prove that \( \angle BAT = \angle ACB \). | Since, \( AC \) is a diameter line, so angle in semi-circle makes an angle \( 90^\circ \).\n\n\( \therefore \angle ABC = 90^\circ \) [by property]\n\nIn \( \triangle ABC \),\n\n\( \angle CAB + \angle ABC + \angle ACB = 180^\circ \) [∵ sum of all interior angles of any triangle is 180^\circ]\n\n\( \Rightarrow \angle C... | |
scpvl_624e257a8c634b57b40556e1e3ddec62 | In figure, tangents \( PQ \) and \( PR \) are drawn to a circle such that \( \angle RPQ = 30^\circ \). A chord \( RS \) is drawn parallel to the tangent \( PQ \). Find the \( \angle RQS \). | Let \( PQ \) and \( PR \) be two tangents drawn from an external point \( P \).\n\n∴ \( PQ = PR \) [the lengths of tangents drawn from an external point to a circle are equal]\n\n⇒ \( \angle PQR = \angle QRP \) [angles opposite to equal sides are equal]\n\nIn \( \triangle PQR \),\n\n\( \angle PQR + \angle QRP + \angl... | |
scpvl_b2b0e82a35654bf49674b4c437ef6532 | In the figure, O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB. | Solution\n\nGiven OT = 13 cm and OE = 5 cm.\n\nSince OE ⟂ AB (radius to tangent), in right-angled \(\triangle OET\):\n\(OT^2 = OE^2 + ET^2\)\n⇒ \(ET^2 = 13^2 - 5^2 = 169 - 25 = 144\)\n⇒ ET = 12 cm.\n\nFrom external point T the two tangents are equal, so both are 12 cm.\nLet PA and QB be the segments on AB; then\n\(TA =... | |
scpvl_765862f9fa124293a498de08cbbf7574 | In figure, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reason for your answer. | Sol. False Given diameter of circle = d ⇒ diagonal of inner square = d. Let side of inner square = x. By Pythagoras, d² = x² + x² = 2x² ⇒ x² = d²/2 ⇒ area of inner square = d²/2. Side of outer square = d ⇒ area = d². Hence area of outer square is not four times area of inner square. | |
scpvl_a1593f50fe674087bb88a15a2d04fa9b | In figure, a square of diagonal 8 cm is inscribed in a circle. Find the area of the shaded region. | Let the side of a square be $a$ and the radius of circle be $r$. \n\nGiven that, length of diagonal of square = 8 cm \n\n$a\sqrt{2} = 8$ \n\n$a = 4\sqrt{2}\text{ cm}$ \n\nNow, \n\nDiagonal of a square = Diameter of a circle \n⇒ Diameter of circle = 8 \n\n⇒ Radius of circle = $r = \tfrac{\text{Diameter}}{2}$ \n\... | |
scpvl_03c91f4d444e40629cd8844b1497c045 | Find the area of the flower bed (with semi-circular ends) shown in figure. | Length and breadth of a circular bed are 38 cm and 10 cm. \nTherefore, the area of rectangle $ACDF = \text{Length}\times\text{Breadth} = 38\times10 = 380\,\text{cm}^2$ \n\nBoth ends of the flower bed are semi-circles. \nTherefore, the radius of a semi-circle = $\tfrac{DF}{2} = \tfrac{10}{2} = 5\,\text{cm}$ \n\nThus... | |
scpvl_c9156dd0e7364b72ac6a71875afc899a | **Question 7:** In the figure, \\( AB \\) is a diameter of the circle, \\( AC = 6 \\text{cm} \\) and \\( BC = 8 \\text{cm} \\). Find the area of the shaded region. (use \\( \pi = 3.14\\))\n\nCircle with diameter AB. Triangle ABC inscribed with AC = 6 cm and BC = 8 cm. Shaded region outside triangle ABC | Given, \\( AC = 6 \\text{cm}\\) and \\( BC = 8 \\text{cm}\\)\n\nWe know that a triangle in a semi-circle with hypotenuse as diameter is a right angled triangle. \nTherefore, \\( \angle C = 90^\circ\\)\n\nIn the right-angled \\(\nabla ACB\\), use the Pythagoras theorem, \n\\[ AB^2 = AC^2 + CB^2 \\\n\\] \n\\[ \\Rig... | |
scpvl_b20a814a502d4edc954820dcd21c24af | Q. 8 Find the area of the shaded field shown in the figure. | **Solution:** In the figure, join \\( ED\\).\n\nFrom the figure, the radius of semi-circle \\( DFE\\), \\( r = 6 - 4 = 2 \\text{m}\\).\n\nNow, \n\nthe area of rectangle \\( ABCD = BC \\times AB = 8 \\times 4 = 32 \\text{m}^2\\)\n\nand \n\nthe area of semi-circle \\( DFE = \\frac{\pi r^2}{2} = \\frac{\pi}{2} (2)^2 = 2... | |
scpvl_6c732a4ce23c443eb9405a2c44677b35 | Q. 11 Find the area of the shaded region in the figure, where arcs drawn with centres \\( A, B, C \\) and \\( D \\) intersect in pairs at mid-point \\( P, Q, R \\) and \\( S \\) of the sides \\( AB, BC, CD \\) and \\( DA \\), respectively of a square \\( ABCD \\). (use \\( \pi = 3.14\\))\n\nSquare ABCD with arcs inters... | **Sol.** Given, side of a square \\( BC = 12 \\text{cm}\\)\n\nSince, \\( Q \\) is a mid-point of \\( BC \\).\n\n\\[ \\therefore \\; \\text{Radius} = BQ = \\frac{12}{2} = 6 \\text{cm}\\]\n\nNow, \n\\[ \\text{area of quadrant } BPQ = \\frac{\pi r^2}{4} = \\frac{3.14 \\times (6)^2}{4} = \\frac{113.04}{4} \\text{cm}^2 \\... | |
scpvl_cd64140a96f4495a8f7d3915a6380dcb | In figure arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region. (use π = 3.14) | Since, ABC is an equilateral triangle.\n\n∴ ∠A = ∠B = ∠C = 60°\nand\nAB = BC = AC = 10 cm\nSo, E, F and D are mid‐points of the sides.\n\n∴ AE = EC = CD = BD = BF = FA = 5 cm\n\nNow, area of sector CDE = (θ π r²)/360 = (60 × 3.14)/360 × (5)²\n= (3.14 × 25)/6 = 78.5/6 = 13.0833 cm²\n\n∴ Area of shaded region = 3 × (Area... | |
scpvl_53c1f0a395f34552bb7fc3851ca463e3 | In figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region. | Given that, radii of each arc (r) = 14 cm.\n\nNow, area of the sector with central ∠P = (∠P/360°) × π r² = (∠P/360°) × π × (14)² cm²\n\nArea of the sector with central ∠Q = (∠Q/360°) × π r² = (∠Q/360°) × π × (14)² cm²\nand area of the sector with central ∠R = (∠R/360°) × π r² = (∠R/360°) × π × (14)² cm²\n\nTherefore, s... | |
scpvl_5595618ff2d54da89a8b899c4ecebfb9 | In figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region. | Given that, radius of each arc (r) = 21 cm\n\nArea of sector with ∠A = (∠A/360°) × π r² = (∠A/360°) × π × (21)² cm²\nArea of sector with ∠B = (∠B/360°) × π r² = (∠B/360°) × π × (21)² cm²\nArea of sector with ∠C = (∠C/360°) × π r² = (∠C/360°) × π × (21)² cm²\nArea of sector with ∠D = (∠D/360°) × π r² = (∠D/360°) × π × (... | |
scpvl_7834788a84e943e4a5e444a0566c5fb1 | In figure, ABCD is a trapezium with AB ∥ DC. AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure. | Sol. Given, AB = 18 cm, DC = 32 cm, height (h) = 14 cm and arc–radius = 7 cm.\n\nSince AB ∥ DC,\n∠A + ∠D = 180° and ∠B + ∠C = 180°.\n\nArea of sector A (and D) = θ/360° × π r^2 = 180°/360° × (22/7) × 7^2 = 77 cm^2.\nSimilarly, area of sector B (and C) = 77 cm^2.\n\nArea of trapezium ABCD = ½ × (AB + DC) × h = ½ × (18 +... | |
scpvl_49eae5e5c8eb45288bb9b5998558bde8 | An archery target has three regions formed by three concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions. | **Solution:**\nLet the diameters of the concentric circles be k, 2k, and 3k. \n∴ Radii of the concentric circles are k/2, k, and 3k/2.\n\n1. Area of inner circle: \n\[ A_1 = \pi \left(\frac{k}{2}\right)^2 = \frac{k^2 \pi}{4} \]\n\n2. Area of middle region: \n\[ A_2 = \pi (k)^2 - \frac{k^2 \pi}{4} = \frac{3k^2 \pi}{4... | |
scpvl_f70cd859bb24419f8441c5f545625d07 | Q. 17 Find the area of the shaded region given in figure.\n\n*Illustration Description:* The figure is a square ABCD with side 14 cm. Inside this square are four semi-circles, each with a radius of 3 cm, arranged such that their diameters lie along the edges of the square. These semi-circles create a smaller square JKL... | Solution:\n\nJoin JK, KL, LM and MJ. \nThere are four equally semi-circles and LMJK formed a square.\n\nTherefore, FH = 14 - (3 + 3) = 8 cm \nSo, the side of square should be 4 cm and radius of semi-circle of both ends are 2 cm each.\n\nTherefore, Area of square JKLM = (4)^2 = 16 cm^2\n\nArea of semi-circle HJM\n\n\... | |
scpvl_e476bd32654e43e5909f96a4487f5888 | A plumbline (sahul) is the combination of (see figure)\n(a) a cone and a cylinder\n(b) a hemisphere and a cone\n(c) frustum of a cone and a cylinder\n(d) sphere and cylinder | Sol. (b)\n\n![Image: Hemisphere and Cone]\nThe figure illustrates a plumbline which is composed of a hemisphere at the top and a cone at the bottom. | |
scpvl_dd255e55d35745428688cf343450f5f5 | The shape of a glass (tumbler) (see figure) is usually in the form of\n(a) a cone\n(b) frustum of a cone\n(c) a cylinder\n(d) a sphere | Sol. (b)\nWe know that, the shape of frustum of a cone is\n\n![Image: Frustum of a Cone]\nThe given figure is usually in the form of frustum of a cone. | |
scpvl_2967bb86f11641d2aceb7bfd8f3b5348 | The shape of a gilli, in the gilli-danda game (see figure) is a combination of\n(a) two cylinders\n(b) a cone and a cylinder\n(c) two cones and a cylinder\n(d) two cylinders and a cone | Sol. (c)\n\n![Image: Cone, Cylinder, Cone]\nThe shape is illustrated as two cones at either end with a cylinder in the middle. | |
scpvl_71f0b8b132d44561b84a684d793cceaf | An ice-cream cone full of ice-cream having radius 5 cm and height 10 cm as shown in figure\n\nCalculate the volume of ice-cream, provided that its 1/6 part is left unfilled with ice-cream. | Sol. Given, ice-cream cone is the combination of a hemisphere and a cone.\nAlso, radius of hemisphere = 5 cm\n\n∴ Volume of hemisphere = (2/3) π r^3 = (2/3) × (22/7) × (5)^3 = 5500/21 = 261.90 cm^3\n\nNow, radius of the cone = 5 cm and height of the cone = 10 – 5 = 5 cm\n\n∴ Volume of the cone = (1/3) π r^2 h\n= (1/3) ... | |
scpvl_c21fab4a82f9429abaecc8c4d7c31569 | For the following distribution,\n\n| Marks | Number of students |\n|----------|--------------------|\n| Below 10 | 3 |\n| Below 20 | 12 |\n| Below 30 | 27 |\n| Below 40 | 57 |\n| Below 50 | 75 |\n| Below 60 | 80 |\n\nthe... | Sol. (c)\n\n| Marks | Number of students | Cumulative frequency |\n|----------|---------------------|----------------------|\n| Below 10 | 3 = 3 | 3 |\n| 10-20 | 12 - 3 = 9 | 12 |\n| 20-30 | 27 - 12 = 15 | 27 |\n| 30-40 | ... | |
scpvl_aff1b7c0d56d4bee8ebe186036a30712 | Consider the data\n\n| Class | Frequency |\n|------------|-----------|\n| 65-85 | 4 |\n| 85-105 | 5 |\n| 105-125 | 13 |\n| 125-145 | 20 |\n| 145-165 | 14 |\n| 165-185 | 7 |\n| 185-205 | 4 |\n\nThe difference of the upper limit of the medi... | Sol. (c)\n\n| Class | Frequency | Cumulative frequency |\n|----------|-----------|----------------------|\n| 65-85 | 4 | 4 |\n| 85-105 | 5 | 9 |\n| 105-125 | 13 | 22 |\n| 125-145 | 20 | 42 |\n| 145-165 | 1... | |
scpvl_a44cb320340a4bb5a91bd5c7b5ef9a0a | The times (in seconds) taken by 150 athletes to run a 110 m hurdle race are tabulated below\n\n| Class | Frequency |\n|----------|-----------|\n| 13.8-14 | 2 |\n| 14-14.2 | 4 |\n| 14.2-14.4| 5 |\n| 14.4-14.6| 71 |\n| 14.6-14.8| 48 |\n| 14.8-15 | 20 |\n\nThe number of a... | Sol. (c) The number of athletes who completed the race in less than 14.6 \n= 2 + 4 + 5 + 71 = 82. | |
scpvl_565e73cc677543a89aad80aec7023a3c | **Q. 11** Consider the following distribution\n\n| Marks obtained | Number of students |\n|----------------------|--------------------|\n| More than or equal to 0 | 63 |\n| More than or equal to 10 | 58 |\n| More than or equal to 20 | 55 |\n| More than or equal to ... | **Sol. (a)**\n\n| Marks obtained | Number of students |\n|----------------|-------------------------|\n| 0-10 | (63 - 58) = 5 |\n| 10-20 | (58 - 55) = 3 |\n| 20-30 | (55 - 51) = 4 |\n| 30-40 | (51 - 48) = 3 |\n| 40-50 | (48 - 42)... | |
scpvl_9f5d1df85c5a4e0a9447b2ebcb3efe69 | A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (see figure). Are the outcomes 1, 2, and 3 equally likely to occur? Give reasons. | No, the outcomes are not equally likely, because 3 contains half part of the total region, so it is more likely than 1 and 2, since 1 and 2 each contain half part of the remaining part of the region.\n\nDescription of the Illustration:\nThe figure accompanying question 6 is a circle divided into four unequal sections. ... | |
scpvl_7a0a883f2b0c4494ab4f6dcdcb4e2f55 | Q. 1 Find the mean of the distribution\n\n| Class | 1-3 | 3-5 | 5-7 | 7-10 |\n|--------|-----|-----|-----|------|\n| Frequency | 9 | 22 | 27 | 17 | | Sol. We first, find the class mark \( x_i \) of each class and then proceed as follows.\n\n| Class | Class marks \( (x_i) \) | Frequency \( (f_i) \) | \( f_i x_i \) |\n|-------|-------------------------|-----------------------|--------------|\n| 1-3 | 2 | 9 | 18 |\n... | |
scpvl_6d7644e3342542e7b9d2e41f448e563a | Q. 2 Calculate the mean of the scores of 20 students in a mathematics test\n\n| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |\n|----------------|-------|-------|-------|-------|-------|\n| Number of students | 2 | 4 | 7 | 6 | 1 | | Sol. We first, find the class mark \( x_i \) of each class and then proceed as follows\n\n| Marks | Class marks \( (x_i) \) | Frequency \( (f_i) \) | \( f_i x_i \) |\n|--------|-------------------------|-----------------------|---------------|\n| 10-20 | 15 | 2 | 30 ... | |
scpvl_70f16ba217ae47b2bfe17175960b54ec | Q. 3 Calculate the mean of the following data\n\n| Class | 4-7 | 8-11 | 12-15 | 16-19 |\n|--------|-----|------|-------|-------|\n| Frequency | 5 | 4 | 9 | 10 | | Sol. Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.\n\nNow, we first find the class mark \( x_i \) of each class and then proceed as follows:\n\n| Class | Class marks \((x_i)\) | Frequency \((f_i)\) | \(f_i x_i\) |\n|-------------|-------... | |
scpvl_db2d6acb5fbf4fb0be8d787690e97c5f | Q. 7 The weights (in kg) of 50 wrestlers are recorded in the following table.\n\n| Weight (in kg) | 100–110 | 110–120 | 120–130 | 130–140 | 140–150 |\n|---------------------|---------|---------|---------|---------|---------|\n| Number of wrestlers | 4 | 14 | 21 | 8 | 3 |\n\nFind the mea... | **Sol.** We first find the class mark \( x_i \) of each class and then proceed as follows:\n\n| Weight (in kg) | Number of wrestlers (\( f_i \)) | Class marks (\( x_i \)) | Deviations \( d_i = x_i - a \) | \( f_i d_i \) |\n|----------------|-----------------------------------|---------------------------|---------------... | |
scpvl_5ed17fd9df3d4eeaaeb8f9c2c7704424 | Q. 8 The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below\n\n| Mileage (\( \mathrm{kmL}^{-1} \)) | 10–12 | 12–14 | 14–16 | 16–18 |\n|------------------------------------|-------|-------|-------|-------|\n| Number of cars | 7 ... | **Sol.**\n\n| Mileage (\( \mathrm{kmL}^{-1} \)) | Class marks (\( x_i \)) | Number of cars (\( f_i \)) | \( f_i x_i \) |\n|------------------------------------|---------------------------|-----------------------------|---------------|\n| 10–12 | 11 | 7 ... | |
scpvl_2c2e778cb2d24326856e9fb0037a32ee | Q. 9 The following is the distribution of weights (in kg) of 40 persons.\n\n| Weight (in kg) | 40–45 | 45–50 | 50–55 | 55–60 | 60–65 | 65–70 | 70–75 | 75–80 |\n|--------------------|-------|-------|-------|-------|-------|-------|-------|-------|\n| Number of persons | 4 | 4 | 13 | 5 | 6 | 5 ... | **Sol.** The cumulative distribution (less than type) table is shown below:\n\n| Weight (in kg) | Cumulative frequency |\n|-----------------|----------------------|\n| Less than 45 | 4 |\n| Less than 50 | 4 + 4 = 8 |\n| Less than 55 | 8 + 13 = 21 |\n| Less than 60 | 2... | |
scpvl_22951f14515c4fb3bae18b7e347070aa | Q. 10 The following table shows the cumulative frequency distribution of marks of 800 students in an examination.\n\n| Marks | Number of students |\n|------------|--------------------|\n| Below 10 | 10 |\n| Below 20 | 50 |\n| Below 30 | 130 |\n| Below 40 | 270... | **Sol.** Here, we observe that 10 students have scored marks below 10 i.e., it lies between class interval 0–10. Similarly, 50 students have scored marks below 20. So, 50 – 10 = 40 students lie in the interval 10–20 and so on. The table of a frequency distribution for the given data is\n\n| Class interval | Number of s... | |
scpvl_97802227c91e4505bb11781f4913f365 | Q. 11 From the frequency distribution table from the following data\n\n| Marks (Out of 90) | Number of candidates |\n|---------------------|----------------------|\n| More than or equal to 80 | 4 |\n| More than or equal to 70 | 6 |\n| More than or equal to 60 | 11 ... | Sol. Here, we observe that, all 34 students have scored marks more than or equal to 0. Since, 32 students have scored marks more than or equal to 10. So, 34 - 32 = 2 students lies in the interval 0-10 and so on. Now, we construct the frequency distribution table.\n\n| Class interval | Number of candidates |\n|----... | |
scpvl_fef072a39b284759a506f73c48e02c93 | Q. 12 Find the unknown entries \( a, b, c, d, e \) and \( f \) in the following distribution of heights of students in a class\n\n| Height (in cm) | Frequency | Cumulative frequency |\n|---------------|-----------|----------------------|\n| 150-155 | 12 | \( a \) |\n| 155-160 | \( b \) |... | | Height (in cm) | Frequency | Cumulative frequency (given) | Cumulative frequency |\n|---------------|-----------|------------------------------|----------------------|\n| 150-155 | 12 | \( a \) | 12 |\n| 155-160 | \( b \) | 25 | \( ... | |
scpvl_1d3ef687643b461f97a5f569e5095663 | Q. 13 The following are the ages of 300 patients getting medical treatment in a hospital on a particular day\n\n| Age (in year) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |\n|--------------------|-------|-------|-------|-------|-------|-------|\n| Number of patients | 60 | 42 | 55 | 70 | 53 | 2... | (i) We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is 0. Similarly, less than 20 include the number of patients which take medical treatment from 0-10 as well as the number of patients which take medical treatment from 10-20. So, the total number of p... | |
scpvl_c9712b7b4be742988305e509ff06a00d | Q. 15 Weekly income of 600 families is tabulated below\n\n| Weekly income (in ₹) | Number of families |\n|----------------------|--------------------|\n| 0-1000 | 250 |\n| 1000-2000 | 190 |\n| 2000-3000 | 100 |\n| 3000-4000 | 40... | First we construct a cumulative frequency table.\n\n| Weekly income (in ₹) | Number of families \( f_i \) | Cumulative frequency \( cf \) |\n|----------------------|------------------------------|-------------------------------|\n| 0-1000 | 250 | 250 |\n|... | |
scpvl_4be75692f47b4b95bc0431be5147366d | Q. 16 The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows\n\n| Speed (in km/h) | Number of players |\n|-----------------|-------------------|\n| 85-100 | 11 |\n| 100-115 | 9 |\n| 115-130 | 8 |... | Sol. First we construct the cumulative frequency table\n\n| Speed (in km/h) | Number of players | Cumulative frequency |\n|-----------------|-------------------|----------------------|\n| 85-100 | 11 | 11 |\n| 100-115 | 9 | 11 + 9 = 20 |\n| 115-... | |
scpvl_c9ede06b00da40df9bc92f7ac49c407a | Q. 17 The monthly income of 100 families are given as below\n\n| Income (in ₹) | Number of families |\n|-----------------|--------------------|\n| 0-5000 | 8 |\n| 5000-10000 | 26 |\n| 10000-15000 | 41 |\n| 15000-20000 | 16 |\n| 200... | Sol. In a given data, the highest frequency is 41, which lies in the interval 10000-15000. \nHere, \( l = 10000, f_m = 41, f_1 = 26, f_2 = 16 \) and \( h = 5000 \)\n\n[\therefore \text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h ]\n\n[= 10000 + \left( \frac{41 - 26}{2 \times 41 - 26 - 16} \r... | |
scpvl_e57032d633ad4289a4865e22dd1666f7 | Q. 18 The weight of coffee in 70 packets are shown in the following table\n\n| Weight (in g) | Number of packets |\n|---------------|-------------------|\n| 200-201 | 12 |\n| 201-202 | 26 |\n| 202-203 | 20 |\n| 203-204 | 9 |\n| 204-205... | Sol. In the given data, the highest frequency is 26, which lies in the interval 201-202 \nHere,\n\n[ l = 201, f_m = 26, f_1 = 12, f_2 = 20 \text{ and (class width)} h = 1 ]\n\n[\therefore \text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h = 201 + \left( \frac{26 - 12}{2 \times 26 - 12 - 20} \... | |
scpvl_16f4b75243cd43898d7dec76415ffc18 | Find the mean marks of students for the following distribution\n\nMarks | Number of students\n0 and above | 80\n10 and above | 77\n20 and above | 72\n30 and above | 65\n40 and above | 55\n50 and above | 43\n60 and above | 28\n70 and above | 16\n80 and above | 10\n90 and above | 8\n100 and above | 0 | Marks | Class marks (xᵢ) | Number of students (Cumulative frequency) | fᵢ | fᵢ·xᵢ\n0–10 | 5 | 80 | 3 | 15\n10–20 | 15 | 77 | 5 | 75\n20–30 | 25 | 72 | 7 ... | |
scpvl_5b05e7279534487f94b9582901bc1d66 | Determine the mean of the following distribution\n\nMarks | Number of students\nBelow 10 | 5\nBelow 20 | 9\nBelow 30 | 17\nBelow 40 | 29\nBelow 50 | 45\nBelow 60 | 60\nBelow 70 | 70\nBelow 80 | 78\nBelow 90 | 83\nBelow 100 | 85 | We compute class frequencies by differencing the cumulative counts:\n\nMarks | fᵢ | Class mark xᵢ | uᵢ=(xᵢ−45)/10 | fᵢ·uᵢ\n0–10 | 5 | 5 | −4 | −20\n10–20 | 4 | 15 | −3 | −12\n20–30 | 8 | 25 | −2 | −16\n30–40 | 12 | 35 | −1 | −... | |
scpvl_b19581c38f7442e9888c37d3ff324a8a | The table below shows the salaries of 280 persons.\n\n| Salary (in ₹ thousand) | Number of persons |\n|------------------------|-------------------|\n| 5-10 | 49 |\n| 10-15 | 133 |\n| 15-20 | 63 |\n| 20-25 |... | First, we construct a cumulative frequency table\n\n| Salary (in ₹ thousand) | Number of persons (f_i) | Cumulative frequency (cf) |\n|------------------------|-------------------------|---------------------------|\n| 5-10 | 49 = f_1 | 49 |\n| 10-15 ... | |
scpvl_2b9de2af439e4f3a94b813c9618ba630 | The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.\n\n| Marks | Frequency |\n|--------|-----------|\n| 20-30 | p |\n| 30-40 | 15 |\n| 40-50 | 25 |\n| 50-60 | 20 |\n| 60-70 | q |\n| 70-80 | 8 |\n| 80-90 | 10 ... | | Marks | Frequency | Cumulative Frequency |\n|--------|-----------|----------------------|\n| 20-30 | p | p |\n| 30-40 | 15 | 15 + p |\n| 40-50 | 25 | 40 + p = cf |\n| 50-60 | 20 = f | 60 + p |\n| 60-70 | q | 60 + p + q ... | |
scpvl_8e001fb411d7446eadbc89111c793ad7 | Size of agricultural holdings in a survey of 200 families is given in the following table:\n\n| Size of agricultural holdings (in hec) | Number of families |\n| -------------------------------------- | ------------------ |\n| 0-5 | 10 |\n| 5-10 ... | Sol.\n\n| Size of agricultural holdings (in hec) | Number of families (fi) | Cumulative frequency |\n| -------------------------------------- | ----------------------- | -------------------- |\n| 0-5 | 10 | 10 |\n| 5-10 ... | |
scpvl_56ff61b6a7504a23aa16f57ea0c8c914 | **Q. 12** The annual rainfall record of a city for 66 days is given in the following table.\n\n| Rainfall (in cm) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |\n|------------------|------|-------|-------|-------|-------|-------|\n| Number of days | 22 | 10 | 8 | 15 | 5 | 6 |\n\nCalculate the me... | **Solution.** We observe that the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10.\n\nSo, the total annual rainfall record of a city for less than 10 cm is \(0 + 22 = 22\) days. Conti... | |
scpvl_a215c99b43184108ae91603061305304 | **Q. 13** The following is the frequency distribution of duration for 100 calls made on a mobile phone.\n\n| Duration (in s) | Number of calls |\n|-----------------|-----------------|\n| 95-125 | 14 |\n| 125-155 | 22 |\n| 155-185 | 28 |\n| 185-215 ... | **Sol.** First, we calculate class marks as follows\n\n| Duration (in s) | Number of calls \( (f_i) \) | Class marks \( (x_i) \) | \( u_i = \frac{x_i - a}{h} \) | \( f_i u_i \) |\n|-----------------|-----------------------------|-------------------------|-------------------------------|---------------|\n| 95-125 ... | |
scpvl_d35c8fce66bb4118b56f936015cc2c0b | Compute up through third differences of the discrete function displayed in the $x_k, y_k$ columns of Table 3.1. (The integer variable $k$ also appears for convenience.)\n\nThe required differences appear in the remaining three columns. Table 3.1 is called a difference table. Its diagonal structure has become a standard... | Table 3.1\n\n| $k$ | $x_k$ | $y_k$ | $\Delta y_k$ | $\Delta^2 y_k$ | $\Delta^3 y_k$ |\n|-----|-------|-------|--------------|----------------|----------------|\n| 0 | 1 | 1 | | | |\n| 1 | 2 | 8 | 7 | | |\n| 2 | 3 ... | |
scpvl_ec68021dfa9148ad8e673cab390547ab | Compute differences of the function displayed in the first two columns of Table 3.3. This may be viewed as a type of “error function,” if one supposes that all its values should be zero but the single 1 is a “unit error.” How does this unit error affect the various differences?\n\nSome of the required differences appea... | The error influences a triangular portion of the difference table, increasing for higher differences and having a binomial coefficient pattern. | |
scpvl_a0df849d63db47d5adf9b1e2da0b70bb | Use Table 4.3 to expand \( k^{(5)} \). | Using row 5 of the table, \n\n\[ k^{(5)} = 24k - 50k^2 + 35k^3 - 10k^4 + k^5. \] | |
scpvl_4ee1c9fdb512465da604d4956934ffe5 | Use Table 4.4 to expand k^5 in factorial polynomials. | Using row 5 of the table,\n\nk^5 = k_{(1)} + 15 k_{(2)} + 25 k_{(3)} + 10 k_{(4)} + k_{(5)}. | |
scpvl_7cefd65d0acd41aeb255de5d0857749b | Use Table 4.3 to express y_k = 2 k^{(3)} − k^{(2)} + 4 k^{(1)} − 7 as a conventional polynomial. | 2k^3 - 7k^2 + 9k - 7 | |
scpvl_c80c40e9bf6d4a39b31f42550f32c5bb | Use Table 4.3 to express y_k = k^{(6)} + k^{(3)} + 1 as a conventional polynomial. | k^6 - 15k^5 + 85k^4 - 224k^3 + 271k^2 - 118k + 1 | |
scpvl_e8d357cf9a60452ab698198ab6bf09d1 | Find the polynomial of degree three which takes the four values listed in the y_k column below at the corresponding arguments x_k.\n\nTable 6.2\nk | x_k | y_k | Δy_k | Δ^2y_k | Δ^3y_k\n0 | 4 | 1 | | | \n1 | 6 | 3 | 2 | | 3 \n2 | 8 | 8 | 5 | 7 | \n3 | 10 | 20 | 1... | Substituting the circled numbers in their places in Newton’s formula,\n\np(x_k) = 1\n + \tfrac12 (x_k − 4)\n + \tfrac{3}{8}(x_k − 4)(x_k − 6)\n + \tfrac{1}{48}(x_k − 4)(x_k − 6)(x_k − 8)\n\nwhich can be simplified to\n\np(x_k) = \frac{1}{24}\bigl[2 x_k^3 − 27 x_k^2 + 142 x_k − 240\bigr]\n\nthough o... | |
scpvl_c29fea107c334e87ab722bf0faa13522 | Apply Newton’s formula to find a polynomial of degree four or less which takes the y_k values of Table 6.3.\n\nTable 6.3\nk | x_k | y_k | Δ | Δ^2 | Δ^3 | Δ^4\n0 | 1 | 1 | | | | \n1 | 2 | −1 | −2 | | 4 | \n2 | 3 | 1 | 2 | −4 | | −8 \n3 | 4 | −1 | −2 | 4 | | 8 \n4... | The needed differences are circled. Substituting the circled entries into their places in Newton’s formula,\n\np_k = 1 − 2k + \tfrac{3}{2} k^{(2)} − \tfrac{5}{6} k^{(3)} + \tfrac{16}{24} k^{(4)},\n\nwhich is also\n\np_k = \tfrac12\,(2 k^4 − 16 k^3 + 40 k^2 − 32 k + 3).\n\nSince k = x_k − 1, this result can also be wri... | |
scpvl_79dcbf259d9b4d93b141378747c425e5 | Find a polynomial of degree four which takes these values.\n\nx_k | 2 | 4 | 6 | 8 | 10\n----+---+---+---+---+---\ny_k | 0 | 0 | 1 | 0 | 0 | (x - 2)(x - 4)/64 [8 - 4(x - 6) + (x - 6)(x - 8)] | |
scpvl_38973d821bc14264a97b161b80e408b7 | Find a polynomial of degree three which takes these values.\n\nx_k | 3 | 4 | 5 | 6\n----+---+---+---+---\ny_k | 6 | 24 | 60 | 120 | 6 + 18(x - 3) + 9(x - 3)(x - 4) + (x - 3)(x - 4)(x - 5) |
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Dataset Card for SCP-VL — Preview
Preview release (v0.1-preview, September 2026). This dataset is an early release and has known quality issues. Problems, images, and solutions are automatically extracted and matched. Information may be missing or corrupted; attached images may be incorrect, incomplete, or unrelated; solutions may contain errors or omit required answers. Further corrections and quality improvements are planned. Do not treat this release as a verified benchmark or its solutions as guaranteed ground truth.
当前为预览版。 数据由自动抽取与匹配流程生成，可能存在题目信息缺失、裁图不完整、错配或多配图片、公式/化学结构丢失、题解错误等问题。后续将逐步修订；使用前请按具体任务进行检查。
Dataset Description
Paper
Dataset Summary
SCP-VL is a multimodal extension of the SCP scientific problem-solution dataset family. This preview contains 41,828 problem-solution pairs, each with at least one attached image and a text solution. Images include scientific diagrams, plots, tables, and other visual material associated with the problems.
The dataset spans physics, chemistry, biology, mathematics, and related subjects. It contains English and Chinese material, with educational levels ranging from secondary-school and competition problems to university-level content. An attached image does not necessarily mean that the problem requires visual reasoning to solve.
This release contains 51,405 image attachments across the 41,828 examples; some images occur in multiple examples. Images are embedded in the Parquet files and can be loaded directly with Hugging Face Datasets. Separate image downloads and local source paths are not required.
GitHub: SCP dataset pipeline
Related text-only dataset: EricLu/SCP-378K
Supported Tasks
Potential research uses include:
- Scientific visual question answering and multimodal reasoning.
- VLM instruction tuning after task-specific validation and filtering.
- Reward-model or verifier research using the extracted reference solutions, subject to their known errors.
Solutions may be short answers rather than complete derivations. Suitability for supervised training or evaluation should be assessed separately for the intended use.
Languages
English and Chinese.
Dataset Structure
Data Fields
The public dataset contains exactly four columns:
| Field | Type | Description |
|---|---|---|
id |
string | Opaque example identifier, such as scpvl_ followed by 32 hexadecimal characters. It does not encode a source title, source path, or page number. |
images |
list of Image |
One or more attached images, embedded as bytes in the Parquet files. Hugging Face Datasets decodes these into a list of PIL images by default. |
problem |
string | Extracted problem text, which may contain mathematical notation and references to figures or tables. |
matched_solution |
string | Automatically matched text solution. It may contain OCR errors, incomplete reasoning, or incorrect results. |
Source-title fields, source identifiers, source filenames, and page metadata are not included. Embedded images have no source paths or filenames in the image feature. Text link destinations and explicit occurrences of the corresponding source-title string have been removed during export. Scientific content and printed text within the crops are otherwise retained; this is not a claim that source material cannot be recognized from its content.
Data Splits
| Split | Examples |
|---|---|
train |
41,828 |
This is a single release split. No separate validation or test split, or contamination-free evaluation partition, is provided.
Loading the Dataset
from datasets import load_dataset
dataset = load_dataset(
"EricLu/SCP-VL",
split="train",
revision="v0.1-preview",
)
example = dataset[0]
print(example["id"])
print(example["problem"])
print(example["matched_solution"])
print(len(example["images"]))
image = example["images"][0] # PIL image
For streaming access:
dataset = load_dataset(
"EricLu/SCP-VL",
split="train",
revision="v0.1-preview",
streaming=True,
)
example = next(iter(dataset))
Relationship to SCP-378K
SCP-378K contains 377,705 text problem-solution pairs. SCP-VL publishes the image-bearing subset recovered through the newer multimodal pipeline as a separate dataset.
The multimodal pipeline reprocessed problems with visual references before the earlier text-only filtering stages. It recovered 46,212 usable-by-model examples, of which 41,828 have attached images and are included here. The remaining 4,384 examples without images are excluded from this release.
SCP-VL is not a disjoint set of new questions relative to SCP-378K: some questions overlap or replace earlier text-only versions. Do not simply concatenate the two datasets without checking overlap. The 417,149-example merged dataset is not included in this repository. SCP-378K's published data remain unchanged.
Dataset Creation
Source Data
The dataset was produced from scientific and educational documents using an automated pipeline. Public records contain the problem, associated visual material, and a matched solution, without source-title metadata.
Extraction and Matching
The pipeline includes:
- Text recognition, segmentation, and extraction of problems and candidate solutions.
- Figure/table detection and image cropping.
- Extraction of visual references from each problem.
- Retrieval of candidate figures within the corresponding document, followed by model-based image matching.
- Solution matching with the selected images supplied to a vision-language model.
- Model-based usability filtering.
This preview uses WeMM-Embedding-9B for figure retrieval and Qwen3.8-27B for the main visual matching and filtering stages. More than one image can be retained for a problem. These automated checks do not guarantee that all retained images or solutions are correct.
Preview Quality Check
A fixed random sample of 100 image-bearing examples, containing 117 images, was reviewed by a Codex assistant through direct inspection of the problem text, images, and solution text. Selected cases were compared with the original document pages.
| Review outcome | Examples |
|---|---|
| Clear issues requiring correction or exclusion | 30 |
| Main answer remained usable in the review, but revisions were needed | 26 |
| No obvious issue found during this review | 44 |
This was an AI-assisted spot check, not independent human expert annotation. Complex derivations were not exhaustively verified. These counts describe the sample; they are not a precise error-rate estimate for the complete dataset. “No obvious issue found” does not mean an example is fully correct.
The identified problems have not been comprehensively corrected in this preview. Apart from removing source-related export metadata and link destinations, this release retains the current extracted problem/image/solution content, including known problematic examples.
Considerations for Using the Data
Known Limitations
- Missing problem information: some examples lack an explicit question, shared instructions, or necessary context from another problem.
- Image matching errors: a problem may include a wrong image or extra images belonging to other problems.
- Incomplete crops: scale bars, numerical labels, or other essential details may be cut off.
- Text extraction errors: mathematical expressions, chemical structures, tables, and symbols may be lost, corrupted, or inconsistently represented in text and images.
- Solution errors: answers may be incomplete, internally inconsistent, or wrong, including mistakes already present in the source material.
- Answer information in images: some crops include worked-example annotations or solution diagrams, which can reveal part of the answer.
- Formatting noise: residual placeholders, literal newline escapes, repeated content, or redundant visual descriptions may remain.
- Uneven coverage: subjects, languages, difficulty levels, and source quality are not uniformly distributed.
- Overlap: overlap with related SCP releases or external training/evaluation datasets has not been eliminated for this release.
Feedback and Updates
Please report issues through the dataset's Community tab, including the example id, the observed problem, and any proposed correction. The opaque ID is intended to support tracking corrections across releases. Future releases may correct or remove examples; IDs of retained examples will be preserved where possible.
Use the v0.1-preview tag or a commit revision for reproducible experiments. The main branch may receive subsequent corrections.
Additional Information
Dataset Curators
The dataset was created as part of research on scientific reasoning and automated problem-solution extraction.
Licensing Information
This dataset is released under the CC-BY-NC-SA 4.0 license, consistent with the SCP-378K release.
Citation Information
If you use this dataset, please cite the SCP pipeline paper and specify the SCP-VL version used:
@misc{lu2025scp116khighqualityproblemsolutiondataset,
title={SCP-116K: A High-Quality Problem-Solution Dataset and a Generalized Pipeline for Automated Extraction in the Higher Education Science Domain},
author={Dakuan Lu and Xiaoyu Tan and Rui Xu and Tianchu Yao and Chao Qu and Wei Chu and Yinghui Xu and Yuan Qi},
year={2025},
eprint={2501.15587},
archivePrefix={arXiv},
primaryClass={cs.CL},
url={https://arxiv.org/abs/2501.15587},
}
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