y = x + 4 Which table gives three values of x and their corresponding values of y for the...
GMAT Algebra : (Alg) Questions
\(\mathrm{y = x + 4}\)
Which table gives three values of x and their corresponding values of y for the given equation?
| x | y |
|---|---|
| 0 | 4 |
| 1 | 5 |
| 2 | 6 |
| x | y |
|---|---|
| 0 | 6 |
| 1 | 5 |
| 2 | 4 |
| x | y |
|---|---|
| 0 | 2 |
| 1 | 1 |
| 2 | 0 |
| x | y |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
1. TRANSLATE the problem requirements
- Given information:
- Equation: \(\mathrm{y = x + 4}\)
- Need to find which table correctly shows x-values and their corresponding y-values
- What this tells us: We need to substitute each x-value from the tables into the equation and check if we get the corresponding y-value shown.
2. SIMPLIFY by testing each table systematically
- Start with Table A and substitute each x-value:
- When \(\mathrm{x = 0}\): \(\mathrm{y = 0 + 4 = 4}\) (matches table)
- When \(\mathrm{x = 1}\): \(\mathrm{y = 1 + 4 = 5}\) (matches table)
- When \(\mathrm{x = 2}\): \(\mathrm{y = 2 + 4 = 6}\) (matches table)
- Since all values match, Table A is correct, but let's verify others are wrong:
3. SIMPLIFY verification for remaining tables
- Table B: When \(\mathrm{x = 0}\), y should be 4, but table shows 6 ✗
- Table C: When \(\mathrm{x = 0}\), y should be 4, but table shows 2 ✗
- Table D: When \(\mathrm{x = 0}\), y should be 4, but table shows 0 ✗
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make basic arithmetic errors when substituting values into \(\mathrm{y = x + 4}\).
For example, they might calculate \(\mathrm{y = 0 + 4 = 6}\) instead of 4, or get confused with the order of operations. These calculation mistakes cause them to incorrectly eliminate the right answer or incorrectly validate a wrong table.
This may lead them to select Choice B, C, or D depending on which arithmetic error they make.
The Bottom Line:
This problem tests whether students can systematically apply substitution to verify equation-table relationships. The key is careful arithmetic - the conceptual part is straightforward, but execution errors in basic addition can derail the entire solution.
| x | y |
|---|---|
| 0 | 4 |
| 1 | 5 |
| 2 | 6 |
| x | y |
|---|---|
| 0 | 6 |
| 1 | 5 |
| 2 | 4 |
| x | y |
|---|---|
| 0 | 2 |
| 1 | 1 |
| 2 | 0 |
| x | y |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |