1. Problem statement: We have three scales with markings 0 to 100 and points a, b, c which are natural numbers representing divisions on the scales. - In Şekil I, two identical objects weigh between points a and b. - In Şekil II, three identical objects weigh between points a and c. - In Şekil III, the weight of object y is shown at position $c + b - a$. We need to find which of the given options (40, 45, 56, 78, 82) can be the weight of y. 2. Understanding the problem: - The scale in Şekil I is divided into equal parts between 0 and 100, with points a and b marking the weights of two identical objects. - Similarly, Şekil II is divided with points a and c marking weights of three identical objects. 3. Let the weight of one identical object be $w$. - From Şekil I, two objects weigh $b - a$ units on the scale. - From Şekil II, three objects weigh $c - a$ units on the scale. 4. Since the scale is linear and the divisions are equal, the weight of one object is: $$w = \frac{b - a}{2} = \frac{c - a}{3}$$ 5. Equate the two expressions for $w$: $$\frac{b - a}{2} = \frac{c - a}{3}$$ Cross-multiplied: $$3(b - a) = 2(c - a)$$ 6. Expand: $$3b - 3a = 2c - 2a$$ Rearranged: $$3b - 3a + 2a = 2c$$ $$3b - a = 2c$$ 7. Solve for $c$: $$c = \frac{3b - a}{2}$$ 8. The weight of object y in Şekil III is at position $c + b - a$. Substitute $c$: $$y = c + b - a = \frac{3b - a}{2} + b - a = \frac{3b - a + 2b - 2a}{2} = \frac{5b - 3a}{2}$$ 9. Since $a$, $b$, and $c$ are natural numbers and the scale max is 100, $y$ must be one of the options given. 10. Test each option $y$ by expressing $y = \frac{5b - 3a}{2}$ and checking if $a$ and $b$ can be natural numbers satisfying $0 < a < b < 100$ and $c$ natural. Try option A) $y=40$: $$40 = \frac{5b - 3a}{2} \Rightarrow 80 = 5b - 3a$$ Try to find integer solutions for $a$, $b$: Rearranged: $$5b = 80 + 3a$$ Try $a=5$: $$5b = 80 + 15 = 95 \Rightarrow b = 19$$ Check $c$: $$c = \frac{3b - a}{2} = \frac{3 \times 19 - 5}{2} = \frac{57 - 5}{2} = 26$$ All are natural numbers and less than 100. Check $w$: $$w = \frac{b - a}{2} = \frac{19 - 5}{2} = 7$$ $$w = \frac{c - a}{3} = \frac{26 - 5}{3} = 7$$ Consistent. Therefore, $y=40$ is possible. Try other options similarly, but since the problem asks which one can be the weight, and we found $40$ works, the answer is A) 40. --- Final answer: **40**