1. The average of $a,b,$ and $e$ is 16, so $$\frac{a+b+e}{3} = 16 \implies a+b+e = 48.$$ 2. The average of $c,d,$ and $e$ is 26, so $$\frac{c+d+e}{3} = 26 \implies c+d+e = 78.$$ 3. The average of $a,b,c,d,$ and $e$ is 20, so $$\frac{a+b+c+d+e}{5} = 20 \implies a+b+c+d+e = 100.$$ 4. Add the first two sums: $(a+b+e) + (c+d+e) = 48 + 78 = 126.$ 5. But $a+b+c+d+2e = 126,$ and from step 3, $a+b+c+d+e=100.$ 6. Subtracting these: $$ (a+b+c+d+2e) - (a+b+c+d+e) = 126 - 100 \implies e = 26.$$ 7. Substitute $e=26$ back into $a+b+e=48$ gives $a+b=48-26=22.$ 8. Substitute $e=26$ into $c+d+e=78$ gives $c+d=78-26=52.$ 9. Using $a+b+c+d+e=100$ with $a+b=22, c+d=52, e=26$ confirms $22 + 52 + 26 = 100.$ 10. From $c+d=52$, we express $d=52 - c.$ 11. Using $a+b=22$, $c+d=52$, and the equation for the sum $a+b+c+d+e=100$, all values are consistent. 12. To find $c$, we use the equality from $a+b+c+d+e=100$ with substitutions already done; thus no direct equation isolates $c$ alone. 13. Since the problem asks for $c$, note from steps above that: $$c + d = 52 \implies d = 52 - c,$$ and from the total sum and given averages, these suffice. 14. To pinpoint numerical values, we use the system: - $a + b = 22$ - $c + d = 52$ - Total sum with $e=26$ 15. Detailed values cannot be derived directly from averages, but the relationships hold; the value of $c$ is hence obtained as part of the system, typically solved if additional info were provided. Problem 2: 1. A positive number is increased by 25%, so new value: $$x + 0.25x = 1.25x.$$ 2. Let the required percentage decrease be $p\%$. Then, decreasing $1.25x$ by $p\\%$ means: $$1.25x \times (1 - \frac{p}{100}) = x.$$ 3. Divide both sides by $x$: $$1.25(1 - \frac{p}{100}) = 1.$$ 4. Simplify: $$1 - \frac{p}{100} = \frac{1}{1.25} = 0.8.$$ 5. Solve for $p$: $$\frac{p}{100} = 1 - 0.8 = 0.2 \implies p = 20.$$ So, decrease by 20%. Problem 3: 1. Points $A, B, C, D, E, F$ are evenly spaced around a circle. There are 6 points total. 2. The angles between adjacent points around the center $O$ are $$\frac{360^\circ}{6} = 60^\circ.$$ 3. To find angle $\angle ACO$, note $O$ is the center and points $A$ and $C$ lie on the circle. 4. The angle at $C$ formed by points $A$ and $O$ typically is 30°, the half of 60°, because $C$ is on the circumference. 5. So, $$\angle ACO = 30^\circ.$$ Problem 4: 1. Rectangle side lengths are positive integers with area 24. 2. Factors of 24 (pairs): $(1,24), (2,12), (3,8), (4,6).$ 3. Perimeter formula: $$P = 2 \times (length + width).$$ 4. Calculate perimeter for each pair: - $(1,24): 2(1+24) = 50$ - $(2,12): 2(2+12) = 28$ - $(3,8): 2(3+8) = 22$ - $(4,6): 2(4+6) = 20$ 5. The perimeters possible are 20, 22, 28, 50. 6. Option E is 36, which is not attainable. 7. Therefore, the perimeter cannot be 36. Problem 5: 1. The operation is defined: $$a \mathbin{\triangledown} b = a + \frac{b}{2},$$ for integers $a,b$ with $a \neq b.$ 2. Given $3 \mathbin{\triangledown} b = -4,$ so: $$3 + \frac{b}{2} = -4.$$ 3. Subtract 3 from both sides: $$\frac{b}{2} = -7.$$ 4. Multiply both sides by 2: $$b = -14.$$ Final answers: - 6: $c$ is part of the system; can be any value satisfying $c+d=52$ with $d$ integer. - 7: 20 - 8: $30^\circ$ - 9: 36 - 10: $-14$