The number of upbeat metropolis comedy in Northwood Movie Festival exists, and its number is greater than 0. The total number of movies in Golden Banana Movie Festival equals 43. The number of calm road movie in Northwood Movie Festival equals 4 times the number of upbeat metropolis comedy in Northwood Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival exists, and its number is greater than 0.

Question: How many upbeat metropolis comedy does Northwood Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Northwood Movie Festival as x; we don't know its value yet but will find it out later. Define calm road movie in Northwood Movie Festival as A; M = x = x; so A = 4 * M = (4) * (x) = 4*x. Define total number of movies in Northwood Movie Festival as R; so R = A + x = 4*x + x = 5*x. Define calm road movie in Golden Banana Movie Festival as s; z = R + x = 5*x + x = 6*x; E = z + A = 6*x + 4*x = 10*x; so s = 3 + E = 3 + 10*x = 10*x + 3. Define total number of movies in Golden Banana Movie Festival as y; so y = s = 10*x + 3. We know y = 43, so we have 10*x + 3 = 43 and we solve it through the following steps: Simplifying the equation: 10*x + 3 = 43. Move all terms to one side: 10*x - 40 = 0. Isolate the term with x: 10*x = 40. Divide both sides by 10: x = 40 / 10. Solution: x = 4. Answer: 4.


The number of adult wolf in Pine Ridge equals the number of adult racoon in Cedar Valley. The number of adult wolf in Maple Creek equals 1 plus the number of adult racoon in Pine Ridge. The number of adult bear in Pine Ridge equals the sum of the number of adult racoon in Pine Ridge, the number of adult fox in Oakridge Riverside, and the number of adult wolf in Maple Creek. The number of adult fox in Oakridge Riverside equals the number of adult wolf in Maple Creek. The number of adult fox in Pine Ridge equals 4 times the number of adult wolf in Cedar Valley. The number of adult racoon in Pine Ridge equals 4. The number of adult fox in Cedar Valley exists, and its number is greater than 0. The number of adult bear in Cedar Valley equals the sum of the number of adult racoon in Cedar Valley and the number of adult fox in Cedar Valley. The number of adult fox in Maple Creek equals the number of adult bear in Cedar Valley. The number of adult wolf in Cedar Valley equals the number of adult racoon in Cedar Valley. The total number of adult animals in Cedar Valley equals 16. The number of adult fox in Cedar Valley exists, and its number is greater than 0. The number of adult racoon in Maple Creek equals the number of adult wolf in Oakridge Riverside. The number of adult wolf in Oakridge Riverside equals 4 times the sum of the number of adult bear in Maple Creek and the number of adult wolf in Maple Creek. The number of adult bear in Oakridge Riverside equals the difference between the number of adult wolf in Maple Creek and the number of adult fox in Oakridge Riverside. The number of adult bear in Maple Creek equals 3 times the number of adult wolf in Maple Creek. The number of adult racoon in Cedar Valley equals 2 times the number of adult fox in Cedar Valley.

Question: How many adult fox does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult fox in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult racoon in Cedar Valley as l; S = x = x; so l = 2 * S = (2) * (x) = 2*x. Define adult wolf in Cedar Valley as C; so C = l = 2*x. Define adult bear in Cedar Valley as d; so d = l + x = 2*x + x = 3*x. Define total number of adult animals in Cedar Valley as u; U = d + x = 3*x + x = 4*x; z = U + C = 4*x + 2*x = 6*x; so u = z + l = 6*x + 2*x = 8*x. We know u = 16, so we have 8*x = 16 and we solve it through the following steps: Simplifying the equation: 8*x = 16. Move all terms to one side: 8*x - 16 = 0. Isolate the term with x: 8*x = 16. Divide both sides by 8: x = 16 / 8. Solution: x = 2. Answer: 2.


The number of elementary school in Oakbridge City equals the total number of schools in Ruby Bay. The number of culinarian school in Ruby Bay equals 3 times the number of elementary school in Ruby Bay. The number of elementary school in Ruby Bay equals 4. The number of culinarian school in Oakbridge City equals 3 times the number of elementary school in Ruby Bay.

Question: What is the total number of schools in Oakbridge City?

Solution: Define elementary school in Ruby Bay as Z; so Z = 4. Define culinarian school in Ruby Bay as D; W = Z = 4; so D = 3 * W = 3 * 4 = 12. Define total number of schools in Ruby Bay as V; so V = Z + D = 4 + 12 = 16. Define elementary school in Oakbridge City as K; so K = V = 16. Define culinarian school in Oakbridge City as w; h = Z = 4; so w = 3 * h = 3 * 4 = 12. Define total number of schools in Oakbridge City as s; so s = K + w = 16 + 12 = 28. Answer: 28.


The number of calm road movie in Festival de Saint-Rivage exists, and its number is greater than 0. The number of upbeat metropolis comedy in Rêves de Belleville equals the sum of the number of calm road movie in Rêves de Belleville and the number of calm road movie in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals the number of calm road movie in Festival de Saint-Rivage. The number of calm road movie in Rêves de Belleville equals the sum of the total number of movies in Festival de Saint-Rivage and the number of calm road movie in Festival de Saint-Rivage. The total number of movies in Rêves de Belleville equals 14.

Question: How many calm road movie does Festival de Saint-Rivage have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Festival de Saint-Rivage as V; so V = x = x. Define total number of movies in Festival de Saint-Rivage as m; so m = x + V = x + x = 2*x. Define calm road movie in Rêves de Belleville as y; so y = m + x = 2*x + x = 3*x. Define upbeat metropolis comedy in Rêves de Belleville as k; so k = y + x = 3*x + x = 4*x. Define total number of movies in Rêves de Belleville as c; so c = y + k = 3*x + 4*x = 7*x. We know c = 14, so we have 7*x = 14 and we solve it through the following steps: Simplifying the equation: 7*x = 14. Move all terms to one side: 7*x - 14 = 0. Isolate the term with x: 7*x = 14. Divide both sides by 7: x = 14 / 7. Solution: x = 2. Answer: 2.


The number of intense detective thriller in Northwood Movie Festival exists, and its number is greater than 0. The total number of movies in Taylor Movie Festival equals 25. The number of upbeat metropolis comedy in Northwood Movie Festival equals 2 times the number of intense detective thriller in Northwood Movie Festival. The number of intense detective thriller in Northwood Movie Festival exists, and its number is greater than 0.

Question: How many intense detective thriller does Northwood Movie Festival have?

Solution: 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 upbeat metropolis comedy in Northwood Movie Festival as M; Y = x = x; so M = 2 * Y = (2) * (x) = 2*x. Define total number of movies in Northwood Movie Festival as d; so d = x + M = x + 2*x = 3*x. Define upbeat metropolis comedy in Taylor Movie Festival as P; n = d + x = 3*x + x = 4*x; p = n + M = 4*x + 2*x = 6*x; so P = 1 + p = 1 + 6*x = 6*x + 1. Define total number of movies in Taylor Movie Festival as B; so B = P = 6*x + 1. We know B = 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.


The number of adult fox in South Zoo equals the total number of adult animals in Jefferson Circus. The number of adult bear in South Zoo equals the sum of the number of adult fox in Jefferson Circus and the total number of adult animals in Jefferson Circus. The number of adult bear in Jefferson Circus equals 4. The number of adult fox in Jefferson Circus equals 2 plus the number of adult bear in Jefferson Circus.

Question: What is the total number of adult animals in South Zoo?

Solution: Define adult bear in Jefferson Circus as U; so U = 4. Define adult fox in Jefferson Circus as l; G = U = 4; so l = 2 + G = 2 + 4 = 6. Define total number of adult animals in Jefferson Circus as p; so p = U + l = 4 + 6 = 10. Define adult fox in South Zoo as Q; so Q = p = 10. Define adult bear in South Zoo as E; so E = l + p = 6 + 10 = 16. Define total number of adult animals in South Zoo as d; so d = E + Q = 16 + 10 = 26. Answer: 26.


The number of private middle school in Clearwater Bay equals the number of private middle school in Oakbridge City. The total number of schools in Clearwater Bay equals 12. The number of private middle school in Oakbridge City exists, and its number is greater than 0. The number of culinarian school in Clearwater Bay equals 4 plus the total number of schools in Oakbridge City. The number of culinarian school in Oakbridge City equals 2 times the number of private middle school in Oakbridge City.

Question: How many private middle school does Oakbridge City have?

Solution: The question is difficult, so we use equations to solve it. Define private middle school 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 z; p = x = x; so z = 2 * p = (2) * (x) = 2*x. Define total number of schools in Oakbridge City as d; so d = x + z = x + 2*x = 3*x. Define private middle school in Clearwater Bay as m; so m = x = x. Define culinarian school in Clearwater Bay as b; w = d = 3*x; so b = 4 + w = 4 + 3*x = 3*x + 4. Define total number of schools in Clearwater Bay as D; so D = m + b = x + 3*x + 4 = 4*x + 4. We know D = 12, so we have 4*x + 4 = 12 and we solve it through the following steps: Simplifying the equation: 4*x + 4 = 12. 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.


The number of culinarian school in Westhaven City exists, and its number is greater than 0. The total number of schools in Evervale City equals 64. The number of private middle school in Westhaven City equals 3 times the number of culinarian school in Westhaven City. The number of culinarian school in Westhaven City exists, and its number is greater than 0.

Question: How many culinarian school does Westhaven City have?

Solution: 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 private middle school in Westhaven City as U; e = x = x; so U = 3 * e = (3) * (x) = 3*x. Define total number of schools in Westhaven City as a; so a = U + x = 3*x + x = 4*x. Define private middle school in Evervale City as O; I = a + x = 4*x + x = 5*x; p = I + U = 5*x + 3*x = 8*x; so O = 2 * p = (2) * (8*x) = 16*x. Define total number of schools in Evervale City as D; so D = O = 16*x. We know D = 64, so we have 16*x = 64 and we solve it through the following steps: Simplifying the equation: 16*x = 64. Move all terms to one side: 16*x - 64 = 0. Isolate the term with x: 16*x = 64. Divide both sides by 16: x = 64 / 16. Solution: x = 4. Answer: 4.


The total number of schools in Westhaven City equals 11. The number of regional medical school in Brightford equals the number of culinarian school in Brightford. The number of culinarian school in Westhaven City equals 1 plus the total number of schools in Brightford. The number of culinarian school in Brightford exists, and its number is greater than 0. The number of regional medical school in Westhaven City equals 1 plus the number of culinarian school in Westhaven City.

Question: How many culinarian school does Brightford have?

Solution: The question is difficult, so we use equations to solve it. Define culinarian school in Brightford as x; we don't know its value yet but will find it out later. Define regional medical school in Brightford as J; so J = x = x. Define total number of schools in Brightford as Y; so Y = x + J = x + x = 2*x. Define culinarian school in Westhaven City as R; E = Y = 2*x; so R = 1 + E = 1 + 2*x = 2*x + 1. Define regional medical school in Westhaven City as T; C = R = 2*x + 1; so T = 1 + C = 1 + 2*x + 1 = 2*x + 2. Define total number of schools in Westhaven City as w; so w = R + T = 2*x + 1 + 2*x + 2 = 4*x + 3. We know w = 11, so we have 4*x + 3 = 11 and we solve it through the following steps: Simplifying the equation: 4*x + 3 = 11. 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.


The number of adult crow in Jefferson Circus exists, and its number is greater than 0. The number of adult crow in Jefferson Circus exists, and its number is greater than 0. The total number of adult animals in Hamilton Farm equals 19. The number of adult parrot in Jefferson Circus equals 3 times the number of adult crow in Jefferson Circus.

Question: How many adult crow does Jefferson Circus have?

Solution: The question is difficult, so we use equations to solve it. Define adult crow in Jefferson Circus as x; we don't know its value yet but will find it out later. Define adult parrot in Jefferson Circus as y; r = x = x; so y = 3 * r = (3) * (x) = 3*x. Define total number of adult animals in Jefferson Circus as l; so l = y + x = 3*x + x = 4*x. Define adult parrot in Hamilton Farm as p; Y = l + x = 4*x + x = 5*x; T = Y + y = 5*x + 3*x = 8*x; so p = 3 + T = 3 + 8*x = 8*x + 3. Define total number of adult animals in Hamilton Farm as g; so g = p = 8*x + 3. We know g = 19, so we have 8*x + 3 = 19 and we solve it through the following steps: Simplifying the equation: 8*x + 3 = 19. Move all terms to one side: 8*x - 16 = 0. Isolate the term with x: 8*x = 16. Divide both sides by 8: x = 16 / 8. Solution: x = 2. Answer: 2.


The number of adult blue jay in Mayer Aquarium exists, and its number is greater than 0. The number of adult owl in Mayer Aquarium equals 4 times the number of adult blue jay in Mayer Aquarium. The total number of adult animals in Jefferson Circus equals 44. The number of adult blue jay in Mayer Aquarium exists, and its number is greater than 0.

Question: How many adult blue jay does Mayer Aquarium have?

Solution: The question is difficult, so we use equations to solve it. Define adult blue jay in Mayer Aquarium as x; we don't know its value yet but will find it out later. Define adult owl in Mayer Aquarium as k; a = x = x; so k = 4 * a = (4) * (x) = 4*x. Define total number of adult animals in Mayer Aquarium as T; so T = k + x = 4*x + x = 5*x. Define adult crow in Jefferson Circus as B; t = T + x = 5*x + x = 6*x; Y = t + k = 6*x + 4*x = 10*x; so B = 4 + Y = 4 + 10*x = 10*x + 4. Define total number of adult animals in Jefferson Circus as n; so n = B = 10*x + 4. We know n = 44, so we have 10*x + 4 = 44 and we solve it through the following steps: Simplifying the equation: 10*x + 4 = 44. Move all terms to one side: 10*x - 40 = 0. Isolate the term with x: 10*x = 40. Divide both sides by 10: x = 40 / 10. Solution: x = 4. Answer: 4.


The number of adult wolf in Jefferson Circus equals 3 plus the number of adult fox in Jefferson Circus. The number of adult fox in Jefferson Circus equals 2. The number of adult fox in Hamilton Farm equals the difference between the number of adult wolf in Jefferson Circus and the number of adult fox in Jefferson Circus. The number of adult wolf in Hamilton Farm equals 2 times the difference between the number of adult wolf in Jefferson Circus and the number of adult fox in Jefferson Circus.

Question: What is the total number of adult animals in Hamilton Farm?

Solution: Define adult fox in Jefferson Circus as T; so T = 2. Define adult wolf in Jefferson Circus as n; x = T = 2; so n = 3 + x = 3 + 2 = 5. Define adult fox in Hamilton Farm as Z; so Z = n - T = 5 - 2 = 3. Define adult wolf in Hamilton Farm as S; f = n - T = 5 - 2 = 3; so S = 2 * f = 2 * 3 = 6. Define total number of adult animals in Hamilton Farm as E; so E = Z + S = 3 + 6 = 9. Answer: 9.


The number of intense detective thriller in Taylor Movie Festival equals 3 times the sum of the total number of movies in Northwood Movie Festival, the number of calm road movie in Northwood Movie Festival, and the number of intense detective thriller in Northwood Movie Festival. The number of intense detective thriller in Northwood Movie Festival equals the number of calm road movie in Northwood Movie Festival. The number of calm road movie in Northwood Movie Festival equals 2. The number of calm road movie in Taylor Movie Festival equals 4 times the number of intense detective thriller in Taylor Movie Festival.

Question: How many calm road movie does Taylor Movie Festival have?

Solution: Define calm road movie in Northwood Movie Festival as D; so D = 2. Define intense detective thriller in Northwood Movie Festival as V; so V = D = 2. Define total number of movies in Northwood Movie Festival as C; so C = V + D = 2 + 2 = 4. Define intense detective thriller in Taylor Movie Festival as U; Q = C + D = 4 + 2 = 6; u = Q + V = 6 + 2 = 8; so U = 3 * u = 3 * 8 = 24. Define calm road movie in Taylor Movie Festival as a; F = U = 24; so a = 4 * F = 4 * 24 = 96. Answer: 96.


The total number of adult animals in Pine Ridge equals 8. The number of adult fox in Cedar Valley equals 2 times the number of adult deer in Cedar Valley. The number of adult deer in Pine Ridge equals the number of adult fox in Cedar Valley. The number of adult deer in Cedar Valley exists, and its number is greater than 0. The number of adult deer in Cedar Valley exists, and its number is greater than 0.

Question: How many adult deer does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult deer in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult fox in Cedar Valley as f; o = x = x; so f = 2 * o = (2) * (x) = 2*x. Define adult deer in Pine Ridge as t; so t = f = 2*x. Define adult fox in Pine Ridge as a; O = t + x = 2*x + x = 3*x; P = O + f = 3*x + 2*x = 5*x; so a = 1 + P = 1 + 5*x = 5*x + 1. Define total number of adult animals in Pine Ridge as j; so j = a + t = 5*x + 1 + 2*x = 7*x + 1. We know j = 8, so we have 7*x + 1 = 8 and we solve it through the following steps: Simplifying the equation: 7*x + 1 = 8. Move all terms to one side: 7*x - 7 = 0. Isolate the term with x: 7*x = 7. Divide both sides by 7: x = 7 / 7. Solution: x = 1. Answer: 1.


The number of public highschool in Hawkesbury equals 4 times the sum of the total number of schools in Evervale City, the number of culinarian school in Evervale City, and the number of public highschool in Evervale City. The number of culinarian school in Evervale City equals 2. The number of public highschool in Evervale City equals 2 plus the number of culinarian school in Evervale City.

Question: What is the total number of schools in Hawkesbury?

Solution: Define culinarian school in Evervale City as s; so s = 2. Define public highschool in Evervale City as U; n = s = 2; so U = 2 + n = 2 + 2 = 4. Define total number of schools in Evervale City as E; so E = s + U = 2 + 4 = 6. Define public highschool in Hawkesbury as k; Q = E + s = 6 + 2 = 8; S = Q + U = 8 + 4 = 12; so k = 4 * S = 4 * 12 = 48. Define total number of schools in Hawkesbury as e; so e = k = 48. Answer: 48.


The number of regional medical school in Brightford equals the sum of the number of private middle school in Brightford, the number of culinarian school in Brightford, and the number of regional medical school in Evervale City. The number of culinarian school in Evervale City equals the sum of the number of private middle school in Evervale City and the number of regional medical school in Brightford. The number of culinarian school in Brightford equals 4 times the number of regional medical school in Evervale City. The number of regional medical school in Evervale City equals 3. The number of private middle school in Brightford equals the number of regional medical school in Evervale City. The number of private middle school in Evervale City equals the number of regional medical school in Brightford.

Question: What is the total number of schools in Brightford?

Solution: Define regional medical school in Evervale City as C; so C = 3. Define culinarian school in Brightford as Z; K = C = 3; so Z = 4 * K = 4 * 3 = 12. Define private middle school in Brightford as A; so A = C = 3. Define regional medical school in Brightford as j; e = A + Z = 3 + 12 = 15; so j = e + C = 15 + 3 = 18. Define total number of schools in Brightford as a; V = Z + j = 12 + 18 = 30; so a = V + A = 30 + 3 = 33. Answer: 33.


The number of calm road movie in Festival de Clairmont equals 4 plus the number of upbeat metropolis comedy in Festival de Clairmont. The number of solemn period drama in Festival Lumière de Valmont equals the sum of the number of upbeat metropolis comedy in Festival Lumière de Valmont, the number of upbeat metropolis comedy in Festival de Clairmont, and the number of calm road movie in Festival Lumière de Valmont. The number of intense detective thriller in Rêves de Belleville equals the sum of the number of solemn period drama in Festival de Clairmont, the number of upbeat metropolis comedy in Festival de Clairmont, and the number of calm road movie in Festival de Clairmont. The number of upbeat metropolis comedy in Rêves de Belleville equals the total number of movies in Cinéma de Montreval. The number of calm road movie in Cinéma de Montreval equals the difference between the number of upbeat metropolis comedy in Festival de Clairmont and the number of solemn period drama in Festival de Clairmont. The number of calm road movie in Rêves de Belleville equals the total number of movies in Festival de Clairmont. The number of calm road movie in Festival Lumière de Valmont equals the total number of movies in Rêves de Belleville. The number of upbeat metropolis comedy in Festival Lumière de Valmont equals 1. The number of upbeat metropolis comedy in Cinéma de Montreval equals 4 plus the number of upbeat metropolis comedy in Festival de Clairmont. The number of solemn period drama in Festival de Clairmont equals 2 plus the number of intense detective thriller in Festival de Clairmont. The number of intense detective thriller in Festival Lumière de Valmont equals 2. The number of intense detective thriller in Festival de Clairmont equals 2. The number of upbeat metropolis comedy in Festival de Clairmont equals the number of intense detective thriller in Festival de Clairmont.

Question: What is the total number of movies in Festival de Clairmont?

Solution: Define intense detective thriller in Festival de Clairmont as s; so s = 2. Define upbeat metropolis comedy in Festival de Clairmont as V; so V = s = 2. Define solemn period drama in Festival de Clairmont as O; n = s = 2; so O = 2 + n = 2 + 2 = 4. Define calm road movie in Festival de Clairmont as D; i = V = 2; so D = 4 + i = 4 + 2 = 6. Define total number of movies in Festival de Clairmont as q; x = V + O = 2 + 4 = 6; Q = x + D = 6 + 6 = 12; so q = Q + s = 12 + 2 = 14. Answer: 14.


The number of regional medical school in Shoreline City exists, and its number is greater than 0. The number of public highschool in Shoreline City equals the number of regional medical school in Shoreline City. The total number of schools in Oakbridge City equals 8. The number of regional medical school in Oakbridge City equals 4 plus the difference between the number of public highschool in Shoreline City and the number of regional medical school in Shoreline City. The number of public highschool in Oakbridge City equals the total number of schools in Shoreline City.

Question: How many regional medical school does Shoreline City have?

Solution: The question is difficult, so we use equations to solve it. Define regional medical school in Shoreline City as x; we don't know its value yet but will find it out later. Define public highschool in Shoreline City as T; so T = x = x. Define total number of schools in Shoreline City as B; so B = T + x = x + x = 2*x. Define public highschool in Oakbridge City as t; so t = B = 2*x. Define regional medical school in Oakbridge City as L; z = T - x = (x) - (x) = 0; so L = 4 + z = 4 + 0 = 4. Define total number of schools in Oakbridge City as A; so A = t + L = 2*x + 4 = 2*x + 4. We know A = 8, so we have 2*x + 4 = 8 and we solve it through the following steps: Simplifying the equation: 2*x + 4 = 8. Move all terms to one side: 2*x - 4 = 0. Isolate the term with x: 2*x = 4. Divide both sides by 2: x = 4 / 2. Solution: x = 2. Answer: 2.


The number of solemn period drama in Festival Lumière de Valmont equals the number of calm road movie in Festival Lumière de Valmont. The total number of movies in Rêves de Belleville equals 10. The number of calm road movie in Festival Lumière de Valmont exists, and its number is greater than 0. The number of solemn period drama in Rêves de Belleville equals the sum of the total number of movies in Festival Lumière de Valmont, the number of solemn period drama in Festival Lumière de Valmont, and the number of calm road movie in Festival Lumière de Valmont. The number of calm road movie in Rêves de Belleville equals the number of solemn period drama in Festival Lumière de Valmont.

Question: How many calm road movie does Festival Lumière de Valmont have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Festival Lumière de Valmont as x; we don't know its value yet but will find it out later. Define solemn period drama in Festival Lumière de Valmont as E; so E = x = x. Define total number of movies in Festival Lumière de Valmont as y; so y = x + E = x + x = 2*x. Define solemn period drama in Rêves de Belleville as n; Q = y + E = 2*x + x = 3*x; so n = Q + x = 3*x + x = 4*x. Define calm road movie in Rêves de Belleville as w; so w = E = x. Define total number of movies in Rêves de Belleville as C; so C = w + n = x + 4*x = 5*x. We know C = 10, so we have 5*x = 10 and we solve it through the following steps: Simplifying the equation: 5*x = 10. 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.


The number of calm road movie in Verdi Movie Festival equals the sum of the number of upbeat metropolis comedy in Verdi Movie Festival and the total number of movies in Golden Banana Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 4 plus the number of calm road movie in Golden Banana Movie Festival. The number of upbeat metropolis comedy in Verdi Movie Festival equals the total number of movies in Golden Banana Movie Festival. The number of calm road movie in Golden Banana Movie Festival exists, and its number is greater than 0. The total number of movies in Verdi Movie Festival equals 18.

Question: How many calm road movie does Golden Banana Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Golden Banana Movie Festival as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Golden Banana Movie Festival as R; g = x = x; so R = 4 + g = 4 + x = x + 4. Define total number of movies in Golden Banana Movie Festival as F; so F = R + x = x + 4 + x = 2*x + 4. Define upbeat metropolis comedy in Verdi Movie Festival as T; so T = F = 2*x + 4. Define calm road movie in Verdi Movie Festival as q; so q = T + F = 2*x + 4 + 2*x + 4 = 4*x + 8. Define total number of movies in Verdi Movie Festival as E; so E = T + q = 2*x + 4 + 4*x + 8 = 6*x + 12. We know E = 18, so we have 6*x + 12 = 18 and we solve it through the following steps: Simplifying the equation: 6*x + 12 = 18. Move all terms to one side: 6*x - 6 = 0. Isolate the term with x: 6*x = 6. Divide both sides by 6: x = 6 / 6. Solution: x = 1. Answer: 1.


The number of adult crow in Jefferson Circus equals the sum of the total number of adult animals in Mayer Aquarium, the number of adult owl in Mayer Aquarium, and the number of adult crow in Mayer Aquarium. The number of adult crow in Mayer Aquarium equals the number of adult owl in Mayer Aquarium. The total number of adult animals in Jefferson Circus equals 13. The number of adult owl in Jefferson Circus exists, and its number is greater than 0. The number of adult owl in Mayer Aquarium equals 3.

Question: How many adult owl does Jefferson Circus have?

Solution: The question is difficult, so we use equations to solve it. Define adult owl in Mayer Aquarium as E; so E = 3. Define adult crow in Mayer Aquarium as Y; so Y = E = 3. Define total number of adult animals in Mayer Aquarium as A; so A = E + Y = 3 + 3 = 6. Define adult crow in Jefferson Circus as o; S = A + E = 6 + 3 = 9; so o = S + Y = 9 + 3 = 12. Define adult owl in Jefferson Circus as x; we don't know its value yet but will find it out later. Define total number of adult animals in Jefferson Circus as R; so R = x + o = x + 12 = x + 12. We know R = 13, so we have x + 12 = 13 and we solve it through the following steps: Simplifying the equation: x + 12 = 13. 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.


The total number of movies in West Sahara Movie Festival equals 32. The number of futuristic sci-fi movie in Golden Banana Movie Festival exists, and its number is greater than 0. The number of futuristic sci-fi movie in Golden Banana Movie Festival exists, and its number is greater than 0. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 4. The number of upbeat metropolis comedy in West Sahara Movie Festival equals the number of futuristic sci-fi movie in West Sahara Movie Festival.

Question: How many futuristic sci-fi movie does Golden Banana Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Golden Banana Movie Festival as D; so D = 4. Define futuristic sci-fi movie in Golden Banana Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in Golden Banana Movie Festival as j; so j = x + D = x + 4 = x + 4. Define futuristic sci-fi movie in West Sahara Movie Festival as E; t = j + x = x + 4 + x = 2*x + 4; so E = t + D = 2*x + 4 + 4 = 2*x + 8. Define upbeat metropolis comedy in West Sahara Movie Festival as m; so m = E = 2*x + 8. Define total number of movies in West Sahara Movie Festival as w; so w = E + m = 2*x + 8 + 2*x + 8 = 4*x + 16. We know w = 32, so we have 4*x + 16 = 32 and we solve it through the following steps: Simplifying the equation: 4*x + 16 = 32. 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.


The number of calm road movie in Festival Lumière de Valmont equals 3. The number of upbeat metropolis comedy in Festival de Saint-Rivage exists, and its number is greater than 0. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 1 plus the number of calm road movie in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals the number of futuristic sci-fi movie in Festival de Clairmont. The number of futuristic sci-fi movie in Festival de Clairmont equals the number of calm road movie in Festival de Saint-Rivage. The number of intense detective thriller in Festival de Saint-Rivage equals 4. The total number of movies in Rêves de Belleville equals 81. The number of intense detective thriller in Festival de Clairmont equals 3. The number of upbeat metropolis comedy in Festival de Clairmont equals the number of intense detective thriller in Festival de Saint-Rivage. 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 calm road movie in Rêves de Belleville equals 3 times the sum of the number of futuristic sci-fi movie in Festival Lumière de Valmont, the number of calm road movie in Festival de Saint-Rivage, and the number of upbeat metropolis comedy in Festival de Saint-Rivage.

Question: How many upbeat metropolis comedy does Festival de Saint-Rivage have?

Solution: 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 C; Y = x = x; so C = 4 * Y = (4) * (x) = 4*x. Define futuristic sci-fi movie in Festival de Clairmont as b; so b = C = 4*x. Define futuristic sci-fi movie in Festival Lumière de Valmont as k; so k = b = 4*x. Define calm road movie in Rêves de Belleville as p; W = k + C = 4*x + 4*x = 8*x; h = W + x = 8*x + x = 9*x; so p = 3 * h = (3) * (9*x) = 27*x. Define total number of movies in Rêves de Belleville as T; so T = p = 27*x. We know T = 81, so we have 27*x = 81 and we solve it through the following steps: Simplifying the equation: 27*x = 81. Move all terms to one side: 27*x - 81 = 0. Isolate the term with x: 27*x = 81. Divide both sides by 27: x = 81 / 27. Solution: x = 3. Answer: 3.


The number of culinarian school in Brightford equals the number of elementary school in Evervale City. The number of regional medical school in Brightford equals 3. The number of culinarian school in Glenfield City equals 4. The number of elementary school in Glenfield City equals the number of regional medical school in Brightford. The number of culinarian school in Westhaven City equals 3 plus the difference between the number of culinarian school in Brightford and the number of elementary school in Evervale City. The number of culinarian school in Evervale City equals the difference between the number of elementary school in Glenfield City and the number of regional medical school in Brightford. The number of regional medical school in Hawkesbury equals the sum of the number of culinarian school in Hawkesbury, the number of culinarian school in Evervale City, and the total number of schools in Evervale City. The number of regional medical school in Evervale City equals 3. The number of elementary school in Westhaven City equals 3 times the sum of the number of culinarian school in Brightford and the number of elementary school in Evervale City. The number of elementary school in Evervale City equals 4. The number of elementary school in Brightford equals the number of culinarian school in Brightford. The number of culinarian school in Hawkesbury equals 3.

Question: What is the total number of schools in Westhaven City?

Solution: Define elementary school in Evervale City as Z; so Z = 4. Define culinarian school in Brightford as I; so I = Z = 4. Define elementary school in Westhaven City as q; d = I + Z = 4 + 4 = 8; so q = 3 * d = 3 * 8 = 24. Define culinarian school in Westhaven City as i; A = I - Z = 4 - 4 = 0; so i = 3 + A = 3 + 0 = 3. Define total number of schools in Westhaven City as e; so e = i + q = 3 + 24 = 27. Answer: 27.


The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 3 plus the total number of movies in Festival de Clairmont. The number of solemn period drama in Festival Lumière de Valmont equals 4 times the number of solemn period drama in Festival de Clairmont. The number of futuristic sci-fi movie in Festival de Clairmont equals 4. The number of solemn period drama in Festival de Clairmont equals 1.

Question: What is the total number of movies in Festival Lumière de Valmont?

Solution: Define futuristic sci-fi movie in Festival de Clairmont as B; so B = 4. Define solemn period drama in Festival de Clairmont as C; so C = 1. Define total number of movies in Festival de Clairmont as x; so x = C + B = 1 + 4 = 5. Define futuristic sci-fi movie in Festival Lumière de Valmont as i; M = x = 5; so i = 3 + M = 3 + 5 = 8. Define solemn period drama in Festival Lumière de Valmont as U; c = C = 1; so U = 4 * c = 4 * 1 = 4. Define total number of movies in Festival Lumière de Valmont as g; so g = U + i = 4 + 8 = 12. Answer: 12.


The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 1 plus the sum of the number of upbeat metropolis comedy in Festival de Clairmont and the number of futuristic sci-fi movie in Festival de Clairmont. The number of upbeat metropolis comedy in Festival Lumière de Valmont equals 4. The number of intense detective thriller in Festival de Saint-Rivage equals the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of solemn period drama in Festival Lumière de Valmont equals the number of upbeat metropolis comedy in Rêves de Belleville. 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 solemn period drama in Festival de Clairmont equals the number of upbeat metropolis comedy in Festival de Clairmont. The number of intense detective thriller in Festival de Clairmont equals 2 times the number of upbeat metropolis comedy in Festival de Clairmont. The number of solemn period drama in Festival de Saint-Rivage equals the number of futuristic sci-fi movie in Rêves de Belleville. The number of futuristic sci-fi movie in Rêves de Belleville equals the difference between the number of futuristic sci-fi movie in Festival de Clairmont and the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of intense detective thriller in Festival Lumière de Valmont equals 1. The number of futuristic sci-fi movie in Festival de Clairmont equals the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 2. The number of upbeat metropolis comedy in Rêves de Belleville equals 4. The number of upbeat metropolis comedy in Festival de Clairmont equals the number of upbeat metropolis comedy in Festival Lumière de Valmont.

Question: What is the total number of movies in Festival de Clairmont?

Solution: Define upbeat metropolis comedy in Festival Lumière de Valmont as j; so j = 4. Define upbeat metropolis comedy in Festival de Clairmont as V; so V = j = 4. Define futuristic sci-fi movie in Festival de Clairmont as b; so b = j = 4. Define intense detective thriller in Festival de Clairmont as a; H = V = 4; so a = 2 * H = 2 * 4 = 8. Define solemn period drama in Festival de Clairmont as v; so v = V = 4. Define total number of movies in Festival de Clairmont as A; o = b + a = 4 + 8 = 12; i = o + V = 12 + 4 = 16; so A = i + v = 16 + 4 = 20. Answer: 20.


The number of futuristic sci-fi movie in Northwood Movie Festival equals 1 plus the sum of the total number of movies in Taylor Movie Festival, the number of futuristic sci-fi movie in Taylor Movie Festival, and the number of intense detective thriller in Taylor Movie Festival. The number of intense detective thriller in Taylor Movie Festival equals 2 times the number of futuristic sci-fi movie in Taylor Movie Festival. The number of futuristic sci-fi movie in Taylor Movie Festival equals 3.

Question: What is the total number of movies in Northwood Movie Festival?

Solution: Define futuristic sci-fi movie in Taylor Movie Festival as D; so D = 3. Define intense detective thriller in Taylor Movie Festival as j; J = D = 3; so j = 2 * J = 2 * 3 = 6. Define total number of movies in Taylor Movie Festival as E; so E = D + j = 3 + 6 = 9. Define futuristic sci-fi movie in Northwood Movie Festival as M; i = E + D = 9 + 3 = 12; d = i + j = 12 + 6 = 18; so M = 1 + d = 1 + 18 = 19. Define total number of movies in Northwood Movie Festival as B; so B = M = 19. Answer: 19.


The number of private middle school in Riverton City equals 2. The number of elementary school in Oakbridge City equals 2 times the sum of the total number of schools in Clearwater Bay and the number of culinarian school in Clearwater Bay. The number of elementary school in Riverton City equals 4. The number of private middle school in Oakbridge City equals 1 plus the difference between the number of elementary school in Oakbridge City and the total number of schools in Clearwater Bay. The number of culinarian school in Clearwater Bay equals 4. The number of culinarian school in Shoreline City equals 2 times the sum of the number of private middle school in Oakbridge City, the number of culinarian school in Clearwater Bay, and the number of culinarian school in Riverton City. The number of elementary school in Shoreline City equals the total number of schools in Clearwater Bay. The number of culinarian school in Riverton City equals 3.

Question: What is the total number of schools in Oakbridge City?

Solution: Define culinarian school in Clearwater Bay as I; so I = 4. Define total number of schools in Clearwater Bay as d; so d = I = 4. Define elementary school in Oakbridge City as G; X = d + I = 4 + 4 = 8; so G = 2 * X = 2 * 8 = 16. Define private middle school in Oakbridge City as p; h = G - d = 16 - 4 = 12; so p = 1 + h = 1 + 12 = 13. Define total number of schools in Oakbridge City as b; so b = p + G = 13 + 16 = 29. Answer: 29.


The number of futuristic sci-fi movie in Festival de Clairmont equals the sum of the total number of movies in Cinéma de Montreval, the number of futuristic sci-fi movie in Festival de Saint-Rivage, and the number of upbeat metropolis comedy in Cinéma de Montreval. The number of intense detective thriller in Festival de Clairmont equals the number of futuristic sci-fi movie 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 Clairmont. The number of upbeat metropolis comedy in Festival de Clairmont equals 4. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 3. The number of upbeat metropolis comedy in Cinéma de Montreval equals 1 plus the total number of movies in Festival de Saint-Rivage.

Question: What is the total number of movies in Rêves de Belleville?

Solution: Define futuristic sci-fi movie in Festival de Saint-Rivage as o; so o = 3. Define total number of movies in Festival de Saint-Rivage as B; so B = o = 3. Define upbeat metropolis comedy in Cinéma de Montreval as m; R = B = 3; so m = 1 + R = 1 + 3 = 4. Define total number of movies in Cinéma de Montreval as M; so M = m = 4. Define futuristic sci-fi movie in Festival de Clairmont as d; e = M + o = 4 + 3 = 7; so d = e + m = 7 + 4 = 11. Define upbeat metropolis comedy in Rêves de Belleville as N; so N = d = 11. Define total number of movies in Rêves de Belleville as O; so O = N = 11. Answer: 11.


The number of elementary school in Westhaven City equals 1 plus the sum of the total number of schools in Hawkesbury, the number of elementary school in Hawkesbury, and the number of private middle school in Hawkesbury. The number of elementary school in Hawkesbury equals 2. The number of private middle school in Hawkesbury equals 4 times the number of elementary school in Hawkesbury.

Question: What is the total number of schools in Westhaven City?

Solution: Define elementary school in Hawkesbury as u; so u = 2. Define private middle school in Hawkesbury as I; e = u = 2; so I = 4 * e = 4 * 2 = 8. Define total number of schools in Hawkesbury as R; so R = I + u = 8 + 2 = 10. Define elementary school in Westhaven City as W; S = R + u = 10 + 2 = 12; x = S + I = 12 + 8 = 20; so W = 1 + x = 1 + 20 = 21. Define total number of schools in Westhaven City as t; so t = W = 21. Answer: 21.


The number of elementary school in Hawkesbury equals 4. The number of public highschool in Brightford equals 2 plus the total number of schools in Westhaven City. The total number of schools in Brightford equals 8. The number of private middle school in Westhaven City equals the sum of the total number of schools in Hawkesbury and the number of regional medical school in Hawkesbury. The number of regional medical school in Hawkesbury exists, and its number is greater than 0.

Question: How many regional medical school does Hawkesbury have?

Solution: The question is difficult, so we use equations to solve it. Define elementary school in Hawkesbury as b; so b = 4. 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 z; so z = x + b = x + 4 = x + 4. Define private middle school in Westhaven City as p; so p = z + x = x + 4 + x = 2*x + 4. Define total number of schools in Westhaven City as a; so a = p = 2*x + 4. Define public highschool in Brightford as k; G = a = 2*x + 4; so k = 2 + G = 2 + 2*x + 4 = 2*x + 6. Define total number of schools in Brightford as v; so v = k = 2*x + 6. We know v = 8, so we have 2*x + 6 = 8 and we solve it through the following steps: Simplifying the equation: 2*x + 6 = 8. Move all terms to one side: 2*x - 2 = 0. Isolate the term with x: 2*x = 2. Divide both sides by 2: x = 2 / 2. Solution: x = 1. Answer: 1.


The number of adult racoon in Oakridge Riverside equals the sum of the total number of adult animals in Beverly Forest, the number of adult deer in Beverly Forest, and the number of adult racoon in Beverly Forest. The number of adult deer in Beverly Forest equals 4. The number of adult racoon in Beverly Forest equals the number of adult deer in Beverly Forest. The number of adult deer in Oakridge Riverside equals the number of adult racoon in Beverly Forest.

Question: What is the total number of adult animals in Oakridge Riverside?

Solution: Define adult deer in Beverly Forest as b; so b = 4. Define adult racoon in Beverly Forest as P; so P = b = 4. Define total number of adult animals in Beverly Forest as Q; so Q = b + P = 4 + 4 = 8. Define adult racoon in Oakridge Riverside as e; D = Q + b = 8 + 4 = 12; so e = D + P = 12 + 4 = 16. Define adult deer in Oakridge Riverside as L; so L = P = 4. Define total number of adult animals in Oakridge Riverside as z; so z = L + e = 4 + 16 = 20. Answer: 20.


The number of culinarian school in Clearwater Bay equals 4 times the sum of the total number of schools in Ruby Bay and the number of elementary school in Clearwater Bay. The total number of schools in Clearwater Bay equals 47. The number of culinarian school in Ruby Bay exists, and its number is greater than 0. The number of elementary school in Clearwater Bay equals 1 plus the sum of the total number of schools in Ruby Bay and the number of culinarian school in Ruby Bay.

Question: How many culinarian school does Ruby Bay have?

Solution: The question is difficult, so we use equations to solve it. 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 v; so v = x = x. Define elementary school in Clearwater Bay as U; c = v + x = x + x = 2*x; so U = 1 + c = 1 + 2*x = 2*x + 1. Define culinarian school in Clearwater Bay as b; a = v + U = x + 2*x + 1 = 3*x + 1; so b = 4 * a = (4) * (3*x + 1) = 12*x + 4. Define total number of schools in Clearwater Bay as s; so s = U + b = 2*x + 1 + 12*x + 4 = 14*x + 5. We know s = 47, so we have 14*x + 5 = 47 and we solve it through the following steps: Simplifying the equation: 14*x + 5 = 47. Move all terms to one side: 14*x - 42 = 0. Isolate the term with x: 14*x = 42. Divide both sides by 14: x = 42 / 14. Solution: x = 3. Answer: 3.


The total number of adult animals in Maple Creek equals 10. The number of adult wolf in Maple Creek exists, and its number is greater than 0. The number of adult fox in Cedar Valley equals 2 times the number of adult wolf in Maple Creek. The number of adult wolf in Maple Creek exists, and its number is greater than 0.

Question: How many adult wolf does Maple Creek have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Maple Creek as x; we don't know its value yet but will find it out later. Define adult fox in Cedar Valley as I; F = x = x; so I = 2 * F = (2) * (x) = 2*x. Define total number of adult animals in Cedar Valley as f; so f = I = 2*x. Define adult fox in Maple Creek as t; C = f + x = 2*x + x = 3*x; Y = C + I = 3*x + 2*x = 5*x; so t = 4 + Y = 4 + 5*x = 5*x + 4. Define total number of adult animals in Maple Creek as E; so E = t + x = 5*x + 4 + x = 6*x + 4. We know E = 10, so we have 6*x + 4 = 10 and we solve it through the following steps: Simplifying the equation: 6*x + 4 = 10. Move all terms to one side: 6*x - 6 = 0. Isolate the term with x: 6*x = 6. Divide both sides by 6: x = 6 / 6. Solution: x = 1. Answer: 1.


The number of adult deer in South Zoo equals 2 plus the number of adult bear in South Zoo. The number of adult bear in South Zoo equals 4. The number of adult deer in Mayer Aquarium equals 4 times the sum of the total number of adult animals in South Zoo, the number of adult bear in South Zoo, and the number of adult deer in South Zoo.

Question: What is the total number of adult animals in Mayer Aquarium?

Solution: Define adult bear in South Zoo as u; so u = 4. Define adult deer in South Zoo as v; z = u = 4; so v = 2 + z = 2 + 4 = 6. Define total number of adult animals in South Zoo as w; so w = v + u = 6 + 4 = 10. Define adult deer in Mayer Aquarium as G; d = w + u = 10 + 4 = 14; L = d + v = 14 + 6 = 20; so G = 4 * L = 4 * 20 = 80. Define total number of adult animals in Mayer Aquarium as J; so J = G = 80. Answer: 80.


The number of culinarian school in Riverton City equals 1. The number of regional medical school in Clearwater Bay equals 1 plus the difference between the total number of schools in Riverton City and the number of culinarian school in Riverton City. The number of culinarian school in Clearwater Bay equals the sum of the total number of schools in Riverton City, the number of regional medical school in Clearwater Bay, and the number of culinarian school in Riverton City.

Question: What is the total number of schools in Clearwater Bay?

Solution: Define culinarian school in Riverton City as B; so B = 1. Define total number of schools in Riverton City as M; so M = B = 1. Define regional medical school in Clearwater Bay as T; z = M - B = 1 - 1 = 0; so T = 1 + z = 1 + 0 = 1. Define culinarian school in Clearwater Bay as k; G = M + T = 1 + 1 = 2; so k = G + B = 2 + 1 = 3. Define total number of schools in Clearwater Bay as N; so N = T + k = 1 + 3 = 4. Answer: 4.


The number of calm road movie in Verdi Movie Festival equals 1. The number of solemn period drama in Taylor Movie Festival equals 1 plus the difference between the total number of movies in Verdi Movie Festival and the number of calm road movie in Verdi Movie Festival. The number of intense detective thriller in Northwood Movie Festival equals the sum of the total number of movies in Taylor Movie Festival, the number of calm road movie in Verdi Movie Festival, and the total number of movies in Verdi Movie Festival.

Question: What is the total number of movies in Northwood Movie Festival?

Solution: Define calm road movie in Verdi Movie Festival as H; so H = 1. Define total number of movies in Verdi Movie Festival as I; so I = H = 1. Define solemn period drama in Taylor Movie Festival as o; X = I - H = 1 - 1 = 0; so o = 1 + X = 1 + 0 = 1. Define total number of movies in Taylor Movie Festival as s; so s = o = 1. Define intense detective thriller in Northwood Movie Festival as d; E = s + H = 1 + 1 = 2; so d = E + I = 2 + 1 = 3. Define total number of movies in Northwood Movie Festival as e; so e = d = 3. Answer: 3.


The number of adult bear in Jefferson Circus equals 3 times the total number of adult animals in Hamilton Farm. The number of adult deer in South Zoo equals 4. The number of adult wolf in Hamilton Farm equals 4 plus the total number of adult animals in South Zoo. The number of adult bear in Hamilton Farm equals the number of adult wolf in Hamilton Farm.

Question: What is the total number of adult animals in Jefferson Circus?

Solution: Define adult deer in South Zoo as q; so q = 4. Define total number of adult animals in South Zoo as Q; so Q = q = 4. Define adult wolf in Hamilton Farm as k; j = Q = 4; so k = 4 + j = 4 + 4 = 8. Define adult bear in Hamilton Farm as E; so E = k = 8. Define total number of adult animals in Hamilton Farm as O; so O = E + k = 8 + 8 = 16. Define adult bear in Jefferson Circus as Y; Z = O = 16; so Y = 3 * Z = 3 * 16 = 48. Define total number of adult animals in Jefferson Circus as e; so e = Y = 48. Answer: 48.


The number of adult eagle in South Zoo equals the number of adult eagle in Bundle Ranch. The number of adult parrot in South Zoo equals 4. The number of adult owl in Bundle Ranch equals the number of adult parrot in South Zoo. The number of adult eagle in Hamilton Farm equals 1. The number of adult owl in Hamilton Farm equals 4 plus the number of adult eagle in Hamilton Farm. The number of adult eagle in Bundle Ranch equals 2 plus the number of adult owl in Bundle Ranch. The number of adult owl in South Zoo equals 1 plus the number of adult owl in Hamilton Farm. The number of adult parrot in Bundle Ranch equals the sum of the number of adult eagle in Bundle Ranch, the number of adult parrot in South Zoo, and the number of adult owl in Bundle Ranch. The number of adult parrot in Hamilton Farm equals the total number of adult animals in South Zoo.

Question: What is the total number of adult animals in Bundle Ranch?

Solution: Define adult parrot in South Zoo as L; so L = 4. Define adult owl in Bundle Ranch as j; so j = L = 4. Define adult eagle in Bundle Ranch as s; p = j = 4; so s = 2 + p = 2 + 4 = 6. Define adult parrot in Bundle Ranch as o; f = s + L = 6 + 4 = 10; so o = f + j = 10 + 4 = 14. Define total number of adult animals in Bundle Ranch as m; M = s + o = 6 + 14 = 20; so m = M + j = 20 + 4 = 24. Answer: 24.


The number of culinarian school in Ruby Bay equals 2 times the sum of the total number of schools in Oakbridge City and the number of elementary school in Oakbridge City. The number of elementary school in Oakbridge City equals 2. The number of elementary school in Ruby Bay equals 2 plus the difference between the number of culinarian school in Ruby Bay and the number of elementary school in Oakbridge City.

Question: What is the total number of schools in Ruby Bay?

Solution: Define elementary school in Oakbridge City as g; so g = 2. Define total number of schools in Oakbridge City as L; so L = g = 2. Define culinarian school in Ruby Bay as E; U = L + g = 2 + 2 = 4; so E = 2 * U = 2 * 4 = 8. Define elementary school in Ruby Bay as Y; X = E - g = 8 - 2 = 6; so Y = 2 + X = 2 + 6 = 8. Define total number of schools in Ruby Bay as F; so F = E + Y = 8 + 8 = 16. Answer: 16.


The number of public highschool in Westhaven City equals 4 times the difference between the number of elementary school in Westhaven City and the number of public highschool in Hawkesbury. The number of regional medical school in Hawkesbury equals 4. The number of public highschool in Hawkesbury equals 4. The number of regional medical school in Westhaven City exists, and its number is greater than 0. The number of elementary school in Westhaven City equals 3 times the number of public highschool in Hawkesbury. The number of culinarian school in Brightford equals the number of public highschool in Hawkesbury. The number of culinarian school in Glenfield City equals the sum of the total number of schools in Hawkesbury and the number of elementary school in Westhaven City. The number of elementary school in Glenfield City equals 2. The total number of schools in Westhaven City equals 46. The number of private middle school in Brightford equals 3.

Question: How many regional medical school does Westhaven City have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Hawkesbury as M; so M = 4. Define elementary school in Westhaven City as D; Q = M = 4; so D = 3 * Q = (3) * (4) = 12. Define regional medical school in Westhaven City as x; we don't know its value yet but will find it out later. Define public highschool in Westhaven City as U; O = D - M = (12) - (4) = 8; so U = 4 * O = (4) * (8) = 32. Define total number of schools in Westhaven City as Z; c = x + D = x + 12 = x + 12; so Z = c + U = x + 12 + 32 = x + 44. We know Z = 46, so we have x + 44 = 46 and we solve it through the following steps: Simplifying the equation: x + 44 = 46. 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.


The number of regional medical school in Brightford equals the sum of the total number of schools in Hawkesbury, the number of culinarian school in Hawkesbury, and the number of regional medical school in Hawkesbury. The number of culinarian school in Brightford exists, and its number is greater than 0. The total number of schools in Brightford equals 11. The number of regional medical school in Hawkesbury equals 2. The number of culinarian school in Hawkesbury equals 2.

Question: How many culinarian school does Brightford have?

Solution: The question is difficult, so we use equations to solve it. Define regional medical school in Hawkesbury as b; so b = 2. Define culinarian school in Hawkesbury as v; so v = 2. Define total number of schools in Hawkesbury as U; so U = v + b = 2 + 2 = 4. Define regional medical school in Brightford as G; t = U + v = 4 + 2 = 6; so G = t + b = 6 + 2 = 8. Define culinarian school in Brightford as x; we don't know its value yet but will find it out later. Define total number of schools in Brightford as n; so n = x + G = x + 8 = x + 8. We know n = 11, so we have x + 8 = 11 and we solve it through the following steps: Simplifying the equation: x + 8 = 11. 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.