Introduction / Context:
We are given a total sum and two comparative conditions to determine three specific fractions. The key is to translate the comparative statements into algebraic relations and solve consistently for the smallest, middle, and greatest fractions that also satisfy the stated sum.
Given Data / Assumptions:
- Total sum of three fractions = 2 11/24 = 59/24.
- (Greatest)/(Smallest) = 7/6.
- This quotient is greater than the middle fraction by 1/3, i.e., Middle = 7/6 − 1/3 = 5/6.
Concept / Approach:
Let the smallest fraction be x. Then the greatest is (7/6)x. From the second condition, the middle is 5/6. Use the sum relation x + (7/6)x + 5/6 = 59/24 and solve for x. Finally, compute the greatest and list the three in decreasing order to match the narrative (greatest, middle, smallest).
Step-by-Step Solution:
Let smallest = x, greatest = (7/6)x, middle = 5/6.Sum: x + (7/6)x + 5/6 = 59/24.Combine: (13/6)x + 5/6 = 59/24 ⇒ multiply by 24: 52x + 20 = 59.Hence 52x = 39 ⇒ x = 39/52 = 3/4.Greatest = (7/6)*(3/4) = 7/8; middle = 5/6; smallest = 3/4.
Verification / Alternative check:
Sum: 7/8 + 5/6 + 3/4 = 21/24 + 20/24 + 18/24 = 59/24. Also (7/8)/(3/4) = (7/8)*(4/3) = 7/6; and 7/6 − 1/3 = 5/6. All conditions satisfied.
Why Other Options Are Wrong:
- 3/5, 4/7, 2/3 and 7/9, 2/3, 3/5: Do not meet both the quotient and sum conditions simultaneously.
- 7/8, 7/9, 7/10: Incorrect middle and smallest; the sum also deviates from 59/24.
Common Pitfalls:
- Misinterpreting “greater than the middle by 1/3” as adding 1/3 to the middle rather than subtracting from the quotient.
- Arithmetic errors in converting 2 11/24 to 59/24.
Final Answer:
7/8, 5/6, 3/4
Discussion & Comments