Answer ✔️
1. Let x be the first even number. Then the next two consecutive even numbers are x+2 and x+4. The sum of these three numbers is x + (x+2) + (x+4) = 3x + 6. We know that this sum is 1524, so we can solve for x: 3x + 6 = 1524, which gives x = 506. Therefore, the three consecutive even numbers are 506, 508, and 510.
2. Let the number be x. We know that 2x + 13 = 38, so we can solve for x: 2x = 25, which gives x = 12.5. Therefore, the number is 12.5.
3. Let the son's age be x. Then the father's age is x + 32. Ten years ago, the father was (x+32)-10 = x+22 years old, and the son was x-10 years old. We know that the father was three times as old as the son at that time, so we can set up an equation: x+22 = 3(x-10). Solving for x gives x = 18, so the son is currently 18 years old and the father is 50 years old.
4. Let the first number be x and the second number be y. We know that x + y = -42 and x - y = 52. Adding these two equations gives 2x = 10, so x = 5. Substituting this value into either equation gives y = -47. Therefore, the two numbers are 5 and -47.
5. Let the number be x. We know that 4x^2 = 1, so x^2 = 1/4. Taking the square root of both sides gives x = ±1/2. Therefore, the number is either 1/2 or -1/2.
6. Let the number be x. We know that 3/4 of the number is equal to 1/10, so we can set up an equation: (3/4)x = 1/10. Solving for x gives x = 4/15. Therefore, the number is 4/15.
7. Let the smaller integer be x. Then the larger integer is x+1. We know that x + (x+1) = 3(x+1 - x), which simplifies to 2x + 1 = 3. Solving for x gives x = 1/2, so the larger integer is x+1 = 3/2. Therefore, the larger number is 3/2.
8. Let the number of sheep be x and the number of hens be y. We know that x + y = 100 and 4x + 2y = 356 (since each sheep has four legs and each hen has two legs). Multiplying the first equation by 2 gives 2x + 2y = 200. Subtracting this equation from the second equation gives 2x = 156, so x = 78. Substituting this value into either equation gives y = 22. Therefore, the farmer has 78 sheep and 22 hens.
9. Let the first number be x and the second number be y. We know that 8x + 5y = 184 and x - y = -3. Solving for x in the second equation gives x = y - 3, which we can substitute into the first equation: 8(y-3) + 5y = 184. Solving for y gives y = 17, so x = 14. Therefore, the two numbers are 14 and 17.
10. Let the width of the field be x. Then the length of the field is x+6. We know that the perimeter is 2(x+x+6) = 2(2x+6) = 4x+12, which is equal to 628. Solving for x gives x = 154, so the width is 154m and the length is 160m.
11. The total surface area of a right circular cylinder is given by 2πr(r+h), where r is the radius of the base and h is the altitude. We know that this surface area is 84π and h = 11cm, so we can set up an equation: 2πr(r+11) = 84π. Simplifying this equation gives r^2 + 11r - 42 = 0. Factoring this equation gives (r+14)(r-3) = 0, so r = -14 (which doesn't make sense in this context) or r = 3. Therefore, the radius of the base is 3cm.
12. Let the amount of land ploughed by the third driver be x hectares. Then the amount ploughed by the second driver is x+2.4 hectares, and the amount ploughed by the first driver is (x+2.4)/2 hectares. We know that the three drivers together ploughed 12.4 hectares, so we can set up an equation: x + (x+2.4) + (x+2.4)/2 = 12.4. Solving for x gives x = 3, so the third driver ploughed 3 hectares, the second driver ploughed 5.4 hectares, and the first driver ploughed 2.7 hectares.
13. Let the son's age be x. Then the man's age is x+18. Three years ago, the sum of their ages was 52, so we can set up an equation: (x-3) + (x+18-3) = 52. Simplifying this equation gives x = 17, so the son is currently 20 years old.
14. Let the amount of work be 1 unit. Adisu can do the work in 15 days, so he can do 1/15 of the work in one day. Similarly, Saron can do the work in 10 days, so he can do 1/10 of the work in one day. Working together, they can do 1/15 + 1/10 = 1/6 of the work in one day. Therefore, they can finish the work in 6 days.
1. Let x be the first even number. Then the next two consecutive even numbers are x+2 and x+4. The sum of these three numbers is x + (x+2) + (x+4) = 3x + 6. We know that this sum is 1524, so we can solve for x: 3x + 6 = 1524, which gives x = 506. Therefore, the three consecutive even numbers are 506, 508, and 510.
2. Let the number be x. We know that 2x + 13 = 38, so we can solve for x: 2x = 25, which gives x = 12.5. Therefore, the number is 12.5.
3. Let the son's age be x. Then the father's age is x + 32. Ten years ago, the father was (x+32)-10 = x+22 years old, and the son was x-10 years old. We know that the father was three times as old as the son at that time, so we can set up an equation: x+22 = 3(x-10). Solving for x gives x = 18, so the son is currently 18 years old and the father is 50 years old.
4. Let the first number be x and the second number be y. We know that x + y = -42 and x - y = 52. Adding these two equations gives 2x = 10, so x = 5. Substituting this value into either equation gives y = -47. Therefore, the two numbers are 5 and -47.
5. Let the number be x. We know that 4x^2 = 1, so x^2 = 1/4. Taking the square root of both sides gives x = ±1/2. Therefore, the number is either 1/2 or -1/2.
6. Let the number be x. We know that 3/4 of the number is equal to 1/10, so we can set up an equation: (3/4)x = 1/10. Solving for x gives x = 4/15. Therefore, the number is 4/15.
7. Let the smaller integer be x. Then the larger integer is x+1. We know that x + (x+1) = 3(x+1 - x), which simplifies to 2x + 1 = 3. Solving for x gives x = 1/2, so the larger integer is x+1 = 3/2. Therefore, the larger number is 3/2.
8. Let the number of sheep be x and the number of hens be y. We know that x + y = 100 and 4x + 2y = 356 (since each sheep has four legs and each hen has two legs). Multiplying the first equation by 2 gives 2x + 2y = 200. Subtracting this equation from the second equation gives 2x = 156, so x = 78. Substituting this value into either equation gives y = 22. Therefore, the farmer has 78 sheep and 22 hens.
9. Let the first number be x and the second number be y. We know that 8x + 5y = 184 and x - y = -3. Solving for x in the second equation gives x = y - 3, which we can substitute into the first equation: 8(y-3) + 5y = 184. Solving for y gives y = 17, so x = 14. Therefore, the two numbers are 14 and 17.
10. Let the width of the field be x. Then the length of the field is x+6. We know that the perimeter is 2(x+x+6) = 2(2x+6) = 4x+12, which is equal to 628. Solving for x gives x = 154, so the width is 154m and the length is 160m.
11. The total surface area of a right circular cylinder is given by 2πr(r+h), where r is the radius of the base and h is the altitude. We know that this surface area is 84π and h = 11cm, so we can set up an equation: 2πr(r+11) = 84π. Simplifying this equation gives r^2 + 11r - 42 = 0. Factoring this equation gives (r+14)(r-3) = 0, so r = -14 (which doesn't make sense in this context) or r = 3. Therefore, the radius of the base is 3cm.
12. Let the amount of land ploughed by the third driver be x hectares. Then the amount ploughed by the second driver is x+2.4 hectares, and the amount ploughed by the first driver is (x+2.4)/2 hectares. We know that the three drivers together ploughed 12.4 hectares, so we can set up an equation: x + (x+2.4) + (x+2.4)/2 = 12.4. Solving for x gives x = 3, so the third driver ploughed 3 hectares, the second driver ploughed 5.4 hectares, and the first driver ploughed 2.7 hectares.
13. Let the son's age be x. Then the man's age is x+18. Three years ago, the sum of their ages was 52, so we can set up an equation: (x-3) + (x+18-3) = 52. Simplifying this equation gives x = 17, so the son is currently 20 years old.
14. Let the amount of work be 1 unit. Adisu can do the work in 15 days, so he can do 1/15 of the work in one day. Similarly, Saron can do the work in 10 days, so he can do 1/10 of the work in one day. Working together, they can do 1/15 + 1/10 = 1/6 of the work in one day. Therefore, they can finish the work in 6 days.