Foundation June 2018 Paper 1 Q15
15 \(3a + b = 7 \qquad\) and \(\qquad 6x + 8y = 40\)
Show that \(\quad 9a + 3b \quad\) has a greater value than \(\quad 3x + 4y\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(3 \times 7\) or 21 or \(40 \div 2\) or 20 | M1 | oe |
| 21 and 20 | A1 | |
| Alternative method 2 – works out and uses correct possible values for \(a\), \(b\), \(x\) and \(y\) | ||
| Substitute values into \(9a + 3b\) that satisfy \(3a + b = 7\) or substitute values into \(3x + 4y\) that satisfy \(6x + 8y = 40\) | M1 | eg \(a = 2\) and \(b = 1\) substituted into \(9a + 3b\) or \(x = 4\) and \(y = 2\) substituted into \(3x + 4y\) |
| 21 and 20 | A1 | Correct evaluation of their expressions with correct values for the letters |
Additional guidance
| Beware 21 or 20 coming from wrong working | |
| Accept either of 21 or 20 seen if there is also an explanation that the other value is one more or one less (as appropriate) than the calculated one | M1A1 |
| Use the scheme that awards the better mark | |
| \(a = 3\) and \(b = -2\) then \(9 \times 3 + 3 \times -2\) or \(x = 0\) and \(y = 5\) then \(3 \times 0 + 4 \times 5\) | M1 |