The number of culinarian school in Brightford equals 2. The number of culinarian school in Glenfield City equals the number of elementary school in Evervale City. The number of culinarian school in Evervale City equals 4 plus the number of elementary school in Evervale City. The number of regional medical school in Evervale City equals 1 plus the total number of schools in Brightford. The number of elementary school in Glenfield City equals 4 plus the number of culinarian school in Brightford. The number of elementary school in Brightford equals 4 times the number of elementary school in Glenfield City. The number of private middle school in Evervale City equals the number of elementary school in Evervale City. The number of private middle school in Brightford equals 2. The number of regional medical school in Glenfield City equals the sum of the number of elementary school in Brightford and the number of elementary school in Glenfield City. The number of private middle school in Glenfield City equals the number of culinarian school in Evervale City. The number of elementary school in Evervale City equals 1. The number of regional medical school in Brightford equals the sum of the number of elementary school in Glenfield City, the number of elementary school in Brightford, and the number of culinarian school in Brightford.

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

Solution: Define culinarian school in Brightford as O; so O = 2. Define elementary school in Glenfield City as E; k = O = 2; so E = 4 + k = 4 + 2 = 6. Define elementary school in Brightford as d; h = E = 6; so d = 4 * h = 4 * 6 = 24. Define private middle school in Brightford as G; so G = 2. Define regional medical school in Brightford as l; z = E + d = 6 + 24 = 30; so l = z + O = 30 + 2 = 32. Define total number of schools in Brightford as j; N = O + l = 2 + 32 = 34; o = N + d = 34 + 24 = 58; so j = o + G = 58 + 2 = 60. Define regional medical school in Evervale City as f; q = j = 60; so f = 1 + q = 1 + 60 = 61. Answer: 61.


The number of adult deer in Mayer Aquarium equals 1. The number of adult racoon in Bundle Ranch exists, and its number is greater than 0. The total number of adult animals in Bundle Ranch equals 23. The number of adult deer in Bundle Ranch equals 4 plus the number of adult bear in Bundle Ranch. The number of adult bear in Bundle Ranch equals 2 times the total number of adult animals in Mayer Aquarium. The number of adult racoon in Mayer Aquarium equals 2 times the number of adult deer in Mayer Aquarium. The number of adult bear in Mayer Aquarium equals the number of adult deer in Mayer Aquarium.

Question: How many adult racoon does Bundle Ranch have?

Solution: The question is difficult, so we use equations to solve it. Define adult deer in Mayer Aquarium as b; so b = 1. Define adult racoon in Mayer Aquarium as W; r = b = 1; so W = 2 * r = (2) * (1) = 2. Define adult bear in Mayer Aquarium as G; so G = b = 1. Define total number of adult animals in Mayer Aquarium as Q; S = b + W = 1 + 2 = 3; so Q = S + G = 3 + 1 = 4. Define adult racoon in Bundle Ranch as x; we don't know its value yet but will find it out later. Define adult bear in Bundle Ranch as H; m = Q = 4; so H = 2 * m = (2) * (4) = 8. Define adult deer in Bundle Ranch as k; Y = H = 8; so k = 4 + Y = 4 + 8 = 12. Define total number of adult animals in Bundle Ranch as R; q = k + x = 12 + x = x + 12; so R = q + H = x + 12 + 8 = x + 20. We know R = 23, so we have x + 20 = 23 and we solve it through the following steps: Simplifying the equation: x + 20 = 23. 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.


The number of solemn period drama in Northwood Movie Festival equals 1. The number of solemn period drama in Golden Banana Movie Festival equals the number of solemn period drama in Northwood Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals 4. The number of futuristic sci-fi movie in Northwood Movie Festival equals 2. The number of intense detective thriller in Golden Banana Movie Festival equals the sum of the total number of movies in Northwood Movie Festival and the number of intense detective thriller in Northwood Movie Festival. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 4. The number of intense detective thriller in Northwood Movie Festival equals the number of futuristic sci-fi movie in Northwood Movie Festival.

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

Solution: Define futuristic sci-fi movie in Northwood Movie Festival as t; so t = 2. Define intense detective thriller in Northwood Movie Festival as G; so G = t = 2. Define solemn period drama in Northwood Movie Festival as R; so R = 1. Define total number of movies in Northwood Movie Festival as d; C = t + G = 2 + 2 = 4; so d = C + R = 4 + 1 = 5. Define intense detective thriller in Golden Banana Movie Festival as i; so i = d + G = 5 + 2 = 7. Define futuristic sci-fi movie in Golden Banana Movie Festival as c; so c = 4. Define solemn period drama in Golden Banana Movie Festival as T; so T = R = 1. Define calm road movie in Golden Banana Movie Festival as A; so A = 4. Define total number of movies in Golden Banana Movie Festival as a; m = c + i = 4 + 7 = 11; v = m + T = 11 + 1 = 12; so a = v + A = 12 + 4 = 16. Answer: 16.


The number of futuristic sci-fi movie in Cinéma de Montreval equals the total number of movies in Festival de Clairmont. The number of solemn period drama in Festival de Clairmont equals the difference between the total number of movies in Festival Lumière de Valmont and the number of solemn period drama in Festival Lumière de Valmont. 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 de Clairmont equals 2 times the total number of movies in Festival Lumière de Valmont. The number of solemn period drama in Cinéma de Montreval equals the sum of the total number of movies in Festival de Clairmont and the total number of movies in Festival Lumière de Valmont. The total number of movies in Cinéma de Montreval equals 12. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals the number of solemn period drama in Festival Lumière de Valmont.

Question: How many solemn period drama does Festival Lumière de Valmont have?

Solution: 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 Festival Lumière de Valmont as L; so L = x = x. Define total number of movies in Festival Lumière de Valmont as T; so T = L + x = x + x = 2*x. Define solemn period drama in Festival de Clairmont as J; so J = T - x = (2*x) - (x) = x. Define futuristic sci-fi movie in Festival de Clairmont as R; G = T = 2*x; so R = 2 * G = (2) * (2*x) = 4*x. Define total number of movies in Festival de Clairmont as V; so V = R + J = 4*x + x = 5*x. Define solemn period drama in Cinéma de Montreval as S; so S = V + T = 5*x + 2*x = 7*x. Define futuristic sci-fi movie in Cinéma de Montreval as N; so N = V = 5*x. Define total number of movies in Cinéma de Montreval as P; so P = N + S = 5*x + 7*x = 12*x. We know P = 12, so we have 12*x = 12 and we solve it through the following steps: Simplifying the equation: 12*x = 12. Move all terms to one side: 12*x - 12 = 0. Isolate the term with x: 12*x = 12. Divide both sides by 12: x = 12 / 12. Solution: x = 1. Answer: 1.


The number of regional medical school in Ruby Bay equals the difference between the total number of schools in Riverton City and the number of culinarian school in Shoreline City. The number of culinarian school in Oakbridge City equals 3. The number of private middle school in Riverton City equals 3 times the total number of schools in Shoreline City. The total number of schools in Ruby Bay equals 45. The number of culinarian school in Shoreline City equals the number of culinarian school in Oakbridge City. The number of regional medical school in Shoreline City equals the number of culinarian school in Shoreline City. The number of private middle school in Shoreline City exists, and its number is greater than 0. The number of elementary school in Riverton City equals the number of private middle school in Riverton City.

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

Solution: The question is difficult, so we use equations to solve it. Define culinarian school in Oakbridge City as r; so r = 3. Define culinarian school in Shoreline City as I; so I = r = 3. Define private middle school in Shoreline City as x; we don't know its value yet but will find it out later. Define regional medical school in Shoreline City as k; so k = I = 3. Define total number of schools in Shoreline City as B; L = I + k = 3 + 3 = 6; so B = L + x = 6 + x = x + 6. Define private middle school in Riverton City as M; J = B = x + 6; so M = 3 * J = (3) * (x + 6) = 3*x + 18. Define elementary school in Riverton City as v; so v = M = 3*x + 18. Define total number of schools in Riverton City as w; so w = M + v = 3*x + 18 + 3*x + 18 = 6*x + 36. Define regional medical school in Ruby Bay as V; so V = w - I = (6*x + 36) - (3) = 6*x + 33. Define total number of schools in Ruby Bay as m; so m = V = 6*x + 33. We know m = 45, so we have 6*x + 33 = 45 and we solve it through the following steps: Simplifying the equation: 6*x + 33 = 45. Move all terms to one side: 6*x - 12 = 0. Isolate the term with x: 6*x = 12. Divide both sides by 6: x = 12 / 6. Solution: x = 2. Answer: 2.


The number of adult racoon in South Zoo equals 4 times the sum of the number of adult wolf in South Zoo and the number of adult racoon in Jefferson Circus. The number of adult deer in Jefferson Circus equals 4 plus the number of adult racoon in Jefferson Circus. The number of adult deer in South Zoo equals 2 times the difference between the number of adult deer in Jefferson Circus and the number of adult racoon in Jefferson Circus. The number of adult racoon in Jefferson Circus exists, and its number is greater than 0. The total number of adult animals in South Zoo equals 124. The number of adult racoon in Jefferson Circus exists, and its number is greater than 0.

Question: How many adult racoon does Jefferson Circus have?

Solution: The question is difficult, so we use equations to solve it. Define adult racoon in Jefferson Circus as x; we don't know its value yet but will find it out later. Define adult deer in Jefferson Circus as E; F = x = x; so E = 4 + F = 4 + x = x + 4. Define adult deer in South Zoo as u; O = E - x = (x + 4) - (x) = 4; so u = 2 * O = (2) * (4) = 8. Define adult wolf in South Zoo as e; j = u + x = 8 + x = x + 8; so e = j + E = x + 8 + x + 4 = 2*x + 12. Define adult racoon in South Zoo as J; q = e + x = 2*x + 12 + x = 3*x + 12; so J = 4 * q = (4) * (3*x + 12) = 12*x + 48. Define total number of adult animals in South Zoo as G; z = J + u = 12*x + 48 + 8 = 12*x + 56; so G = z + e = 12*x + 56 + 2*x + 12 = 14*x + 68. We know G = 124, so we have 14*x + 68 = 124 and we solve it through the following steps: Simplifying the equation: 14*x + 68 = 124. Move all terms to one side: 14*x - 56 = 0. Isolate the term with x: 14*x = 56. Divide both sides by 14: x = 56 / 14. Solution: x = 4. Answer: 4.


The number of calm road movie in West Sahara Movie Festival equals 2. The number of solemn period drama in Golden Banana Movie Festival equals 2. The number of solemn period drama in Verdi Movie Festival equals 4 times the number of calm road movie in Taylor Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival equals the sum of the total number of movies in West Sahara Movie Festival, the number of solemn period drama in West Sahara Movie Festival, and the number of calm road movie in West Sahara Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 3 plus the total number of movies in West Sahara Movie Festival. The number of solemn period drama in Taylor Movie Festival equals 3 times the total number of movies in West Sahara Movie Festival. The number of solemn period drama in West Sahara Movie Festival equals 3. The number of calm road movie in Golden Banana Movie Festival equals the number of solemn period drama in West Sahara Movie Festival. The number of calm road movie in Taylor Movie Festival equals 1. The number of upbeat metropolis comedy in Verdi Movie Festival equals 1 plus the sum of the number of calm road movie in West Sahara Movie Festival, the number of solemn period drama in Taylor Movie Festival, and the number of solemn period drama in Golden Banana Movie Festival. The number of calm road movie in Verdi Movie Festival equals 2 plus the number of calm road movie in Taylor Movie Festival.

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

Solution: Define solemn period drama in West Sahara Movie Festival as R; so R = 3. Define calm road movie in West Sahara Movie Festival as i; so i = 2. Define total number of movies in West Sahara Movie Festival as k; so k = R + i = 3 + 2 = 5. Define intense detective thriller in Golden Banana Movie Festival as x; L = k + R = 5 + 3 = 8; so x = L + i = 8 + 2 = 10. Define upbeat metropolis comedy in Golden Banana Movie Festival as X; Q = k = 5; so X = 3 + Q = 3 + 5 = 8. Define calm road movie in Golden Banana Movie Festival as E; so E = R = 3. Define solemn period drama in Golden Banana Movie Festival as g; so g = 2. Define total number of movies in Golden Banana Movie Festival as y; t = g + x = 2 + 10 = 12; m = t + E = 12 + 3 = 15; so y = m + X = 15 + 8 = 23. Answer: 23.


The number of adult deer in South Zoo equals 2 times the number of adult bear in South Zoo. The number of adult fox in Mayer Aquarium equals the total number of adult animals in Hamilton Farm. The number of adult fox in Hamilton Farm equals the number of adult deer in South Zoo. The number of adult deer in Mayer Aquarium equals the number of adult fox in Mayer Aquarium. The number of adult bear in South Zoo equals 4. The total number of adult animals in Mayer Aquarium equals 48. The number of adult deer in Hamilton Farm exists, and its number is greater than 0. The number of adult bear in Hamilton Farm equals the total number of adult animals in South Zoo.

Question: How many adult deer does Hamilton Farm have?

Solution: The question is difficult, so we use equations to solve it. Define adult bear in South Zoo as Q; so Q = 4. Define adult deer in South Zoo as s; O = Q = 4; so s = 2 * O = (2) * (4) = 8. Define total number of adult animals in South Zoo as N; so N = Q + s = 4 + 8 = 12. Define adult fox in Hamilton Farm as m; so m = s = 8. Define adult bear in Hamilton Farm as K; so K = N = 12. Define adult deer in Hamilton Farm as x; we don't know its value yet but will find it out later. Define total number of adult animals in Hamilton Farm as e; n = K + m = 12 + 8 = 20; so e = n + x = 20 + x = x + 20. Define adult fox in Mayer Aquarium as b; so b = e = x + 20. Define adult deer in Mayer Aquarium as L; so L = b = x + 20. Define total number of adult animals in Mayer Aquarium as Z; so Z = b + L = x + 20 + x + 20 = 2*x + 40. We know Z = 48, so we have 2*x + 40 = 48 and we solve it through the following steps: Simplifying the equation: 2*x + 40 = 48. Move all terms to one side: 2*x - 8 = 0. Isolate the term with x: 2*x = 8. Divide both sides by 2: x = 8 / 2. Solution: x = 4. Answer: 4.


The number of adult owl in Oakridge Riverside equals 3 times the sum of the number of adult crow in Oakridge Riverside and the number of adult owl in Cedar Valley. The number of adult owl in Cedar Valley exists, and its number is greater than 0. The number of adult blue jay in Cedar Valley equals the number of adult crow in Oakridge Riverside. The number of adult owl in Cedar Valley exists, and its number is greater than 0. The number of adult eagle in Cedar Valley equals 4 times the number of adult crow in Beverly Forest. The number of adult parrot in Beverly Forest equals 2 plus the sum of the number of adult owl in Beverly Forest and the number of adult owl in Oakridge Riverside. The number of adult blue jay in Beverly Forest equals 52. The number of adult eagle in Oakridge Riverside equals the sum of the number of adult parrot in Beverly Forest, the number of adult owl in Beverly Forest, the total number of adult animals in Beverly Forest, and the number of adult eagle in Cedar Valley. The number of adult parrot in Oakridge Riverside equals the number of adult crow in Oakridge Riverside. The number of adult crow in Oakridge Riverside equals 1 plus the number of adult owl in Cedar Valley. The number of adult parrot in Cedar Valley equals the sum of the number of adult crow in Cedar Valley, the number of adult eagle in Cedar Valley, and the number of adult crow in Beverly Forest. The number of adult crow in Beverly Forest equals 4. The number of adult blue jay in Oakridge Riverside equals the number of adult owl in Cedar Valley. The number of adult crow in Cedar Valley equals 4. The number of adult blue jay in Beverly Forest equals 2 times the number of adult parrot in Beverly Forest.

Question: How many adult owl does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult owl in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult crow in Oakridge Riverside as p; F = x = x; so p = 1 + F = 1 + x = x + 1. Define adult owl in Oakridge Riverside as v; t = p + x = x + 1 + x = 2*x + 1; so v = 3 * t = (3) * (2*x + 1) = 6*x + 3. Define adult owl in Beverly Forest as J; h = v + x = 6*x + 3 + x = 7*x + 3; e = h + p = 7*x + 3 + x + 1 = 8*x + 4; so J = 3 + e = 3 + 8*x + 4 = 8*x + 7. Define adult parrot in Beverly Forest as s; l = J + v = 8*x + 7 + 6*x + 3 = 14*x + 10; so s = 2 + l = 2 + 14*x + 10 = 14*x + 12. Define adult blue jay in Beverly Forest as S; k = s = 14*x + 12; so S = 2 * k = (2) * (14*x + 12) = 28*x + 24. We know S = 52, so we have 28*x + 24 = 52 and we solve it through the following steps: Simplifying the equation: 28*x + 24 = 52. Move all terms to one side: 28*x - 28 = 0. Isolate the term with x: 28*x = 28. Divide both sides by 28: x = 28 / 28. Solution: x = 1. Answer: 1.


The number of upbeat metropolis comedy in Rêves de Belleville equals the number of intense detective thriller in Rêves de Belleville. The number of intense detective thriller 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 2 times the total number of movies in Rêves de Belleville. The number of intense detective thriller in Rêves de Belleville equals 4. The number of intense detective thriller in Festival de Clairmont equals the number of intense detective thriller in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 4 times the total number of movies in Festival de Clairmont. The number of intense detective thriller in Festival de Saint-Rivage equals 4 plus the total number of movies in Festival Lumière de Valmont.

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

Solution: Define intense detective thriller in Rêves de Belleville as M; so M = 4. Define upbeat metropolis comedy in Rêves de Belleville as z; so z = M = 4. Define total number of movies in Rêves de Belleville as Z; so Z = z + M = 4 + 4 = 8. Define upbeat metropolis comedy in Festival Lumière de Valmont as s; i = Z = 8; so s = 2 * i = 2 * 8 = 16. Define intense detective thriller in Festival Lumière de Valmont as p; so p = Z = 8. Define total number of movies in Festival Lumière de Valmont as b; so b = s + p = 16 + 8 = 24. Define intense detective thriller in Festival de Saint-Rivage as f; r = b = 24; so f = 4 + r = 4 + 24 = 28. Define intense detective thriller in Festival de Clairmont as t; so t = f = 28. Define total number of movies in Festival de Clairmont as k; so k = t = 28. Define upbeat metropolis comedy in Festival de Saint-Rivage as X; c = k = 28; so X = 4 * c = 4 * 28 = 112. Answer: 112.


The number of adult wolf in Hamilton Farm equals 2 plus the difference between the total number of adult animals in Jefferson Circus and the number of adult racoon in Jefferson Circus. The number of adult deer in Jefferson Circus equals 1 plus the sum of the total number of adult animals in South Zoo and the number of adult wolf in South Zoo. The number of adult racoon in Hamilton Farm equals the number of adult racoon in Jefferson Circus. The number of adult racoon in Jefferson Circus equals 3 times the number of adult deer in Jefferson Circus. The number of adult wolf in South Zoo equals 4.

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

Solution: Define adult wolf in South Zoo as o; so o = 4. Define total number of adult animals in South Zoo as L; so L = o = 4. Define adult deer in Jefferson Circus as G; t = L + o = 4 + 4 = 8; so G = 1 + t = 1 + 8 = 9. Define adult racoon in Jefferson Circus as P; l = G = 9; so P = 3 * l = 3 * 9 = 27. Define total number of adult animals in Jefferson Circus as m; so m = G + P = 9 + 27 = 36. Define adult wolf in Hamilton Farm as W; B = m - P = 36 - 27 = 9; so W = 2 + B = 2 + 9 = 11. Define adult racoon in Hamilton Farm as E; so E = P = 27. Define total number of adult animals in Hamilton Farm as Q; so Q = E + W = 27 + 11 = 38. Answer: 38.


The number of calm road movie in Northwood Movie Festival equals 2. The number of calm road movie in Golden Banana Movie Festival equals the difference between the number of solemn period drama in Taylor Movie Festival and the total number of movies in Verdi Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals the number of calm road movie in Golden Banana Movie Festival. The number of calm road movie in Taylor Movie Festival equals the sum of the total number of movies in Verdi Movie Festival, the total number of movies in Northwood Movie Festival, the number of calm road movie in Northwood Movie Festival, and the number of calm road movie in Verdi Movie Festival. The number of calm road movie in Verdi Movie Festival equals the total number of movies in Northwood Movie Festival. The number of solemn period drama in Taylor Movie Festival equals the sum of the number of calm road movie in Taylor Movie Festival and the total number of movies in Northwood Movie Festival. The number of solemn period drama in Verdi Movie Festival equals the total number of movies in Northwood Movie Festival.

Question: How many solemn period drama does Golden Banana Movie Festival have?

Solution: Define calm road movie in Northwood Movie Festival as o; so o = 2. Define total number of movies in Northwood Movie Festival as a; so a = o = 2. Define solemn period drama in Verdi Movie Festival as G; so G = a = 2. Define calm road movie in Verdi Movie Festival as R; so R = a = 2. Define total number of movies in Verdi Movie Festival as V; so V = G + R = 2 + 2 = 4. Define calm road movie in Taylor Movie Festival as d; n = V + a = 4 + 2 = 6; v = n + o = 6 + 2 = 8; so d = v + R = 8 + 2 = 10. Define solemn period drama in Taylor Movie Festival as r; so r = d + a = 10 + 2 = 12. Define calm road movie in Golden Banana Movie Festival as N; so N = r - V = 12 - 4 = 8. Define solemn period drama in Golden Banana Movie Festival as l; so l = N = 8. Answer: 8.


The number of adult blue jay in South Zoo exists, and its number is greater than 0. The number of adult owl in Mayer Aquarium equals 2. The number of adult blue jay in Hamilton Farm equals 4 times the number of adult owl in Hamilton Farm. The number of adult owl in South Zoo equals 3 plus the number of adult crow in South Zoo. The number of adult crow in Mayer Aquarium equals 2 times the sum of the number of adult crow in Hamilton Farm, the total number of adult animals in Hamilton Farm, and the number of adult crow in South Zoo. The number of adult crow in Hamilton Farm equals the total number of adult animals in South Zoo. The total number of adult animals in Hamilton Farm equals 23. The number of adult owl in Hamilton Farm equals 2 plus the number of adult crow in South Zoo. The number of adult crow in South Zoo equals 1.

Question: How many adult blue jay does South Zoo have?

Solution: The question is difficult, so we use equations to solve it. Define adult crow in South Zoo as j; so j = 1. Define adult owl in South Zoo as F; c = j = 1; so F = 3 + c = 3 + 1 = 4. Define adult owl in Hamilton Farm as n; l = j = 1; so n = 2 + l = 2 + 1 = 3. Define adult blue jay in South Zoo as x; we don't know its value yet but will find it out later. Define adult blue jay in Hamilton Farm as i; M = n = 3; so i = 4 * M = (4) * (3) = 12. Define total number of adult animals in South Zoo as p; I = j + x = 1 + x = x + 1; so p = I + F = x + 1 + 4 = x + 5. Define adult crow in Hamilton Farm as T; so T = p = x + 5. Define total number of adult animals in Hamilton Farm as b; P = T + i = x + 5 + 12 = x + 17; so b = P + n = x + 17 + 3 = x + 20. We know b = 23, so we have x + 20 = 23 and we solve it through the following steps: Simplifying the equation: x + 20 = 23. 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.


The total number of movies in Golden Banana Movie Festival equals 336. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals the number of calm road movie in Golden Banana Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals the total number of movies in Taylor Movie Festival. The number of upbeat metropolis comedy in Taylor Movie Festival equals 4 times the number of futuristic sci-fi movie in Taylor Movie Festival. The number of calm road movie in Northwood Movie Festival exists, and its number is greater than 0. The number of futuristic sci-fi movie in Taylor Movie Festival equals 2 times the total number of movies in Northwood Movie Festival. The number of calm road movie in Taylor Movie Festival equals 4 times the number of upbeat metropolis comedy in Taylor Movie Festival.

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

Solution: The question is difficult, so we use equations to solve it. Define calm road 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 G; so G = x = x. Define futuristic sci-fi movie in Taylor Movie Festival as V; I = G = x; so V = 2 * I = (2) * (x) = 2*x. Define upbeat metropolis comedy in Taylor Movie Festival as o; Q = V = 2*x; so o = 4 * Q = (4) * (2*x) = 8*x. Define calm road movie in Taylor Movie Festival as d; s = o = 8*x; so d = 4 * s = (4) * (8*x) = 32*x. Define total number of movies in Taylor Movie Festival as a; g = d + o = 32*x + 8*x = 40*x; so a = g + V = 40*x + 2*x = 42*x. Define calm road movie in Golden Banana Movie Festival as S; so S = a = 42*x. Define futuristic sci-fi movie in Golden Banana Movie Festival as Y; so Y = S = 42*x. Define total number of movies in Golden Banana Movie Festival as N; so N = S + Y = 42*x + 42*x = 84*x. We know N = 336, so we have 84*x = 336 and we solve it through the following steps: Simplifying the equation: 84*x = 336. Move all terms to one side: 84*x - 336 = 0. Isolate the term with x: 84*x = 336. Divide both sides by 84: x = 336 / 84. Solution: x = 4. Answer: 4.


The number of upbeat metropolis comedy in Golden Banana Movie Festival equals the sum of the number of calm road movie in Golden Banana Movie Festival, the number of solemn period drama in Golden Banana Movie Festival, the number of upbeat metropolis comedy in Verdi Movie Festival, and the number of solemn period drama in Verdi Movie Festival. The number of upbeat metropolis comedy in Verdi Movie Festival equals 4 times the number of calm road movie 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 Verdi Movie Festival and the number of solemn period drama in Golden Banana Movie Festival. The number of solemn period drama in Verdi Movie Festival equals the sum of the number of solemn period drama in Golden Banana Movie Festival and the number of calm road movie in Verdi Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 30. The number of calm road movie in Verdi Movie Festival exists, and its number is greater than 0. The number of solemn period drama in Golden Banana Movie Festival equals 2 plus the difference between the number of upbeat metropolis comedy in Verdi Movie Festival and the number of calm road movie in Verdi Movie Festival.

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

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Verdi Movie Festival as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Verdi Movie Festival as C; l = x = x; so C = 4 * l = (4) * (x) = 4*x. Define solemn period drama in Golden Banana Movie Festival as u; o = C - x = (4*x) - (x) = 3*x; so u = 2 + o = 2 + 3*x = 3*x + 2. Define solemn period drama in Verdi Movie Festival as T; so T = u + x = 3*x + 2 + x = 4*x + 2. Define calm road movie in Golden Banana Movie Festival as A; m = T - u = (4*x + 2) - (3*x + 2) = x; so A = 2 * m = (2) * (x) = 2*x. Define upbeat metropolis comedy in Golden Banana Movie Festival as d; B = A + u = 2*x + 3*x + 2 = 5*x + 2; Y = B + C = 5*x + 2 + 4*x = 9*x + 2; so d = Y + T = 9*x + 2 + 4*x + 2 = 13*x + 4. We know d = 30, so we have 13*x + 4 = 30 and we solve it through the following steps: Simplifying the equation: 13*x + 4 = 30. Move all terms to one side: 13*x - 26 = 0. Isolate the term with x: 13*x = 26. Divide both sides by 13: x = 26 / 13. Solution: x = 2. Answer: 2.


The number of adult parrot in Mayer Aquarium equals the total number of adult animals in Jefferson Circus. The number of adult owl in Bundle Ranch equals the total number of adult animals in South Zoo. The number of adult crow in Mayer Aquarium equals 2 plus the total number of adult animals in Jefferson Circus. The number of adult crow in South Zoo equals the number of adult crow in Jefferson Circus. The number of adult owl in South Zoo equals the total number of adult animals in Mayer Aquarium. The number of adult crow in Bundle Ranch equals 4. The number of adult crow in Jefferson Circus equals 3.

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

Solution: Define adult crow in Jefferson Circus as b; so b = 3. Define total number of adult animals in Jefferson Circus as X; so X = b = 3. Define adult crow in Mayer Aquarium as m; U = X = 3; so m = 2 + U = 2 + 3 = 5. Define adult parrot in Mayer Aquarium as S; so S = X = 3. Define total number of adult animals in Mayer Aquarium as v; so v = S + m = 3 + 5 = 8. Define adult crow in South Zoo as d; so d = b = 3. Define adult owl in South Zoo as J; so J = v = 8. Define total number of adult animals in South Zoo as Z; so Z = d + J = 3 + 8 = 11. Define adult owl in Bundle Ranch as V; so V = Z = 11. Define adult crow in Bundle Ranch as u; so u = 4. Define total number of adult animals in Bundle Ranch as a; so a = u + V = 4 + 11 = 15. Answer: 15.


The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 4 times the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 2. The number of upbeat metropolis comedy in Festival Lumière de Valmont equals 3 times the difference between the total number of movies in Rêves de Belleville and the number of upbeat metropolis comedy in Rêves de Belleville. The number of intense detective thriller in Festival de Saint-Rivage equals the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of intense detective thriller in Rêves de Belleville equals the number of upbeat metropolis comedy in Rêves de Belleville. The number of upbeat metropolis comedy in Rêves de Belleville equals 3 times the total number of movies in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 2.

Question: How many upbeat metropolis comedy does Festival Lumière de Valmont have?

Solution: Define upbeat metropolis comedy in Festival de Saint-Rivage as t; so t = 2. Define intense detective thriller in Festival de Saint-Rivage as K; so K = t = 2. Define futuristic sci-fi movie in Festival de Saint-Rivage as Z; H = t = 2; so Z = 4 * H = 4 * 2 = 8. Define total number of movies in Festival de Saint-Rivage as z; O = t + Z = 2 + 8 = 10; so z = O + K = 10 + 2 = 12. Define upbeat metropolis comedy in Rêves de Belleville as Q; c = z = 12; so Q = 3 * c = 3 * 12 = 36. Define intense detective thriller in Rêves de Belleville as j; so j = Q = 36. Define total number of movies in Rêves de Belleville as W; so W = Q + j = 36 + 36 = 72. Define upbeat metropolis comedy in Festival Lumière de Valmont as T; N = W - Q = 72 - 36 = 36; so T = 3 * N = 3 * 36 = 108. Answer: 108.


The number of culinarian school in Glenfield City equals the difference between the number of public highschool in Evervale City and the number of public highschool in Hawkesbury. The number of public highschool in Hawkesbury equals 3 times the number of elementary school in Evervale City. The number of elementary school in Evervale City equals 3. The number of culinarian school in Hawkesbury equals 3 times the sum of the number of public highschool in Hawkesbury and the number of elementary school in Evervale City. The number of culinarian school in Evervale City equals the sum of the total number of schools in Glenfield City, the number of public highschool in Evervale City, and the number of culinarian school in Hawkesbury. The number of public highschool in Evervale City equals 1 plus the sum of the number of culinarian school in Hawkesbury and the number of public highschool in Hawkesbury.

Question: How many culinarian school does Evervale City have?

Solution: Define elementary school in Evervale City as v; so v = 3. Define public highschool in Hawkesbury as t; j = v = 3; so t = 3 * j = 3 * 3 = 9. Define culinarian school in Hawkesbury as B; Y = t + v = 9 + 3 = 12; so B = 3 * Y = 3 * 12 = 36. Define public highschool in Evervale City as Z; e = B + t = 36 + 9 = 45; so Z = 1 + e = 1 + 45 = 46. Define culinarian school in Glenfield City as I; so I = Z - t = 46 - 9 = 37. Define total number of schools in Glenfield City as u; so u = I = 37. Define culinarian school in Evervale City as z; r = u + Z = 37 + 46 = 83; so z = r + B = 83 + 36 = 119. Answer: 119.


The number of upbeat metropolis comedy in Rêves de Belleville equals 2. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals the number of upbeat metropolis comedy in Festival de Clairmont. The number of upbeat metropolis comedy in Festival de Clairmont equals 4 times the number of solemn period drama in Festival de Clairmont. The number of solemn period drama in Rêves de Belleville equals the sum of the total number of movies in Festival de Saint-Rivage, the total number of movies in Rêves de Belleville, the number of intense detective thriller in Cinéma de Montreval, and the number of upbeat metropolis comedy in Festival de Clairmont. The number of solemn period drama in Cinéma de Montreval equals 4. The number of solemn period drama in Festival de Clairmont equals the number of intense detective thriller in Cinéma de Montreval. The number of intense detective thriller in Festival de Clairmont equals 3 plus the number of intense detective thriller in Cinéma de Montreval. The number of intense detective thriller in Rêves de Belleville equals 3 times the total number of movies in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Cinéma de Montreval equals 4. The number of futuristic sci-fi movie in Rêves de Belleville equals 3. The number of upbeat metropolis comedy in Cinéma de Montreval equals 4 times the number of intense detective thriller in Rêves de Belleville. The number of solemn period drama in Festival de Saint-Rivage equals the total number of movies in Festival de Clairmont. The number of intense detective thriller in Cinéma de Montreval equals 2.

Question: How many intense detective thriller does Rêves de Belleville have?

Solution: Define intense detective thriller in Cinéma de Montreval as N; so N = 2. Define solemn period drama in Festival de Clairmont as v; so v = N = 2. Define intense detective thriller in Festival de Clairmont as Y; s = N = 2; so Y = 3 + s = 3 + 2 = 5. Define upbeat metropolis comedy in Festival de Clairmont as q; a = v = 2; so q = 4 * a = 4 * 2 = 8. Define total number of movies in Festival de Clairmont as r; J = v + Y = 2 + 5 = 7; so r = J + q = 7 + 8 = 15. Define solemn period drama in Festival de Saint-Rivage as I; so I = r = 15. Define upbeat metropolis comedy in Festival de Saint-Rivage as o; so o = q = 8. Define total number of movies in Festival de Saint-Rivage as Q; so Q = I + o = 15 + 8 = 23. Define intense detective thriller in Rêves de Belleville as X; i = Q = 23; so X = 3 * i = 3 * 23 = 69. Answer: 69.


The number of adult blue jay in Jefferson Circus equals the total number of adult animals in Hamilton Farm. The number of adult parrot in Mayer Aquarium equals 4. The number of adult parrot in Hamilton Farm equals the sum of the total number of adult animals in Mayer Aquarium and the number of adult eagle in Mayer Aquarium. The number of adult eagle in Mayer Aquarium equals 4. The number of adult crow in Hamilton Farm equals the sum of the total number of adult animals in South Zoo and the number of adult eagle in Mayer Aquarium. The number of adult eagle in South Zoo equals 1 plus the number of adult eagle in Mayer Aquarium.

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

Solution: Define adult parrot in Mayer Aquarium as A; so A = 4. Define adult eagle in Mayer Aquarium as Q; so Q = 4. Define total number of adult animals in Mayer Aquarium as a; so a = Q + A = 4 + 4 = 8. Define adult parrot in Hamilton Farm as C; so C = a + Q = 8 + 4 = 12. Define adult eagle in South Zoo as i; q = Q = 4; so i = 1 + q = 1 + 4 = 5. Define total number of adult animals in South Zoo as y; so y = i = 5. Define adult crow in Hamilton Farm as N; so N = y + Q = 5 + 4 = 9. Define total number of adult animals in Hamilton Farm as Y; so Y = C + N = 12 + 9 = 21. Define adult blue jay in Jefferson Circus as G; so G = Y = 21. Define total number of adult animals in Jefferson Circus as u; so u = G = 21. Answer: 21.


The number of private middle school in Evervale City equals the number of regional medical school in Brightford. The number of regional medical school in Westhaven City equals 4 plus the number of culinarian school in Westhaven City. The number of culinarian school in Brightford equals 2 times the number of private middle school in Evervale City. The number of private middle school in Brightford equals 3. The number of culinarian school in Westhaven City equals 2 plus the sum of the number of culinarian school in Brightford and the number of regional medical school in Brightford. The number of regional medical school in Evervale City equals 3. The number of regional medical school in Brightford exists, and its number is greater than 0. The number of private middle school in Westhaven City equals 2 plus the sum of the number of culinarian school in Brightford and the number of regional medical school in Brightford. The total number of schools in Westhaven City equals 46. The number of culinarian school in Evervale City equals the sum of the total number of schools in Westhaven City and the total number of schools in Evervale City.

Question: How many regional medical school does Brightford have?

Solution: The question is difficult, so we use equations to solve it. Define regional medical school in Brightford as x; we don't know its value yet but will find it out later. Define private middle school in Evervale City as l; so l = x = x. Define culinarian school in Brightford as Z; s = l = x; so Z = 2 * s = (2) * (x) = 2*x. Define culinarian school in Westhaven City as A; M = Z + x = 2*x + x = 3*x; so A = 2 + M = 2 + 3*x = 3*x + 2. Define private middle school in Westhaven City as P; J = Z + x = 2*x + x = 3*x; so P = 2 + J = 2 + 3*x = 3*x + 2. Define regional medical school in Westhaven City as O; W = A = 3*x + 2; so O = 4 + W = 4 + 3*x + 2 = 3*x + 6. Define total number of schools in Westhaven City as u; h = P + O = 3*x + 2 + 3*x + 6 = 6*x + 8; so u = h + A = 6*x + 8 + 3*x + 2 = 9*x + 10. We know u = 46, so we have 9*x + 10 = 46 and we solve it through the following steps: Simplifying the equation: 9*x + 10 = 46. Move all terms to one side: 9*x - 36 = 0. Isolate the term with x: 9*x = 36. Divide both sides by 9: x = 36 / 9. Solution: x = 4. Answer: 4.


The number of adult racoon in Hamilton Farm equals the sum of the number of adult deer in South Zoo and the number of adult racoon in South Zoo. The number of adult racoon in Jefferson Circus equals 3 plus the difference between the total number of adult animals in South Zoo and the number of adult racoon in South Zoo. The number of adult racoon in South Zoo equals 1. The number of adult fox in Hamilton Farm equals the sum of the total number of adult animals in Jefferson Circus, the total number of adult animals in South Zoo, and the number of adult deer in South Zoo. The number of adult deer in South Zoo equals 2.

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

Solution: Define adult deer in South Zoo as r; so r = 2. Define adult racoon in South Zoo as q; so q = 1. Define total number of adult animals in South Zoo as U; so U = q + r = 1 + 2 = 3. Define adult racoon in Jefferson Circus as H; e = U - q = 3 - 1 = 2; so H = 3 + e = 3 + 2 = 5. Define total number of adult animals in Jefferson Circus as O; so O = H = 5. Define adult fox in Hamilton Farm as p; W = O + U = 5 + 3 = 8; so p = W + r = 8 + 2 = 10. Define adult racoon in Hamilton Farm as K; so K = r + q = 2 + 1 = 3. Define total number of adult animals in Hamilton Farm as w; so w = K + p = 3 + 10 = 13. Answer: 13.


The number of adult wolf in Oakridge Riverside exists, and its number is greater than 0. The number of adult fox in Beverly Forest equals 4 times the total number of adult animals in Pine Ridge. The number of adult fox in Oakridge Riverside equals 3. The number of adult racoon in Cedar Valley equals 3 times the number of adult deer in Pine Ridge. The number of adult racoon in Pine Ridge equals 3 plus the number of adult wolf in Oakridge Riverside. The total number of adult animals in Cedar Valley equals 80. The number of adult deer in Cedar Valley equals 3 plus the total number of adult animals in Beverly Forest. The number of adult deer in Pine Ridge equals the number of adult racoon in Pine Ridge.

Question: How many adult wolf does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define adult racoon in Pine Ridge as J; g = x = x; so J = 3 + g = 3 + x = x + 3. Define adult deer in Pine Ridge as i; so i = J = x + 3. Define total number of adult animals in Pine Ridge as S; so S = i + J = x + 3 + x + 3 = 2*x + 6. Define adult fox in Beverly Forest as d; V = S = 2*x + 6; so d = 4 * V = (4) * (2*x + 6) = 8*x + 24. Define total number of adult animals in Beverly Forest as R; so R = d = 8*x + 24. Define adult deer in Cedar Valley as s; M = R = 8*x + 24; so s = 3 + M = 3 + 8*x + 24 = 8*x + 27. Define adult racoon in Cedar Valley as f; F = i = x + 3; so f = 3 * F = (3) * (x + 3) = 3*x + 9. Define total number of adult animals in Cedar Valley as A; so A = s + f = 8*x + 27 + 3*x + 9 = 11*x + 36. We know A = 80, so we have 11*x + 36 = 80 and we solve it through the following steps: Simplifying the equation: 11*x + 36 = 80. Move all terms to one side: 11*x - 44 = 0. Isolate the term with x: 11*x = 44. Divide both sides by 11: x = 44 / 11. Solution: x = 4. Answer: 4.


The number of futuristic sci-fi movie in Verdi Movie Festival equals 2 plus the number of intense detective thriller in Northwood Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals 2. The number of solemn period drama in Northwood Movie Festival equals the number of intense detective thriller in West Sahara Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival equals 4 times the number of solemn period drama in West Sahara Movie Festival. The number of solemn period drama in Taylor Movie Festival exists, and its number is greater than 0. The number of intense detective thriller in Taylor Movie Festival equals the number of futuristic sci-fi movie in Verdi Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 3 plus the sum of the total number of movies in Verdi Movie Festival and the number of intense detective thriller in Northwood Movie Festival. The number of futuristic sci-fi movie in Northwood Movie Festival equals 2 times the number of upbeat metropolis comedy in West Sahara Movie Festival. The number of intense detective thriller in West Sahara Movie Festival equals 1 plus the number of solemn period drama in West Sahara Movie Festival. The number of intense detective thriller in Northwood Movie Festival equals 4. The number of upbeat metropolis comedy in Northwood Movie Festival equals 2 plus the number of solemn period drama in Northwood Movie Festival. The number of upbeat metropolis comedy in Taylor Movie Festival equals 1 plus the sum of the number of upbeat metropolis comedy in Golden Banana Movie Festival and the number of futuristic sci-fi movie in Verdi Movie Festival. The number of solemn period drama in West Sahara Movie Festival equals 3 times the number of upbeat metropolis comedy in West Sahara Movie Festival. The total number of movies in Taylor Movie Festival equals 29. The number of solemn period drama in Golden Banana Movie Festival equals 2 plus the number of solemn period drama in Taylor Movie Festival.

Question: How many solemn period drama does Taylor Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define intense detective thriller in Northwood Movie Festival as Q; so Q = 4. Define futuristic sci-fi movie in Verdi Movie Festival as f; P = Q = 4; so f = 2 + P = 2 + 4 = 6. Define total number of movies in Verdi Movie Festival as V; so V = f = 6. Define upbeat metropolis comedy in Golden Banana Movie Festival as Y; s = V + Q = 6 + 4 = 10; so Y = 3 + s = 3 + 10 = 13. Define upbeat metropolis comedy in Taylor Movie Festival as t; k = Y + f = 13 + 6 = 19; so t = 1 + k = 1 + 19 = 20. Define intense detective thriller in Taylor Movie Festival as r; so r = f = 6. Define solemn period drama 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 B; R = x + r = x + 6 = x + 6; so B = R + t = x + 6 + 20 = x + 26. We know B = 29, so we have x + 26 = 29 and we solve it through the following steps: Simplifying the equation: x + 26 = 29. 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.


The number of adult wolf in Mayer Aquarium exists, and its number is greater than 0. The total number of adult animals in Mayer Aquarium equals 94. The number of adult fox in Mayer Aquarium equals 3 times the sum of the total number of adult animals in Hamilton Farm and the number of adult fox in Jefferson Circus. The number of adult wolf in Hamilton Farm equals 2. The number of adult fox in Hamilton Farm equals the total number of adult animals in Jefferson Circus. The number of adult fox in Jefferson Circus equals 3 times the number of adult wolf in Jefferson Circus. The number of adult wolf in Jefferson Circus equals 4.

Question: How many adult wolf does Mayer Aquarium have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Jefferson Circus as B; so B = 4. Define adult fox in Jefferson Circus as U; Z = B = 4; so U = 3 * Z = (3) * (4) = 12. Define total number of adult animals in Jefferson Circus as i; so i = U + B = 12 + 4 = 16. Define adult fox in Hamilton Farm as g; so g = i = 16. Define adult wolf in Hamilton Farm as S; so S = 2. Define total number of adult animals in Hamilton Farm as K; so K = g + S = 16 + 2 = 18. Define adult fox in Mayer Aquarium as k; R = K + U = 18 + 12 = 30; so k = 3 * R = (3) * (30) = 90. Define adult wolf in Mayer Aquarium as x; we don't know its value yet but will find it out later. Define total number of adult animals in Mayer Aquarium as N; so N = k + x = 90 + x = x + 90. We know N = 94, so we have x + 90 = 94 and we solve it through the following steps: Simplifying the equation: x + 90 = 94. 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.


The number of adult parrot in Oakridge Riverside equals the sum of the total number of adult animals in Maple Creek and the number of adult crow in Beverly Forest. The number of adult eagle in Maple Creek equals 2 plus the number of adult blue jay in Cedar Valley. The number of adult blue jay in Beverly Forest equals 2. The number of adult crow in Beverly Forest equals 3 times the number of adult eagle in Oakridge Riverside. The number of adult blue jay in Cedar Valley equals 3 plus the difference between the number of adult crow in Beverly Forest and the number of adult eagle in Oakridge Riverside. The number of adult eagle in Oakridge Riverside equals 3. The number of adult eagle in Beverly Forest equals 3. The number of adult parrot in Maple Creek equals 4. The number of adult crow in Maple Creek equals the sum of the number of adult blue jay in Cedar Valley, the number of adult parrot in Maple Creek, and the number of adult crow in Beverly Forest. The number of adult crow in Oakridge Riverside equals the number of adult blue jay in Beverly Forest.

Question: What is the total number of adult animals in Maple Creek?

Solution: Define adult eagle in Oakridge Riverside as n; so n = 3. Define adult crow in Beverly Forest as p; s = n = 3; so p = 3 * s = 3 * 3 = 9. Define adult blue jay in Cedar Valley as v; w = p - n = 9 - 3 = 6; so v = 3 + w = 3 + 6 = 9. Define adult eagle in Maple Creek as S; j = v = 9; so S = 2 + j = 2 + 9 = 11. Define adult parrot in Maple Creek as B; so B = 4. Define adult crow in Maple Creek as O; u = v + B = 9 + 4 = 13; so O = u + p = 13 + 9 = 22. Define total number of adult animals in Maple Creek as y; m = S + B = 11 + 4 = 15; so y = m + O = 15 + 22 = 37. Answer: 37.


The number of adult bear in Cedar Valley equals the number of adult wolf in Cedar Valley. The number of adult wolf in Oakridge Riverside equals 4 times the total number of adult animals in Cedar Valley. The number of adult deer in Pine Ridge equals 2 times the number of adult bear in Cedar Valley. The number of adult wolf in Pine Ridge equals 2. The number of adult deer in Oakridge Riverside equals the number of adult wolf in Cedar Valley. The number of adult bear in Oakridge Riverside equals 4. The number of adult deer in Cedar Valley equals 3 times the difference between the number of adult bear in Cedar Valley and the number of adult wolf in Cedar Valley. The number of adult wolf in Cedar Valley equals 2.

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

Solution: Define adult wolf in Cedar Valley as a; so a = 2. Define adult bear in Cedar Valley as h; so h = a = 2. Define adult deer in Cedar Valley as B; d = h - a = 2 - 2 = 0; so B = 3 * d = 3 * 0 = 0. Define total number of adult animals in Cedar Valley as Q; S = B + a = 0 + 2 = 2; so Q = S + h = 2 + 2 = 4. Define adult wolf in Oakridge Riverside as y; e = Q = 4; so y = 4 * e = 4 * 4 = 16. Define adult deer in Oakridge Riverside as r; so r = a = 2. Define adult bear in Oakridge Riverside as A; so A = 4. Define total number of adult animals in Oakridge Riverside as c; W = r + y = 2 + 16 = 18; so c = W + A = 18 + 4 = 22. Answer: 22.


The number of adult parrot in Bundle Ranch equals the number of adult eagle in Bundle Ranch. The number of adult parrot in Jefferson Circus equals the number of adult eagle in Hamilton Farm. The number of adult blue jay in Hamilton Farm equals 2. The number of adult eagle in Bundle Ranch equals the total number of adult animals in Hamilton Farm. The number of adult eagle in Jefferson Circus equals the total number of adult animals in Bundle Ranch. The number of adult blue jay in Jefferson Circus equals the total number of adult animals in Hamilton Farm. The number of adult eagle in Hamilton Farm equals 3 plus the number of adult blue jay in Hamilton Farm.

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

Solution: Define adult blue jay in Hamilton Farm as w; so w = 2. Define adult eagle in Hamilton Farm as a; G = w = 2; so a = 3 + G = 3 + 2 = 5. Define total number of adult animals in Hamilton Farm as r; so r = w + a = 2 + 5 = 7. Define adult eagle in Bundle Ranch as h; so h = r = 7. Define adult parrot in Bundle Ranch as c; so c = h = 7. Define adult parrot in Jefferson Circus as q; so q = a = 5. Define total number of adult animals in Bundle Ranch as U; so U = c + h = 7 + 7 = 14. Define adult blue jay in Jefferson Circus as Z; so Z = r = 7. Define adult eagle in Jefferson Circus as v; so v = U = 14. Define total number of adult animals in Jefferson Circus as X; K = q + Z = 5 + 7 = 12; so X = K + v = 12 + 14 = 26. Answer: 26.


The number of solemn period drama in Northwood Movie Festival equals the sum of the total number of movies in Verdi Movie Festival, the number of solemn period drama in Verdi Movie Festival, and the number of upbeat metropolis comedy in Verdi Movie Festival. The number of calm road movie in Verdi Movie Festival equals the number of upbeat metropolis comedy in Verdi Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals 2 plus the number of solemn period drama in Northwood Movie Festival. The number of calm road movie in West Sahara Movie Festival equals the number of solemn period drama in Verdi Movie Festival. The number of solemn period drama in Verdi Movie Festival equals 3 times the number of calm road movie in Verdi Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival equals 4. The number of upbeat metropolis comedy in Verdi Movie Festival equals 3.

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

Solution: Define upbeat metropolis comedy in Verdi Movie Festival as I; so I = 3. Define calm road movie in Verdi Movie Festival as h; so h = I = 3. Define solemn period drama in Verdi Movie Festival as v; q = h = 3; so v = 3 * q = 3 * 3 = 9. Define total number of movies in Verdi Movie Festival as H; n = h + v = 3 + 9 = 12; so H = n + I = 12 + 3 = 15. Define solemn period drama in Northwood Movie Festival as N; M = H + v = 15 + 9 = 24; so N = M + I = 24 + 3 = 27. Define upbeat metropolis comedy in West Sahara Movie Festival as E; B = N = 27; so E = 2 + B = 2 + 27 = 29. Define calm road movie in West Sahara Movie Festival as S; so S = v = 9. Define total number of movies in West Sahara Movie Festival as P; so P = S + E = 9 + 29 = 38. Answer: 38.


The number of adult eagle in South Zoo equals 1 plus the total number of adult animals in Hamilton Farm. The number of adult blue jay in Hamilton Farm equals the number of adult eagle in Hamilton Farm. The number of adult eagle in Jefferson Circus equals 1 plus the number of adult eagle in South Zoo. The number of adult owl in Jefferson Circus equals the number of adult eagle in Hamilton Farm. The number of adult blue jay in Jefferson Circus equals the number of adult blue jay in South Zoo. The number of adult blue jay in South Zoo equals 2. The number of adult eagle in Mayer Aquarium equals the number of adult crow in Hamilton Farm. The number of adult eagle in Hamilton Farm equals 3 times the number of adult crow in South Zoo. The number of adult crow in South Zoo equals the number of adult crow in Jefferson Circus. The number of adult owl in Mayer Aquarium equals 1 plus the sum of the number of adult blue jay in South Zoo and the number of adult blue jay in Jefferson Circus. The number of adult owl in Hamilton Farm equals 2. The number of adult crow in Hamilton Farm equals the sum of the number of adult eagle in Hamilton Farm, the number of adult owl in Hamilton Farm, and the number of adult crow in Jefferson Circus. The number of adult crow in Jefferson Circus equals 3. The number of adult crow in Mayer Aquarium equals 3 times the number of adult eagle in Hamilton Farm. The number of adult owl in South Zoo equals 3 times the number of adult owl in Jefferson Circus. The number of adult blue jay in Mayer Aquarium equals 4 times the number of adult eagle in South Zoo.

Question: How many adult eagle does South Zoo have?

Solution: Define adult crow in Jefferson Circus as H; so H = 3. Define adult crow in South Zoo as I; so I = H = 3. Define adult eagle in Hamilton Farm as a; V = I = 3; so a = 3 * V = 3 * 3 = 9. Define adult owl in Hamilton Farm as d; so d = 2. Define adult blue jay in Hamilton Farm as B; so B = a = 9. Define adult crow in Hamilton Farm as t; Q = a + d = 9 + 2 = 11; so t = Q + H = 11 + 3 = 14. Define total number of adult animals in Hamilton Farm as w; q = d + a = 2 + 9 = 11; z = q + B = 11 + 9 = 20; so w = z + t = 20 + 14 = 34. Define adult eagle in South Zoo as W; o = w = 34; so W = 1 + o = 1 + 34 = 35. Answer: 35.


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 Cinéma de Montreval equals the number of calm road movie in Festival de Clairmont. The number of calm road movie in Cinéma de Montreval equals 2 times the sum of the number of calm road movie in Festival de Clairmont and the number of intense detective thriller in Festival de Clairmont. The number of solemn period drama in Cinéma de Montreval equals the total number of movies in Festival de Clairmont. The number of calm road movie in Festival de Clairmont equals 3. The number of intense detective thriller in Festival de Clairmont equals 3.

Question: What is the total number of movies in Cinéma de Montreval?

Solution: Define intense detective thriller in Festival de Clairmont as o; so o = 3. Define calm road movie in Festival de Clairmont as D; so D = 3. Define solemn period drama in Festival de Clairmont as t; A = o = 3; so t = 2 + A = 2 + 3 = 5. Define total number of movies in Festival de Clairmont as W; f = D + t = 3 + 5 = 8; so W = f + o = 8 + 3 = 11. Define solemn period drama in Cinéma de Montreval as y; so y = W = 11. Define calm road movie in Cinéma de Montreval as l; R = D + o = 3 + 3 = 6; so l = 2 * R = 2 * 6 = 12. Define intense detective thriller in Cinéma de Montreval as K; so K = D = 3. Define total number of movies in Cinéma de Montreval as U; d = K + l = 3 + 12 = 15; so U = d + y = 15 + 11 = 26. Answer: 26.


The number of private middle school in Evervale City equals 3 plus the difference between the number of private middle school in Westhaven City and the number of regional medical school in Westhaven City. The number of regional medical school in Brightford equals 3. The number of elementary school in Glenfield City equals 3 times the total number of schools in Brightford. The number of private middle school in Brightford equals 4. The number of elementary school in Brightford equals 2 times the number of elementary school in Evervale City. The number of elementary school in Evervale City equals 4. The number of regional medical school in Westhaven City equals 2 plus the number of elementary school in Brightford. The number of private middle school in Westhaven City equals the number of regional medical school in Westhaven City. The number of culinarian school in Evervale City equals the sum of the number of elementary school in Evervale City, the number of private middle school in Glenfield City, and the total number of schools in Brightford. The number of private middle school in Glenfield City equals 1 plus the total number of schools in Westhaven City. The number of culinarian school in Brightford equals 3 plus the total number of schools in Westhaven City. The number of culinarian school in Westhaven City equals the number of private middle school in Westhaven City. The number of elementary school in Westhaven City equals the difference between the number of private middle school in Westhaven City and the number of elementary school in Brightford.

Question: How many culinarian school does Brightford have?

Solution: Define elementary school in Evervale City as t; so t = 4. Define elementary school in Brightford as w; V = t = 4; so w = 2 * V = 2 * 4 = 8. Define regional medical school in Westhaven City as L; G = w = 8; so L = 2 + G = 2 + 8 = 10. Define private middle school in Westhaven City as b; so b = L = 10. Define culinarian school in Westhaven City as M; so M = b = 10. Define elementary school in Westhaven City as O; so O = b - w = 10 - 8 = 2. Define total number of schools in Westhaven City as j; i = L + O = 10 + 2 = 12; s = i + b = 12 + 10 = 22; so j = s + M = 22 + 10 = 32. Define culinarian school in Brightford as E; y = j = 32; so E = 3 + y = 3 + 32 = 35. Answer: 35.


The number of adult fox in Mayer Aquarium equals the number of adult bear in Hamilton Farm. The number of adult bear in Hamilton Farm equals 4 plus the sum of the total number of adult animals in South Zoo and the number of adult wolf in South Zoo. The number of adult wolf in Hamilton Farm equals 2. The number of adult bear in Mayer Aquarium equals the total number of adult animals in Hamilton Farm. The number of adult fox in South Zoo equals 2 times the number of adult wolf in South Zoo. The number of adult wolf in South Zoo equals 4.

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

Solution: Define adult wolf in South Zoo as n; so n = 4. Define adult fox in South Zoo as X; q = n = 4; so X = 2 * q = 2 * 4 = 8. Define total number of adult animals in South Zoo as p; so p = X + n = 8 + 4 = 12. Define adult bear in Hamilton Farm as i; D = p + n = 12 + 4 = 16; so i = 4 + D = 4 + 16 = 20. Define adult wolf in Hamilton Farm as a; so a = 2. Define total number of adult animals in Hamilton Farm as m; so m = i + a = 20 + 2 = 22. Define adult bear in Mayer Aquarium as U; so U = m = 22. Define adult fox in Mayer Aquarium as Z; so Z = i = 20. Define total number of adult animals in Mayer Aquarium as O; so O = U + Z = 22 + 20 = 42. Answer: 42.


The number of adult owl in Bundle Ranch equals 4 times the number of adult parrot in Bundle Ranch. The number of adult parrot in South Zoo equals 2 plus the total number of adult animals in Mayer Aquarium. The number of adult parrot in Bundle Ranch equals 3. The number of adult owl in South Zoo equals the sum of the number of adult parrot in Bundle Ranch, the total number of adult animals in Bundle Ranch, and the number of adult crow in Mayer Aquarium. The number of adult crow in Mayer Aquarium equals 2 times the total number of adult animals in Bundle Ranch.

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

Solution: Define adult parrot in Bundle Ranch as o; so o = 3. Define adult owl in Bundle Ranch as t; y = o = 3; so t = 4 * y = 4 * 3 = 12. Define total number of adult animals in Bundle Ranch as h; so h = o + t = 3 + 12 = 15. Define adult crow in Mayer Aquarium as k; a = h = 15; so k = 2 * a = 2 * 15 = 30. Define total number of adult animals in Mayer Aquarium as C; so C = k = 30. Define adult owl in South Zoo as d; T = o + h = 3 + 15 = 18; so d = T + k = 18 + 30 = 48. Define adult parrot in South Zoo as f; Q = C = 30; so f = 2 + Q = 2 + 30 = 32. Define total number of adult animals in South Zoo as E; so E = f + d = 32 + 48 = 80. Answer: 80.


The number of elementary school in Clearwater Bay equals 4. The number of private middle school in Shoreline City equals 4 times the number of public highschool in Shoreline City. The number of public highschool in Clearwater Bay equals the total number of schools in Oakbridge City. The number of public highschool in Shoreline City equals 2. The number of elementary school in Oakbridge City equals 2. The number of private middle school in Oakbridge City equals 3 times the sum of the total number of schools in Shoreline City and the number of public highschool in Shoreline City.

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

Solution: Define public highschool in Shoreline City as f; so f = 2. Define private middle school in Shoreline City as J; l = f = 2; so J = 4 * l = 4 * 2 = 8. Define total number of schools in Shoreline City as k; so k = f + J = 2 + 8 = 10. Define private middle school in Oakbridge City as m; R = k + f = 10 + 2 = 12; so m = 3 * R = 3 * 12 = 36. Define elementary school in Oakbridge City as O; so O = 2. Define total number of schools in Oakbridge City as T; so T = O + m = 2 + 36 = 38. Define public highschool in Clearwater Bay as o; so o = T = 38. Define elementary school in Clearwater Bay as a; so a = 4. Define total number of schools in Clearwater Bay as A; so A = o + a = 38 + 4 = 42. Answer: 42.


The number of intense detective thriller in Festival de Clairmont equals 4. The number of intense detective thriller in Festival de Saint-Rivage equals 3 plus the total number of movies in Festival de Clairmont. The number of futuristic sci-fi movie in Festival de Clairmont equals 4 plus the total number of movies in Rêves de Belleville. The number of intense detective thriller in Rêves de Belleville equals 1 plus the number of futuristic sci-fi movie in Rêves de Belleville. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals the number of futuristic sci-fi movie in Festival de Clairmont. The number of futuristic sci-fi movie in Rêves de Belleville equals 4.

Question: What is the total number of movies in Festival de Saint-Rivage?

Solution: Define futuristic sci-fi movie in Rêves de Belleville as f; so f = 4. Define intense detective thriller in Rêves de Belleville as L; W = f = 4; so L = 1 + W = 1 + 4 = 5. Define total number of movies in Rêves de Belleville as P; so P = L + f = 5 + 4 = 9. Define futuristic sci-fi movie in Festival de Clairmont as s; h = P = 9; so s = 4 + h = 4 + 9 = 13. Define futuristic sci-fi movie in Festival de Saint-Rivage as I; so I = s = 13. Define intense detective thriller in Festival de Clairmont as e; so e = 4. Define total number of movies in Festival de Clairmont as Q; so Q = e + s = 4 + 13 = 17. Define intense detective thriller in Festival de Saint-Rivage as a; X = Q = 17; so a = 3 + X = 3 + 17 = 20. Define total number of movies in Festival de Saint-Rivage as J; so J = I + a = 13 + 20 = 33. Answer: 33.


The number of intense detective thriller in Verdi Movie Festival equals 2. The number of calm road movie in Golden Banana Movie Festival equals 2 plus the number of calm road movie in Northwood Movie Festival. The number of calm road movie in Verdi Movie Festival equals 2. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 4 times the number of futuristic sci-fi movie in Northwood Movie Festival. The number of futuristic sci-fi movie in Northwood Movie Festival equals 3 times the sum of the number of intense detective thriller in Golden Banana Movie Festival, the number of calm road movie in Northwood Movie Festival, and the number of calm road movie in Golden Banana Movie Festival. The number of futuristic sci-fi movie in Verdi Movie Festival equals 1. The total number of movies in Golden Banana Movie Festival equals 152. The number of intense detective thriller in Golden Banana Movie Festival exists, and its number is greater than 0. The number of calm road movie in Northwood Movie Festival equals 4 times the number of futuristic sci-fi movie in Verdi Movie Festival. The number of intense detective thriller in Northwood Movie Festival equals 1.

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

Solution: The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Verdi Movie Festival as b; so b = 1. Define calm road movie in Northwood Movie Festival as N; B = b = 1; so N = 4 * B = (4) * (1) = 4. Define calm road movie in Golden Banana Movie Festival as G; u = N = 4; so G = 2 + u = 2 + 4 = 6. Define intense detective thriller in Golden Banana Movie Festival as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Northwood Movie Festival as D; h = x + N = x + 4 = x + 4; j = h + G = x + 4 + 6 = x + 10; so D = 3 * j = (3) * (x + 10) = 3*x + 30. Define futuristic sci-fi movie in Golden Banana Movie Festival as E; T = D = 3*x + 30; so E = 4 * T = (4) * (3*x + 30) = 12*x + 120. Define total number of movies in Golden Banana Movie Festival as I; r = G + E = 6 + 12*x + 120 = 12*x + 126; so I = r + x = 12*x + 126 + x = 13*x + 126. We know I = 152, so we have 13*x + 126 = 152 and we solve it through the following steps: Simplifying the equation: 13*x + 126 = 152. Move all terms to one side: 13*x - 26 = 0. Isolate the term with x: 13*x = 26. Divide both sides by 13: x = 26 / 13. Solution: x = 2. Answer: 2.


The number of culinarian school in Riverton City equals 3 times the total number of schools in Shoreline City. The number of elementary school in Oakbridge City equals the sum of the total number of schools in Riverton City and the number of culinarian school in Riverton City. The number of regional medical school in Oakbridge City equals 1. The number of private middle school in Ruby Bay equals 4 plus the total number of schools in Oakbridge City. The number of regional medical school in Shoreline City exists, and its number is greater than 0. The total number of schools in Clearwater Bay equals 29. The number of public highschool in Clearwater Bay equals the total number of schools in Ruby Bay.

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 Oakbridge City as o; so o = 1. Define regional medical school in Shoreline City as x; we don't know its value yet but will find it out later. Define total number of schools in Shoreline City as F; so F = x = x. Define culinarian school in Riverton City as q; a = F = x; so q = 3 * a = (3) * (x) = 3*x. Define total number of schools in Riverton City as h; so h = q = 3*x. Define elementary school in Oakbridge City as n; so n = h + q = 3*x + 3*x = 6*x. Define total number of schools in Oakbridge City as k; so k = n + o = 6*x + 1 = 6*x + 1. Define private middle school in Ruby Bay as p; V = k = 6*x + 1; so p = 4 + V = 4 + 6*x + 1 = 6*x + 5. Define total number of schools in Ruby Bay as B; so B = p = 6*x + 5. Define public highschool in Clearwater Bay as O; so O = B = 6*x + 5. Define total number of schools in Clearwater Bay as b; so b = O = 6*x + 5. We know b = 29, so we have 6*x + 5 = 29 and we solve it through the following steps: Simplifying the equation: 6*x + 5 = 29. 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 racoon in Cedar Valley equals 3 plus the sum of the number of adult deer in Pine Ridge and the number of adult deer in Beverly Forest. The number of adult deer in Beverly Forest equals 4 times the total number of adult animals in Pine Ridge. The number of adult bear in Pine Ridge equals 2. The number of adult bear in Beverly Forest equals 3. The number of adult deer in Pine Ridge equals 4 times the number of adult racoon in Pine Ridge. The number of adult bear in Cedar Valley equals 1. The number of adult deer in Cedar Valley equals 3. The number of adult racoon in Beverly Forest equals the difference between the total number of adult animals in Pine Ridge and the number of adult bear in Pine Ridge. The number of adult racoon in Pine Ridge equals the number of adult bear in Beverly Forest.

Question: What is the total number of adult animals in Beverly Forest?

Solution: Define adult bear in Pine Ridge as O; so O = 2. Define adult bear in Beverly Forest as D; so D = 3. Define adult racoon in Pine Ridge as L; so L = D = 3. Define adult deer in Pine Ridge as d; e = L = 3; so d = 4 * e = 4 * 3 = 12. Define total number of adult animals in Pine Ridge as n; S = L + O = 3 + 2 = 5; so n = S + d = 5 + 12 = 17. Define adult racoon in Beverly Forest as s; so s = n - O = 17 - 2 = 15. Define adult deer in Beverly Forest as b; K = n = 17; so b = 4 * K = 4 * 17 = 68. Define total number of adult animals in Beverly Forest as U; o = s + D = 15 + 3 = 18; so U = o + b = 18 + 68 = 86. Answer: 86.


The number of private middle school in Clearwater Bay equals the number of elementary school in Clearwater Bay. The number of elementary school in Clearwater Bay equals the difference between the total number of schools in Shoreline City and the number of public highschool in Ruby Bay. The number of elementary school in Shoreline City exists, and its number is greater than 0. The number of culinarian school in Riverton City equals 2 times the number of private middle school in Clearwater Bay. The number of public highschool in Ruby Bay equals 4. The number of regional medical school in Shoreline City equals the number of culinarian school in Ruby Bay. The number of private middle school in Oakbridge City equals the total number of schools in Riverton City. The number of culinarian school in Ruby Bay equals 3 times the number of public highschool in Ruby Bay. The total number of schools in Oakbridge City equals 22.

Question: How many elementary school does Shoreline City have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Ruby Bay as t; so t = 4. Define culinarian school in Ruby Bay as Z; a = t = 4; so Z = 3 * a = (3) * (4) = 12. Define regional medical school in Shoreline City as D; so D = Z = 12. Define elementary school in Shoreline City as x; we don't know its value yet but will find it out later. Define total number of schools in Shoreline City as G; so G = D + x = 12 + x = x + 12. Define elementary school in Clearwater Bay as p; so p = G - t = (x + 12) - (4) = x + 8. Define private middle school in Clearwater Bay as h; so h = p = x + 8. Define culinarian school in Riverton City as I; c = h = x + 8; so I = 2 * c = (2) * (x + 8) = 2*x + 16. Define total number of schools in Riverton City as A; so A = I = 2*x + 16. Define private middle school in Oakbridge City as M; so M = A = 2*x + 16. Define total number of schools in Oakbridge City as j; so j = M = 2*x + 16. We know j = 22, so we have 2*x + 16 = 22 and we solve it through the following steps: Simplifying the equation: 2*x + 16 = 22. Move all terms to one side: 2*x - 6 = 0. Isolate the term with x: 2*x = 6. Divide both sides by 2: x = 6 / 2. Solution: x = 3. Answer: 3.


The number of upbeat metropolis comedy in Rêves de Belleville equals 4. The number of calm road movie in Rêves de Belleville equals 3 times the number of upbeat metropolis comedy in Rêves de Belleville. The number of upbeat metropolis comedy in Cinéma de Montreval equals 3 times the total number of movies in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Rêves de Belleville equals 4. 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 calm road movie in Festival Lumière de Valmont equals 2 plus the total number of movies in Rêves de Belleville.

Question: What is the total number of movies in Cinéma de Montreval?

Solution: Define upbeat metropolis comedy in Rêves de Belleville as D; so D = 4. Define calm road movie in Rêves de Belleville as q; U = D = 4; so q = 3 * U = 3 * 4 = 12. Define futuristic sci-fi movie in Rêves de Belleville as H; so H = 4. Define total number of movies in Rêves de Belleville as u; X = D + q = 4 + 12 = 16; so u = X + H = 16 + 4 = 20. Define calm road movie in Festival Lumière de Valmont as g; M = u = 20; so g = 2 + M = 2 + 20 = 22. Define solemn period drama in Festival Lumière de Valmont as A; so A = D = 4. Define total number of movies in Festival Lumière de Valmont as N; so N = g + A = 22 + 4 = 26. Define upbeat metropolis comedy in Cinéma de Montreval as t; J = N = 26; so t = 3 * J = 3 * 26 = 78. Define total number of movies in Cinéma de Montreval as K; so K = t = 78. Answer: 78.


The number of calm road movie in Festival de Saint-Rivage equals 3 times the number of upbeat metropolis comedy in Cinéma de Montreval. The number of upbeat metropolis comedy in Cinéma de Montreval equals the number of calm road movie in Cinéma de Montreval. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 3. The number of intense detective thriller in Festival de Saint-Rivage equals the difference between the total number of movies in Cinéma de Montreval and the number of upbeat metropolis comedy in Cinéma de Montreval. The number of intense detective thriller in Cinéma de Montreval equals 4 times the number of calm road movie in Cinéma de Montreval. The number of calm road movie in Cinéma de Montreval equals 4.

Question: What is the total number of movies in Festival de Saint-Rivage?

Solution: Define calm road movie in Cinéma de Montreval as e; so e = 4. Define upbeat metropolis comedy in Cinéma de Montreval as S; so S = e = 4. Define intense detective thriller in Cinéma de Montreval as Z; J = e = 4; so Z = 4 * J = 4 * 4 = 16. Define total number of movies in Cinéma de Montreval as p; M = S + e = 4 + 4 = 8; so p = M + Z = 8 + 16 = 24. Define intense detective thriller in Festival de Saint-Rivage as w; so w = p - S = 24 - 4 = 20. Define calm road movie in Festival de Saint-Rivage as s; I = S = 4; so s = 3 * I = 3 * 4 = 12. Define upbeat metropolis comedy in Festival de Saint-Rivage as X; so X = 3. Define total number of movies in Festival de Saint-Rivage as r; n = X + s = 3 + 12 = 15; so r = n + w = 15 + 20 = 35. Answer: 35.


The number of intense detective thriller in Verdi Movie Festival equals the sum of the number of solemn period drama in Verdi Movie Festival, the total number of movies in Golden Banana Movie Festival, and the number of solemn period drama 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 Verdi Movie Festival equals the number of solemn period drama in Verdi Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals 4 plus the number of intense detective thriller in Golden Banana Movie Festival. The number of solemn period drama in Verdi Movie Festival equals the sum of the total number of movies in Golden Banana Movie Festival, the number of intense detective thriller in Golden Banana Movie Festival, and the number of solemn period drama in Golden Banana Movie Festival.

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

Solution: Define intense detective thriller in Golden Banana Movie Festival as j; so j = 4. Define solemn period drama in Golden Banana Movie Festival as C; I = j = 4; so C = 4 + I = 4 + 4 = 8. Define total number of movies in Golden Banana Movie Festival as Y; so Y = C + j = 8 + 4 = 12. Define solemn period drama in Verdi Movie Festival as L; H = Y + j = 12 + 4 = 16; so L = H + C = 16 + 8 = 24. Define calm road movie in Verdi Movie Festival as f; so f = L = 24. Define intense detective thriller in Verdi Movie Festival as p; y = L + Y = 24 + 12 = 36; so p = y + C = 36 + 8 = 44. Define total number of movies in Verdi Movie Festival as h; U = f + L = 24 + 24 = 48; so h = U + p = 48 + 44 = 92. Answer: 92.


The number of adult blue jay in Maple Creek equals 3 times the number of adult eagle in Maple Creek. The number of adult owl in Maple Creek equals the number of adult eagle in Maple Creek. The number of adult eagle in Oakridge Riverside equals the sum of the total number of adult animals in Maple Creek, the number of adult eagle in Maple Creek, the number of adult owl in Maple Creek, and the number of adult blue jay in Maple Creek. The number of adult eagle in Maple Creek exists, and its number is greater than 0. The number of adult owl in Oakridge Riverside equals 2 times the number of adult eagle in Oakridge Riverside. The total number of adult animals in Oakridge Riverside equals 60.

Question: How many adult eagle does Maple Creek have?

Solution: The question is difficult, so we use equations to solve it. Define adult eagle in Maple Creek as x; we don't know its value yet but will find it out later. Define adult owl in Maple Creek as D; so D = x = x. Define adult blue jay in Maple Creek as i; h = x = x; so i = 3 * h = (3) * (x) = 3*x. Define total number of adult animals in Maple Creek as P; C = D + x = x + x = 2*x; so P = C + i = 2*x + 3*x = 5*x. Define adult eagle in Oakridge Riverside as f; b = P + x = 5*x + x = 6*x; L = b + D = 6*x + x = 7*x; so f = L + i = 7*x + 3*x = 10*x. Define adult owl in Oakridge Riverside as c; t = f = 10*x; so c = 2 * t = (2) * (10*x) = 20*x. Define total number of adult animals in Oakridge Riverside as K; so K = c + f = 20*x + 10*x = 30*x. We know K = 60, so we have 30*x = 60 and we solve it through the following steps: Simplifying the equation: 30*x = 60. Move all terms to one side: 30*x - 60 = 0. Isolate the term with x: 30*x = 60. Divide both sides by 30: x = 60 / 30. Solution: x = 2. Answer: 2.


The number of adult wolf in Jefferson Circus equals 4 times the total number of adult animals in Bundle Ranch. The number of adult bear in Mayer Aquarium equals the total number of adult animals in Jefferson Circus. The number of adult bear in Jefferson Circus exists, and its number is greater than 0. The number of adult wolf in Bundle Ranch equals 1. The number of adult wolf in South Zoo equals the sum of the total number of adult animals in Mayer Aquarium, the number of adult wolf in Bundle Ranch, and the number of adult bear in Jefferson Circus. The number of adult fox in South Zoo equals the number of adult wolf in Bundle Ranch. The total number of adult animals in South Zoo equals 14.

Question: How many adult bear does Jefferson Circus have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Bundle Ranch as E; so E = 1. Define total number of adult animals in Bundle Ranch as T; so T = E = 1. Define adult wolf in Jefferson Circus as p; z = T = 1; so p = 4 * z = (4) * (1) = 4. Define adult bear 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 j; so j = p + x = 4 + x = x + 4. Define adult bear in Mayer Aquarium as r; so r = j = x + 4. Define total number of adult animals in Mayer Aquarium as h; so h = r = x + 4. Define adult wolf in South Zoo as Z; P = h + E = x + 4 + 1 = x + 5; so Z = P + x = x + 5 + x = 2*x + 5. Define adult fox in South Zoo as B; so B = E = 1. Define total number of adult animals in South Zoo as D; so D = B + Z = 1 + 2*x + 5 = 2*x + 6. We know D = 14, so we have 2*x + 6 = 14 and we solve it through the following steps: Simplifying the equation: 2*x + 6 = 14. Move all terms to one side: 2*x - 8 = 0. Isolate the term with x: 2*x = 8. Divide both sides by 2: x = 8 / 2. Solution: x = 4. Answer: 4.


The number of adult blue jay in Jefferson Circus equals the sum of the total number of adult animals in South Zoo and the number of adult eagle in South Zoo. The number of adult eagle in South Zoo equals 4 plus the total number of adult animals in Hamilton Farm. The number of adult parrot in Jefferson Circus equals 3. The total number of adult animals in Mayer Aquarium equals 39. The number of adult blue jay in Hamilton Farm exists, and its number is greater than 0. The number of adult parrot in South Zoo equals 3 plus the number of adult eagle in South Zoo. The number of adult crow in South Zoo equals 3 times the number of adult eagle in South Zoo. The number of adult blue jay in Mayer Aquarium equals the number of adult blue jay in Jefferson Circus.

Question: How many adult blue jay does Hamilton Farm have?

Solution: The question is difficult, so we use equations to solve it. Define adult blue jay in Hamilton Farm as x; we don't know its value yet but will find it out later. Define total number of adult animals in Hamilton Farm as f; so f = x = x. Define adult eagle in South Zoo as z; a = f = x; so z = 4 + a = 4 + x = x + 4. Define adult crow in South Zoo as y; I = z = x + 4; so y = 3 * I = (3) * (x + 4) = 3*x + 12. Define adult parrot in South Zoo as P; e = z = x + 4; so P = 3 + e = 3 + x + 4 = x + 7. Define total number of adult animals in South Zoo as U; H = z + P = x + 4 + x + 7 = 2*x + 11; so U = H + y = 2*x + 11 + 3*x + 12 = 5*x + 23. Define adult blue jay in Jefferson Circus as q; so q = U + z = 5*x + 23 + x + 4 = 6*x + 27. Define adult blue jay in Mayer Aquarium as T; so T = q = 6*x + 27. Define total number of adult animals in Mayer Aquarium as L; so L = T = 6*x + 27. We know L = 39, so we have 6*x + 27 = 39 and we solve it through the following steps: Simplifying the equation: 6*x + 27 = 39. Move all terms to one side: 6*x - 12 = 0. Isolate the term with x: 6*x = 12. Divide both sides by 6: x = 12 / 6. Solution: x = 2. Answer: 2.


The number of adult eagle in South Zoo equals 4 plus the total number of adult animals in Jefferson Circus. The number of adult blue jay in Jefferson Circus equals 4 plus the total number of adult animals in Mayer Aquarium. The number of adult crow in Bundle Ranch equals 3. The number of adult owl in Mayer Aquarium equals the total number of adult animals in Bundle Ranch. The number of adult blue jay in Hamilton Farm equals 3 plus the total number of adult animals in South Zoo. The number of adult parrot in Mayer Aquarium equals the total number of adult animals in Bundle Ranch.

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

Solution: Define adult crow in Bundle Ranch as W; so W = 3. Define total number of adult animals in Bundle Ranch as v; so v = W = 3. Define adult parrot in Mayer Aquarium as L; so L = v = 3. Define adult owl in Mayer Aquarium as f; so f = v = 3. Define total number of adult animals in Mayer Aquarium as T; so T = L + f = 3 + 3 = 6. Define adult blue jay in Jefferson Circus as V; P = T = 6; so V = 4 + P = 4 + 6 = 10. Define total number of adult animals in Jefferson Circus as O; so O = V = 10. Define adult eagle in South Zoo as a; A = O = 10; so a = 4 + A = 4 + 10 = 14. Define total number of adult animals in South Zoo as S; so S = a = 14. Define adult blue jay in Hamilton Farm as s; e = S = 14; so s = 3 + e = 3 + 14 = 17. Define total number of adult animals in Hamilton Farm as I; so I = s = 17. Answer: 17.


The number of elementary school in Riverton City equals 3 plus the number of regional medical school in Riverton City. The number of elementary school in Shoreline City equals 3 times the sum of the number of regional medical school in Ruby Bay and the number of elementary school in Ruby Bay. The number of elementary school in Ruby Bay equals the difference between the number of public highschool in Ruby Bay and the number of public highschool in Riverton City. The number of regional medical school in Ruby Bay equals the total number of schools in Riverton City. The number of public highschool in Shoreline City equals the sum of the number of regional medical school in Shoreline City, the total number of schools in Riverton City, and the number of regional medical school in Ruby Bay. The number of regional medical school in Riverton City equals 4. The number of public highschool in Ruby Bay equals the difference between the total number of schools in Riverton City and the number of public highschool in Riverton City. The number of regional medical school in Shoreline City equals the number of public highschool in Riverton City. The number of public highschool in Riverton City equals the number of regional medical school in Riverton City.

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

Solution: Define regional medical school in Riverton City as M; so M = 4. Define elementary school in Riverton City as y; s = M = 4; so y = 3 + s = 3 + 4 = 7. Define public highschool in Riverton City as N; so N = M = 4. Define total number of schools in Riverton City as m; d = y + M = 7 + 4 = 11; so m = d + N = 11 + 4 = 15. Define public highschool in Ruby Bay as T; so T = m - N = 15 - 4 = 11. Define elementary school in Ruby Bay as o; so o = T - N = 11 - 4 = 7. Define regional medical school in Ruby Bay as Z; so Z = m = 15. Define total number of schools in Ruby Bay as u; K = o + Z = 7 + 15 = 22; so u = K + T = 22 + 11 = 33. Answer: 33.


The number of solemn period drama in Verdi Movie Festival equals the sum of the total number of movies in Taylor Movie Festival and the number of calm road movie in Golden Banana Movie Festival. The number of intense detective thriller in Taylor Movie Festival equals 2. The number of upbeat metropolis comedy in Verdi Movie Festival equals 3. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 4. The number of solemn period drama in Golden Banana Movie Festival equals the number of solemn period drama in Northwood Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival equals the number of upbeat metropolis comedy in Golden Banana Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals the total number of movies in Northwood Movie Festival. The number of solemn period drama 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 the number of upbeat metropolis comedy in Golden Banana Movie Festival. The number of upbeat metropolis comedy in Taylor Movie Festival equals the total number of movies in Golden Banana Movie Festival. The number of calm road movie in Verdi Movie Festival equals the difference between the number of solemn period drama in Northwood Movie Festival and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival equals 4. The number of solemn period drama in Taylor Movie Festival equals 4.

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

Solution: Define upbeat metropolis comedy in Golden Banana Movie Festival as g; so g = 4. Define upbeat metropolis comedy in Northwood Movie Festival as C; so C = g = 4. Define calm road movie in Northwood Movie Festival as b; so b = g = 4. Define solemn period drama in Northwood Movie Festival as E; so E = b = 4. Define total number of movies in Northwood Movie Festival as Q; q = E + b = 4 + 4 = 8; so Q = q + C = 8 + 4 = 12. Define calm road movie in Golden Banana Movie Festival as B; so B = Q = 12. Define intense detective thriller in Golden Banana Movie Festival as J; so J = 4. Define solemn period drama in Golden Banana Movie Festival as z; so z = E = 4. Define total number of movies in Golden Banana Movie Festival as c; L = z + B = 4 + 12 = 16; A = L + g = 16 + 4 = 20; so c = A + J = 20 + 4 = 24. Define upbeat metropolis comedy in Taylor Movie Festival as v; so v = c = 24. Answer: 24.


The number of intense detective thriller in Northwood Movie Festival equals 1. The total number of movies in Verdi Movie Festival equals 119. The number of upbeat metropolis comedy in Taylor Movie Festival equals 1 plus the number of solemn period drama in Taylor Movie Festival. The number of solemn period drama in Northwood Movie Festival equals the total number of movies in Northwood Movie Festival. The number of upbeat metropolis comedy in Verdi Movie Festival equals 2 times the sum of the number of upbeat metropolis comedy in Northwood Movie Festival and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival equals 4 plus the number of upbeat metropolis comedy in Taylor Movie Festival. The number of calm road movie in Northwood Movie Festival equals 1. The number of calm road movie in Golden Banana Movie Festival equals 4 times the sum of the total number of movies in Northwood Movie Festival and the total number of movies in Golden Banana Movie Festival. The number of intense detective thriller in Verdi Movie Festival equals 4 times the sum of the number of upbeat metropolis comedy in Verdi Movie Festival and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of solemn period drama in Golden Banana Movie Festival equals the number of intense detective thriller in Verdi Movie Festival. The number of solemn period drama in Taylor Movie Festival exists, and its number is greater than 0. The number of calm road movie in Taylor Movie Festival equals the number of upbeat metropolis comedy in Northwood Movie Festival. The number of calm road movie in Verdi Movie Festival equals the number of upbeat metropolis comedy in Northwood Movie Festival.

Question: How many solemn period drama does Taylor Movie Festival have?

Solution: 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 upbeat metropolis comedy in Taylor Movie Festival as T; N = x = x; so T = 1 + N = 1 + x = x + 1. Define upbeat metropolis comedy in Northwood Movie Festival as s; p = T = x + 1; so s = 4 + p = 4 + x + 1 = x + 5. Define upbeat metropolis comedy in Verdi Movie Festival as y; P = s + T = x + 5 + x + 1 = 2*x + 6; so y = 2 * P = (2) * (2*x + 6) = 4*x + 12. Define calm road movie in Verdi Movie Festival as t; so t = s = x + 5. Define intense detective thriller in Verdi Movie Festival as J; Z = y + T = 4*x + 12 + x + 1 = 5*x + 13; so J = 4 * Z = (4) * (5*x + 13) = 20*x + 52. Define total number of movies in Verdi Movie Festival as f; Y = t + y = x + 5 + 4*x + 12 = 5*x + 17; so f = Y + J = 5*x + 17 + 20*x + 52 = 25*x + 69. We know f = 119, so we have 25*x + 69 = 119 and we solve it through the following steps: Simplifying the equation: 25*x + 69 = 119. Move all terms to one side: 25*x - 50 = 0. Isolate the term with x: 25*x = 50. Divide both sides by 25: x = 50 / 25. Solution: x = 2. Answer: 2.


The number of adult owl in South Zoo equals the sum of the number of adult crow in South Zoo, the number of adult parrot in South Zoo, and the number of adult crow in Jefferson Circus. The number of adult owl in Jefferson Circus equals 4 times the difference between the number of adult parrot in South Zoo and the number of adult crow in South Zoo. The number of adult crow in South Zoo exists, and its number is greater than 0. The total number of adult animals in South Zoo equals 69. The number of adult parrot in South Zoo equals 4 times the number of adult crow in Jefferson Circus. The number of adult crow in Jefferson Circus equals 3. The number of adult eagle in South Zoo equals the number of adult owl in Jefferson Circus. The number of adult parrot in Jefferson Circus equals 2 times the sum of the number of adult owl in Jefferson Circus, the number of adult parrot in South Zoo, and the number of adult eagle in South Zoo. The number of adult eagle in Jefferson Circus equals 2 times the number of adult parrot in South Zoo.

Question: How many adult crow does South Zoo have?

Solution: The question is difficult, so we use equations to solve it. Define adult crow in Jefferson Circus as h; so h = 3. Define adult parrot in South Zoo as R; Z = h = 3; so R = 4 * Z = (4) * (3) = 12. Define adult crow in South Zoo as x; we don't know its value yet but will find it out later. Define adult owl in South Zoo as V; D = x + R = x + 12 = x + 12; so V = D + h = x + 12 + 3 = x + 15. Define adult owl in Jefferson Circus as G; Y = R - x = (12) - (x) = 12 - x; so G = 4 * Y = (4) * (12 - x) = 48 - 4*x. Define adult eagle in South Zoo as i; so i = G = 48 - 4*x. Define total number of adult animals in South Zoo as I; q = V + x = x + 15 + x = 2*x + 15; U = q + R = 2*x + 15 + 12 = 2*x + 27; so I = U + i = 2*x + 27 + 48 - 4*x = 75 - 2*x. We know I = 69, so we have 75 - 2*x = 69 and we solve it through the following steps: Simplifying the equation: 75 - 2*x = 69. Move all terms to one side: 6 - 2*x = 0. Isolate the term with x: -2*x = -6. Divide both sides by -2: x = -6 / -2. Solution: x = 3. Answer: 3.