Out of six objects, A, B, C, D, E, and F, D is twice as heavy as B. C's weight is half that of E. B and E together are one and a half times as heavy as A. A is one and a half times as heavy as D. F is one and a half times as heavy as C. Who is the heaviest among them?
E
Understanding the Weight Comparison Problem
We are given six objects: A, B, C, D, E, and F. We have several clues relating their weights. Our goal is to find the heaviest among these six objects. Let's denote the weight of each object by its letter (e.g., weight of A is A).
Setting Up Equations from Weight Relationships
We can write down the given relationships as mathematical equations:
- D is twice as heavy as B: \( D = 2B \)
- C's weight is half that of E: \( C = \frac{1}{2}E \)
- B and E together are one and a half times as heavy as A: \( B + E = 1.5A \)
- A is one and a half times as heavy as D: \( A = 1.5D \)
- F is one and a half times as heavy as C: \( F = 1.5C \)
Solving for Relative Weights
To compare the weights and find the heaviest object, it's helpful to express all weights in terms of a single object's weight, say A.
From \( A = 1.5D \), we can write \( A = \frac{3}{2}D \). This gives us \( D = \frac{2}{3}A \).
From \( D = 2B \) and \( D = \frac{2}{3}A \), we have \( 2B = \frac{2}{3}A \), which simplifies to \( B = \frac{1}{3}A \).
Now we use the equation \( B + E = 1.5A \). Substitute the value of B: \( \frac{1}{3}A + E = 1.5A \) \( E = 1.5A - \frac{1}{3}A \) \( E = \frac{3}{2}A - \frac{1}{3}A \) To subtract, we find a common denominator (6): \( E = \frac{9}{6}A - \frac{2}{6}A \) \( E = \frac{7}{6}A \)
Next, use \( C = \frac{1}{2}E \). Substitute the value of E: \( C = \frac{1}{2} \times \left(\frac{7}{6}A\right) \) \( C = \frac{7}{12}A \)
Finally, use \( F = 1.5C \). Substitute the value of C: \( F = 1.5 \times \left(\frac{7}{12}A\right) \) \( F = \frac{3}{2} \times \left(\frac{7}{12}A\right) \) \( F = \frac{21}{24}A \) \( F = \frac{7}{8}A \)
Comparing the Weights of Objects A, B, C, D, E, F
Now we have the weight of each object expressed in terms of the weight of A:
| Object | Weight (in terms of A) |
|---|---|
| A | \( 1A \) |
| B | \( \frac{1}{3}A \) |
| C | \( \frac{7}{12}A \) |
| D | \( \frac{2}{3}A \) |
| E | \( \frac{7}{6}A \) |
| F | \( \frac{7}{8}A \) |
To find the heaviest object, we need to compare the coefficients of A: \( 1, \frac{1}{3}, \frac{7}{12}, \frac{2}{3}, \frac{7}{6}, \frac{7}{8} \).
Let's convert these fractions to have a common denominator, say 24:
- A: \( 1 = \frac{24}{24} \)
- B: \( \frac{1}{3} = \frac{8}{24} \)
- C: \( \frac{7}{12} = \frac{14}{24} \)
- D: \( \frac{2}{3} = \frac{16}{24} \)
- E: \( \frac{7}{6} = \frac{28}{24} \)
- F: \( \frac{7}{8} = \frac{21}{24} \)
Comparing the numerators (24, 8, 14, 16, 28, 21), the largest value is 28. This corresponds to the object E.
Therefore, E is the heaviest object among A, B, C, D, E, and F.
Revision Table: Summarizing Weight Relationships
| Relationship | Equation |
|---|---|
| D is twice as heavy as B | \( D = 2B \) |
| C's weight is half that of E | \( C = \frac{1}{2}E \) |
| B and E together are one and a half times as heavy as A | \( B + E = 1.5A \) |
| A is one and a half times as heavy as D | \( A = 1.5D \) |
| F is one and a half times as heavy as C | \( F = 1.5C \) |
Additional Information: Solving Weight Puzzles
Weight comparison problems like this are common in logical reasoning and quantitative aptitude tests. The key to solving them is systematically converting the given word relationships into mathematical equations. Once you have the equations, you can use substitution to express all variables in terms of a single variable. This allows for direct comparison of their values or coefficients to determine the relative weights, and thus, find the heaviest or lightest object. Always double-check your algebraic steps to ensure accuracy. Practice with similar problems helps improve speed and accuracy in solving such weight puzzles.
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