Higher June 2024 Paper 5 Q7
7 A bottle contains \(1\frac{3}{4}\) litres of cordial.
To make orange squash, 1 part of this cordial is mixed with 7 parts of water.
Cups that can hold \(\frac{1}{6}\) of a litre are completely filled with this orange squash.
Work out the maximum number of cups that can be filled from the bottle of cordial.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 84 with correct working | 6 | B1 for \(\frac{1}{8}\) or 8 soi | Correct working requires evidence of at least B1M3 (could be done in stages) or other alternate correct approach leading to 84 accept use of equivalent decimals throughout |
| M4 for \(\frac{7}{4} \times \frac{6}{1} \times \frac{8}{1}\) oe or better | e.g. M4 for 1750 ÷ 1000 × 6 × 8 | ||
| or M3 for \(\frac{7}{4} \times \frac{6}{1}\) oe isw or better \(\frac{7}{4} \times \frac{8}{1}\) oe isw or better or 1750 × 6 × 8 oe or better | M3 implied by \(\frac{42}{4}\) oe or 10.5 nfww or \(\frac{56}{4}\) oe or 14 nfww oe e.g. M3 for 1750 ÷ 1000 × 6 [or × 8] oe If \(\frac{1}{7}\) or 7 used as ratio then max mark is M3 for \(\frac{7}{4} \times \frac{6}{1}\) oe isw (leads to answer 73.5) | ||
| or M2 for \(\frac{7}{4} \div \frac{1}{6}\) oe or \(\frac{7}{4} \div \frac{1}{8}\) oe | or for equivalent improper fraction to \(\frac{7}{4}\) M2 oe for both decimal values correct e.g. 1.75 ÷ 0.167 or 1.75 ÷ 0.125, For M2, allow error in decimal e.g. 0.160 for 0.167 if 1 ÷ 6 method shown | ||
| or for \(\frac{1}{6} \times \frac{1}{8}\) oe or better or 1750 × 6 oe or 1750 × 8 oe or better | Accept 8 × 6 | ||
| or M1 for \(1\frac{3}{4} \div \frac{1}{6}\) oe or \(1\frac{3}{4} \div \frac{1}{8}\) oe | Allow M1 for \(1.75 \div \frac{1}{6}\) or \(1.75 \div \frac{1}{8}\) | ||
| If 0 or 1 scored, instead award SC2 for answer 84 If 0 scored SC1 for \(\frac{7}{4} \times 7\) oe seen | Implied by \(\frac{49}{4}\) oe seen , 12.25 seen See AG | ||
Additional guidance: Question 7
| Example | Response | Comment |
|---|---|---|
| Example A | 1.75 × 8 = 14 14 × 6 = 84 | Concise complete method with decimals award 6 marks |
| Example B | 1 : 7 \(7 \times 1\frac{3}{4} = 12\frac{1}{4}\), \(12\frac{1}{4} + 1\frac{3}{4} = 15\), \(15 \div 0.6 = 25\) | B1 earned when they add \(12\frac{1}{4}\) and \(1\frac{3}{4}\) as ×8 implied and \(\frac{7}{4} \times 8\) oe earns M3 There is an arithmetic error but does not affect the method Division by 0.6 is incorrect, should be \(\frac{1}{6}\) Award B1M3 |
| Example C | \(1.75 \div 0.125 = 10 + 6 = 16\) \(16 \times 6 = 96\) | 0.125 implies B1 \(1.75 \div 0.125 \times 6\) is equivalent to M4. The only error is 16 should be 14 which is arithmetic. Method is fine Award B1M4 |
| Example D | \(7 \times 1\frac{3}{4} = 12\frac{1}{4}\) \(12\frac{1}{4} \times 6 = 73.5\) 73 cups | B0 as 7 used not 8 and is not recovered \(1\frac{3}{4} \times 6\) is embedded within lines 1 and 2 and scores M3. Award B0M3 |
| Example E | 1750 [ml] × 8 = 14000 14000 × 6 = 84000 84000 cups | B1 for 8 in line 1 Works in ml and earns M3 for 1750 × 8 × 6 there is no divide by 1000 Award B1M3 |
| Example F | 1750 ml × 8 = 14000 1000 ÷ 6 = 160 14000 ÷ 160 = 87.5 = 87 cups | B1 for 8 in line 1 Works in ml and earns M4 for a correct method 1750 ÷ 1000 × 8 × 6 oe, the only error is using 160 but this comes from 1000 ÷ 6 which is correct Award B1M4 |
| Example G | \(\frac{7}{4} \div \frac{8}{1}\), \(\frac{7}{4} \times \frac{1}{8} = \frac{7}{32}\) \(\frac{7}{32} \times \frac{1}{6} = \frac{7}{192}\), \(\frac{7}{4} \times \frac{192}{7} = \frac{192}{4}\) \(192 \div 4 = 48\) | B1 for 8 in line 1 There is an error in dividing \(\frac{7}{4}\) by \(\frac{8}{1}\) After this error the remaining steps imply \(\frac{1}{6} \times \frac{1}{8}\) and earn M2 Award B1M2 |
| Example H | \(\frac{1}{6} \times \frac{1}{8} = \frac{1}{48}\) [ litres cordial for 1 cup] \(\frac{7}{4}\) [litres] = 48 + 36 \(\frac{7}{4} = \frac{84}{48}\) 84 cups | Non-standard approach using ratio after \(\frac{1}{6} \times \frac{1}{8} = \frac{1}{48}\) and finding an equivalent fraction over 48 resulting in 84 cups Award 6 marks for convincing alternate approach |