Higher June 2017 Paper 2 Q15
15 Sami is trying to work out the exact value of \(y\) using Pythagoras’ theorem.
Not drawn accurately
Here is her working.
\[\begin{aligned} (2y)^2 &= 6^2 + 8^2 \\ 2y^2 &= 36 + 64 \\ 2y^2 &= 100 \\ y^2 &= 100 \div 2 \\ y^2 &= 50 \\ y &= \sqrt{50} \end{aligned}\](a) What error has she made in her working? [1 mark]
(b) Kai works out that \(y = 5\)
Mel says,
“\(y\) cannot be 5 because the hypotenuse should be the longest side and the other sides are longer than 5 cm”
Is Mel correct?
Tick a box.
- Yes
- No
Give a reason for your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Identifies error in working | B1 | eg \(2y^2\) should be \(4y^2\) 2 should be 4 2 should be squared Should have worked out \((2y)^2\) but has only worked out \(y^2\) |
Additional guidance
| Answer may be seen next to Sami’s method below the diagram | |
| Adding brackets around \(2y\) to Sami’s working in line 2 (working lines may be blank) | B1 |
| Showing the error being corrected eg1 \(\;(2y)^2 = 100\) and \(2y = 10\) eg2 \(\;4y^2 = 36 + 64\) | B1 B1 |
| She hasn’t squared the bracket | B1 |
| Has only squared \(y\) | B1 |
| The brackets have been left out | B1 |
| \((2y)^2\) is not equal to \(2y^2\) | B1 |
| Should have square rooted 100 before dividing by 2 because the \(2y\) should not have been taken out of the bracket | B1 |
| Should have square rooted 100 before dividing by 2 (could be referring to working from line 3 to line 4) | B0 |
| Line 2 is wrong (has not identified which part of line 2 is wrong) | B0 |
| Answer should be \(y = 5\) (has not shown what the error is) | B0 |
| Ignore non-contradictory work if correct response seen |
| Answer | Mark | Comments |
|---|---|---|
| No and valid reason | B1 | eg No and the hypotenuse is 10 No and \(2y\) is 10 No and if you double \(y\) it is more than 8 |
Additional guidance
| Valid reason must be for Mel’s argument | |
| Neither box ticked with valid reason can score B1 if decision in words eg \(2y\) is 10 so Mel is wrong | B1 |
| No and she didn’t double it to 10 | B1 |
| No and she didn’t double \(y\) | B0 |
| No and she has to double 5 which makes it 10 | B1 |
| No and she has to double 5 | B0 |
| No and the hypotenuse is \(2y\) so that’s more than 8 | B1 |
| No and the hypotenuse is \(2y\) | B0 |
| No and the hypotenuse is the longest side | B0 |
| No and \(y\) is 5 | B0 |
| No and if you double \(y\) it is more than 6 and 8 | B1 |
| No and if you double \(y\) it is more than 6 | B0 |
| Yes and valid reason | B0 |