Intermediate June 2023 Unit 2 Q16
16.
(a) A number is decreased by 5% of its value.
This is done 4 times in total. Each time, the value decreases by 5%.
Circle the multiplier that you would use to find the value after the 4 decreases. [1]
This is done 4 times in total. Each time, the value decreases by 5%.
Circle the multiplier that you would use to find the value after the 4 decreases. [1]
× 0·05\(^4\) × 0·95\(^4\) × 0·20 × 1·05\(^4\) × 0·04\(^5\)
(b) A number has been decreased by 17% to give an answer of 3569.
What was the original number? [3]
What was the original number? [3]
| Scheme | Marks | Comments |
|---|---|---|
| × 0·95\(^4\) | B1 |
| Scheme | Marks | Comments |
|---|---|---|
| Sight of 0·83 OR 83% | B1 | Allow (100 –17 =) 83 |
| \(\dfrac{3569}{0{\cdot}83}\) or \(\dfrac{3569}{83}\) × 100 or equivalent | M1 | FT ‘their 1 – 0·17’ provided <1 or ‘their 100% – 17%’ provided < 100%. |
| = 4300 | A1 | Award B1M1A1 for an embedded answer (e.g. 0·83 × 4300 = 3569 or \(\dfrac{3569}{4300}\) × 100 = 83), BUT only B1M1A0 if contradicted by stating original amount ≠ 4300. Unsupported 4300 is awarded B1M1A1. |