Out of a group of three numbers, the second number is three times the first and also twice the third. If the sum of the three numbers is 55, what is the largest number in the group?

Introduction / Context:
This algebraic word problem describes relationships among three numbers and their total sum. The second number is related both to the first and to the third, and we must use these relationships to form equations. The goal is to find which number is the largest and its value.

Given Data / Assumptions:
Let the three numbers be x (first), y (second) and z (third).The second number y is three times the first: y = 3x.The second number y is also twice the third: y = 2z.The sum x + y + z equals 55.

Concept / Approach:
We express all three numbers in terms of a single variable. From y = 3x and y = 2z, we can write y and z in terms of x, then substitute into the sum equation. This produces a single equation in x, which we solve. After that, we compute y and z and identify the largest number among them.

Step-by-Step Solution:
Step 1: From y = 3x, we have y expressed in terms of x.Step 2: From y = 2z, solve for z in terms of x: y = 2z implies 3x = 2z, so z = (3x) / 2.Step 3: Write the sum equation: x + y + z = 55.Step 4: Substitute y = 3x and z = 3x / 2 into the sum: x + 3x + 3x / 2 = 55.Step 5: Combine like terms: x + 3x = 4x, so we have 4x + 3x / 2 = 55.Step 6: Express 4x as 8x / 2 to get a common denominator: 8x / 2 + 3x / 2 = 55.Step 7: Add the fractions: (11x / 2) = 55.Step 8: Multiply both sides by 2: 11x = 110, so x = 10.Step 9: Then y = 3x = 3 × 10 = 30 and z = 3x / 2 = 30 / 2 = 15.

Verification / Alternative check:
Check the relationships: y is three times x (30 = 3 × 10) and y is twice z (30 = 2 × 15). The sum x + y + z = 10 + 30 + 15 = 55 matches the given total. These confirm that the numbers 10, 30 and 15 satisfy all conditions and that 30 is indeed the largest.

Why Other Options Are Wrong:
Options such as 26, 29 or 32 do not satisfy the equations when treated as the largest number y. For instance, if y were 26, then x would be about 26 / 3 and would not combine with a consistent z to give a sum of 55 while preserving the relationships. Only y = 30 satisfies all conditions.

Common Pitfalls:
Students may confuse which number is twice the other or incorrectly set up the relationships (for example, writing x = 3y instead of y = 3x). Fraction handling in the step z = 3x / 2 also sometimes causes errors. Always clearly assign variables and recheck that each equation reflects the wording of the problem.

Final Answer:
The largest number in the group is 30.