P4 HCF and LCM: Paper 2 Intermediate
This paper drills short division for two and three numbers, then uses the rule that the product equals H.C.F. times L.C.M. to recover an unknown value, and settles the bounds on an H.C.F.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: P4 HCF and LCM |
| Level | P4 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 12. | Dividing both by 12 leaves 3 and 4, which share no factor other than 1, so the H.C.F. is 12. |
| 2 | 15. | All three divide by 15, and 3, 4 and 5 share no common factor. |
| 3 | 36. | 12 times 18 divided by the H.C.F. 6 gives 36. |
| 4 | 72. | The L.C.M. of 8 and 12 is 24, and pairing that with 18 gives 72. |
| 5 | 24. | Short division is left with 2 and 3, and the factors taken out multiply to 24. |
| 6 | 60. | The L.C.M. of 10 and 15 is 30, and 30 with 20 gives 60. |
| 7 | 18. | The product equals H.C.F. times L.C.M.: 216 divided by 12 is 18. |
| 8 | 14. | Dividing both by 14 leaves 3 and 4, so the H.C.F. is 14. |
| 9 | 72. | The H.C.F. is 12, and 24 times 36 divided by 12 is 72. |
| 10 | Not always: H.C.F.(9, 9) is 9, equal to the number itself; the H.C.F. never exceeds the smaller number. | The H.C.F. is strictly smaller than both numbers only when the two numbers differ. |
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