S5 Maths: Variation: Paper 3 (Challenge)
Ten mixed items on joint and partial variation plus percentage change.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Maths: Variation |
| Level | Challenge |
| # | Answer | One-line note |
|---|---|---|
| 1 | 60 | k = 24 / 12 = 2, so z = 2 x 5 x 6 = 60. |
| 2 | 27 | k = 8 x 3 / 4 = 6, so z = 6 x 9 / 2 = 27. |
| 3 | 35 | Subtracting gives 3b = 12, so b = 4 and a = 3; y = 3 + 4 x 8 = 35. |
| 4 | $9,800 | The rate is (8200 – 5800) / 120 = $20 per item and the basic wage is $1,800, so 400 items pay 1800 + 8000 = $9,800. |
| 5 | 10 | k = 6 / 3 = 2, so z = 2 x 5 = 10. |
| 6 | 13.5 | k = 4 x 3 / 4 = 3, so y = 3 x 9 / 2 = 13.5. |
| 7 | 108 | k = 32 / 8 = 4, so V = 4 x 27 = 108. |
| 8 | 6 | The product xy is fixed at 24, and 4x = 24 gives x = 6. |
| 9 | 44% | (1.2)^2 = 1.44, so y increases by 44%. |
| 10 | It should be k = zy / x = 3, so z = 6 when x = 6 and y = 3 | For z = kx / y we get k = zy / x; swapping x and y gives the wrong constant. |
Write the full relation before substituting so numerators and denominators stay in place.Tutor Registration →
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