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The total number of adult animals in Bundle Ranch equals 8. The number of adult owl in Jefferson Circus equals the number of adult parrot in Jefferson Circus. The number of adult crow in Bundle Ranch equals 2 times the total number of adult animals in Mayer Aquarium. The number of adult parrot in Jefferson Circus exists, and its number is greater than 0. The number of adult parrot in Mayer Aquarium equals the total number of adult animals in Jefferson Circus.
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How many adult parrot does Jefferson Circus have?
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The question is difficult, so we use equations to solve it. Define adult parrot in Jefferson Circus as x; we don't know its value yet but will find it out later. Define adult owl in Jefferson Circus as b; so b = x = x. Define total number of adult animals in Jefferson Circus as D; so D = x + b = x + x = 2*x. Define adult parrot in Mayer Aquarium as d; so d = D = 2*x. Define total number of adult animals in Mayer Aquarium as K; so K = d = 2*x. Define adult crow in Bundle Ranch as s; R = K = 2*x; so s = 2 * R = (2) * (2*x) = 4*x. Define total number of adult animals in Bundle Ranch as z; so z = s = 4*x. We know z = 8, so we have 4*x = 8 and we solve it through the following steps: Simplifying the equation: 4*x = 8. Move all terms to one side: 4*x - 8 = 0. Isolate the term with x: 4*x = 8. Divide both sides by 4: x = 8 / 4. Solution: x = 2. Answer: 2.
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crazy_zootopia
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zero_context
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The number of adult deer in Oakridge Riverside exists, and its number is greater than 0. The number of adult bear in Cedar Valley equals 3. The number of adult racoon in Oakridge Riverside equals the total number of adult animals in Maple Creek. The number of adult bear in Maple Creek equals 4 plus the total number of adult animals in Cedar Valley. The total number of adult animals in Oakridge Riverside equals 8.
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How many adult deer does Oakridge Riverside have?
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The question is difficult, so we use equations to solve it. Define adult bear in Cedar Valley as i; so i = 3. Define total number of adult animals in Cedar Valley as l; so l = i = 3. Define adult bear in Maple Creek as a; n = l = 3; so a = 4 + n = 4 + 3 = 7. Define total number of adult animals in Maple Creek as A; so A = a = 7. Define adult racoon in Oakridge Riverside as p; so p = A = 7. Define adult deer in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define total number of adult animals in Oakridge Riverside as h; so h = p + x = 7 + x = x + 7. We know h = 8, so we have x + 7 = 8 and we solve it through the following steps: Simplifying the equation: x + 7 = 8. Move all terms to one side: x - 1 = 0. Isolate the term with x: 1*x = 1. Divide both sides by 1: x = 1 / 1. Solution: x = 1. Answer: 1.
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crazy_zootopia
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zero_context
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The number of adult racoon in Oakridge Riverside equals 4 plus the number of adult racoon in Beverly Forest. The number of adult deer in Maple Creek equals the number of adult wolf in Oakridge Riverside. The number of adult racoon in Beverly Forest exists, and its number is greater than 0. The number of adult deer in Oakridge Riverside equals the number of adult wolf in Oakridge Riverside. The total number of adult animals in Oakridge Riverside equals 22. The number of adult wolf in Maple Creek equals 4 times the number of adult wolf in Beverly Forest. The number of adult deer in Beverly Forest equals the sum of the number of adult racoon in Oakridge Riverside, the total number of adult animals in Oakridge Riverside, and the number of adult racoon in Beverly Forest. The number of adult racoon in Maple Creek equals the number of adult wolf in Beverly Forest. The number of adult wolf in Beverly Forest equals 2. The number of adult wolf in Oakridge Riverside equals the sum of the number of adult racoon in Oakridge Riverside and the number of adult racoon in Beverly Forest.
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How many adult racoon does Beverly Forest have?
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The question is difficult, so we use equations to solve it. Define adult racoon in Beverly Forest as x; we don't know its value yet but will find it out later. Define adult racoon in Oakridge Riverside as K; i = x = x; so K = 4 + i = 4 + x = x + 4. Define adult wolf in Oakridge Riverside as Y; so Y = K + x = x + 4 + x = 2*x + 4. Define adult deer in Oakridge Riverside as g; so g = Y = 2*x + 4. Define total number of adult animals in Oakridge Riverside as U; l = g + Y = 2*x + 4 + 2*x + 4 = 4*x + 8; so U = l + K = 4*x + 8 + x + 4 = 5*x + 12. We know U = 22, so we have 5*x + 12 = 22 and we solve it through the following steps: Simplifying the equation: 5*x + 12 = 22. Move all terms to one side: 5*x - 10 = 0. Isolate the term with x: 5*x = 10. Divide both sides by 5: x = 10 / 5. Solution: x = 2. Answer: 2.
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zero_context
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The number of adult bear in Maple Creek equals 2 plus the difference between the number of adult bear in Cedar Valley and the number of adult racoon in Cedar Valley. The number of adult racoon in Cedar Valley exists, and its number is greater than 0. The number of adult bear in Cedar Valley equals the number of adult racoon in Cedar Valley. The total number of adult animals in Maple Creek equals 8. The number of adult racoon in Maple Creek equals 3 plus the number of adult racoon in Cedar Valley.
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How many adult racoon does Cedar Valley have?
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The question is difficult, so we use equations to solve it. Define adult racoon in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult racoon in Maple Creek as l; n = x = x; so l = 3 + n = 3 + x = x + 3. Define adult bear in Cedar Valley as Q; so Q = x = x. Define adult bear in Maple Creek as O; y = Q - x = (x) - (x) = 0; so O = 2 + y = 2 + 0 = 2. Define total number of adult animals in Maple Creek as g; so g = O + l = 2 + x + 3 = x + 5. We know g = 8, so we have x + 5 = 8 and we solve it through the following steps: Simplifying the equation: x + 5 = 8. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.
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The number of adult racoon in Oakridge Riverside exists, and its number is greater than 0. The total number of adult animals in Cedar Valley equals 30. The number of adult racoon in Pine Ridge equals the number of adult racoon in Oakridge Riverside. The number of adult deer in Oakridge Riverside equals 4 times the number of adult racoon in Oakridge Riverside. The number of adult racoon in Cedar Valley equals the number of adult deer in Cedar Valley. The number of adult deer in Pine Ridge equals 2 times the sum of the number of adult racoon in Oakridge Riverside, the total number of adult animals in Oakridge Riverside, and the number of adult deer in Oakridge Riverside. The number of adult deer in Cedar Valley equals the total number of adult animals in Oakridge Riverside.
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How many adult racoon does Oakridge Riverside have?
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The question is difficult, so we use equations to solve it. Define adult racoon in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define adult deer in Oakridge Riverside as M; p = x = x; so M = 4 * p = (4) * (x) = 4*x. Define total number of adult animals in Oakridge Riverside as t; so t = M + x = 4*x + x = 5*x. Define adult deer in Cedar Valley as P; so P = t = 5*x. Define adult racoon in Cedar Valley as u; so u = P = 5*x. Define total number of adult animals in Cedar Valley as Z; so Z = P + u = 5*x + 5*x = 10*x. We know Z = 30, so we have 10*x = 30 and we solve it through the following steps: Simplifying the equation: 10*x = 30. Move all terms to one side: 10*x - 30 = 0. Isolate the term with x: 10*x = 30. Divide both sides by 10: x = 30 / 10. Solution: x = 3. Answer: 3.
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The number of adult crow in Mayer Aquarium equals 2 plus the number of adult eagle in Mayer Aquarium. The number of adult crow in Jefferson Circus equals 3 times the number of adult crow in Hamilton Farm. The number of adult parrot in Mayer Aquarium equals the number of adult crow in Mayer Aquarium. The number of adult crow in Hamilton Farm equals 1. The number of adult eagle in Hamilton Farm equals 3 times the number of adult eagle in Mayer Aquarium. The number of adult eagle in Mayer Aquarium equals 4 times the number of adult eagle in Jefferson Circus. The number of adult eagle in Jefferson Circus exists, and its number is greater than 0. The total number of adult animals in Mayer Aquarium equals 52. The number of adult parrot in Hamilton Farm equals the total number of adult animals in Mayer Aquarium. The number of adult parrot in Jefferson Circus equals 2 times the total number of adult animals in Mayer Aquarium.
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How many adult eagle does Jefferson Circus have?
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The question is difficult, so we use equations to solve it. Define adult eagle in Jefferson Circus as x; we don't know its value yet but will find it out later. Define adult eagle in Mayer Aquarium as p; D = x = x; so p = 4 * D = (4) * (x) = 4*x. Define adult crow in Mayer Aquarium as N; G = p = 4*x; so N = 2 + G = 2 + 4*x = 4*x + 2. Define adult parrot in Mayer Aquarium as m; so m = N = 4*x + 2. Define total number of adult animals in Mayer Aquarium as K; V = N + p = 4*x + 2 + 4*x = 8*x + 2; so K = V + m = 8*x + 2 + 4*x + 2 = 12*x + 4. We know K = 52, so we have 12*x + 4 = 52 and we solve it through the following steps: Simplifying the equation: 12*x + 4 = 52. Move all terms to one side: 12*x - 48 = 0. Isolate the term with x: 12*x = 48. Divide both sides by 12: x = 48 / 12. Solution: x = 4. Answer: 4.
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zero_context
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The number of adult wolf in Pine Ridge equals the sum of the number of adult deer in Maple Creek and the number of adult bear in Maple Creek. The number of adult bear in Maple Creek equals 1 plus the number of adult deer in Maple Creek. The number of adult bear in Cedar Valley equals the number of adult bear in Maple Creek. The total number of adult animals in Cedar Valley equals 8. The number of adult bear in Pine Ridge equals 4. The number of adult deer in Maple Creek exists, and its number is greater than 0. The number of adult deer in Cedar Valley equals the total number of adult animals in Maple Creek.
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How many adult deer does Maple Creek have?
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The question is difficult, so we use equations to solve it. Define adult deer in Maple Creek as x; we don't know its value yet but will find it out later. Define adult bear in Maple Creek as K; d = x = x; so K = 1 + d = 1 + x = x + 1. Define adult bear in Cedar Valley as M; so M = K = x + 1. Define total number of adult animals in Maple Creek as V; so V = x + K = x + x + 1 = 2*x + 1. Define adult deer in Cedar Valley as s; so s = V = 2*x + 1. Define total number of adult animals in Cedar Valley as W; so W = s + M = 2*x + 1 + x + 1 = 3*x + 2. We know W = 8, so we have 3*x + 2 = 8 and we solve it through the following steps: Simplifying the equation: 3*x + 2 = 8. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of public highschool in Oakbridge City exists, and its number is greater than 0. The total number of schools in Oakbridge City equals 13. The number of culinarian school in Oakbridge City equals 2 plus the total number of schools in Riverton City. The number of culinarian school in Riverton City equals 4. The number of public highschool in Riverton City equals 4.
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How many public highschool does Oakbridge City have?
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The question is difficult, so we use equations to solve it. Define public highschool in Riverton City as a; so a = 4. Define culinarian school in Riverton City as l; so l = 4. Define total number of schools in Riverton City as R; so R = a + l = 4 + 4 = 8. Define public highschool in Oakbridge City as x; we don't know its value yet but will find it out later. Define culinarian school in Oakbridge City as f; I = R = 8; so f = 2 + I = 2 + 8 = 10. Define total number of schools in Oakbridge City as p; so p = x + f = x + 10 = x + 10. We know p = 13, so we have x + 10 = 13 and we solve it through the following steps: Simplifying the equation: x + 10 = 13. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.
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The number of culinarian school in Hawkesbury equals 2 plus the number of regional medical school in Brightford. The number of culinarian school in Brightford equals 2. The number of regional medical school in Evervale City equals the sum of the number of private middle school in Hawkesbury, the number of private middle school in Evervale City, the number of regional medical school in Brightford, and the total number of schools in Hawkesbury. The number of private middle school in Brightford equals 2 times the difference between the number of culinarian school in Brightford and the number of culinarian school in Evervale City. The number of culinarian school in Evervale City equals 4. The number of private middle school in Evervale City equals 2 times the sum of the number of regional medical school in Brightford, the total number of schools in Hawkesbury, and the number of culinarian school in Hawkesbury. The number of private middle school in Hawkesbury equals 2 plus the number of culinarian school in Hawkesbury. The number of regional medical school in Hawkesbury exists, and its number is greater than 0. The number of regional medical school in Brightford equals 2. The total number of schools in Hawkesbury equals 14.
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How many regional medical school does Hawkesbury have?
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The question is difficult, so we use equations to solve it. Define regional medical school in Brightford as Y; so Y = 2. Define culinarian school in Hawkesbury as T; C = Y = 2; so T = 2 + C = 2 + 2 = 4. Define private middle school in Hawkesbury as I; P = T = 4; so I = 2 + P = 2 + 4 = 6. Define regional medical school in Hawkesbury as x; we don't know its value yet but will find it out later. Define total number of schools in Hawkesbury as Q; E = T + x = 4 + x = x + 4; so Q = E + I = x + 4 + 6 = x + 10. We know Q = 14, so we have x + 10 = 14 and we solve it through the following steps: Simplifying the equation: x + 10 = 14. Move all terms to one side: x - 4 = 0. Isolate the term with x: 1*x = 4. Divide both sides by 1: x = 4 / 1. Solution: x = 4. Answer: 4.
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The number of public highschool in Brightford equals 2 times the number of elementary school in Brightford. The number of elementary school in Glenfield City equals 2. The number of private middle school in Evervale City equals the total number of schools in Brightford. The number of elementary school in Brightford equals 2 times the number of private middle school in Glenfield City. The number of private middle school in Brightford equals the number of elementary school in Brightford. The total number of schools in Brightford equals 24. The number of public highschool in Evervale City equals 2. The number of private middle school in Glenfield City exists, and its number is greater than 0.
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How many private middle school does Glenfield City have?
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The question is difficult, so we use equations to solve it. Define private middle school in Glenfield City as x; we don't know its value yet but will find it out later. Define elementary school in Brightford as B; Q = x = x; so B = 2 * Q = (2) * (x) = 2*x. Define public highschool in Brightford as s; p = B = 2*x; so s = 2 * p = (2) * (2*x) = 4*x. Define private middle school in Brightford as n; so n = B = 2*x. Define total number of schools in Brightford as l; g = n + s = 2*x + 4*x = 6*x; so l = g + B = 6*x + 2*x = 8*x. We know l = 24, so we have 8*x = 24 and we solve it through the following steps: Simplifying the equation: 8*x = 24. Move all terms to one side: 8*x - 24 = 0. Isolate the term with x: 8*x = 24. Divide both sides by 8: x = 24 / 8. Solution: x = 3. Answer: 3.
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The total number of schools in Brightford equals 36. The number of private middle school in Evervale City exists, and its number is greater than 0. The number of elementary school in Glenfield City equals the sum of the number of elementary school in Evervale City and the number of public highschool in Evervale City. The number of public highschool in Evervale City equals 3. The number of private middle school in Glenfield City equals the number of private middle school in Evervale City. The number of private middle school in Brightford equals the number of elementary school in Brightford. The number of elementary school in Evervale City equals 2. The number of elementary school in Brightford equals 3 times the number of private middle school in Evervale City. The number of public highschool in Glenfield City equals the number of public highschool in Evervale City. The number of public highschool in Brightford equals 2 times the number of private middle school in Brightford.
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How many private middle school does Evervale City have?
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The question is difficult, so we use equations to solve it. Define private middle school in Evervale City as x; we don't know its value yet but will find it out later. Define elementary school in Brightford as M; O = x = x; so M = 3 * O = (3) * (x) = 3*x. Define private middle school in Brightford as j; so j = M = 3*x. Define public highschool in Brightford as b; m = j = 3*x; so b = 2 * m = (2) * (3*x) = 6*x. Define total number of schools in Brightford as d; r = b + j = 6*x + 3*x = 9*x; so d = r + M = 9*x + 3*x = 12*x. We know d = 36, so we have 12*x = 36 and we solve it through the following steps: Simplifying the equation: 12*x = 36. Move all terms to one side: 12*x - 36 = 0. Isolate the term with x: 12*x = 36. Divide both sides by 12: x = 36 / 12. Solution: x = 3. Answer: 3.
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The number of private middle school in Ruby Bay equals the sum of the number of elementary school in Ruby Bay and the number of private middle school in Shoreline City. The number of private middle school in Shoreline City equals 2 plus the number of elementary school in Shoreline City. The number of elementary school in Ruby Bay equals 1 plus the number of private middle school in Shoreline City. The total number of schools in Ruby Bay equals 14. The number of elementary school in Shoreline City exists, and its number is greater than 0.
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How many elementary school does Shoreline City have?
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The question is difficult, so we use equations to solve it. Define elementary school in Shoreline City as x; we don't know its value yet but will find it out later. Define private middle school in Shoreline City as N; F = x = x; so N = 2 + F = 2 + x = x + 2. Define elementary school in Ruby Bay as m; T = N = x + 2; so m = 1 + T = 1 + x + 2 = x + 3. Define private middle school in Ruby Bay as J; so J = m + N = x + 3 + x + 2 = 2*x + 5. Define total number of schools in Ruby Bay as p; so p = m + J = x + 3 + 2*x + 5 = 3*x + 8. We know p = 14, so we have 3*x + 8 = 14 and we solve it through the following steps: Simplifying the equation: 3*x + 8 = 14. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of private middle school in Glenfield City equals 4 plus the number of private middle school in Evervale City. The number of elementary school in Evervale City equals 1 plus the number of elementary school in Hawkesbury. The number of regional medical school in Glenfield City equals 2 times the number of private middle school in Evervale City. The number of regional medical school in Hawkesbury equals 2. The number of private middle school in Evervale City exists, and its number is greater than 0. The number of elementary school in Glenfield City equals 2. The total number of schools in Glenfield City equals 12. The number of elementary school in Hawkesbury equals 3. The number of regional medical school in Evervale City equals the difference between the number of private middle school in Glenfield City and the number of private middle school in Hawkesbury. The number of private middle school in Hawkesbury equals 2.
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How many private middle school does Evervale City have?
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The question is difficult, so we use equations to solve it. Define elementary school in Glenfield City as b; so b = 2. Define private middle school in Evervale City as x; we don't know its value yet but will find it out later. Define regional medical school in Glenfield City as Y; l = x = x; so Y = 2 * l = (2) * (x) = 2*x. Define private middle school in Glenfield City as A; R = x = x; so A = 4 + R = 4 + x = x + 4. Define total number of schools in Glenfield City as D; W = A + Y = x + 4 + 2*x = 3*x + 4; so D = W + b = 3*x + 4 + 2 = 3*x + 6. We know D = 12, so we have 3*x + 6 = 12 and we solve it through the following steps: Simplifying the equation: 3*x + 6 = 12. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of public highschool in Westhaven City equals the total number of schools in Evervale City. The total number of schools in Westhaven City equals 7. The number of private middle school in Evervale City equals 2 plus the total number of schools in Glenfield City. The number of elementary school in Glenfield City equals 2. The number of private middle school in Westhaven City exists, and its number is greater than 0.
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How many private middle school does Westhaven City have?
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The question is difficult, so we use equations to solve it. Define elementary school in Glenfield City as v; so v = 2. Define total number of schools in Glenfield City as j; so j = v = 2. Define private middle school in Evervale City as P; h = j = 2; so P = 2 + h = 2 + 2 = 4. Define total number of schools in Evervale City as G; so G = P = 4. Define public highschool in Westhaven City as A; so A = G = 4. Define private middle school in Westhaven City as x; we don't know its value yet but will find it out later. Define total number of schools in Westhaven City as I; so I = x + A = x + 4 = x + 4. We know I = 7, so we have x + 4 = 7 and we solve it through the following steps: Simplifying the equation: x + 4 = 7. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.
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The number of elementary school in Shoreline City equals the number of regional medical school in Shoreline City. The number of public highschool in Ruby Bay equals 4 times the number of elementary school in Ruby Bay. The number of elementary school in Ruby Bay exists, and its number is greater than 0. The number of elementary school in Oakbridge City equals 3 plus the sum of the number of public highschool in Ruby Bay and the number of elementary school in Ruby Bay. The total number of schools in Oakbridge City equals 21. The number of regional medical school in Shoreline City equals the sum of the total number of schools in Oakbridge City and the number of elementary school in Oakbridge City. The number of regional medical school in Oakbridge City equals the number of public highschool in Ruby Bay. The number of regional medical school in Ruby Bay equals the difference between the number of regional medical school in Shoreline City and the number of elementary school in Ruby Bay.
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How many elementary school does Ruby Bay have?
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The question is difficult, so we use equations to solve it. Define elementary school in Ruby Bay as x; we don't know its value yet but will find it out later. Define public highschool in Ruby Bay as I; S = x = x; so I = 4 * S = (4) * (x) = 4*x. Define elementary school in Oakbridge City as Z; O = I + x = 4*x + x = 5*x; so Z = 3 + O = 3 + 5*x = 5*x + 3. Define regional medical school in Oakbridge City as w; so w = I = 4*x. Define total number of schools in Oakbridge City as E; so E = w + Z = 4*x + 5*x + 3 = 9*x + 3. We know E = 21, so we have 9*x + 3 = 21 and we solve it through the following steps: Simplifying the equation: 9*x + 3 = 21. Move all terms to one side: 9*x - 18 = 0. Isolate the term with x: 9*x = 18. Divide both sides by 9: x = 18 / 9. Solution: x = 2. Answer: 2.
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The number of public highschool in Clearwater Bay equals 4 times the number of private middle school in Ruby Bay. The number of public highschool in Shoreline City equals the number of public highschool in Clearwater Bay. The number of elementary school in Shoreline City equals 4. The number of elementary school in Ruby Bay equals the number of public highschool in Ruby Bay. The number of elementary school in Clearwater Bay equals the number of private middle school in Ruby Bay. The number of private middle school in Shoreline City equals 1. The number of private middle school in Clearwater Bay equals 1 plus the number of elementary school in Clearwater Bay. The number of private middle school in Ruby Bay exists, and its number is greater than 0. The total number of schools in Clearwater Bay equals 25. The number of public highschool in Ruby Bay equals 3.
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How many private middle school does Ruby Bay have?
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The question is difficult, so we use equations to solve it. Define private middle school in Ruby Bay as x; we don't know its value yet but will find it out later. Define public highschool in Clearwater Bay as u; b = x = x; so u = 4 * b = (4) * (x) = 4*x. Define elementary school in Clearwater Bay as v; so v = x = x. Define private middle school in Clearwater Bay as f; L = v = x; so f = 1 + L = 1 + x = x + 1. Define total number of schools in Clearwater Bay as i; G = f + v = x + 1 + x = 2*x + 1; so i = G + u = 2*x + 1 + 4*x = 6*x + 1. We know i = 25, so we have 6*x + 1 = 25 and we solve it through the following steps: Simplifying the equation: 6*x + 1 = 25. Move all terms to one side: 6*x - 24 = 0. Isolate the term with x: 6*x = 24. Divide both sides by 6: x = 24 / 6. Solution: x = 4. Answer: 4.
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The number of regional medical school in Westhaven City equals the number of regional medical school in Evervale City. The number of culinarian school in Westhaven City exists, and its number is greater than 0. The number of culinarian school in Evervale City equals the total number of schools in Westhaven City. The total number of schools in Evervale City equals 11. The number of regional medical school in Evervale City equals 1 plus the number of culinarian school in Westhaven City.
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How many culinarian school does Westhaven City have?
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The question is difficult, so we use equations to solve it. Define culinarian school in Westhaven City as x; we don't know its value yet but will find it out later. Define regional medical school in Evervale City as K; u = x = x; so K = 1 + u = 1 + x = x + 1. Define regional medical school in Westhaven City as z; so z = K = x + 1. Define total number of schools in Westhaven City as Q; so Q = x + z = x + x + 1 = 2*x + 1. Define culinarian school in Evervale City as l; so l = Q = 2*x + 1. Define total number of schools in Evervale City as S; so S = l + K = 2*x + 1 + x + 1 = 3*x + 2. We know S = 11, so we have 3*x + 2 = 11 and we solve it through the following steps: Simplifying the equation: 3*x + 2 = 11. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of regional medical school in Clearwater Bay equals the number of private middle school in Clearwater Bay. The total number of schools in Ruby Bay equals 19. The number of private middle school in Clearwater Bay exists, and its number is greater than 0. The number of private middle school in Ruby Bay equals the total number of schools in Clearwater Bay. The number of regional medical school in Ruby Bay equals 3 plus the total number of schools in Clearwater Bay.
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How many private middle school does Clearwater Bay have?
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The question is difficult, so we use equations to solve it. Define private middle school in Clearwater Bay as x; we don't know its value yet but will find it out later. Define regional medical school in Clearwater Bay as q; so q = x = x. Define total number of schools in Clearwater Bay as Y; so Y = x + q = x + x = 2*x. Define regional medical school in Ruby Bay as C; G = Y = 2*x; so C = 3 + G = 3 + 2*x = 2*x + 3. Define private middle school in Ruby Bay as j; so j = Y = 2*x. Define total number of schools in Ruby Bay as V; so V = j + C = 2*x + 2*x + 3 = 4*x + 3. We know V = 19, so we have 4*x + 3 = 19 and we solve it through the following steps: Simplifying the equation: 4*x + 3 = 19. Move all terms to one side: 4*x - 16 = 0. Isolate the term with x: 4*x = 16. Divide both sides by 4: x = 16 / 4. Solution: x = 4. Answer: 4.
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The number of regional medical school in Oakbridge City equals 4 plus the total number of schools in Ruby Bay. The number of regional medical school in Riverton City equals the sum of the number of elementary school in Oakbridge City and the number of elementary school in Ruby Bay. The number of elementary school in Oakbridge City equals the sum of the number of regional medical school in Oakbridge City and the total number of schools in Ruby Bay. The number of elementary school in Ruby Bay exists, and its number is greater than 0. The total number of schools in Riverton City equals 13.
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How many elementary school does Ruby Bay have?
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The question is difficult, so we use equations to solve it. Define elementary school in Ruby Bay as x; we don't know its value yet but will find it out later. Define total number of schools in Ruby Bay as c; so c = x = x. Define regional medical school in Oakbridge City as K; E = c = x; so K = 4 + E = 4 + x = x + 4. Define elementary school in Oakbridge City as G; so G = K + c = x + 4 + x = 2*x + 4. Define regional medical school in Riverton City as F; so F = G + x = 2*x + 4 + x = 3*x + 4. Define total number of schools in Riverton City as D; so D = F = 3*x + 4. We know D = 13, so we have 3*x + 4 = 13 and we solve it through the following steps: Simplifying the equation: 3*x + 4 = 13. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of public highschool in Oakbridge City equals the number of culinarian school in Oakbridge City. The number of culinarian school in Ruby Bay exists, and its number is greater than 0. The number of culinarian school in Oakbridge City equals 1. The number of public highschool in Ruby Bay equals 2 times the total number of schools in Oakbridge City. The total number of schools in Ruby Bay equals 8.
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How many culinarian school does Ruby Bay have?
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The question is difficult, so we use equations to solve it. Define culinarian school in Oakbridge City as T; so T = 1. Define public highschool in Oakbridge City as Q; so Q = T = 1. Define total number of schools in Oakbridge City as I; so I = T + Q = 1 + 1 = 2. Define public highschool in Ruby Bay as o; n = I = 2; so o = 2 * n = (2) * (2) = 4. Define culinarian school in Ruby Bay as x; we don't know its value yet but will find it out later. Define total number of schools in Ruby Bay as s; so s = x + o = x + 4 = x + 4. We know s = 8, so we have x + 4 = 8 and we solve it through the following steps: Simplifying the equation: x + 4 = 8. Move all terms to one side: x - 4 = 0. Isolate the term with x: 1*x = 4. Divide both sides by 1: x = 4 / 1. Solution: x = 4. Answer: 4.
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The number of culinarian school in Brightford equals the number of regional medical school in Brightford. The number of elementary school in Brightford equals 4. The number of regional medical school in Glenfield City equals 2 times the number of elementary school in Westhaven City. The number of culinarian school in Westhaven City equals 1. The number of regional medical school in Westhaven City equals the number of regional medical school in Glenfield City. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of culinarian school in Glenfield City equals the sum of the number of regional medical school in Glenfield City and the number of elementary school in Westhaven City. The number of regional medical school in Brightford equals the number of elementary school in Brightford. The number of elementary school in Westhaven City equals 4. The total number of schools in Glenfield City equals 22.
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How many elementary school does Glenfield City have?
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The question is difficult, so we use equations to solve it. Define elementary school in Westhaven City as b; so b = 4. Define regional medical school in Glenfield City as P; c = b = 4; so P = 2 * c = (2) * (4) = 8. Define culinarian school in Glenfield City as k; so k = P + b = 8 + 4 = 12. Define elementary school in Glenfield City as x; we don't know its value yet but will find it out later. Define total number of schools in Glenfield City as H; z = P + x = 8 + x = x + 8; so H = z + k = x + 8 + 12 = x + 20. We know H = 22, so we have x + 20 = 22 and we solve it through the following steps: Simplifying the equation: x + 20 = 22. Move all terms to one side: x - 2 = 0. Isolate the term with x: 1*x = 2. Divide both sides by 1: x = 2 / 1. Solution: x = 2. Answer: 2.
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The total number of schools in Ruby Bay equals 32. The number of private middle school in Clearwater Bay exists, and its number is greater than 0. The number of regional medical school in Clearwater Bay equals 2 times the sum of the number of culinarian school in Shoreline City, the total number of schools in Ruby Bay, and the total number of schools in Shoreline City. The number of private middle school in Ruby Bay equals the number of regional medical school in Ruby Bay. The number of culinarian school in Ruby Bay equals 2 times the number of regional medical school in Ruby Bay. The number of culinarian school in Shoreline City equals 3. The number of private middle school in Shoreline City equals 2. The number of culinarian school in Clearwater Bay equals 2. The number of regional medical school in Shoreline City equals 4 times the number of culinarian school in Clearwater Bay. The number of regional medical school in Ruby Bay equals 4 times the number of private middle school in Clearwater Bay.
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How many private middle school does Clearwater Bay have?
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The question is difficult, so we use equations to solve it. Define private middle school in Clearwater Bay as x; we don't know its value yet but will find it out later. Define regional medical school in Ruby Bay as k; g = x = x; so k = 4 * g = (4) * (x) = 4*x. Define culinarian school in Ruby Bay as f; V = k = 4*x; so f = 2 * V = (2) * (4*x) = 8*x. Define private middle school in Ruby Bay as C; so C = k = 4*x. Define total number of schools in Ruby Bay as R; U = C + k = 4*x + 4*x = 8*x; so R = U + f = 8*x + 8*x = 16*x. We know R = 32, so we have 16*x = 32 and we solve it through the following steps: Simplifying the equation: 16*x = 32. Move all terms to one side: 16*x - 32 = 0. Isolate the term with x: 16*x = 32. Divide both sides by 16: x = 32 / 16. Solution: x = 2. Answer: 2.
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The number of elementary school in Evervale City equals 4. The number of culinarian school in Evervale City exists, and its number is greater than 0. The total number of schools in Glenfield City equals 17. The number of public highschool in Glenfield City equals the total number of schools in Evervale City. The number of culinarian school in Glenfield City equals the sum of the total number of schools in Evervale City and the number of culinarian school in Evervale City. The number of culinarian school in Hawkesbury equals 1. The number of elementary school in Hawkesbury equals 2.
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How many culinarian school does Evervale City have?
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The question is difficult, so we use equations to solve it. Define elementary school in Evervale City as G; so G = 4. Define culinarian school in Evervale City as x; we don't know its value yet but will find it out later. Define total number of schools in Evervale City as t; so t = x + G = x + 4 = x + 4. Define culinarian school in Glenfield City as J; so J = t + x = x + 4 + x = 2*x + 4. Define public highschool in Glenfield City as v; so v = t = x + 4. Define total number of schools in Glenfield City as f; so f = J + v = 2*x + 4 + x + 4 = 3*x + 8. We know f = 17, so we have 3*x + 8 = 17 and we solve it through the following steps: Simplifying the equation: 3*x + 8 = 17. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of regional medical school in Glenfield City exists, and its number is greater than 0. The number of regional medical school in Glenfield City exists, and its number is greater than 0. The number of culinarian school in Westhaven City equals 3 plus the number of culinarian school in Hawkesbury. The number of private middle school in Glenfield City equals 3 plus the number of regional medical school in Glenfield City. The number of regional medical school in Westhaven City equals 66. The number of culinarian school in Hawkesbury equals 3. The number of private middle school in Hawkesbury equals 3 times the number of private middle school in Glenfield City. The number of regional medical school in Hawkesbury equals the sum of the number of culinarian school in Hawkesbury and the number of culinarian school in Westhaven City.
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How many regional medical school does Glenfield City have?
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The question is difficult, so we use equations to solve it. Define regional medical school in Glenfield City as x; we don't know its value yet but will find it out later. Define private middle school in Glenfield City as y; L = x = x; so y = 3 + L = 3 + x = x + 3. Define private middle school in Hawkesbury as r; f = y = x + 3; so r = 3 * f = (3) * (x + 3) = 3*x + 9. Define regional medical school in Westhaven City as A; u = r + x = 3*x + 9 + x = 4*x + 9; g = u + y = 4*x + 9 + x + 3 = 5*x + 12; so A = 3 * g = (3) * (5*x + 12) = 15*x + 36. We know A = 66, so we have 15*x + 36 = 66 and we solve it through the following steps: Simplifying the equation: 15*x + 36 = 66. Move all terms to one side: 15*x - 30 = 0. Isolate the term with x: 15*x = 30. Divide both sides by 15: x = 30 / 15. Solution: x = 2. Answer: 2.
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The number of regional medical school in Riverton City exists, and its number is greater than 0. The number of public highschool in Riverton City equals the sum of the number of public highschool in Shoreline City, the total number of schools in Riverton City, and the number of private middle school in Shoreline City. The number of private middle school in Riverton City equals 3. The total number of schools in Shoreline City equals 18. The number of regional medical school in Clearwater Bay equals the number of public highschool in Riverton City. The number of public highschool in Shoreline City equals 4 times the number of regional medical school in Riverton City. The number of private middle school in Shoreline City equals the number of regional medical school in Riverton City. The number of regional medical school in Shoreline City equals 4 times the number of private middle school in Shoreline City. The number of private middle school in Clearwater Bay equals 3. The number of public highschool in Clearwater Bay equals 1 plus the total number of schools in Riverton City.
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How many regional medical school does Riverton City have?
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The question is difficult, so we use equations to solve it. Define regional medical school in Riverton City as x; we don't know its value yet but will find it out later. Define public highschool in Shoreline City as l; y = x = x; so l = 4 * y = (4) * (x) = 4*x. Define private middle school in Shoreline City as t; so t = x = x. Define regional medical school in Shoreline City as h; R = t = x; so h = 4 * R = (4) * (x) = 4*x. Define total number of schools in Shoreline City as n; d = t + h = x + 4*x = 5*x; so n = d + l = 5*x + 4*x = 9*x. We know n = 18, so we have 9*x = 18 and we solve it through the following steps: Simplifying the equation: 9*x = 18. Move all terms to one side: 9*x - 18 = 0. Isolate the term with x: 9*x = 18. Divide both sides by 9: x = 18 / 9. Solution: x = 2. Answer: 2.
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The number of regional medical school in Riverton City equals the number of public highschool in Riverton City. The number of regional medical school in Shoreline City equals the sum of the total number of schools in Riverton City and the number of public highschool in Riverton City. The number of private middle school in Ruby Bay equals the total number of schools in Shoreline City. The number of public highschool in Riverton City exists, and its number is greater than 0. The total number of schools in Ruby Bay equals 6.
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How many public highschool does Riverton City have?
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The question is difficult, so we use equations to solve it. Define public highschool in Riverton City as x; we don't know its value yet but will find it out later. Define regional medical school in Riverton City as R; so R = x = x. Define total number of schools in Riverton City as r; so r = R + x = x + x = 2*x. Define regional medical school in Shoreline City as a; so a = r + x = 2*x + x = 3*x. Define total number of schools in Shoreline City as u; so u = a = 3*x. Define private middle school in Ruby Bay as k; so k = u = 3*x. Define total number of schools in Ruby Bay as J; so J = k = 3*x. We know J = 6, so we have 3*x = 6 and we solve it through the following steps: Simplifying the equation: 3*x = 6. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of intense detective thriller in Festival de Clairmont exists, and its number is greater than 0. The total number of movies in Festival de Clairmont equals 13. The number of futuristic sci-fi movie in Festival de Clairmont equals 2 plus the total number of movies in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 4. The number of intense detective thriller in Festival Lumière de Valmont equals 4.
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How many intense detective thriller does Festival de Clairmont have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Festival Lumière de Valmont as a; so a = 4. Define futuristic sci-fi movie in Festival Lumière de Valmont as l; so l = 4. Define total number of movies in Festival Lumière de Valmont as R; so R = a + l = 4 + 4 = 8. Define intense detective thriller in Festival de Clairmont as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Festival de Clairmont as f; I = R = 8; so f = 2 + I = 2 + 8 = 10. Define total number of movies in Festival de Clairmont as p; so p = x + f = x + 10 = x + 10. We know p = 13, so we have x + 10 = 13 and we solve it through the following steps: Simplifying the equation: x + 10 = 13. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.
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The number of solemn period drama in Verdi Movie Festival equals 2 plus the number of intense detective thriller in Golden Banana Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals 2. The number of intense detective thriller in Taylor Movie Festival equals the sum of the number of calm road movie in Verdi Movie Festival, the number of calm road movie in Taylor Movie Festival, the number of intense detective thriller in Golden Banana Movie Festival, and the total number of movies in Verdi Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals 2 times the difference between the number of solemn period drama in Golden Banana Movie Festival and the number of solemn period drama in Taylor Movie Festival. The number of solemn period drama in Taylor Movie Festival equals 4. The number of calm road movie in Taylor Movie Festival equals 2 times the sum of the number of intense detective thriller in Golden Banana Movie Festival, the total number of movies in Verdi Movie Festival, and the number of solemn period drama in Verdi Movie Festival. The number of calm road movie in Verdi Movie Festival equals 2 plus the number of solemn period drama in Verdi Movie Festival. The number of intense detective thriller in Verdi Movie Festival exists, and its number is greater than 0. The number of intense detective thriller in Golden Banana Movie Festival equals 2. The total number of movies in Verdi Movie Festival equals 14.
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How many intense detective thriller does Verdi Movie Festival have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Golden Banana Movie Festival as Y; so Y = 2. Define solemn period drama in Verdi Movie Festival as T; C = Y = 2; so T = 2 + C = 2 + 2 = 4. Define calm road movie in Verdi Movie Festival as I; P = T = 4; so I = 2 + P = 2 + 4 = 6. Define intense detective thriller in Verdi Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in Verdi Movie Festival as Q; E = T + x = 4 + x = x + 4; so Q = E + I = x + 4 + 6 = x + 10. We know Q = 14, so we have x + 10 = 14 and we solve it through the following steps: Simplifying the equation: x + 10 = 14. Move all terms to one side: x - 4 = 0. Isolate the term with x: 1*x = 4. Divide both sides by 1: x = 4 / 1. Solution: x = 4. Answer: 4.
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The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 2 times the number of futuristic sci-fi movie in Golden Banana Movie Festival. The number of solemn period drama in West Sahara Movie Festival equals 2. The number of solemn period drama in Taylor Movie Festival equals the total number of movies in Golden Banana Movie Festival. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 2 times the number of upbeat metropolis comedy in West Sahara Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals the number of futuristic sci-fi movie in Golden Banana Movie Festival. The total number of movies in Golden Banana Movie Festival equals 24. The number of futuristic sci-fi movie in Taylor Movie Festival equals 2. The number of upbeat metropolis comedy in West Sahara Movie Festival exists, and its number is greater than 0.
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How many upbeat metropolis comedy does West Sahara Movie Festival have?
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The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in West Sahara Movie Festival as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Golden Banana Movie Festival as B; Q = x = x; so B = 2 * Q = (2) * (x) = 2*x. Define upbeat metropolis comedy in Golden Banana Movie Festival as s; p = B = 2*x; so s = 2 * p = (2) * (2*x) = 4*x. Define solemn period drama in Golden Banana Movie Festival as n; so n = B = 2*x. Define total number of movies in Golden Banana Movie Festival as l; g = n + s = 2*x + 4*x = 6*x; so l = g + B = 6*x + 2*x = 8*x. We know l = 24, so we have 8*x = 24 and we solve it through the following steps: Simplifying the equation: 8*x = 24. Move all terms to one side: 8*x - 24 = 0. Isolate the term with x: 8*x = 24. Divide both sides by 8: x = 24 / 8. Solution: x = 3. Answer: 3.
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The total number of movies in Golden Banana Movie Festival equals 36. The number of solemn period drama in Taylor Movie Festival exists, and its number is greater than 0. The number of futuristic sci-fi movie in West Sahara Movie Festival equals the sum of the number of futuristic sci-fi movie in Taylor Movie Festival and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of upbeat metropolis comedy in Taylor Movie Festival equals 3. The number of solemn period drama in West Sahara Movie Festival equals the number of solemn period drama in Taylor Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals the number of futuristic sci-fi movie in Golden Banana Movie Festival. The number of futuristic sci-fi movie in Taylor Movie Festival equals 2. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 3 times the number of solemn period drama in Taylor Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals the number of upbeat metropolis comedy in Taylor Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 2 times the number of solemn period drama in Golden Banana Movie Festival.
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How many solemn period drama does Taylor Movie Festival have?
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The question is difficult, so we use equations to solve it. Define solemn period drama in Taylor Movie Festival as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Golden Banana Movie Festival as M; O = x = x; so M = 3 * O = (3) * (x) = 3*x. Define solemn period drama in Golden Banana Movie Festival as j; so j = M = 3*x. Define upbeat metropolis comedy in Golden Banana Movie Festival as b; m = j = 3*x; so b = 2 * m = (2) * (3*x) = 6*x. Define total number of movies in Golden Banana Movie Festival as d; r = b + j = 6*x + 3*x = 9*x; so d = r + M = 9*x + 3*x = 12*x. We know d = 36, so we have 12*x = 36 and we solve it through the following steps: Simplifying the equation: 12*x = 36. Move all terms to one side: 12*x - 36 = 0. Isolate the term with x: 12*x = 36. Divide both sides by 12: x = 36 / 12. Solution: x = 3. Answer: 3.
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The number of solemn period drama in Festival de Saint-Rivage equals the sum of the number of upbeat metropolis comedy in Festival de Saint-Rivage and the number of solemn period drama in Cinéma de Montreval. The number of solemn period drama in Cinéma de Montreval equals 2 plus the number of upbeat metropolis comedy in Cinéma de Montreval. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 1 plus the number of solemn period drama in Cinéma de Montreval. The total number of movies in Festival de Saint-Rivage equals 14. The number of upbeat metropolis comedy in Cinéma de Montreval exists, and its number is greater than 0.
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How many upbeat metropolis comedy does Cinéma de Montreval have?
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The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Cinéma de Montreval as x; we don't know its value yet but will find it out later. Define solemn period drama in Cinéma de Montreval as N; F = x = x; so N = 2 + F = 2 + x = x + 2. Define upbeat metropolis comedy in Festival de Saint-Rivage as m; T = N = x + 2; so m = 1 + T = 1 + x + 2 = x + 3. Define solemn period drama in Festival de Saint-Rivage as J; so J = m + N = x + 3 + x + 2 = 2*x + 5. Define total number of movies in Festival de Saint-Rivage as p; so p = m + J = x + 3 + 2*x + 5 = 3*x + 8. We know p = 14, so we have 3*x + 8 = 14 and we solve it through the following steps: Simplifying the equation: 3*x + 8 = 14. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of calm road movie in West Sahara Movie Festival equals 4 plus the number of upbeat metropolis comedy in Taylor Movie Festival. The number of solemn period drama in Taylor Movie Festival equals 1 plus the number of solemn period drama in Verdi Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals 2 times the number of upbeat metropolis comedy in Taylor Movie Festival. The number of calm road movie in Verdi Movie Festival equals 2. The number of upbeat metropolis comedy in Taylor Movie Festival exists, and its number is greater than 0. The number of solemn period drama in West Sahara Movie Festival equals 2. The total number of movies in West Sahara Movie Festival equals 12. The number of solemn period drama in Verdi Movie Festival equals 3. The number of calm road movie in Taylor Movie Festival equals the difference between the number of calm road movie in West Sahara Movie Festival and the number of upbeat metropolis comedy in Verdi Movie Festival. The number of upbeat metropolis comedy in Verdi Movie Festival equals 2.
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How many upbeat metropolis comedy does Taylor Movie Festival have?
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The question is difficult, so we use equations to solve it. Define solemn period drama in West Sahara Movie Festival as b; so b = 2. Define upbeat metropolis comedy in Taylor Movie Festival as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in West Sahara Movie Festival as Y; l = x = x; so Y = 2 * l = (2) * (x) = 2*x. Define calm road movie in West Sahara Movie Festival as A; R = x = x; so A = 4 + R = 4 + x = x + 4. Define total number of movies in West Sahara Movie Festival as D; W = A + Y = x + 4 + 2*x = 3*x + 4; so D = W + b = 3*x + 4 + 2 = 3*x + 6. We know D = 12, so we have 3*x + 6 = 12 and we solve it through the following steps: Simplifying the equation: 3*x + 6 = 12. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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The number of solemn period drama in Northwood Movie Festival equals the total number of movies in Taylor Movie Festival. The total number of movies in Northwood Movie Festival equals 7. The number of solemn period drama in Taylor Movie Festival equals 2 plus the total number of movies in West Sahara Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals 2. The number of futuristic sci-fi movie in Northwood Movie Festival exists, and its number is greater than 0.
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How many futuristic sci-fi movie does Northwood Movie Festival have?
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The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in West Sahara Movie Festival as v; so v = 2. Define total number of movies in West Sahara Movie Festival as j; so j = v = 2. Define solemn period drama in Taylor Movie Festival as P; h = j = 2; so P = 2 + h = 2 + 2 = 4. Define total number of movies in Taylor Movie Festival as G; so G = P = 4. Define solemn period drama in Northwood Movie Festival as A; so A = G = 4. Define futuristic sci-fi movie in Northwood Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in Northwood Movie Festival as I; so I = x + A = x + 4 = x + 4. We know I = 7, so we have x + 4 = 7 and we solve it through the following steps: Simplifying the equation: x + 4 = 7. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.
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The number of calm road movie in Cinéma de Montreval equals the number of upbeat metropolis comedy in Cinéma de Montreval. The number of calm road movie in Festival de Saint-Rivage equals 4 times the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage exists, and its number is greater than 0. The number of calm road movie in Festival de Clairmont equals 3 plus the sum of the number of calm road movie in Festival de Saint-Rivage and the number of upbeat metropolis comedy in Festival de Saint-Rivage. The total number of movies in Festival de Clairmont equals 21. The number of upbeat metropolis comedy in Cinéma de Montreval equals the sum of the total number of movies in Festival de Clairmont and the number of calm road movie in Festival de Clairmont. The number of upbeat metropolis comedy in Festival de Clairmont equals the number of calm road movie in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals the difference between the number of upbeat metropolis comedy in Cinéma de Montreval and the number of upbeat metropolis comedy in Festival de Saint-Rivage.
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How many upbeat metropolis comedy does Festival de Saint-Rivage have?
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The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define calm road movie in Festival de Saint-Rivage as I; S = x = x; so I = 4 * S = (4) * (x) = 4*x. Define calm road movie in Festival de Clairmont as Z; O = I + x = 4*x + x = 5*x; so Z = 3 + O = 3 + 5*x = 5*x + 3. Define upbeat metropolis comedy in Festival de Clairmont as w; so w = I = 4*x. Define total number of movies in Festival de Clairmont as E; so E = w + Z = 4*x + 5*x + 3 = 9*x + 3. We know E = 21, so we have 9*x + 3 = 21 and we solve it through the following steps: Simplifying the equation: 9*x + 3 = 21. Move all terms to one side: 9*x - 18 = 0. Isolate the term with x: 9*x = 18. Divide both sides by 9: x = 18 / 9. Solution: x = 2. Answer: 2.
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The number of solemn period drama in Rêves de Belleville equals 4 times the number of futuristic sci-fi movie in Festival de Saint-Rivage. The number of solemn period drama in Cinéma de Montreval equals the number of solemn period drama in Rêves de Belleville. The number of upbeat metropolis comedy in Cinéma de Montreval equals 4. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals the number of solemn period drama in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Rêves de Belleville equals the number of futuristic sci-fi movie in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Cinéma de Montreval equals 1. The number of futuristic sci-fi movie in Rêves de Belleville equals 1 plus the number of upbeat metropolis comedy in Rêves de Belleville. The number of futuristic sci-fi movie in Festival de Saint-Rivage exists, and its number is greater than 0. The total number of movies in Rêves de Belleville equals 25. The number of solemn period drama in Festival de Saint-Rivage equals 3.
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How many futuristic sci-fi movie does Festival de Saint-Rivage have?
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The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define solemn period drama in Rêves de Belleville as u; b = x = x; so u = 4 * b = (4) * (x) = 4*x. Define upbeat metropolis comedy in Rêves de Belleville as v; so v = x = x. Define futuristic sci-fi movie in Rêves de Belleville as f; L = v = x; so f = 1 + L = 1 + x = x + 1. Define total number of movies in Rêves de Belleville as i; G = f + v = x + 1 + x = 2*x + 1; so i = G + u = 2*x + 1 + 4*x = 6*x + 1. We know i = 25, so we have 6*x + 1 = 25 and we solve it through the following steps: Simplifying the equation: 6*x + 1 = 25. Move all terms to one side: 6*x - 24 = 0. Isolate the term with x: 6*x = 24. Divide both sides by 6: x = 24 / 6. Solution: x = 4. Answer: 4.
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The number of calm road movie in Northwood Movie Festival equals the number of calm road movie in Taylor Movie Festival. The number of intense detective thriller in Northwood Movie Festival exists, and its number is greater than 0. The number of intense detective thriller in Taylor Movie Festival equals the total number of movies in Northwood Movie Festival. The total number of movies in Taylor Movie Festival equals 11. The number of calm road movie in Taylor Movie Festival equals 1 plus the number of intense detective thriller in Northwood Movie Festival.
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How many intense detective thriller does Northwood Movie Festival have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Northwood Movie Festival as x; we don't know its value yet but will find it out later. Define calm road movie in Taylor Movie Festival as K; u = x = x; so K = 1 + u = 1 + x = x + 1. Define calm road movie in Northwood Movie Festival as z; so z = K = x + 1. Define total number of movies in Northwood Movie Festival as Q; so Q = x + z = x + x + 1 = 2*x + 1. Define intense detective thriller in Taylor Movie Festival as l; so l = Q = 2*x + 1. Define total number of movies in Taylor Movie Festival as S; so S = l + K = 2*x + 1 + x + 1 = 3*x + 2. We know S = 11, so we have 3*x + 2 = 11 and we solve it through the following steps: Simplifying the equation: 3*x + 2 = 11. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of solemn period drama in Rêves de Belleville equals the number of calm road movie in Rêves de Belleville. The total number of movies in Festival de Saint-Rivage equals 19. The number of calm road movie in Rêves de Belleville exists, and its number is greater than 0. The number of calm road movie in Festival de Saint-Rivage equals the total number of movies in Rêves de Belleville. The number of solemn period drama in Festival de Saint-Rivage equals 3 plus the total number of movies in Rêves de Belleville.
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How many calm road movie does Rêves de Belleville have?
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The question is difficult, so we use equations to solve it. Define calm road movie in Rêves de Belleville as x; we don't know its value yet but will find it out later. Define solemn period drama in Rêves de Belleville as q; so q = x = x. Define total number of movies in Rêves de Belleville as Y; so Y = x + q = x + x = 2*x. Define solemn period drama in Festival de Saint-Rivage as C; G = Y = 2*x; so C = 3 + G = 3 + 2*x = 2*x + 3. Define calm road movie in Festival de Saint-Rivage as j; so j = Y = 2*x. Define total number of movies in Festival de Saint-Rivage as V; so V = j + C = 2*x + 2*x + 3 = 4*x + 3. We know V = 19, so we have 4*x + 3 = 19 and we solve it through the following steps: Simplifying the equation: 4*x + 3 = 19. Move all terms to one side: 4*x - 16 = 0. Isolate the term with x: 4*x = 16. Divide both sides by 4: x = 16 / 4. Solution: x = 4. Answer: 4.
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The number of upbeat metropolis comedy in Festival de Clairmont equals 4 plus the total number of movies in Festival de Saint-Rivage. The number of calm road movie in Festival Lumière de Valmont equals the sum of the number of calm road movie in Festival de Clairmont and the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of calm road movie in Festival de Clairmont equals the sum of the number of upbeat metropolis comedy in Festival de Clairmont and the total number of movies in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage exists, and its number is greater than 0. The total number of movies in Festival Lumière de Valmont equals 13.
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How many upbeat metropolis comedy does Festival de Saint-Rivage have?
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The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define total number of movies in Festival de Saint-Rivage as c; so c = x = x. Define upbeat metropolis comedy in Festival de Clairmont as K; E = c = x; so K = 4 + E = 4 + x = x + 4. Define calm road movie in Festival de Clairmont as G; so G = K + c = x + 4 + x = 2*x + 4. Define calm road movie in Festival Lumière de Valmont as F; so F = G + x = 2*x + 4 + x = 3*x + 4. Define total number of movies in Festival Lumière de Valmont as D; so D = F = 3*x + 4. We know D = 13, so we have 3*x + 4 = 13 and we solve it through the following steps: Simplifying the equation: 3*x + 4 = 13. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of futuristic sci-fi movie in Festival de Clairmont equals the number of intense detective thriller in Festival de Clairmont. The number of intense detective thriller in Festival de Saint-Rivage exists, and its number is greater than 0. The number of intense detective thriller in Festival de Clairmont equals 1. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 2 times the total number of movies in Festival de Clairmont. The total number of movies in Festival de Saint-Rivage equals 8.
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How many intense detective thriller does Festival de Saint-Rivage have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Festival de Clairmont as T; so T = 1. Define futuristic sci-fi movie in Festival de Clairmont as Q; so Q = T = 1. Define total number of movies in Festival de Clairmont as I; so I = T + Q = 1 + 1 = 2. Define futuristic sci-fi movie in Festival de Saint-Rivage as o; n = I = 2; so o = 2 * n = (2) * (2) = 4. Define intense detective thriller in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define total number of movies in Festival de Saint-Rivage as s; so s = x + o = x + 4 = x + 4. We know s = 8, so we have x + 4 = 8 and we solve it through the following steps: Simplifying the equation: x + 4 = 8. Move all terms to one side: x - 4 = 0. Isolate the term with x: 1*x = 4. Divide both sides by 1: x = 4 / 1. Solution: x = 4. Answer: 4.
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The number of upbeat metropolis comedy in Golden Banana Movie Festival equals the number of calm road movie in Golden Banana Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival equals 4. The number of calm road movie in West Sahara Movie Festival equals 2 times the number of intense detective thriller in Northwood Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival equals 1. The number of calm road movie in Northwood Movie Festival equals the number of calm road movie in West Sahara Movie Festival. The number of intense detective thriller in West Sahara Movie Festival exists, and its number is greater than 0. The number of upbeat metropolis comedy in West Sahara Movie Festival equals the sum of the number of calm road movie in West Sahara Movie Festival and the number of intense detective thriller in Northwood Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals the number of intense detective thriller in Golden Banana Movie Festival. The number of intense detective thriller in Northwood Movie Festival equals 4. The total number of movies in West Sahara Movie Festival equals 22.
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How many intense detective thriller does West Sahara Movie Festival have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Northwood Movie Festival as b; so b = 4. Define calm road movie in West Sahara Movie Festival as P; c = b = 4; so P = 2 * c = (2) * (4) = 8. Define upbeat metropolis comedy in West Sahara Movie Festival as k; so k = P + b = 8 + 4 = 12. Define intense detective thriller in West Sahara Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in West Sahara Movie Festival as H; z = P + x = 8 + x = x + 8; so H = z + k = x + 8 + 12 = x + 20. We know H = 22, so we have x + 20 = 22 and we solve it through the following steps: Simplifying the equation: x + 20 = 22. Move all terms to one side: x - 2 = 0. Isolate the term with x: 1*x = 2. Divide both sides by 1: x = 2 / 1. Solution: x = 2. Answer: 2.
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The total number of movies in Festival de Saint-Rivage equals 32. The number of intense detective thriller in Rêves de Belleville exists, and its number is greater than 0. The number of solemn period drama in Rêves de Belleville equals 2 times the sum of the number of intense detective thriller in Cinéma de Montreval, the total number of movies in Festival de Saint-Rivage, and the total number of movies in Cinéma de Montreval. The number of calm road movie in Festival de Saint-Rivage equals the number of solemn period drama in Festival de Saint-Rivage. The number of intense detective thriller in Festival de Saint-Rivage equals 2 times the number of solemn period drama in Festival de Saint-Rivage. The number of intense detective thriller in Cinéma de Montreval equals 3. The number of calm road movie in Cinéma de Montreval equals 2. The number of calm road movie in Rêves de Belleville equals 2. The number of solemn period drama in Cinéma de Montreval equals 4 times the number of calm road movie in Rêves de Belleville. The number of solemn period drama in Festival de Saint-Rivage equals 4 times the number of intense detective thriller in Rêves de Belleville.
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How many intense detective thriller does Rêves de Belleville have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Rêves de Belleville as x; we don't know its value yet but will find it out later. Define solemn period drama in Festival de Saint-Rivage as k; g = x = x; so k = 4 * g = (4) * (x) = 4*x. Define intense detective thriller in Festival de Saint-Rivage as f; V = k = 4*x; so f = 2 * V = (2) * (4*x) = 8*x. Define calm road movie in Festival de Saint-Rivage as C; so C = k = 4*x. Define total number of movies in Festival de Saint-Rivage as R; U = C + k = 4*x + 4*x = 8*x; so R = U + f = 8*x + 8*x = 16*x. We know R = 32, so we have 16*x = 32 and we solve it through the following steps: Simplifying the equation: 16*x = 32. Move all terms to one side: 16*x - 32 = 0. Isolate the term with x: 16*x = 32. Divide both sides by 16: x = 32 / 16. Solution: x = 2. Answer: 2.
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The number of intense detective thriller in Taylor Movie Festival equals 4. The number of upbeat metropolis comedy in Taylor Movie Festival exists, and its number is greater than 0. The total number of movies in West Sahara Movie Festival equals 17. The number of futuristic sci-fi movie in West Sahara Movie Festival equals the total number of movies in Taylor Movie Festival. The number of intense detective thriller in West Sahara Movie Festival equals the sum of the total number of movies in Taylor Movie Festival and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of upbeat metropolis comedy in Verdi Movie Festival equals 1. The number of intense detective thriller in Verdi Movie Festival equals 2.
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How many upbeat metropolis comedy does Taylor Movie Festival have?
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The question is difficult, so we use equations to solve it. Define intense detective thriller in Taylor Movie Festival as G; so G = 4. Define upbeat metropolis comedy in Taylor Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in Taylor Movie Festival as t; so t = x + G = x + 4 = x + 4. Define intense detective thriller in West Sahara Movie Festival as J; so J = t + x = x + 4 + x = 2*x + 4. Define futuristic sci-fi movie in West Sahara Movie Festival as v; so v = t = x + 4. Define total number of movies in West Sahara Movie Festival as f; so f = J + v = 2*x + 4 + x + 4 = 3*x + 8. We know f = 17, so we have 3*x + 8 = 17 and we solve it through the following steps: Simplifying the equation: 3*x + 8 = 17. Move all terms to one side: 3*x - 9 = 0. Isolate the term with x: 3*x = 9. Divide both sides by 3: x = 9 / 3. Solution: x = 3. Answer: 3.
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The number of calm road movie in West Sahara Movie Festival exists, and its number is greater than 0. The number of calm road movie in West Sahara Movie Festival exists, and its number is greater than 0. The number of intense detective thriller in Northwood Movie Festival equals 3 plus the number of solemn period drama in Verdi Movie Festival. The number of solemn period drama in West Sahara Movie Festival equals 3 plus the number of calm road movie in West Sahara Movie Festival. The number of calm road movie in Northwood Movie Festival equals 66. The number of solemn period drama in Verdi Movie Festival equals 3. The number of intense detective thriller in Verdi Movie Festival equals 3 times the number of solemn period drama in West Sahara Movie Festival. The number of calm road movie in Verdi Movie Festival equals the sum of the number of solemn period drama in Verdi Movie Festival and the number of intense detective thriller in Northwood Movie Festival.
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How many calm road movie does West Sahara Movie Festival have?
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The question is difficult, so we use equations to solve it. Define calm road movie in West Sahara Movie Festival as x; we don't know its value yet but will find it out later. Define solemn period drama in West Sahara Movie Festival as y; L = x = x; so y = 3 + L = 3 + x = x + 3. Define intense detective thriller in Verdi Movie Festival as r; f = y = x + 3; so r = 3 * f = (3) * (x + 3) = 3*x + 9. Define calm road movie in Northwood Movie Festival as A; u = r + x = 3*x + 9 + x = 4*x + 9; g = u + y = 4*x + 9 + x + 3 = 5*x + 12; so A = 3 * g = (3) * (5*x + 12) = 15*x + 36. We know A = 66, so we have 15*x + 36 = 66 and we solve it through the following steps: Simplifying the equation: 15*x + 36 = 66. Move all terms to one side: 15*x - 30 = 0. Isolate the term with x: 15*x = 30. Divide both sides by 15: x = 30 / 15. Solution: x = 2. Answer: 2.
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The number of solemn period drama in Festival Lumière de Valmont exists, and its number is greater than 0. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals the sum of the number of futuristic sci-fi movie in Cinéma de Montreval, the total number of movies in Festival Lumière de Valmont, and the number of solemn period drama in Cinéma de Montreval. The number of calm road movie in Festival Lumière de Valmont equals 3. The total number of movies in Cinéma de Montreval equals 18. The number of calm road movie in Rêves de Belleville equals the number of futuristic sci-fi movie in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Cinéma de Montreval equals 4 times the number of solemn period drama in Festival Lumière de Valmont. The number of solemn period drama in Cinéma de Montreval equals the number of solemn period drama in Festival Lumière de Valmont. The number of calm road movie in Cinéma de Montreval equals 4 times the number of solemn period drama in Cinéma de Montreval. The number of solemn period drama in Rêves de Belleville equals 3. The number of futuristic sci-fi movie in Rêves de Belleville equals 1 plus the total number of movies in Festival Lumière de Valmont.
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How many solemn period drama does Festival Lumière de Valmont have?
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The question is difficult, so we use equations to solve it. Define solemn period drama in Festival Lumière de Valmont as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Cinéma de Montreval as l; y = x = x; so l = 4 * y = (4) * (x) = 4*x. Define solemn period drama in Cinéma de Montreval as t; so t = x = x. Define calm road movie in Cinéma de Montreval as h; R = t = x; so h = 4 * R = (4) * (x) = 4*x. Define total number of movies in Cinéma de Montreval as n; d = t + h = x + 4*x = 5*x; so n = d + l = 5*x + 4*x = 9*x. We know n = 18, so we have 9*x = 18 and we solve it through the following steps: Simplifying the equation: 9*x = 18. Move all terms to one side: 9*x - 18 = 0. Isolate the term with x: 9*x = 18. Divide both sides by 9: x = 18 / 9. Solution: x = 2. Answer: 2.
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The number of calm road movie in Festival Lumière de Valmont equals the number of futuristic sci-fi movie in Festival Lumière de Valmont. The number of calm road movie in Cinéma de Montreval equals the sum of the total number of movies in Festival Lumière de Valmont and the number of futuristic sci-fi movie in Festival Lumière de Valmont. The number of solemn period drama in Festival de Saint-Rivage equals the total number of movies in Cinéma de Montreval. The number of futuristic sci-fi movie in Festival Lumière de Valmont exists, and its number is greater than 0. The total number of movies in Festival de Saint-Rivage equals 6.
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How many futuristic sci-fi movie does Festival Lumière de Valmont have?
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The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Festival Lumière de Valmont as x; we don't know its value yet but will find it out later. Define calm road movie in Festival Lumière de Valmont as R; so R = x = x. Define total number of movies in Festival Lumière de Valmont as r; so r = R + x = x + x = 2*x. Define calm road movie in Cinéma de Montreval as a; so a = r + x = 2*x + x = 3*x. Define total number of movies in Cinéma de Montreval as u; so u = a = 3*x. Define solemn period drama in Festival de Saint-Rivage as k; so k = u = 3*x. Define total number of movies in Festival de Saint-Rivage as J; so J = k = 3*x. We know J = 6, so we have 3*x = 6 and we solve it through the following steps: Simplifying the equation: 3*x = 6. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.
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