The number of regional medical school in Riverton City equals 2 plus the difference between the total number of schools in Ruby Bay and the number of regional medical school in Ruby Bay. The total number of schools in Oakbridge City equals 22. The number of elementary school in Oakbridge City equals 3 times the sum of the total number of schools in Clearwater Bay and the number of regional medical school in Ruby Bay. The number of regional medical school in Ruby Bay exists, and its number is greater than 0. The number of elementary school in Clearwater Bay equals 2 times the total number of schools in Riverton City. The number of regional medical school in Oakbridge City equals the total number of schools in Clearwater Bay.

Question: How many regional medical school does Ruby Bay have?

Solution: The question is difficult, so we use equations to solve it. Define regional medical school in Ruby Bay as x; we don't know its value yet but will find it out later. Define total number of schools in Ruby Bay as d; so d = x = x. Define regional medical school in Riverton City as h; Y = d - x = (x) - (x) = 0; so h = 2 + Y = 2 + 0 = 2. Define total number of schools in Riverton City as N; so N = h = 2. Define elementary school in Clearwater Bay as D; z = N = 2; so D = 2 * z = (2) * (2) = 4. Define total number of schools in Clearwater Bay as B; so B = D = 4. Define elementary school in Oakbridge City as g; i = B + x = 4 + x = x + 4; so g = 3 * i = (3) * (x + 4) = 3*x + 12. Define regional medical school in Oakbridge City as C; so C = B = 4. Define total number of schools in Oakbridge City as c; so c = g + C = 3*x + 12 + 4 = 3*x + 16. We know c = 22, so we have 3*x + 16 = 22 and we solve it through the following steps: Simplifying the equation: 3*x + 16 = 22. Move all terms to one side: 3*x - 6 = 0. Isolate the term with x: 3*x = 6. Divide both sides by 3: x = 6 / 3. Solution: x = 2. Answer: 2.


The number of intense detective thriller in Rêves de Belleville equals the number of intense detective thriller in Festival de Clairmont. The total number of movies in Festival de Clairmont equals 16. The number of futuristic sci-fi movie in Rêves de Belleville equals 3. The number of calm road movie in Festival de Clairmont equals 4 plus the difference between the number of calm road movie in Rêves de Belleville and the number of calm road movie in Festival de Saint-Rivage. The number of calm road movie in Festival de Saint-Rivage equals 2. The number of intense detective thriller in Festival de Clairmont exists, and its number is greater than 0. The number of calm road movie in Rêves de Belleville equals 4 plus the total number of movies in Festival de Saint-Rivage. The number of intense detective thriller in Festival de Saint-Rivage equals 3 plus the number of calm road movie in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 1.

Question: How many intense detective thriller does Festival de Clairmont have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Festival de Saint-Rivage as R; so R = 2. Define upbeat metropolis comedy in Festival de Saint-Rivage as h; so h = 1. Define intense detective thriller in Festival de Saint-Rivage as z; Z = R = 2; so z = 3 + Z = 3 + 2 = 5. Define total number of movies in Festival de Saint-Rivage as N; H = z + R = 5 + 2 = 7; so N = H + h = 7 + 1 = 8. Define calm road movie in Rêves de Belleville as G; k = N = 8; so G = 4 + k = 4 + 8 = 12. Define calm road movie in Festival de Clairmont as s; m = G - R = (12) - (2) = 10; so s = 4 + m = 4 + 10 = 14. Define intense detective thriller in Festival de Clairmont as x; we don't know its value yet but will find it out later. Define total number of movies in Festival de Clairmont as B; so B = x + s = x + 14 = x + 14. We know B = 16, so we have x + 14 = 16 and we solve it through the following steps: Simplifying the equation: x + 14 = 16. Move all terms to one side: x - 2 = 0. Isolate the term with x: 1*x = 2. Divide both sides by 1: x = 2 / 1. Solution: x = 2. Answer: 2.


The total number of schools in Oakbridge City equals 49. The number of private middle school in Ruby Bay equals 1 plus the number of private middle school in Riverton City. The number of private middle school in Riverton City exists, and its number is greater than 0. The number of elementary school in Oakbridge City equals 2 times the sum of the number of private middle school in Ruby Bay and the number of private middle school in Riverton City. The number of regional medical school in Oakbridge City equals 4 times the number of private middle school in Ruby Bay. The number of private middle school in Oakbridge City equals the sum of the number of regional medical school in Oakbridge City, the number of private middle school in Riverton City, the number of elementary school in Oakbridge City, and the number of private middle school in Ruby Bay.

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

Solution: The question is difficult, so we use equations to solve it. Define private middle school in Riverton City as x; we don't know its value yet but will find it out later. Define private middle school in Ruby Bay as o; r = x = x; so o = 1 + r = 1 + x = x + 1. Define elementary school in Oakbridge City as L; v = o + x = x + 1 + x = 2*x + 1; so L = 2 * v = (2) * (2*x + 1) = 4*x + 2. Define regional medical school in Oakbridge City as j; K = o = x + 1; so j = 4 * K = (4) * (x + 1) = 4*x + 4. Define private middle school in Oakbridge City as J; u = j + x = 4*x + 4 + x = 5*x + 4; y = u + L = 5*x + 4 + 4*x + 2 = 9*x + 6; so J = y + o = 9*x + 6 + x + 1 = 10*x + 7. Define total number of schools in Oakbridge City as k; h = j + J = 4*x + 4 + 10*x + 7 = 14*x + 11; so k = h + L = 14*x + 11 + 4*x + 2 = 18*x + 13. We know k = 49, so we have 18*x + 13 = 49 and we solve it through the following steps: Simplifying the equation: 18*x + 13 = 49. Move all terms to one side: 18*x - 36 = 0. Isolate the term with x: 18*x = 36. Divide both sides by 18: x = 36 / 18. Solution: x = 2. Answer: 2.


The number of culinarian school in Brightford equals 4 plus the difference between the total number of schools in Evervale City and the total number of schools in Hawkesbury. The number of culinarian school in Hawkesbury equals 2 times the sum of the total number of schools in Glenfield City and the number of private middle school in Glenfield City. The number of public highschool in Evervale City equals 2 times the total number of schools in Hawkesbury. The number of private middle school in Glenfield City equals 4. The number of elementary school in Brightford equals 4.

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

Solution: Define private middle school in Glenfield City as t; so t = 4. Define total number of schools in Glenfield City as A; so A = t = 4. Define culinarian school in Hawkesbury as W; g = A + t = 4 + 4 = 8; so W = 2 * g = 2 * 8 = 16. Define total number of schools in Hawkesbury as I; so I = W = 16. Define public highschool in Evervale City as l; u = I = 16; so l = 2 * u = 2 * 16 = 32. Define total number of schools in Evervale City as O; so O = l = 32. Define culinarian school in Brightford as R; k = O - I = 32 - 16 = 16; so R = 4 + k = 4 + 16 = 20. Define elementary school in Brightford as p; so p = 4. Define total number of schools in Brightford as T; so T = p + R = 4 + 20 = 24. Answer: 24.


The number of adult crow in Mayer Aquarium equals the sum of the number of adult crow in Hamilton Farm, the number of adult blue jay in Hamilton Farm, and the total number of adult animals in Hamilton Farm. The number of adult owl in Bundle Ranch equals 1 plus the number of adult crow in Mayer Aquarium. The number of adult owl in Mayer Aquarium equals the sum of the total number of adult animals in Hamilton Farm and the number of adult crow in Hamilton Farm. The number of adult blue jay in Hamilton Farm equals 2 plus the number of adult crow in Hamilton Farm. The number of adult crow in Hamilton Farm exists, and its number is greater than 0. The number of adult owl in Hamilton Farm equals 1 plus the number of adult crow in Hamilton Farm. The number of adult blue jay in Bundle Ranch equals 4. The total number of adult animals in Mayer Aquarium equals 35.

Question: How many adult crow does Hamilton Farm have?

Solution: The question is difficult, so we use equations to solve it. Define adult crow in Hamilton Farm as x; we don't know its value yet but will find it out later. Define adult blue jay in Hamilton Farm as n; G = x = x; so n = 2 + G = 2 + x = x + 2. Define adult owl in Hamilton Farm as b; O = x = x; so b = 1 + O = 1 + x = x + 1. Define total number of adult animals in Hamilton Farm as z; v = n + b = x + 2 + x + 1 = 2*x + 3; so z = v + x = 2*x + 3 + x = 3*x + 3. Define adult owl in Mayer Aquarium as T; so T = z + x = 3*x + 3 + x = 4*x + 3. Define adult crow in Mayer Aquarium as f; K = x + n = x + x + 2 = 2*x + 2; so f = K + z = 2*x + 2 + 3*x + 3 = 5*x + 5. Define total number of adult animals in Mayer Aquarium as D; so D = T + f = 4*x + 3 + 5*x + 5 = 9*x + 8. We know D = 35, so we have 9*x + 8 = 35 and we solve it through the following steps: Simplifying the equation: 9*x + 8 = 35. Move all terms to one side: 9*x - 27 = 0. Isolate the term with x: 9*x = 27. Divide both sides by 9: x = 27 / 9. Solution: x = 3. Answer: 3.


The number of adult blue jay in Mayer Aquarium equals 4 plus the number of adult parrot in South Zoo. The total number of adult animals in Jefferson Circus equals 32. The number of adult crow in Jefferson Circus equals the number of adult parrot in Jefferson Circus. The number of adult eagle in Bundle Ranch exists, and its number is greater than 0. The number of adult parrot in South Zoo equals 3 times the sum of the number of adult eagle in South Zoo and the number of adult eagle in Hamilton Farm. The number of adult parrot in Jefferson Circus equals 1 plus the difference between the number of adult blue jay in Mayer Aquarium and the number of adult eagle in Bundle Ranch. The number of adult eagle in South Zoo equals the number of adult eagle in Hamilton Farm. The number of adult eagle in Hamilton Farm equals 1 plus the number of adult eagle in Bundle Ranch. The number of adult parrot in Mayer Aquarium equals the sum of the number of adult parrot in Jefferson Circus, the number of adult eagle in South Zoo, and the total number of adult animals in Hamilton Farm.

Question: How many adult eagle does Bundle Ranch have?

Solution: The question is difficult, so we use equations to solve it. Define adult eagle in Bundle Ranch as x; we don't know its value yet but will find it out later. Define adult eagle in Hamilton Farm as W; K = x = x; so W = 1 + K = 1 + x = x + 1. Define adult eagle in South Zoo as o; so o = W = x + 1. Define adult parrot in South Zoo as m; D = o + W = x + 1 + x + 1 = 2*x + 2; so m = 3 * D = (3) * (2*x + 2) = 6*x + 6. Define adult blue jay in Mayer Aquarium as T; e = m = 6*x + 6; so T = 4 + e = 4 + 6*x + 6 = 6*x + 10. Define adult parrot in Jefferson Circus as Y; j = T - x = (6*x + 10) - (x) = 5*x + 10; so Y = 1 + j = 1 + 5*x + 10 = 5*x + 11. Define adult crow in Jefferson Circus as C; so C = Y = 5*x + 11. Define total number of adult animals in Jefferson Circus as r; so r = Y + C = 5*x + 11 + 5*x + 11 = 10*x + 22. We know r = 32, so we have 10*x + 22 = 32 and we solve it through the following steps: Simplifying the equation: 10*x + 22 = 32. Move all terms to one side: 10*x - 10 = 0. Isolate the term with x: 10*x = 10. Divide both sides by 10: x = 10 / 10. Solution: x = 1. Answer: 1.


The number of adult bear in Bundle Ranch equals the difference between the total number of adult animals in Mayer Aquarium and the number of adult wolf in Jefferson Circus. The number of adult deer in Hamilton Farm equals 1. The number of adult deer in Jefferson Circus equals the total number of adult animals in Hamilton Farm. The number of adult racoon in Mayer Aquarium equals the number of adult deer in Jefferson Circus. The number of adult wolf in Jefferson Circus equals the number of adult deer in Hamilton Farm. The number of adult bear in Mayer Aquarium equals the sum of the total number of adult animals in Jefferson Circus, the number of adult wolf in Jefferson Circus, and the total number of adult animals in Hamilton Farm.

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

Solution: Define adult deer in Hamilton Farm as e; so e = 1. Define adult wolf in Jefferson Circus as O; so O = e = 1. Define total number of adult animals in Hamilton Farm as C; so C = e = 1. Define adult deer in Jefferson Circus as l; so l = C = 1. Define total number of adult animals in Jefferson Circus as U; so U = l + O = 1 + 1 = 2. Define adult bear in Mayer Aquarium as k; Q = U + O = 2 + 1 = 3; so k = Q + C = 3 + 1 = 4. Define adult racoon in Mayer Aquarium as u; so u = l = 1. Define total number of adult animals in Mayer Aquarium as W; so W = k + u = 4 + 1 = 5. Define adult bear in Bundle Ranch as D; so D = W - O = 5 - 1 = 4. Define total number of adult animals in Bundle Ranch as b; so b = D = 4. Answer: 4.