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More questions on this topic: IGCSE Number: Rounding (144)
26 Karim rounds a number, \(n\), to 1 decimal place.
The result is 23.8
Complete the error interval for \(n\). (2)
\[\ldots\ldots \leqslant n \lt \ldots\ldots\]| Answer | Mark | Mark scheme |
|---|---|---|
| 23.75 and 23.85 | B1 | for 23.75 in the correct position |
| B1 | for 23.85 in the correct position |
Accept \(23.84\dot{9}\) or 23.8499(9…)
20
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.265(0780261) | M1 | for 12.57 or 47.42 or 0.2 or 0.26 or 0.27 or 0.3 or \(\dfrac{1257}{4742}\) oe |
| A1 | for 0.265(0780261) |
Answer must be given to at least 3 decimal places rounded or truncated
Check first 3 decimal places only
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.27 | B1 | for 0.27 or ft from part (a) provided answer to (a) has at least 3 sf |
10
| Answer | Mark | Mark scheme |
|---|---|---|
| 24 000 | B1 |
| Answer | Mark | Mark scheme |
|---|---|---|
| 600 | M1 | for 6 or 100 or 98 used in a calculation |
| A1 | for 600 (from \(6 \times 100\)) or 588 (from \(6 \times 98\)) or 590 (from \(5.9 \times 100\)) or 588.6 (from \(6 \times 98.1\)) |
578.2 from \(5.9 \times 98\) scores M1A0
M0A0 for working with unrounded figures with answer rounded to 600
ISW rounding of an estimated answer e.g. rounding 588 to 600
Subtraction of additional products after a rounded calculation may score the M mark for 6 or 100 or 98, but will not score the A mark unless one of the acceptable answers is reached.
5 Karim rounds a number, \(n\), to 1 decimal place.
The result is 23.8
Complete the error interval for \(n\). (2)
\[\ldots\ldots \leqslant n \lt \ldots\ldots\]| Answer | Mark | Mark scheme |
|---|---|---|
| 23.75 and 23.85 | B1 | for 23.75 in the correct position |
| B1 | for 23.85 in the correct position |
Accept \(23.84\dot{9}\) or 23.8499(9…)
22 A car travelled 4.96 miles at an average speed of 30.4 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated time | P1 | for rounding of distance = 5 (miles) or speed = 30 (mph) |
| P1 | (dep) for using time = distance/speed eg \(5 \div 30\) or for a complete process, eg \(30 \div 60\ (= 0.5)\) and \(5 \div \text{``}0.5\text{''}\) or \(30 \div 5\ (= 6)\) and \(60 \div \text{``}6\text{''}\) or \(4.96 \times \dfrac{60}{30}\) | |
| A1 | for a correct answer following through their correct rounded distance and/or speed |
| Answer | Mark | Mark scheme |
|---|---|---|
| Overestimate with reason | C1 | ft from (a) for decision with correct reasoning, eg overestimate as dividing a larger number by a smaller number or overestimate as miles rounded up and speed rounded down |
Ft the rounding and process from (a)
Must relate to estimation and not rounding of their final answer and they must have a final answer to part (a)
5 A car travelled 4.96 miles at an average speed of 30.4 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated time | P1 | for rounding of distance = 5 (miles) or speed = 30 (mph) |
| P1 | (dep) for using time = distance/speed eg \(5 \div 30\) or for a complete process, eg \(30 \div 60\ (= 0.5)\) and \(5 \div \text{``}0.5\text{''}\) or \(30 \div 5\ (= 6)\) and \(60 \div \text{``}6\text{''}\) or \(4.96 \times \dfrac{60}{30}\) | |
| A1 | for a correct answer following through their correct rounded distance and/or speed |
| Answer | Mark | Mark scheme |
|---|---|---|
| Overestimate with reason | C1 | ft from (a) for decision with correct reasoning, eg overestimate as dividing a larger number by a smaller number or overestimate as miles rounded up and speed rounded down |
Ft the rounding and process from (a)
Must relate to estimation and not rounding of their final answer and they must have a final answer to part (a)
24 Sophie drives a distance of 513 kilometres on a motorway in France.
She pays 0.81 euros for every 10 kilometres she drives.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated value | P1 | for using a value rounded to 1sf in a calculation eg \(500 \div 10\) or \(500 \times 0.8\) or \(510 \times 0.8\) or \(513 \times 0.8\) or \(500 \times 0.81\) |
| P1 | for a full process to find the total amount eg \(500 \div 10 \times 0.8\ (= 40)\) or \(510 \div 10 \times 0.8\ (= 40.8)\) or \(500 \div 10 \times 0.81\ (= 40.5)\) or [distance] \(\div\, 10 \times\) [amount] oe | |
| A1 | for a correct answer following through their correct rounded value(s) |
Their rounded value must be used in a calculation
Rounding may occur after a correct process, eg
\(513 \div 10 = 51.3 \approx 50\) and \(50 \times 0.81\)
\(513 \div 10 = 51.3 \approx 51\) and \(51 \times 0.8\)
scores P1P1
Accept 0.81 rounded to 0.80 for this mark
Condone 0.81 rounded to 1 for this mark.
Where [distance] is their rounded 513 or 513 and [amount] is their rounded 0.81 or 0.81
Accept \(513 \div 10 \times 0.81\) for this mark.
Do not award this mark if 0.81 is rounded to 1
| Answer | Mark | Mark scheme |
|---|---|---|
| underestimate with reason | C1 | ft from (a) eg underestimate as numbers rounded down |
Must relate to estimation and not rounding of their final answer and they must have a final answer to part (a)
19
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.517(0189759) | M1 | for any correct partial calculation, eg 40.113 or 6.333(482454) or 12.25 or answer of 0.51 or 0.52 or digits 517… |
| A1 | for 0.517(…) |
Answer must be given to at least 3 decimal places rounded or truncated.
Check first 3 significant figures only.
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.52 | B1 | for 0.52 or ft their answer to part (a) correctly rounded to 2 sf, provided part (a) has at least 3 sf |
Do not accept trailing 0, eg 0.520
9 A number \(N\) is rounded to 2 significant figures.
The result is 7.3
Leila says,
“The value of \(N\) cannot be greater than 7.349 because 7.350 would round up to 7.4”
| Answer | Mark | Mark scheme |
|---|---|---|
| 7.25 | B1 | oe |
| Answer | Mark | Mark scheme |
|---|---|---|
| No with supporting reason | C1 | No with supporting reason Acceptable reasons: No, the 9 should be recurring No, 7.3491 is greater No, there are numbers between 7.349 and 7.350 Not acceptable reasons: Yes with any reason given Leila is correct … 7.350 would round down She has rounded to 3sf and should only round to 2sf 7.349 rounds to 7.4 |
5 Sophie drives a distance of 513 kilometres on a motorway in France.
She pays 0.81 euros for every 10 kilometres she drives.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated value | P1 | for using a value rounded to 1sf in a calculation eg \(500 \div 10\) or \(500 \times 0.8\) or \(510 \times 0.8\) or \(513 \times 0.8\) or \(500 \times 0.81\) |
| P1 | for a full process to find the total amount eg \(500 \div 10 \times 0.8\) oe (= 40) or \(510 \div 10 \times 0.8\) oe (= 40.8) or \(500 \div 10 \times 0.81\ (= 40.5)\) or \([\text{distance}] \div 10 \times [\text{amount}]\) oe | |
| A1 | for a correct answer following through their correct rounded value(s) |
Their rounded value must be used in a calculation
Rounding may occur after a correct process, eg
\(513 \div 10 = 51.3 \approx 50\) and \(50 \times 0.81\)
\(513 \div 10 = 51.3 \approx 51\) and \(51 \times 0.8\)
scores P1P1
Accept 0.81 rounded to 0.80 for this mark
Condone 0.81 rounded to 1 for this mark
Where [distance] is their rounded 513 or 513 and [amount] is their rounded 0.81 or 0.81
Accept \(513 \div 10 \times 0.81\) for this mark
Do not award this mark if 0.81 is rounded to 1
| Answer | Mark | Mark scheme |
|---|---|---|
| underestimate with reason | C1 | ft from (a) eg underestimate as numbers rounded down |
Must relate to estimation and not rounding of their final answer and they must have a final answer to part (a)
1 Write the number 18 475 correct to the nearest thousand. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 18 000 | B1 | cao |
26 Work out an estimate for \(\quad \dfrac{5.7 \times 8.2}{0.26}\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 160 to 200 | M1 | rounds one figure appropriately (6, 8, 0.25 or 0.3) |
| M1 | (dep) for carrying out an accurate calculation using 0.25 or 0.3 eg \(6 \div 0.3 = 20\), \(8 \div 0.25 = 32\), \(6 \div 0.25 = 24\) or digits 16 | |
| A1 | Answer in the range 160 to 200 from appropriate rounding |
Do not award any marks for an accurate calculation if then rounded
18
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.882(0861678…) | B2 | 1.882(0861678…) |
| (B1 | for 16.6 or 8.82 or \(\dfrac{830}{441}\) or 1.88) |
Condone \(1.\dot{8}82(0861678\ldots)\) for both marks
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.88 | B1 | for 1.88 or ft their answer to part (a) correctly rounded to 2 dp, providing part (a) has at least 3 dp |
Condone \(1.\dot{8}8\)
Do not accept trailing 0, eg 1.880
2 Write 8061 correct to the nearest hundred. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8100 | B1 | cao |
26 A number, \(d\), is rounded to 1 decimal place.
The result is 12.7
Complete the error interval for \(d\). (2)
\(\ldots\ldots\ldots \leqslant d \lt \ldots\ldots\ldots\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 12.65, 12.75 | B1 | for 12.65 in correct position |
| B1 | for 12.75 in correct position |
Accept \(12.74\dot{9}\) or 12.7499(…)
11 A total of 700 tickets were on sale for a football match.
452 of the tickets were sold.
For a different football match,
297 tickets were sold for £9.50 each.
399 tickets were sold for £19.50 each.
| Answer | Mark | Mark scheme |
|---|---|---|
| 248 | P1 | for \(700 - 452\) |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 11000 | P1 | for evidence of rounding values to 1 significant figure, eg 300 or 400 or 10 or 9 or 20 |
| P1 | (dep on P1) for beginning a process to work with ticket sales, eg. \(300 \times 10\ (= 3000)\) or \(290 \times 10\ (= 2900)\) or \(297 \times 10\ (= 2970)\) or \(300 \times 9\ (= 2700)\) or \(300 \times 9.5\ (= 2850)\) or \(290 \times 9\ (= 2610)\) or \(297 \times 9\ (= 2673)\) OR \(400 \times 20\ (= 8000)\) or \(390 \times 20\ (= 7800)\) or \(399 \times 20\ (= 7980)\) or \(400 \times 19.5\ (= 7800)\) or \(400 \times 19\ (= 7600)\) | |
| A1 | for using correct values giving an answer in the range 10 200 to 11 000 from calculations using their rounded values |
Note: not \(290 \times 9.5\ (= 2755)\)
or \(297 \times 9.5\ (= 2821.5)\)
Note: not \(390 \times 19\ (= 7410)\)
or \(390 \times 19.5\ (= 7605)\)
or \(399 \times 19\ (= 7581)\)
or \(399 \times 19.5\ (= 7780.5)\)
Award 0 marks for an answer in the range with no supportive working
| Answer | Mark | Mark scheme |
|---|---|---|
| Overestimate with reason | C1 | (dep on P2 in (b)) for overestimate and reason, eg (ft from (b)) true total amount of money paid will be less as all values were rounded up |
Must relate to estimation and not to rounding of their final answer
and they must have a final answer to part (b)
7 A number, \(d\), is rounded to 1 decimal place.
The result is 12.7
Complete the error interval for \(d\).
\(\ldots\ldots \leqslant d \lt \ldots\ldots\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 12.65, 12.75 | B1 | for 12.65 in correct position |
| B1 | for 12.75 in correct position |
Accept \(12.74\dot{9}\) or 12.7499(…)
1 Write 6184 correct to the nearest hundred. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6200 | B1 | for 6200 (accept in words) |
17
| Answer | Mark | Mark scheme |
|---|---|---|
| 87 600 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 13.524 | M1 | for 33.81 or 2.5 or \(\dfrac{3381}{250}\) or digits 13524 |
| A1 | cao |
4 Write 19.4949 correct to the nearest whole number. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 19 | B1 | cao |
23 The length of a football pitch is 90 metres, correct to the nearest metre.
Complete the error interval for the length of the football pitch. (2)
.................... m \(\leqslant\) length \(\lt\) .................... m
| Answer | Mark | Mark scheme |
|---|---|---|
| 89.5 and 90.5 | B1 | for 89.5 in the correct position |
| B1 | for 90.5 in the correct position |
Accept \(90.4\dot{9}\) or 90.499(…)
18
| Answer | Mark | Mark scheme |
|---|---|---|
| 2500 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.09 | B1 | cao |
15
Given that \(2.96 \times 3.2 = 9.472\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 180 | M1 | rounds one figure appropriately 92 to 90 or 100 or 1.63 to 2 or 1.5 or 1.6 or 1.7 |
| A1 | for 180 \((= 90 \times 2)\) or 135 \((= 90 \times 1.5)\) or 144 \((= 90 \times 1.6)\) or 153 \((= 90 \times 1.7)\) or 200 \((= 100 \times 2)\) or 150 \((= 100 \times 1.5)\) or 160 \((= 100 \times 1.6)\) or 170 \((= 100 \times 1.7)\) or 163 \((= 100 \times 1.63)\) or 184 \((= 92 \times 2)\) or 138 \((= 92 \times 1.5)\) or 147.2 \((= 92 \times 1.6)\) or 156.4 \((= 92 \times 1.7)\) |
Answer of 149.96 \((92 \times 1.63)\) gets M0A0
Answer with no working gets M0A0
Ignore further rounding of their result
| Answer | Mark | Mark scheme |
|---|---|---|
| 947.2 | B1 | cao |
3 The length of a football pitch is 90 metres, correct to the nearest metre.
Complete the error interval for the length of the football pitch. (2)
\(\ldots\ldots\ldots\ \text{m} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{m}\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 89.5 and 90.5 | B1 | for 89.5 in the correct position |
| B1 | for 90.5 in the correct position |
Accept \(90.4\dot{9}\) or 90.499(…)
1 Write 1476 to the nearest 10 (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1480 | B1 | cao |
27 Freddie measured the length of a pencil as 7.2 cm correct to 1 decimal place.
Complete the error interval for the length, \(p\) cm, of the pencil. (2)
.................... \(\leqslant p \lt\) ....................
| Answer | Mark | Mark scheme |
|---|---|---|
| 7.15 and 7.25 | B1 | for 7.15 as the lower bound |
| B1 | for 7.25 as the upper bound |
Accept \(7.24\dot{9}\) oe or 7.2499… oe
12
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.844977205 | M1 | for 3.403(940887) or 3.717(526059) or 2.014(944168) or 1.84(...) or 1.8(...) |
| A1 | for 1.844(977205) |
Accept consistent use of a comma to indicate a decimal point
Answer must be given to at least 3 decimal places rounded or truncated
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.84 | B1 | for 1.84 or ft from (a) provided answer to (a) has at least 3 dp |
10 The diagram shows a shape made from 6 identical squares.
The total area of the shape is 5406 cm2
| Answer | Mark | Mark scheme |
|---|---|---|
| 30 | P1 | for a start to the process, eg \(5406 \div 6\ (= 901)\) or \(5400 \div 6\ (= 900)\) or \(5000 \div 6\ (= 833.33..)\) or \(5410 \div 6\ (= 901.66..)\) |
| P1 | for a process to find the length of one side, eg \(\sqrt{\text{``}901\text{''}}\) or \(\sqrt{\text{``}900\text{''}}\) or \(\sqrt{\text{``}833.33..\text{''}}\) or \(\sqrt{\text{``}901.66..\text{''}}\) | |
| A1 | for 30 |
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | for a correct explanation based on their working in (a), eg underestimate because I rounded the total area down |
Must be based on the use of a rounded value in a calculation
4 Write 327 correct to the nearest ten. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 330 | B1 | cao |
23
| Answer | Mark | Mark scheme |
|---|---|---|
| 157.668(255) | M1 | for 836.4 or 5.304(809139) or 28.141 or a truncated or rounded version of 157.668255 to no less than 3 sf |
| A1 | for 157.668(255) |
Answer must be given to at least 3 decimal places rounded or truncated
Accept a clear indication of the decimal point.
Check first 3 decimal places only
| Answer | Mark | Mark scheme |
|---|---|---|
| 157.7 | B1 | ft from part (a) provided answer to (a) has at least 5 sf |
22 The length of a pencil is 128 mm correct to the nearest millimetre.
Complete the error interval for the length of the pencil. (2)
\(\ldots\ldots\ldots\) mm \(\leqslant\) length \(\lt\) \(\ldots\ldots\ldots\) mm
| Answer | Mark | Mark scheme |
|---|---|---|
| 127.5 and 128.5 | B1 | for 127.5 in the correct position |
| B1 | for 128.5 in the correct position |
Accept \(128.4\dot{9}\) or 128.499...
12 Rehan is on holiday in the USA.
He has $200 to spend on clothes.
Rehan buys
1 pair of trainers costing $60
3 T-shirts costing $25 each.
He also wants to buy a jacket costing $80
The trainers cost $60
The exchange rate is $1 = £0.749
Rehan says,
“The trainers cost less than £40”
Rehan is wrong.
| Answer | Mark | Mark scheme |
|---|---|---|
| No from correct figures | P1 | for first step in process to solve the problem, eg find cost of 3 T-shirts, \(25 \times 3\ (= 75)\) or eg find remaining money after just one purchase, eg \(200 - 60\ (= 140)\) or \(200 - 25\ (= 175)\) |
| P1 | for process to find total cost of trainers and T-shirts, eg \(60 + \text{``}75\text{''}\ (= 135)\) or find total cost including cost of jacket, eg. \(60 + \text{``}75\text{''} + 80\ (= 215)\) or find the change after buying all 4 items, eg. \(200 - 60 - 3 \times 25\ (= 65)\) oe | |
| C1 | for No from correct figures Acceptable examples No, needs 215 No, only has 65 left No, needs 15 more Not acceptable examples Yes .... |
Award this mark for addition of 2 or more items or for subtraction of one item or more from 200 eg \(200 - 50\ (= 150)\) etc.
Figures can be given without units ($)
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | P1 | for a start to a method, eg. approximating 0.749 to 0.7, 0.74, 0.75 or 0.8 |
| C1 | for explanation Acceptable examples \(0.7 \times 60 = 42\) [is an underestimate] \(0.74 \times 60 = 44.4(0)\) [is an underestimate] Not acceptable examples \(0.75 \times 60 = 45\) [is an overestimate] \(0.8 \times 60 = 48\) [is an overestimate] |
For full marks, any calculations must be correct.
No statement in words is needed.
2 The length of a pencil is 128 mm correct to the nearest millimetre.
Complete the error interval for the length of the pencil. (2)
\(\ldots\ldots\ldots\) mm \(\leqslant\) length \(\lt \) \(\ldots\ldots\ldots\) mm
| Answer | Mark | Mark scheme |
|---|---|---|
| 127.5 and 128.5 | B1 | for 127.5 in the correct position |
| B1 | for 128.5 in the correct position |
Accept \(128.4\dot{9}\) or 128.499…
2
| Answer | Mark | Mark scheme |
|---|---|---|
| 157.668(255) | M1 | for 836.4 or 5.304(809139) or 28.141 or a truncated or rounded version of 157.668255 to no less than 3 sf |
| A1 | for 157.668(255) |
Answer must be given to at least 3 decimal places rounded or truncated
Accept a clear indication of the decimal point.
Check first 3 decimal places only
| Answer | Mark | Mark scheme |
|---|---|---|
| 157.7 | B1 | ft from part (a) provided answer to (a) has at least 5 sf |
2 Write 4.58 correct to 1 decimal place. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4.6 | B1 | cao |
2 Write the number 8375 correct to the nearest thousand. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8000 | B1 | cao |
11 The table shows the costs of sending a parcel by the Express service and by the Rapid service.
| Type of service | Cost |
|---|---|
| Express | £15.25 |
| Rapid | £35.38 |
Brendan has to send 12 parcels.
It will be cheaper to send the parcels by the Express service than by the Rapid service.
Luke wants to send 21 parcels by the Express service.
He does the calculation 20 × £15 = £300 to estimate the cost.
| Answer | Mark | Mark scheme |
|---|---|---|
| 241.56 | P1 | for difference for 1 parcel eg \(35.38 - 15.25\ (= 20.13)\) OR for total cost for 12 parcels by either service eg \(35.38 \times 12\ (= 424.56)\) or \(15.25 \times 12\ (= 183)\) |
| P1 | for a complete process eg \(\text{``}20.13\text{''} \times 12\) or \(\text{``}424.56\text{''} - \text{``}183\text{''}\) | |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | for explanation Acceptable examples both figures rounded down (refers to both figures) 20 is less than 21 and 15 is less than 15.25 Not acceptable examples both figures rounded (up); rounded down either 20 is less than 21 or 15 is less than 15.25 (refers to just one figure) the cost is 320.25 (more than 300); multiplying with bigger numbers |
8
\((2.3)^6 = 148\) correct to 3 significant figures.
| Answer | Mark | Mark scheme |
|---|---|---|
| 75 to 81 | B2 | for answer in the range 75 to 81 |
| (B1 | for 60 or 100 or 6000 or 6400 or \(\sqrt{64 \times 100}\)) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.000 148 | B1 | for 0.000 148 oe |
Can use standard form
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{1}{25}\) | B1 | for \(\dfrac{1}{25}\) or 0.04 |
7 Work out \(\sqrt{\dfrac{2.5 \times \sin 43^\circ}{8.2^2 - 50.5}}\)
Give your answer correct to 3 significant figures. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.319 | M1 | for partial method eg 1.70(499…) or 16.74 or \(\dfrac{837}{50}\) or 0.101(8516...) or 0.102 or 0.32 |
| A1 | for 0.319(1419…) |
Accept 0.319 or better. Condone incorrect digits after the 0.319; isw incorrect rounding if 0.319(1419…) is shown in working.
1 Write 478 to the nearest hundred. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 500 | B1 | cao |
24 A plane travels at a speed of 213 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| 16 to 20 | P1 | for using \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\), eg \(\dfrac{1}{200}\) or \(\dfrac{1}{213}\) or for 1 hour \(= 60 \times 60\ (= 3600)\) seconds |
| P1 | complete process, eg \(\dfrac{1}{200} \times 60 \times 60\) oe or \(\dfrac{1}{213} \times 60 \times 60\) | |
| A1 | for answer in range 16 to 20 |
Calculation could be done in stages.
| Answer | Mark | Mark scheme |
|---|---|---|
| decision with reason | C1 | (dep on correct use of \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\)) for reason related to their response to part(a), eg overestimate as speed rounded down |
20
| Answer | Mark | Mark scheme |
|---|---|---|
| 7360 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.1077981356 | B2 | for 0.1077(981…) |
| (B1 | for 5.74(45626…) or 53.29 or 0.11 or 0.107 or 0.108 ) |
Answer must be given to at least 4 decimal places rounded or truncated
Accept a clear indication of the decimal point.
Check first four decimal places only
9 Martin truncates the number \(N\) to 1 digit.
The result is 7
Write down the error interval for \(N\). (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(7 \leqslant N \lt 8\) | M1 | for identifying the key numbers 7 and 8 |
| A1 | cao |
Ignore any inequality symbols used at this stage
Accept 7.9 (recurring) for 8 as shown by 7.999 or 7.9… or recurring notation (or words)
5 A plane travels at a speed of 213 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| 16 to 20 | P1 | for using \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\), eg \(\dfrac{1}{200}\) or \(\dfrac{1}{213}\) or for 1 hour \(= 60 \times 60\ (= 3600)\) seconds |
| P1 | complete process, eg \(\dfrac{1}{200} \times 60 \times 60\) oe or \(\dfrac{1}{213} \times 60 \times 60\) oe | |
| A1 | for answer in range 16 to 20 |
Calculation could be done in stages.
| Answer | Mark | Mark scheme |
|---|---|---|
| decision with reason | C1 | (dep on correct use of \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\)) for reason related to their response to part(a), eg overestimate as speed rounded down |
4 Write 7829 to the nearest 1000 (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8000 | B1 | cao |
1
| Answer | Mark | Mark scheme |
|---|---|---|
| 7360 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.1077981356 | B2 | for 0.1077(981…) |
| (B1 | for 5.74(45626…) or 53.29 or 0.11 or 0.107 or 0.108 ) |
Answer must be given to at least 4 decimal places rounded or truncated
Accept a clear indication of the decimal point.
Check first four decimal places only
22 A cycle race across America is 3069.25 miles in length.
Juan knows his average speed for his previous races is 15.12 miles per hour.
For the next race across America he will cycle for 8 hours per day.
Juan trains for the race.
The average speed he can cycle at increases.
It is now 16.27 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated value | P1 | for using a rounded value in a correct process eg \(3000 \div 15\) or \(15 \times 8\) or \(20 \times 8\) |
| P1 | for a full process to find the number of days eg \(\text{``}3000\text{''} \div \text{``}15\text{''} \div \text{``}10\text{''}\ (= 20)\) or \(\text{``}3000\text{''} \div \text{``}15\text{''} \div 8\ (= 25)\) | |
| A1 | for a correct answer following through their rounded values |
Their rounded value must be used in a calculation
Rounding may appear after a correct process
eg \(15.12 \times 8 = 120.96 \approx 100\)
followed by eg \(3069.25 \div 100\)
Accept \(3069.25 \div 15.12 \div 8\) oe
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | eg less days required or it doesn’t affect the answer because I would still round 16.27 down to 15 (or up to 20) |
Refers to time taken
4 A cycle race across America is 3069.25 miles in length.
Juan knows his average speed for his previous races is 15.12 miles per hour.
For the next race across America he will cycle for 8 hours per day.
Juan trains for the race.
The average speed he can cycle at increases.
It is now 16.27 miles per hour.
| Answer | Mark | Mark scheme |
|---|---|---|
| Estimated value | P1 | for using a rounded value in a correct process eg \(3000 \div 15\) or \(15 \times 8\) or \(20 \times 8\) |
| P1 | for a full process to find the number of days eg \(\text{``}3000\text{''} \div \text{``}15\text{''} \div \text{``}10\text{''}\ (= 20)\) or \(\text{``}3000\text{''} \div \text{``}15\text{''} \div 8\ (= 25)\) | |
| A1 | for a correct answer following through their rounded values |
Their rounded value must be used in a calculation
Rounding may appear after a correct process
eg \(15.12 \times 8 = 120.96 \approx 100\)
followed by eg \(3069.25 \div 100\)
Accept \(3069.25 \div 15.12 \div 8\) oe
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | eg less days required or it doesn’t affect the answer because I would still round 16.27 down to 15 (or up to 20) |
Refers to time taken
3 Write the number 2538 correct to the nearest hundred. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2500 | B1 | cao |
2 Write 1.59 correct to 1 decimal place. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.6 | B1 | cao |
1 Write 6324 correct to the nearest thousand. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6000 | B1 | cao |
Accept 6 thousand or six thousand
23 Use your calculator to work out \(\sqrt{\dfrac{\sin 25^\circ + \sin 40^\circ}{\cos 25^\circ - \cos 40^\circ}}\)
| Answer | Mark | Notes |
|---|---|---|
| 2.7560… | M1 | for 1.0654(059…), 0.1402(633…), 7.5957(541…), 2.756 truncated or rounded to no less than 2dp |
| A1 | for 2.7560(….) |
| Answer | Mark | Notes |
|---|---|---|
| 2.76 | B1 | for 2.76 ft from (a) |
20 Ami and Josh use a calculator to work out \(\dfrac{595}{4.08^2 + 5.3}\)
Ami’s answer is 27.1115
Josh’s answer is 271.115
One of these answers is correct.
Use approximations to find out which answer is correct. (3)
| Answer | Mark | Notes |
|---|---|---|
| Ami with estimate | M2 | for an approximate calculation eg \(\dfrac{600}{16+5}\) or \(\dfrac{600}{21}\) or \(\dfrac{600}{20}\) or \(\dfrac{600}{20+5}\) or \(\dfrac{600}{25}\) or \(\dfrac{600}{25+5}\) or \(\dfrac{600}{30}\) or \(\dfrac{595}{20}\) |
| (M1 | for using 600 or 5 or 4) | |
| C1 | Ami’s answer /27.1115 is closest with accurately calculated figure from approximation |
16 Berenika wants to buy 35 T-shirts.
Each T-shirt costs £5.80
Berenika does the calculation \(40 \times 6 = 240\) to estimate the cost of 35 T-shirts.
There is a special offer.
| T-shirts £5.80 each. Buy 30 or more T-shirts. Get 10% off the total cost. |
| Answer | Mark | Notes |
|---|---|---|
| Explanation | C1 | eg States over-estimated for both values |
| Answer | Mark | Notes |
|---|---|---|
| 182.7(0) | P1 | for a process to find 10% of a value stated in the question eg \(\dfrac{10}{100} \times 5.80\ (= 0.58)\) or \(\dfrac{10}{100} \times 35\ (= 3.5)\) oe or \(35 \times 5.80\ (= 203)\), allow \(30 \times 5.80\ (= 174)\) or \(35 \times\) [reduced price] |
| P1 | for a process to find 90% of a value stated in the question eg \(35 - \text{``}3.5\text{''}\ (= 31.5)\) or \(0.9 \times 5.80\ (= 5.22)\) oe or \(\dfrac{10}{100} \times \text{``}203\text{''}\ (= 20.3)\) or \(\dfrac{10}{100} \times \text{``}174\text{''}\ (= 17.4)\) oe | |
| P1 | for a complete process to find actual cost of 35 eg \(0.9 \times 5.80 \times 35\) oe | |
| A1 | cao SC B2 156.6(0) |
8 Use your calculator to work out \(\sqrt{\dfrac{\sin 25^\circ + \sin 40^\circ}{\cos 25^\circ - \cos 40^\circ}}\)
| Answer | Mark | Notes |
|---|---|---|
| 2.7560… | M1 | for 1.0654(059…), 0.1402(633…), 7.5957(541…), 2.756 truncated or rounded to no less than 2dp |
| A1 | for 2.7560(….) |
| Answer | Mark | Notes |
|---|---|---|
| 2.76 | B1 | for 2.76 ft from (a) |
1 Write 3758 correct to the nearest 1000 (1)
| Answer | Mark | Notes |
|---|---|---|
| 4000 | B1 | for 4000 |
15 A cone has a volume of 98 cm3.
The radius of the cone is 5.13 cm.
John uses a calculator to work out the height of the cone to 2 decimal places.
| Answer | Mark | Notes |
|---|---|---|
| 3.5 to 4.5 | M1 | substitution into formula \(\dfrac{1}{3}\pi r^2 h\) of chosen values for \(r\) and \(V\) (accept \(r = 5.13\) and \(V = 98\)) and starts rearrangement e.g. multiplies by 3, divides by \(\pi\) or divides by \(r^2\) (both sides) |
| M1 | uses estimates in calculation e.g. \(\dfrac{3 \times 100}{3 \times 25}\) (or in rearranged formula) or \(\dfrac{12}{\pi}\) | |
| A1 | arrives at a single value from estimate in the range 3.5 to 4.5 |
| Answer | Mark | Notes |
|---|---|---|
| more | C1 | ft e.g. more since number in numerator goes up; numbers in denominator go down. |
2 Write 7.264 51 correct to 3 decimal places. (1)
| Answer | Mark | Notes |
|---|---|---|
| 7.265 | B1 | cao |
2 Write 56.78 correct to one significant figure. (1)
| Answer | Mark | Notes |
|---|---|---|
| 60 | B1 | cao |
1
| Answer | Mark | Notes |
|---|---|---|
| 6500 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 168 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| Arrow at 7.2 | 1 | B1 cao |
1
| Answer | Mark | Notes |
|---|---|---|
| 5906 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7200 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 28 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 3.67, 5.08, 5.45, 6.03 | 1 | B1 cao |
5 There are 892 litres of oil in Mr Aston’s oil tank.
He uses 18.7 litres of oil each day.
Estimate the number of days it will take him to use all the oil in the tank. (2)
| Answer | Mark | Notes |
|---|---|---|
| 44 - 56 | 2 | B2 for 44 - 56 (B1 for 1000 or 900 or 20 or 18 or 19, unless it is clear these have not come from estimation) |
1
| Answer | Mark | Notes |
|---|---|---|
| 4726 in words | 1 | B1 accept 4 thousand 7 hundred and twenty six |
| Answer | Mark | Notes |
|---|---|---|
| 5 000 000 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 3600 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7.04, 7.47, 7.58, 7.6, 7.69 ordered | 1 | B1 cao |
3
| Answer | Mark | Notes |
|---|---|---|
| 7400 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 6402 in words | 1 | B1 for eg six thousand four hundred and two |
| Answer | Mark | Notes |
|---|---|---|
| 54 000 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 13 | 1 | B1 cao |
3 Use your calculator to work out \(\dfrac{\sqrt{70.25}}{4.2 - 2.37}\)
| Answer | Mark | Notes |
|---|---|---|
| 4.58006(9567) | 2 | M1 for 1.83 or 8.381(527307) or 4.6 or 4.58 or 4.580 or 4.5801 A1 for 4.58006(9567) |
| Answer | Mark | Notes |
|---|---|---|
| 4.5801 | 1 | B1 ft provided at least 5 decimal places in (a) |
1 The table shows the heights of six mountains.
| Mountain | Height (metres) |
|---|---|
| Diran | 7266 |
| Lhotse | 8516 |
| K12 | 7428 |
| Cho Oyu | 8188 |
| Jannu | 7711 |
| Makalu | 8485 |
The mountain K3 is eight thousand and fifty one metres high.
| Answer | Mark | Notes |
|---|---|---|
| Lhotse | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 8200 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| Seven thousand four hundred and twenty-eight | 1 | B1 for any unambiguous answer |
| Answer | Mark | Notes |
|---|---|---|
| 8051 | 1 | B1 cao |
*13
| Competition a prize every 2014 seconds |
In a competition, a prize is won every 2014 seconds.
Work out an estimate for the number of prizes won in 24 hours.
You must show your working. (4)
| Answer | Mark | Notes |
|---|---|---|
| Answer in range 35 – 50 | 4 | M1 for a method to either find the exact or approximate number of seconds in one day, e.g. \(24 \times 60 \times 60\) (=86400) or the number of minutes in 2014 seconds, e.g. \(2014 \div 60\) or \(2000 \div 60\) (≈30) M1 for a correct method to find the number of prizes; eg. ‘\(24 \times 60 \times 60\)’ \(\div 2014\) oe or \(60 \div \text{``}30\text{''} \times 24\) oe B1 for rounding at least one appropriate value in the working to 1 sf, e.g. 24 rounded to 20 or 2014 rounded to 2000 or 86400 rounded to 90000 C1 (dep on M2) for answer in 35 – 50 clearly identified |
5 28569 people watch a football match.
5619 of the 28569 people are female.
| Answer | Mark | Notes |
|---|---|---|
| 28 600 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 20 000 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 22 950 | 1 | B1 cao |
22
| Answer | Mark | Notes |
|---|---|---|
| 34.81 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| (i) 35.1606.... | 2 | B1 for 35.1606(7977...) |
| (ii) 35.2 | B1 ft from (i) provided (i) has more than one decimal place |
1
| Answer | Mark | Notes |
|---|---|---|
| 18.75 | 2 | M1 for 84 or 4.48 or \(\dfrac{112}{25}\) or 18.7 or 18.8 or 19 or 20 or \(\dfrac{75}{4}\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 20 | 1 | B1 for 20 or ft from their answer to (a) provided (a) is written to 2 or more significant figures |
1
| Answer | Mark | Notes |
|---|---|---|
| 3502 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| Two thousand and nineteen | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7 tens | 1 | B1 for 7 tens or 70 accept in words |
| Answer | Mark | Notes |
|---|---|---|
| 6700 | 1 | B1 cao |
17 A ticket for a seat at a school play costs £2.95
There are 21 rows of seats.
There are 39 seats in each row.
The school will sell all the tickets.
Work out an estimate for the total money the school will get. (3)
| Answer | Mark | Notes |
|---|---|---|
| 2400 | 3 | B1 for one of 20, 40, 3 or 300 M1 for “20” × “40” × “3” or “20” × “40” × “300” (values do not need to be rounded) A1 for answer in range 2280 − 2520 SC : Award B3 for an answer of 2400 if no working seen NB. An answer of 2416.05 implies B0 M1 A1 |
8 Margaret has some goats.
The goats produce an average total of 21.7 litres of milk per day for 280 days.
Margaret sells the milk in \(\dfrac{1}{2}\) litre bottles.
Work out an estimate for the total number of bottles that Margaret will be able to fill with the milk.
You must show clearly how you got your estimate. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{20 \times 300}{0.5}\) | 12000 | 3 | B1 for 20 or 300 used M1 for \(\text{``}20\text{''} \times \text{``}300\text{''}\) or \(\dfrac{\text{``}20\text{''}}{0.5}\) or \(\dfrac{\text{``}300\text{''}}{0.5}\), values do not need to be rounded A1 for answer in the range 11200 –13200 SC B3 for 12000 with or without working |
1
| Answer | Mark | Notes |
|---|---|---|
| 2085 | 1 | B1 for 2085 or 2,085 |
| Answer | Mark | Notes |
|---|---|---|
| five thousand, one hundred and eight | 1 | B1 for five thousand, one hundred (and) eight |
| Answer | Mark | Notes |
|---|---|---|
| 9 tenths | 1 | B1 for 9 tenths oe, eg \(\dfrac{9}{10}\) or \(\dfrac{90}{100}\) |
| Answer | Mark | Notes |
|---|---|---|
| 160 | 1 | B1 cao |
17
| Answer | Mark | Notes |
|---|---|---|
| (i) 12.978(61279…) (ii) 13 | 2 | (i) B1 for 12.978(…..) (ii) B1 for 13 or ft from a(i) [Note: An answer of 13.0 gets B0] |
| Answer | Mark | Notes |
|---|---|---|
| 100000 | 1 | B1 cao |
1
| Answer | Mark | Notes |
|---|---|---|
| 7025 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| Nine thousand four hundred (and) fifty | 1 | B1 for Nine thousand four hundred (and) fifty Accept ‘nine thousand’ written as ‘9 thousand’, ‘four hundred’ written as ‘4 hundred’ and ‘fifty’ written as ‘5 tens’ or any mixture of these. Do not penalise poor spelling. |
| Answer | Mark | Notes |
|---|---|---|
| 29 | 1 | B1 for 29 or twenty nine or 29.0 |
| Answer | Mark | Notes |
|---|---|---|
| 7000 | 1 | B1 for 7000 or 7 thousand or seven thousand |
5 Work out an estimate for \(\dfrac{31 \times 9.87}{0.509}\) (3)
| Answer | Mark | Notes |
|---|---|---|
| 600 | 3 | (M2 for 300 ÷ 0.5 or 60 × 10 or 30 × 20) M1 for at least two of 30, 10 and 0.5 or sight of 300 or 60 or 20 A1 for 600 – 620 but not 601.1(198428…) OR (M2 for 310 ÷ 0.5 or 62 × 10 or 31 × 20) M1 for at least two of 31, 10 and 0.5 or sight of 310 or 62 or 20 A1 for 600 – 620 but not 601.1(198428…) |
2
| Answer | Mark | Notes |
|---|---|---|
| 507 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 40 | 1 | B1 for 40 or forty or 4 tens (do not accept an answer of “tens”) |
| Answer | Mark | Notes |
|---|---|---|
| 6000 | 1 | B1 for 6000 or six (6) thousand |
2
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{546.7}{12.5} =\) | 43.736 | 2 | B2 for 43.736 (B1 for 546.7 or \(\frac{5467}{10}\) or \(\frac{5467}{125}\) or 12.5 or \(\frac{25}{2}\) or 43.7 or 43.8 or 43.73 or 43.74 or 40 or 44) |
| Answer | Mark | Notes |
|---|---|---|
| 40 | 1 | B1 for 40 or ft from their answer to (a) provided (a) is written to 2 or more significant figures |
1
| Answer | Mark | Notes |
|---|---|---|
| 4216 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| eight thousand | 1 | B1 for eight thousand or 8000 |
| Answer | Mark | Notes |
|---|---|---|
| 3570 | 1 | B1 cao |
23
Complete the error interval. [2 marks]
\(\ldots\ldots\ldots\ \text{cm} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{cm}\)
Kim says that the total length of three of these fence panels must be less than 6.2 m
Show that Kim is correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 155 cm \(\leqslant\) length \(\lt\) 165 cm | B2 | oe B1 155 or 165 in correct position SC1 165 cm \(\leqslant\) length \(\lt\) 155 cm |
Accept \(164.\dot{9}\) for 165
Accept eg 155.0 for 155
| Answer | Mark | Comments |
|---|---|---|
| 2.05 or \(6.2 \div 3\) or \(2.0\dot{6}\) | M1 | oe eg in cm |
| \(2.05 \times 3\) and 6.15 or \(205 \times 3\) and 6.15 or \(205 \times 3\) and 615 and 620 or \(2.0\dot{6}\) and 2.05 or \(206.\dot{6}\) and 205 | A1 | accept eg \(2.05 + 2.05 + 2.05\) for \(2.05 \times 3\) |
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Accept rounding or truncating of \(2.0\dot{6}\) or \(206.\dot{6}\) to at least 4 sf Accept rounding or truncating of \(2.0\dot{6}\) to 3 sf only if \(6.2 \div 3\) is seen Accept rounding or truncating of \(206.\dot{6}\) to 3 sf only if \(620 \div 3\) is seen | |
| Accept \(2.04\dot{9}\) for 2.05 | |
| Ignore any reference to lower bounds | |
| Ignore reference to units | |
| 2.05 or 205 may be embedded eg \(2.05 + 2.06 + 2.09 = 6.2\) with no further work or explanation | M1A0 |
| \(2.06 + 2.06 + 2.08 = 6.2\) with no further work or explanation \(2.06 + 2.06 + 2.06 = 6.18\) with a clear reference to 2.05 and a full explanation | M0A0 M1A1 |
14
You must show your working. [3 marks]
Tick one box.
Give a reason for your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 2 or 4 or 7 or 49 | M1 | |
| 2 and 4 and 7 | M1 | accept 49 for 7 |
| 57 with 2 and 4 and 7 seen | A1 | accept 49 for 7 condone 60 with 2 and 4 and 7 seen |
| Ignore multiplication and/or addition of the given numbers | |
| 57 from multiplication and addition of values other than 2, 4 and 7 | M0M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| Overestimate and correct reason or Underestimate and used 1 and 3 and 6 in part (a) and correct reason | B1 | oe eg every number rounded up |
| Every number (was over .5 so) had to be rounded up | B1 |
| I rounded them up | B1 |
| (All) the numbers were closer to the higher number | B1 |
| I rounded up | B0 |
| The number was bigger than the real answer | B0 |
5
Complete the error interval. [2 marks]
\(\ldots\ldots\ldots\ \text{cm} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{cm}\)
Kim says that the total length of three of these fence panels must be less than 6.2 m
Show that Kim is correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 155 cm \(\leqslant\) length \(\lt\) 165 cm | B2 | oe B1 155 or 165 in correct position SC1 165 cm \(\leqslant\) length \(\lt\) 155 cm |
Accept \(164.\dot{9}\) for 165
Accept eg 155.0 for 155
| Answer | Mark | Comments |
|---|---|---|
| 2.05 or \(6.2 \div 3\) or \(2.0\dot{6}\) | M1 | oe eg in cm |
| \(2.05 \times 3\) and 6.15 or \(205 \times 3\) and 6.15 or \(205 \times 3\) and 615 and 620 or \(2.0\dot{6}\) and 2.05 or \(206.\dot{6}\) and 205 | A1 | accept eg \(2.05 + 2.05 + 2.05\) for \(2.05 \times 3\) |
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Accept rounding or truncating of \(2.0\dot{6}\) or \(206.\dot{6}\) to at least 4 sf Accept rounding or truncating of \(2.0\dot{6}\) to 3 sf only if \(6.2 \div 3\) is seen Accept rounding or truncating of \(206.\dot{6}\) to 3 sf only if \(620 \div 3\) is seen | |
| Accept \(2.04\dot{9}\) for 2.05 | |
| Ignore any reference to lower bounds | |
| Ignore reference to units | |
| 2.05 or 205 may be embedded eg \(2.05 + 2.06 + 2.09 = 6.2\) with no further work or explanation | M1A0 |
| \(2.06 + 2.06 + 2.08 = 6.2\) with no further work or explanation \(2.06 + 2.06 + 2.06 = 6.18\) with a clear reference to 2.05 and a full explanation | M0A0 M1A1 |
24
there are 912 students
the ratio of students to teachers is \(\quad\) 15.2 : 1
The number of students stays the same.
The number of teachers increases by 2
Work out the new ratio of students to teachers.
Give your answer in the form \(\quad n : 1\) [3 marks]
72 students want to go on the trip.
Lexi tries to work out how many teachers are needed.
| \(72 \div 10 = 7.2\) 7 teachers are needed. |
What is wrong with her answer? [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(912 \div 15.2\) or 60 | M1 | oe implied by 62 |
| \(912 \div (\text{their } 60 + 2)\) or \(912 \div 62\) or [14.7, 14.71] | M1dep | oe |
| [14.7, 14.71] : 1 or 15 : 1 with M2 awarded | A1 | oe eg \(\dfrac{456}{31} : 1\) |
M1 may be awarded for correct work, with no answer or incorrect answer, even if this is seen amongst multiple attempts
Do not allow misreads for 15.2
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation | B1 | eg she needs to round up |
| She rounded down | B1 |
| Needs 8 (teachers) | B1 |
| Need (one) more | B1 |
| Only 70 students could go | B1 |
| 2 students do not have a teacher | B1 |
| There are students without a teacher | B1 |
| Have groups of 8 with 9 teachers | B0 |
| 8 | B0 |
| Cannot have 0.2 of a teacher | B0 |
13
estimate the value of \(\quad \dfrac{\sqrt{401} + 1.9^3}{\cos 58.7^\circ}\)
You must show your working. [3 marks]
Not drawn accurately
Sam attempts to find the value of \(\cos x\) using the cosine rule.
Here is his working.
| \(5^2 = 3^2 + 7^2 + 2 \times 3 \times 7 \times \cos x\) \(5^2 = 10^2 + 42 \times \cos x\) \(25 = 142 \cos x\) Therefore \(\quad \cos x = \dfrac{25}{142}\) |
Identify two errors Sam has made. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt{400}\) or 20 or \(2^3\) or 8 or \(\cos 60^\circ\) or 0.5 | M1 | oe |
| 20 and 8 and 0.5 | M1 | |
| 56 with correct values | A1 |
| 56 seen then rounded to 60 | M1M1A1 |
| 28 rounded to 30 then \(\dfrac{30}{0.5}\) leading to 60 as final answer | M1M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Any two of It should be \(-2 \times 3 \times 7 \times \cos x\) \(3^2 + 7^2\) is not \(10^2\) \(10^2 + 42 \cos x\) is not \(142 \cos x\) | B2 | oe B1 any one error identified SC1 answer should be \(\cos x = \dfrac{33}{42}\) oe or answer should be \(\cos x = -\dfrac{33}{42}\) oe |
| All three errors identified | B2 |
| Three errors identified with one or two correct | B1 |
| The second + should be \(-\) | B1 |
| You have to rearrange (to make \(\cos x\) the subject) | B0 |
10
Give your answer to 2 decimal places. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 37.5 or \(37\dfrac{1}{2}\) | B1 | accept \(\dfrac{75}{2}\) |
| 37 | B0 |
| 37.5 followed by 37 or 38 on the answer line | B0 |
| 38 | B0 |
| 37.50 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 0.47 | B2 | B1 0.46 or 0.469 or 0.4688 or 0.46875 or 0.47 with one or more trailing 0s B1 their 3dp or more value correctly rounded to 2dp |
7
there are 912 students
the ratio of students to teachers is \(\quad\) 15.2 : 1
The number of students stays the same.
The number of teachers increases by 2
Work out the new ratio of students to teachers.
Give your answer in the form \(\quad n : 1\) [3 marks]
72 students want to go on the trip.
Lexi tries to work out how many teachers are needed.
| \(72 \div 10 = 7.2\) 7 teachers are needed. |
What is wrong with her answer? [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(912 \div 15.2\) or 60 | M1 | oe implied by 62 |
| \(912 \div (\text{their } 60 + 2)\) or \(912 \div 62\) or [14.7, 14.71] | M1dep | oe |
| [14.7, 14.71] : 1 or 15 : 1 with M2 awarded | A1 | oe eg \(\dfrac{456}{31} : 1\) |
M1 may be awarded for correct work, with no answer or incorrect answer, even if this is seen amongst multiple attempts
Do not allow misreads for 15.2
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation | B1 | eg she needs to round up |
| She rounded down | B1 |
| Needs 8 (teachers) | B1 |
| Need (one) more | B1 |
| Only 70 students could go | B1 |
| 2 students do not have a teacher | B1 |
| There are students without a teacher | B1 |
| Have groups of 8 with 9 teachers | B0 |
| 8 | B0 |
| Cannot have 0.2 of a teacher | B0 |
20 The attendance for a rugby match is 8400 people to the nearest 100
| Answer | Mark | Comments |
|---|---|---|
| 8350 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 8449 | B1 |
15
| A | B |
|---|---|
| \(1.92^7 + 6.9^3\) | \(5 \times \sqrt[3]{1\,000\,350}\) |
Use approximations to show that \(\quad\) answer to A \(\lt\) answer to B [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 14 and 15 | B1 | either order |
| Ignore incorrect calculations | |
| Answer \(14^2\) and \(15^2\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(2^7\) or 128 or \(7^3\) or 343 or \((5 \times) \sqrt[3]{1\,000\,000}\) or \((5 \times)\ 100\) or 500 | M1 | |
| At least two of 128, 343 and 500 or 471 | A1 | |
| 471 and 500 | A1 |
4 The attendance for a rugby match is 8400 people to the nearest 100
| Answer | Mark | Comments |
|---|---|---|
| 8350 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 8449 | B1 |
19 The length of a line is 8 cm to the nearest centimetre.
Complete the error interval. [2 marks]
\(\ldots\ldots\ldots\ \text{cm} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{cm}\)
| Answer | Mark | Comments |
|---|---|---|
| \(7.5\ \text{cm} \leqslant \text{length} \lt 8.5\ \text{cm}\) | B2 | B1 one length correct in correct position SC1 \(8.5\ \text{cm} \leqslant \text{length} \lt 7.5\ \text{cm}\) |
Accept \(8.4\dot{9}\) for 8.5
9 Connor is estimating the answer to \(\quad 385 + 1479\)
Work out his estimate. [2 marks]
“My estimate will be more than the exact answer.”
How does he know this without working out the exact answer? [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 390 or 1480 | M1 | |
| 1870 | A1 |
| Answer only | M1A1 |
| Ignore calculation of 1864 but 1870 only coming from incorrect rounding of 1864 is M0 |
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation referring to both of the original numbers being rounded up | B1 | eg he rounded each number up or each rounded number is bigger than the actual number or 390 is bigger than 385 and 1480 is bigger than 1479 |
| Ignore irrelevant, non-contradictory statements | |
| Ignore 1864 alongside a correct explanation | |
| Incorrect rounding or values seen in this part even alongside a correct explanation | B0 |
| The numbers are rounded up (to the nearest 10) | B1 |
| 385 became 390 and 1479 became 1480 | B1 |
| One number increased 5 and the other 1 (so it will be 6 bigger) | B1 |
| 385 became 390 and 1479 became 1500 | B0 |
| The numbers are rounded to the nearest 10 | B0 |
| It’s rounding so the answer is bigger | B0 |
| 1870 is bigger than 1864 | B0 |
24 Millie is estimating the value of \(\quad \dfrac{1}{\left(\sqrt[3]{8.34}\right)^2 \times 10.21}\)
She rounds each decimal number to 1 significant figure.
You must show your working. [2 marks]
“My estimate must be more than the exact value.”
Without working out the exact value, give a reason how she can know this. [1 mark]| Answer | Mark | Comments |
|---|---|---|
| 8 or 10 | M1 | 8 may be implied by \(2^2\) or 4 |
| 8 and 10 and \(\dfrac{1}{40}\) or 0.025 | A1 | 8 may be implied by \(2^2\) or 4 accept 0.03 with \(\dfrac{1}{40}\) or 0.025 seen |
| Do not allow exact calculations for M1A1 eg \(4.113 = 4\) and \(10.21 = 10\) and \(\dfrac{1}{40}\) | M1A0 |
| \(\dfrac{1}{40}\) or 0.025 with 8 or 10 seen (8 may be implied by \(2^2\) or 4) | M1A0 |
| \(\dfrac{1}{40}\) or 0.025 without 8 or 10 seen (8 may be implied by \(2^2\) or 4) | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation | B1 | eg both numbers have been rounded down |
| Ignore irrelevant reasons alongside a correct reason, unless contradictory | |
| Ignore a calculation using exact values alongside a correct reason eg 0.025 is greater than 0.0238… and both numbers rounded down | B1 |
| 0.025 is greater than 0.0238… | B0 |
| The denominator is smaller | B1 |
| The denominator using the exact values is bigger | B1 |
| (Decimals) rounded down | B1 |
| Because 8.34 is more than 8 and 10.21 is more than 10 | B1 |
| One is divided by less (with answer more) | B1 |
| Estimating rounds the numbers down which makes the denominator less | B1 |
| Estimating rounds the numbers down which makes it less | B0 |
| Because it rounds up | B0 |
| Because she rounded each number to one significant figure | B0 |
| The numbers get rounded up so more than the exact value | B0 |
| Rounded up when estimating | B0 |
| Removing the decimals makes the number bigger | B0 |
11 Millie is estimating the value of \(\quad \dfrac{1}{\left(\sqrt[3]{8.34}\right)^2 \times 10.21}\)
She rounds each decimal number to 1 significant figure.
You must show your working. [2 marks]
“My estimate must be more than the exact value.”
Without working out the exact value, give a reason how she can know this. [1 mark]| Answer | Mark | Comments |
|---|---|---|
| 8 or 10 | M1 | 8 may be implied by \(2^2\) or 4 |
| 8 and 10 and \(\dfrac{1}{40}\) or 0.025 | A1 | 8 may be implied by \(2^2\) or 4 accept 0.03 with \(\dfrac{1}{40}\) or 0.025 seen |
| Do not allow exact calculations for M1A1 eg \(4.113 = 4\) and \(10.21 = 10\) and \(\dfrac{1}{40}\) | M1A0 |
| \(\dfrac{1}{40}\) or 0.025 with 8 or 10 seen (8 may be implied by \(2^2\) or 4) | M1A0 |
| \(\dfrac{1}{40}\) or 0.025 without 8 or 10 seen (8 may be implied by \(2^2\) or 4) | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| Valid explanation | B1 | eg both numbers have been rounded down |
| Ignore irrelevant reasons alongside a correct reason, unless contradictory | |
| Ignore a calculation using exact values alongside a correct reason eg 0.025 is greater than 0.0238… and both numbers rounded down | B1 |
| 0.025 is greater than 0.0238… | B0 |
| The denominator is smaller | B1 |
| The denominator using the exact values is bigger | B1 |
| (Decimals) rounded down | B1 |
| Because 8.34 is more than 8 and 10.21 is more than 10 | B1 |
| One is divided by less (with answer more) | B1 |
| Estimating rounds the numbers down which makes the denominator less | B1 |
| Estimating rounds the numbers down which makes it less | B0 |
| Because it rounds up | B0 |
| Because she rounded each number to one significant figure | B0 |
| The numbers get rounded up so more than the exact value | B0 |
| Rounded up when estimating | B0 |
| Removing the decimals makes the number bigger | B0 |
21
The cube root of \(k\) is 3, to the nearest whole number.
Work out the largest possible value of \(k\). [2 marks]
She says,
“The only possible value of \(x\) is 10”
Give a reason why she is not correct. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Correct evaluation of the cube root of an integer \([40, 50]\) or correct evaluation of the cube of a decimal or fraction \((3, 3.5]\) | M1 | eg \(\sqrt[3]{40} = 3.4\) or \(40 \to 3.4\) eg \(3.5^3 = 42.8\) or \(3.5 \to 42.8\) |
| 42 | A1 | SC1 answer given as \(\sqrt[3]{42}\) |
| Up to M1 may be awarded for correct work with no answer, or incorrect answer, even if this is seen amongst multiple attempts | |||||||||||||||||||||||
| Condone eg \(40 = 3.4\) or \(\sqrt{40} = 3.4\) to mean \(\sqrt[3]{40} = 3.4\) | |||||||||||||||||||||||
| Answer only 42 | M1A1 | ||||||||||||||||||||||
| Must select 42 as final answer for M1A1 ie 42 as the last in a list with a blank answer line is not enough for A1 unless 42 selected | |||||||||||||||||||||||
| If \(\sqrt[3]{42}\) or \(3.5^3\) is evaluated then it must be correct to award the A1 for 42 | |||||||||||||||||||||||
| NB 42 only from incorrect method eg listing multiples of 3 or \(42 \div 3\) seen or 42 is divisible by 3 as the working | M0A0 | ||||||||||||||||||||||
Acceptable values for cube roots of integers in range
| |||||||||||||||||||||||
Examples of cubes of numbers in range with their acceptable values
|
| Answer | Mark | Comments |
|---|---|---|
| Valid response that indicates there is one (negative) answer missing | B1 | eg −10 (is also an answer) or there is a negative value as well or square roots have two answers or answer is 10 and −10 |
| \(-10 \times -10\;(= 100)\) | B1 |
| Another number can square to make 100 (implies exactly two) | B1 |
| She has forgotten the other value (implies exactly two) | B1 |
| There is another value it could be (implies exactly two) | B1 |
| It could be a different number (implies exactly two) | B1 |
| It could be negative (bod means 10 could be −10) | B1 |
| \(-10^2\;(= 100)\) (condone missing brackets around −10) | B1 |
| \(\pm\sqrt{100}\) | B1 |
| Indication that there might be more than two possible values for \(x\) eg There are other possible numbers eg There could be other values eg Other numbers square to make 100 eg She hasn’t included negatives | B0 B0 B0 B0 |
| Repeating the question eg There is more than 1 possible value eg 10 is not the only possible value eg More than 1 number works | B0 B0 B0 |
| A partially correct statement eg \(x\) could be negative or decimal eg \(-10 \times -10 = -100\) eg \(x^2 = -10\) | B0 B0 B0 |
7
The cube root of \(k\) is 3, to the nearest whole number.
Work out the largest possible value of \(k\). [2 marks]
She says,
“The only possible value of \(x\) is 10”
Give a reason why she is not correct. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Correct evaluation of the cube root of an integer [40, 50] or correct evaluation of the cube of a decimal or fraction (3, 3.5] | M1 | eg \(\sqrt[3]{40} = 3.4\) or \(40 \to 3.4\) eg \(3.5^3 = 42.8\) or \(3.5 \to 42.8\) |
| 42 | A1 | SC1 answer given as \(\sqrt[3]{42}\) |
| Up to M1 may be awarded for correct work with no answer, or incorrect answer, even if this is seen amongst multiple attempts | |||||||||||||||||||||||||
| Condone eg \(40 = 3.4\) or \(\sqrt{40} = 3.4\) to mean \(\sqrt[3]{40} = 3.4\) | |||||||||||||||||||||||||
| Answer only 42 | M1A1 | ||||||||||||||||||||||||
| Must select 42 as final answer for M1A1 ie 42 as the last in a list with a blank answer line is not enough for A1 unless 42 selected | |||||||||||||||||||||||||
| If \(\sqrt[3]{42}\) or \(3.5^3\) is evaluated then it must be correct to award the A1 for 42 | |||||||||||||||||||||||||
| NB 42 only from incorrect method eg listing multiples of 3 or \(42 \div 3\) seen or 42 is divisible by 3 as the working | M0A0 | ||||||||||||||||||||||||
Acceptable values for cube roots of integers in range
| |||||||||||||||||||||||||
Examples of cubes of numbers in range with their acceptable values
|
| Answer | Mark | Comments |
|---|---|---|
| Valid response that indicates there is one (negative) answer missing | B1 | eg \(-10\) (is also an answer) or there is a negative value as well or square roots have two answers or answer is 10 and \(-10\) |
| \(-10 \times -10\ (= 100)\) | B1 |
| Another number can square to make 100 (implies exactly two) | B1 |
| She has forgotten the other value (implies exactly two) | B1 |
| There is another value it could be (implies exactly two) | B1 |
| It could be a different number (implies exactly two) | B1 |
| It could be negative (bod means 10 could be \(-10\)) | B1 |
| \(-10^2\ (= 100)\) (condone missing brackets around \(-10\)) | B1 |
| \(\pm\sqrt{100}\) | B1 |
| Indication that there might be more than two possible values for \(x\) eg There are other possible numbers eg There could be other values eg Other numbers square to make 100 eg She hasn’t included negatives | B0 B0 B0 B0 |
| Repeating the question eg There is more than 1 possible value eg 10 is not the only possible value eg More than 1 number works | B0 B0 B0 |
| A partially correct statement eg \(x\) could be negative or decimal eg \(-10 \times -10 = -100\) eg \(x^2 = -10\) | B0 B0 B0 |
29 The mass of a baby is 3.6 kilograms to 1 decimal place.
What is the error interval for the mass in kilograms?
Tick one box. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(3.55 \leqslant \text{mass} \lt 3.65\) | B1 |
5
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{13}{9}\) | B1 | oe improper fraction |
| \(\dfrac{13}{9}\) in working with a decimal on answer line | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 0.4375 | B1 | accept .4375 |
| \(7 \div 16\) with incorrect or no decimal | B0 |
| 0.4375 in working with 0.437 or 0.438 or 0.43 or 0.44 or 0.4 on answer line | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 2.8 | B1 |
| 2.80 | B0 |
3 By rounding each number to the nearest 10, estimate the value of \(31 \times 18\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 30 | B1 | |
| 20 | B1 | |
| (30 and 20 and) 600 | B1ft | ft their 30 \(\times\) their 20 with B1B0 or B0B1 SC1 (\(31 \times 18 =\)) 558, answer 560 |
| Answer 600 with no working | B1B1B1 |
| Answer 558 with neither 30 nor 20 seen | B0B0B0 |
| \(30 \times 18\) with answer 540 \(31 \times 20\) with answer 620 and answer 600 (ignore further work) \(31 \times 20\) with answer 600 | B1B0B1ft B0B1B1ft B0B1B0ft |
22 The number of people living in a town is 47 000 to the nearest 1000
Which one of these is a possible number of people living in the town?
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 46 500 | B1 |
20 Freddie tries to work out \(\quad \dfrac{29.15 + 83.47}{9.82}\)
His answer is 37.65
By rounding each number to the nearest 10, show that his answer is incorrect. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 30 or 80 or 10 | M1 | |
| \(\dfrac{30 + 80}{10}\) or \(\dfrac{110}{10}\) or \(\dfrac{112.62}{10}\) or 11.262 | M1dep | |
| 11 with 30, 80 and 10 seen | A1 |
| 11 with no working | M0M0A0 |
7 Use approximations to estimate the answer to \(\quad \dfrac{\sqrt{97} + 2.014^3}{0.49}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(\sqrt{97}\) =) \(\sqrt{100}\) or 10 or (\(2.014^3\) =) \(2^3\) or 8 or (0.49 =) 0.5 or \(\dfrac{1}{2}\) | M1 | |
| (\(\sqrt{97}\) =) \(\sqrt{100}\) or 10 and (\(2.014^3\) =) \(2^3\) or 8 and (0.49 =) 0.5 or \(\dfrac{1}{2}\) | M1 | \(\dfrac{10 + 8}{0.5}\) or \(\dfrac{18}{0.5}\) scores M2 |
| 36 | A1 |
1 What is 6.2819 to 2 decimal places?
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 6.28 | B1 |
13 By rounding each number to the nearest 10, estimate the value of \(\quad 262 \div 19.8\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| (262 rounded to) 260 or (19.8 rounded to) 20 or \(26 \div 2\) | M1 | |
| 13 | A1 |
| 13 embedded eg \(260 \div 13 = 20\) | M1A0 |
| Beware, 13 may not get full marks eg \(262 \div 20 = 13.1\), answer 13 | M1A0 |
| \(300 \div 20\) | M1A0 |
13 The triangular cross section of a prism is an isosceles right-angled triangle.
The volume of the prism is 102 cm3
Use approximations to estimate the value of \(x\).
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| One of \((102 \to)\ 100\) \((8.14 \to)\ 8\) | M1 | |
| their \(100 = 0.5 \times x^2 \times\) their 8 or \((x^2 =)\) their \(100 \div 8 \times 2\) or \((x^2 =)\ 100 \div\) their \(8 \times 2\) or 25 or their \(8 \times 5 \times 5 \times 0.5 = 100\) or \(8 \times 5 \times 5 \times 0.5 =\) their 100 | M1dep | oe must have used at least one correct 1 sf value |
| 5 with M2 seen | A1 |
If working is done with approximations and with the given values ignore the working with the given values and mark the working with approximations
6
Give your answer as a decimal.
Write down your full calculator display. [2 marks]
Check by rounding each of 9.75, 1.875 and 6.4 to the nearest whole number.
You must show your working. [3 marks]
Tick a box.
| Answer | Mark | Comments |
|---|---|---|
| 494.325 or \(\dfrac{19\,773}{40}\) or \(494\dfrac{13}{40}\) or 40.96 or \(\dfrac{1024}{25}\) or \(40\dfrac{24}{25}\) or 535.29 or 535.3 or \(\dfrac{107\,057}{200}\) or \(535\dfrac{57}{200}\) | M1 | |
| 535.285 | A1 |
| Ignore any subsequent truncation or rounding if 535.285 seen in working | M1A1 |
| Answer | Mark | Comments |
|---|---|---|
| \(10^3\) and 2 and \(6^2\) and 536 and indicates Sensible | B3ft | ft correct decision for comparing 536 with their 535.285 B2 \(10^3\) and 2 and \(6^2\) seen B1 any two of 10, 2 and 6 seen allow 1000 to imply 10 or \(10^3\) and 36 to imply 6 or \(6^2\) for B1 or B2 only |
Students must give the correct ft decision for part (a) for B3
Correct decision for their (a) should be Sensible if their 535.285 is 530 or 540 to 2 sf. Otherwise they should indicate Not sensible
Condone eg 10.00 for 10 etc
2 The height of a tree is 12 metres, correct to the nearest metre.
Circle the error interval. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(11.5\text{ m} \leqslant \text{height} < 12.5\text{ m}\) | B1 |
18 There are 240 cows on a farm.
number of bulls : number of cows = 1 : 30
How many bulls are there? [1 mark]
the 240 cows produce milk for 10 months each year
each cow produces an average of 25 litres of milk per day.
Estimate the total milk production, in litres, of the 240 cows in one year.
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 8 | B1 |
| Ignore mention of bulls or cows eg condone 8 cows | B1 |
| Condone an answer of 8 : 240 | B1 |
| 8 : 240 followed by 1 : 30 | B0 |
| 8 : 30 | B0 |
| Do not accept 8 from an incorrect method eg \(240 \div 31 = 7.7\ldots\) and answer 8 | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \([28, 31] \times 10\) or [280, 310] | M1 | appropriate days in 10-month year |
| their \([280, 310] \times 25\) or [7000, 7750] or their \([280, 310] \times 240\) or [67 200, 74 400] | M1dep | litres per year per cow milkings per year for 240 cows |
| their \([7000, 7750] \times 240\) or their \([67\,200, 74\,400] \times 25\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 2 | ||
| \(25 \times 240\) or 6000 | M1 | litres per day for 240 cows may be seen embedded in a product eg \(25 \times 10 \times 240\) |
| their \(6000 \times [28, 31]\) or [168 000, 186 000] or \(25 \times 240\) or 6000 and \([28, 31] \times 10\) or [280, 310] | M1dep | litres per month for 240 cows litres per day for 240 cows and appropriate days in 10-month year |
| their \([168\,000, 186\,000] \times 10\) or \(25 \times 240 \times [28, 31] \times 10\) or their \(6000 \times\) their [280, 310] | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 3 | ||
| \([28, 31] \times 25\) or [700, 775] | M1 | litres per month per cow |
| their \([700, 775] \times 10\) or [7000, 7750] or their \([700, 775] \times 240\) or [168 000, 186 000] | M1dep | litres per year per cow litres per month for 240 cows |
| their \([7000, 7750] \times 240\) or their \([168\,000, 186\,000] \times 10\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 4 | ||
| \([28, 31] \times 240\) or [6720, 7440] | M1 | milkings per month for 240 cows |
| their \([6720, 7440] \times 10\) or [67 200, 74 400] or their \([6720, 7440] \times 25\) or [168 000, 186 000] | M1dep | milkings per year for 240 cows litres per month for 240 cows |
| their \([67\,200, 74\,400] \times 25\) or their \([168\,000, 186\,000] \times 10\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Use the scheme that awards the most marks and ignore choice | |
| A value in the range [280, 310] may come from subtracting two months from a year eg uses 303 (may come from 365 – 31 – 31) | M1 |
| The special case allows 2 marks for those using 12 months or using [336, 372] days | |
| Allow consistent use of approximations to 1 sf throughout (this leads to an answer in the given range) ie \(30 \times 10 \times 30 \times 200 = 1\,800\,000\) | M3A1 |
| Mark inconsistent use of approximations to 1sf as the scheme | |
| Their final answer must be in range and correct for their product but may be given to 1 or 2 sf eg 280 days: \(28 \times 10 \times 25 \times 240 = 1\,680\,000\) 300 days: \(30 \times 10 \times 25 \times 240 = 1\,800\,000\) 310 days: \(31 \times 10 \times 25 \times 240 = 1\,860\,000\) 303 days: \(303 \times 25 \times 240 = 1\,818\,000\) 304 days: \(304 \times 25 \times 240 = 1\,824\,000\) 305 days: \(305 \times 25 \times 240 = 1\,830\,000\) | M3A1 |
| eg 12 months of 28 days: \(28 \times 12 \times 25 \times 240 = 2\,016\,000\) 12 months of 30 days: \(30 \times 12 \times 25 \times 240 = 2\,160\,000\) 12 months of 31 days: \(31 \times 12 \times 25 \times 240 = 2\,232\,000\) 365 days: \(365 \times 25 \times 240 = 2\,190\,000\) 366 days: \(366 \times 25 \times 240 = 2\,196\,000\) | SC2 |
16 Circle the number that is closest to the value of \(\quad 29^3\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 27 000 | B1 |
14
Give your answer as a decimal.
Write down your full calculator display. [1 mark]
Use approximations to decide.
You must show your working. [3 marks]
Tick a box.
| Answer | Mark | Comments |
|---|---|---|
| 2950.2745(00…) | B1 |
| 2’950.2745 or 2,950.2745 | B1 |
| 2.950.2745 | B0 |
| Allow correct rounding or truncation once full value seen |
| Answer | Mark | Comments |
|---|---|---|
| 10 or \(10^2\) or 100 or 30 | M1 | |
| \(10 \times 10 \times 30\) or \(10^2 \times 30\) or \(100 \times 30\) | M1dep | |
| \(10 \times 10 \times 30 = 3000\) and Sensible or \(10^2 \times 30 = 3000\) and Sensible or \(100 \times 30 = 3000\) and Sensible | A1ft | ft their answer to part (a) for the decision |
| 3000 (and Sensible) with no working | M0M0A0 |
| Their decision must be based on part (a) unless new calculation shown in part (b) | |
| \(10^2 \times 30 = 3000\) and \(10^2 \times 29 = 2900\) and Sensible | M1M1A1 |
| \(10^2 \times 30\) and \(10^2 \times 29\) and Sensible | M1M1A0 |
| \(10^2 \times 29 = 2900\) and Sensible | M1M0A0 |
| ft should be Sensible if their part (a) is 3000 to 1sf or vice versa | |
| eg (a) 295.02745 (b) \(10 \times 10 \times 30 = 3000\) and Not sensible | (B0)M1M1A1ft |
4 Circle the number that is closest in value to \(\quad \dfrac{9.8}{0.0195}\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 500 | B1 |
24 Three whole numbers are each rounded to the nearest 10
The sum of the rounded numbers is 70
Work out the maximum possible sum for the original three numbers. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Three whole numbers that each are less than 80 and have units digit 4 or States that each number must have units digit 4 | M1 | |
| 82 | A1 | |
| Alternative method 2 | ||
| Correctly evaluated trial for three whole numbers, none of which are a multiple of 10, and that, when rounded, total 70 | M1 | eg \(33 + 33 + 13 = 79\) |
| 82 | A1 | |
| \(39 + 33 + 13 = 85\) \((40 + 30 + 10 = 80)\) | M0 |
| Beware 82 from incorrect values, eg \(39 + 24 + 19 = 82\) | M0A0 |
| Ignore incorrectly evaluated trials that do not solely lead to the answer |
13
| Answer | Mark | Comments |
|---|---|---|
| 752 951 or 752951 or 752,951 | B1 | Allow commas even if positioning incorrect eg 75,2951 or 752’951 B1 |
| 752.951 | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 20 000 and 400 and 10 and 800 000 and Yes | B3ft | ft correct decision for their answer to (a) oe decision eg it is sensible B2 20 000 and 400 and 10 B1 20 000 or 400 or 10 |
| 800 000 (and Yes) with no other values | B0 |
| If answer to (a) is 800 000 to 1sf then the correct ft decision in (b) is Yes eg1 (a) 770 000 (b) decision should be Yes eg2 (a) 1762 (b) decision should be No eg3 (a) 752.951 (b) allow decision to be Yes or No |
12 Use approximations to 1 significant figure to estimate the value of
\[\frac{0.526 \times 39.6^2}{\sqrt{97.65}}\]You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Any two of 0.5, 40 and 100 | M1 | 1600 implies 40 10 implies 100 |
| (\(40^2 =\)) 1600 or (\(0.5 \times 40^2 =\)) 800 or (\(\sqrt{100} =\)) 10 | M1 | |
| 80 with correct working | A1 |
| \(\dfrac{0.5 \times 1600}{\sqrt{100}}\) or \(\dfrac{0.5 \times 40^2}{10}\) or \(\dfrac{1 \times 1600}{10}\) or \(\dfrac{800}{\sqrt{100}}\) or \(\dfrac{800}{10}\) | M1M1 |
| 80 with no or incorrect working, eg attempt at actual calculation and then rounding to 80 | M0M0A0 |
| Condone 0.50(0) for 0.5, 40.0(0) for 40 and 100.0(0) for 100 etc | |
| Rounding 0.526 to 1, but otherwise correct, with answer 160 | M1M1A0 |
7 By rounding each number to the nearest 10,
estimate the answer to \(\quad \dfrac{61 \times 47}{102}\)
You must show your working. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Any two of 60, 50 and 100 | M1 | \(\dfrac{60 \times 50}{100}\) 60 and 50 may be implied by 3000 |
| 30 | A1 |
| 30 with no working | M0A0 |
| 28.1… (from original values) and then rounded to 30 | M0A0 |
5
Write down your full calculator display. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 26.47640(….) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 26.5 | B1ft | Correct or ft provided their answer to (a) is given to more than 1 dp |
| 8.88326612 in (a) and 8.9 in (b) | B1ft |
| 8.88326612 in (a) and 26.5 in (b) | B1 |
| 26.50 | B0 |
18 Calculate.
\[\frac{4.5^3}{1.8} + \frac{3.6}{2.4^2}\]Give your answer correct to 3 significant figures. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 51.3 | 2 | B1 for 51.2[5] or \(\dfrac{205}{4}\) or \(51\dfrac{1}{4}\) If 0 scored SC1 for their answer to more than 3 figures correctly rounded to 3 s.f. | |
15
Heidi works out \(\dfrac{5}{8} \times N\).
Ben works out \(0.63 \times N\).
Tick (✓) the correct statement.
Give a reason why you ticked the statement. [2]
Heidi’s answer is 314.1592654 cm\(^2\).
Ben correctly works out the area of the same circle using the same formula.
Ben’s answer is bigger than Heidi’s answer.
Show how this can happen. [2]
| Answer | Marks | Part marks and guidance | |||||||
|---|---|---|---|---|---|---|---|---|---|
Correct row only
| 2 | B1 for correct statement only selected | Accept any clear indication | ||||||
| and \(\left[\dfrac{5}{8} =\right]\) 0.625 | B1 for correct conversion isw or correctly finding 0.63 and \(\dfrac{5}{8}\) of the same amount isw | For second B1 Mark converted value(s) on the answer line first If insufficient converted value(s) on line, mark converted value(s) in body Accept both converted to comparable form e.g. 63% and 62.5% or \(\dfrac{630}{1000}\) and \(\dfrac{625}{1000}\). Condone \(\dfrac{63}{100}\) and \(\dfrac{62.5}{100}\) condone rounding/truncation to a level to compare quantities | |||||||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(k \times 10^2\) where \(k\) is one from 3.1415927, 3.141593, 3.1416, 3.142 leading to 314.15927, 314.1593, 314.16, 314.2 | 2 | B1 for \(k \times 10^2\) where \(k\) is one from 3.1415927, 3.141593, 3.1416, 3.142 | For 1 or 2 marks
e.g 314.1592654 = 314.2 [to 1dp] |
11 A solid steel cube has a mass of 1.05 kg.
The density of the steel is 7.9 g/cm\(^3\).
Orla says,
The length of one side of the cube is approximately 10 cm.
Use estimation to decide if Orla’s statement is reasonable.
Show how you decide. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Mass = 1050 or 1000 | B1 | For B1 accept 1050 or 1000 seen | |
| Volume = their mass ÷ 8 oe | M2 | M1 for their mass ÷ 7.9 oe | For M2 or M1 allow use of 1050 or 1000 or their mass ÷ 8, their mass could be in kg |
| Length = \(\sqrt[3]{\textit{their} \text{ volume}}\) | M1 | or [Volume = ]10 × 10 × 10 = 1000 | If cube root not shown allow implied by 5 nfww after 125 shown |
| Unreasonable oe and length = 5 [cm] | A1 | Dep on B1M2M1 or incorrect and shows [Volume = ]1000 and 125 | For A1 If 5 [cm] is shown in working it does not need to be stated in conclusion |
| OR | OR | ||
| [Volume =] 10 × 10 × 10 or 10\(^3\) | M1 | Accept Volume = 1000 | |
| Mass = (10 × 10 × 10) × 8 oe | M2 | M1 for (10 × 10 × 10) × 7.9 oe | |
| Their mass converted to kg or 1050 [g] or 1000 [g] | B1 | ||
| Unreasonable oe and either 8 kg with 1 kg or [1.05 kg] or 8000 g with 1050 g or 1000 g | A1 | Dep on M1M2B1 | For A1 values could be shown in working and do not need to be stated in conclusion See Appendix 1 for alternate method |
A further possibility is to show that the density of the 10 cm cube is different to 7.9 g/cm\(^3\)
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Mass = 1050 or 1000 or 0.0079 or 0.008 | B1 | For B1 accept 1050 or 1000 or 0.0079 or 0.008 seen | |
| [Volume = ]10 × 10 × 10 or 10\(^3\) | M1 | Accept Volume = 1000 | |
| 1000 ÷ (10 × 10 × 10) oe or 1[kg] ÷ (10 × 10 × 10) oe | M2 | M1 for 1050 ÷ (10 × 10 × 10) oe or 1.05[kg] ÷ (10 × 10 × 10) oe | |
| Unreasonable oe and [density =] 1 oe or [density = ]0.001 and 0.0079 or 0.008 | A1 | Dep on M1M2B1 | For A1 , [density =] 1 could be shown in working and does not need to be stated in conclusion |
10 By writing each number correct to 1 significant figure, find an estimate for this calculation.
\[\frac{3.81 \times 19.8}{2.1}\][3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{4 \times 20}{2}\) and answer 40 | 3 | B2 for \(\dfrac{4 \times 20}{2}\) or B1 for two of 4, 20 or 2 seen If 0 scored and no workings seen instead award SC1 for answer of 40. | Condone trailing zeros |
2
The decorator
Work out how much the decorator charges for painting an area of 74 m\(^2\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 4400 cao | 1 | ||
| (ii) 5.37 cao | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 520 | 2 | M1 for 80 [\(\times\) 65] or 8 [\(\times\) 65] If 0 scored, SC1 for answer 481 or 455 | Accept money as e.g. 520.0 or 520.00 |
1 Calculate.
\[\frac{4.5^3}{1.8} + \frac{3.6}{2.4^2}\]Give your answer correct to 3 significant figures. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 51.3 | 2 | B1 for 51.2[5] or \(\frac{205}{4}\) or \(51\frac{1}{4}\) If 0 scored SC1 for their answer to more than 3 figures correctly rounded to 3 s.f. | |
12 A housing estate is built on a triangular piece of land.
Not to scale
There are 3951 people living on the estate.
Work out an estimate of the population density of the estate in people per km2. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Answer 800 with 2, 5 and 4000 seen | 4 | B1 for at least two of the values 5, 2 and 4000 M2 for \(\frac{\textit{their } 4000}{0.5 \times \textit{their } 5 \times \textit{their } 2}\) oe or M1 for \(\frac{1}{2} \times \textit{their } 5 \times \textit{their } 2\) | For B1 condone trailing zeros For M2 and M1 their values can be 3951, 5.03, 1.96 or other rounded relatable values |
8
[1]
Give your answer correct to 3 significant figures. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\times \qquad +\) | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.16 | 3 | B2 for 3.158…or 3.159 or B1 for 249.41 or 9.9764 If 0 scored, instead award SC1 for their answer to more than 3 figures correctly rounded to 3 sf | |
6
Their method is to:
The student writes
\[\dfrac{14 + 1}{5} = \dfrac{15}{5} = 3.\]What error has the student made in using their method? [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(5\ [\times]\ 70\) or \(5\ [\times]\ 0.7\) | M1 | Not from \(73 \times 4.9\) and rounded to 360 or truncated 350 Accept e.g. 70+70+70+70+70 for M1 Condone units included in product Condone answer 350p with £ crossed through or £3.50p | |
| and 3.5[0] | A1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Recognition that the figures are not all to the same number of sig fig e.g. They’re not all the same accuracy oe | 1 | See appendix Allow complete and correct example(s) | |
Question 6b
| Response | Mark | |
|---|---|---|
| 13.7 rounded to the same number of significant figures would be 10 | Mentions “same number of sig fig” and the correct number chosen with correct example | 1 |
| 1.28 and 5.099 have not been rounded to the same number of sig fig as 13.7 | Correct, but only because of the mention of 13.7 (because they have been rounded to the same number of sig fig as each other) | 1 |
| They should have rounded 13.7 to 10 | Accept correct example as illustrating the error | 1 |
| They should have rounded 1.28 to 1.3 and 5.099 to 5.1 | Accept the two correct examples as illustrating the error | 1 |
| They should have rounded 1.28 to 1.30 and 5.099 to 5.1 | Condone the use of 1.30 for 1.3 | 1 |
| He should have rounded 13.7 to the first significant figure | Condone “first” for “1” | 1 |
| They should have rounded 1.28 to 1.3 | Recognises only one of the incorrectly rounded figures. If using 1.28 and 5.099 must describe both | 0 |
| The numerator is bigger than the denominator | Not a comment on the method | 0 |
| He did not round each number to its correct significant figure | A comment about the method BUT does not mention the same number of significant figures | 0 |
| They did not round each number to the correct amount of significant figures | A comment about the method BUT does not mention the same number of significant figures | 0 |
| 1.28 rounded to the same number of significant figures would be 2 | Does mention same number of sig fig but the wrong figure chosen and example is incorrect | 0 |
| They have rounded 13 wrong | Not explained what “wrong” means | 0 |
| He didn’t round 1.28 and 5.099 down properly | Does not explain the error. Might just be suggesting an arithmetic error | 0 |
19 Work out.
\[\sqrt[3]{\frac{19.5^4 - 18^2}{1.45}}\]Write your answer correct to 4 significant figures. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 46.34 | 3 | B2 for 46.33[7…] or B1 for 99 493[.836…] or 144 266[.0625] If 0 scored SC1 for their answer to more than 4 figures correctly rounded to 4 s.f. | for B1 accept these numbers rot to at least integers |
16 The table shows how much Amaya earns per hour.
| Work done on Monday to Friday | £20.15 per hour |
| Work done on Saturday or Sunday | £30.23 per hour |
One week Amaya works for \(40\frac{1}{3}\) hours between Monday and Friday and then for \(4\frac{1}{4}\) hours on Saturday.
I will earn at least £900 for my work this week.
By rounding each value to the nearest integer, use estimation to show that Amaya may be correct. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Three of 20[.00], 30[.00], 40 and 4 seen | B1 | B1 seen as rounding | |
| \(40 \times 20\) | M1 | for their 40 x their 20 | their 40 can be \(40\frac{1}{3}\), \(40.\dot{3}\), 40.33[3…] or 41 their 20 can be 20.15 or 21 their 4 can be \(4\frac{1}{4}\), 4.25 or 5 their 30 can be 30.23 or 31 |
| \(4 \times 30\) | M1 | for their 4 x their 30 | |
| 920 or 800 and 120 | A2 | A2 dep on B1M1M1 or A1 dep on B1M1 for 800 or 120 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The calculation is an underestimate | 1 | All values were rounded down oe Mark best response as long as not contradictory or incorrect See Appendix 1 | |
APPENDIX 1 Question 16(b)
We want the candidate to say WHY they are certain of the figure being at least £900.
If TOTAL is used then this MUST be £920. IF rounded figures are used then all four must be rounded down correctly (20,30,40,4)
| Response | Mark | |
|---|---|---|
| 1 | It is an Underestimate | 1 |
| 2 | All Values were rounded down | 1 |
| 3 | This is correct because at minimum she will make 920 (‘at minimum’ implies rounding and is correct) | 1 |
| 4 | The values are rounded down meaning she earnt more than this | 1 |
| 5 | (40 x 20) + (4 x 30) already equal over £900 | 1 |
| 6 | The underestimation is over £900 | 1 |
| 7 | If you round the earnings you get over £900 (this is true for rounding both up and down) | 1BOD |
| 8 | Because in my working not adding the minutes, it’s still over £900 as it’s £928.80 (incorrect) | 0 |
| 9 | Because I estimated it meaning she’d probably got more (probably does not mean certain) | 0 |
| 10 | As it has gone over £900 (not describing rounding or underestimate) | 0 |
| 11 | This is correct because she will make 920 (the candidate is not telling us WHY they are certain) | 0 |
| 12 | Because £20.15x40=£806 and £30.23 x4=120.92 + them = 926.92 not including the minimum (All rounded values not used) | 0 |
| 13 | If you added everything without a fraction it adds up to almost £900 (incorrect) | 0 |
| 14 | Because it works out that all of her work will add to over 900 (no reference to underestimate) | 0 |
| 15 | Because it’s over 900 (not enough, need to say WHY it is certainly over £900) | 0 |
| 16 | The estimation is over £900 (not enough, need to say WHY estimation is certainly over £900) | 0 |
| 17 | Because I rounded down (not enough- needs to be e.g. rounded all numbers down) | 0 |
6 A bronze ornament has a volume of \(198\,\text{cm}^3\).
The density of bronze is \(8.9\,\text{g/cm}^3\).
By rounding each value correct to one significant figure, work out an estimate for the mass of the bronze ornament. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1800 final answer and 200 and 9 shown | 3 | M2 for 200 × 9 or M1 for their 200 × their 9 or B1 for 200 and 9 seen | For M1 e.g. uses 198 and 8.9 or uses incorrectly rounded values |
1 Work out.
\[\sqrt[3]{\frac{19.5^4 - 18^2}{1.45}}\]Write your answer correct to 4 significant figures. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 46.34 | 3 | B2 for 46.33[7…] or B1 for 99 493[.836…] or 144 266[.0625] If 0 scored SC1 for their answer to more than 4 figures correctly rounded to 4 s.f. | for B1 accept these numbers rot to at least integers |
23 A number, \(r\), is 6.2 when rounded correct to 2 significant figures.
A number, \(h\), is 6.2 when truncated to 1 decimal place.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Any value of \(r\) in \(6.15 \leqslant r < 6.2\) | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Any value of \(h\) in \(6.25 \leqslant h < 6.3\) | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Any pair of values where \(r > h\) and \(6.2 < r < 6.25\) and \(6.2 \leqslant h < 6.25\) | 1 | ||
9
Find the cost of five of these apples. [2]
Give your answer in pounds. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [£]2.25 | 2 | M1 for \(1.80 \div 4\) [\(\times 5\)] oe or for \(180 \div 4\) [\(\times 5\)] oe If 0 scored SC1 for answer of 225 on answer line | Condone 225p as final answer in working space or answer line if £ sign is crossed out for 2 marks. |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| £5.60 | 3 | B2 for 560 or 5.84 or 5.6 OR M2 for \(8 \times\) their 0.7 or \(70 \times\) their 8 or \(0.7 \times\) their 8 or B1 for one of 8 or 70 or 0.7 | Isw rounding after 5.60 3 marks their 0.7 can be 70, 73, 0.73 only their 8 can be 8.2 or 8.5 only Condone trailing 0’s after rounding e.g. 8.0 |
2
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3600 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [£] 3.29 | 1 | Condone 329p but not £329p or £3.29p Do not condone extra zeros | |
2 By writing each number correct to 1 significant figure, find an estimate for this calculation.
\[\frac{486}{\sqrt{101.2}}\][2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 50 with correct 500 and \(\sqrt{100}\) or 10 shown nfww | 2 | M1 for one of 500 or \(\sqrt{100}\) or 10 | |
19 Calculate.
\[\sqrt{5.2^2 - 4.8 \times -6.3}\]Give your answer correct to 3 significant figures. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.57 | 2 | B1 for 7.56[8…] or \(\dfrac{2\sqrt{358}}{5}\) or 7.570 If 0 scored SC1 for their positive answer to more than 3 figures correctly rounded to 3 s.f. | Must see unrounded value |
17 The number of people watching a football match is 30 000, correct to the nearest thousand.
Show that fewer people could be watching the rugby match than the football match. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 29 500 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair of values in which rugby < football from these ranges rugby: 30 450 to 30 498 and football: 30 451 to 30 499 | 2 | B1 for one value for rugby or football in these ranges rugby from 30 450 to 30 498 or football from 30 451 to 30 499 | Error interval alone scores 0 Only mark candidate’s chosen value(s) Rugby < football does not need to be stated If a pair of values and rugby or football not stated assume the value on the left or below another is rugby For B1 if one or multiple values seen and no clear choice/designation all must be in range 30 451 to 30 498 |
8 A block made of iron is in the shape of a cuboid.
The block is 3.1 cm by 4.9 cm by 2.2 cm.
The density of iron is 7.87 g / cm3.
Sam thinks that the mass of the block is about 2.4 kg.
Use estimation to decide if Sam’s answer is reasonable.
Show how you decide. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [Vol of block ≈ 3 × 5 × 2 = ] 30 or [3 × 4.9 × 2 =] 29[.4] or [ 3 × 4.9 × 2 =] 31 or [ 3 × 5 × 2.2 = ] 33 | M2 | M1 for 3.1 × 4.9 × 2.2 or with one rounded value[s] | Throughout question if the candidate attempts to convert to different units this must be done correctly for method marks |
| their 30 × their 8 or 2400 ÷ their 30 or 2400 ÷ their 8 | M1 | Dep on at least M1 or for 2400 and their 8 used If M0M0, SC1 for two of 3, 5, 2 and 8 used | their 8 is 7.87, 7.9 or 8 Accept 2.4 ÷ their 30 their 8 is 7.87, 7.9 or 8 If these calculations are not shown they may be implied by correct figures for the answers |
| 240 or 80 or 300 | A1 | Accept e.g. 0.24 kg Accept e.g. 0.08 kg/cm3 | |
| Not reasonable oe and 2400 g or 0.24[0] kg shown oe or Not reasonable oe and 80 or 0.08 kg/cm3 shown or Not reasonable oe and 300 and 30 shown | A1 | Dep on M2M1A1 | Allow e.g. wrong/not correct for not reasonable Accept implicit comparison with 2.4 kg if 2400 g or 0.24 kg not shown e.g. it should be 10 times heavier |
1 Calculate.
\[\sqrt{5.2^2 - 4.8 \times -6.3}\]Give your answer correct to 3 significant figures. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.57 | 2 | B1 for 7.56[8…] or \(\frac{2\sqrt{358}}{5}\) or 7.570 If 0 scored SC1 for their positive answer to more than 3 figures correctly rounded to 3 s.f. | Must see unrounded value |
9
Do you think Jamie has used their calculator correctly?
Explain why.
.................... because ........................................ [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 240 with working | 2 | B1 for rounded value of 80 or 3 If 0 scored SC1 for 80.0 and 3.0 and 240.0 as answer | For 2 marks 80 and 3 must be seen |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| No oe and [because] the answer should be close to 240 oe | 1 | Eg No his answer is ten times too big No the decimal point is in the wrong place See AG | |
Appendix Q9b
| Response | Mark | |
|---|---|---|
| No | ||
| 1 | The answer should be 247.38 | 1 |
| 2 | The decimal point should be between the 7 and 3 | 1 |
| 3 | The number is a lot higher than my estimate. I think she placed a decimal point wrong (The first part is not enough without seeing what their estimate is, ‘placed decimal point wrong’ scores the mark) | 1 |
| 4 | His answer should be in 200s not 2000s he must have placed decimal incorrectly | 1 |
| 5 | There should be 2 numbers behind the decimal place and not one | 1 |
| 6 | Because if you do \(80 \times 3\) you’ll get 240 (implies the answer is 10 times too big) | 1 |
| 7 | it should be closer to 240 (This is the correct estimate, so no need for calculation to be shown here) | 1 |
| 8 | His answer was in thousands which is too large (with it should be in the hundreds would have scored the mark) | 0 |
| 9 | Has not put decimals in | 0 |
| 10 | His answer is too high and shouldn’t have a decimal at the end (First part of answer is not enough to score and ‘shouldn’t have a decimal at the end’ is incorrect) | 0 |
| 11 | The answer that Jamie got is bigger than the estimate by 1 significant figure | 0 |
| 12 | There is only one two digit number so it cannot be above 1000 | 0 |
| 13 | It would be less than 2473.8 due to \(80 \times 3\) (if the candidate had written = 240, this would be enough) | 0 |
| 14 | his answer is significantly larger | 0 |
| 15 | there is no way it can be higher than 300 (not enough as we need to see the calculation of their ‘estimate’ of 300) | 0 |
| 16 | The place value is not in the correct position (‘place value’ not an equivalent for decimal point) | 0 |
20 Force is measured in newtons (N).
A force of 198.5 N is applied to a rectangular surface of length 4.9 cm and width 4.1 cm.
Work out an estimate of the pressure, in N/cm\(^2\), applied to this rectangular surface.
[The formula for pressure is: \(\text{Pressure} = \dfrac{\text{Force}}{\text{Area}}\)] [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 nfww | 4 | M1 for \(5 \times 4\) M1 for 200 or 199 used M1dep for their 200 \(\div\) their area, dep on first M1 | nfww for 4 marks no errors in calculating values and at least one of 5, 4, 200 or 199 used Allow for 20 or for \(4.9 \times 4.1\) [20.09] or with one unrounded value [19.6 or 20.5] Allow for \(198.5 \div (4.9 \times 4.1)\) |
10 Force is measured in newtons (N).
A force of 198.5 N is applied to a rectangular surface of length 4.9 cm and width 4.1 cm.
Work out an estimate of the pressure, in N/cm², applied to this rectangular surface.
[The formula for pressure is: Pressure \(= \dfrac{\text{Force}}{\text{Area}}\)] [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 nfww | 4 | nfww for 4 marks no errors in calculating values and at least one of 5, 4, 200 or 199 used | |
| M1 for 5 × 4 | Allow for 20 or for 4.9 × 4.1 [20.09] or with one unrounded value [19.6 or 20.5] | ||
| M1 for 200 or 199 used | |||
| M1dep for their 200 ÷ their area, dep on first M1 | Allow for 198.5 ÷ (4.9 × 4.1) | ||
5
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 560 cao | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 439 000 cao | 1 | ||
1 Calculate.
Write your answer correct to 3 significant figures. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 68 921 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.86 | 3 | B2 for 2.85[7…] OR B1 for 66.95 or 8.2 or 8.16[4…] and B1 for their answer written to more than 3 figures correctly rounded to 3 sf | |
16 The height, \(h\), of a lorry is 4.3 metres, correct to 1 decimal place.
Complete the error interval for the height, \(h\).
.................. \(\leqslant h \lt\) .................. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4.25 4.35 | 2 | B1 for each or for correct answers reversed | |
4 Nina estimates the value of \(\sqrt{\left(\dfrac{3.93 \times 393}{0.546 \times 220}\right)^3}\) by rounding each number to 1 significant figure.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| e.g. \(\sqrt{\left(\frac{4 \times 400}{0.5 \times 200}\right)^3} = \sqrt{16^3} = 64\) | 3 | B2 for 4, 400, 0.5 and 200 or B1 for at least two of 4, 400, 0.5 and 200 | For full marks, at least one of these intermediate steps leading to 64 must be seen \(\sqrt{16^3}\) or \(4^3\) or \(\sqrt{4096}\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 38.7 to 38.9 | 4 | B2 for 46.1 to 46.11 or 17.89 to 17.9 or B1 for 12.8 to 12.9 or 3.57 to 3.6[0] or 2125 to 2126 and M1 for (64 – their 46.1 to 46.11) ÷ their 46.1 to 46.11 [× 100] oe | Accept 39 with correct working |
15 Angie is planning a presentation evening.
She writes down her costs and income.
| Costs | Income |
|---|---|
| 10 staff each working 6 hours at £8 per hour Food: 60 meals at £8.95 each Prizes: 12 prizes at £19.99 each | 60 guests each paying £5 Sponsorship £1000 |
Angie thinks she will make a small profit.
Use estimation to decide if Angie is correct.
Show all of your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 × 6 × 8 | M1 | May be implied by 480 | |
| Makes use of rounding 8.95 to 9 or 19.99 to 20 | M1 | May be implied by use in a calculation but must be used correctly | |
| 60 × 9 oe or 12 × 20 oe | M1 | Alternative Method M1 for 60 × 9 oe or 10 × 20 oe | |
| 1260 | A1 | A1 for 1220 | |
| 1000 + 60 × 5 oe | M1 | May be implied by 1300 | |
| Correct decision for 1300 and their estimated costs | B1 | Their estimated costs means 8.95 and 19.99 not used | |
3
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 33 000 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 30 000 | 1 | ||
1
\(\dfrac{3}{4} + \dfrac{1}{6}\)
Give your answer in its simplest form. [2]
\(\dfrac{39.6 \times 20.2}{\sqrt{99.2}} \approx 80\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{11}{12}\) final answer | 2 | M1 for \(\frac{11k}{12k}\) or \(\frac{9k}{12k}\) [+] \(\frac{2k}{12k}\) | \(k\) = positive integer |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{40 \times 20}{\sqrt{100}}\) | M1 | For M1 allow 2 correct If more than one attempt, mark the better oe allow cancelling in fraction with 10 and 40 or 20 | |
| 800 ÷ 10 = 80 | A2 | A1 for 800 or 10 | |
26
Complete the error interval for \(g\).
\(\ldots\ldots\ldots\ldots \leqslant g \lt \ldots\ldots\ldots\ldots\) [2]
Complete the error interval for \(h\).
\(3 \leqslant h \lt \ldots\ldots\ldots\ldots\) [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4.045 and 4.055 | 2 | B1 for each or for both correct but reversed | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4 cao | 1 | Do not accept 4.0 | |
14
| Answer | Marks | Part marks and guidance |
|---|---|---|
| (i) 360 cao | 1 | |
| (ii) 356.1 cao | 1 |
| Answer | Marks | Part marks and guidance |
|---|---|---|
| (i) 4 | 1 | Do not accept \(3^4\) |
| (ii) 8 | 1 | Do not accept \(6^8\) |
2 By rounding each value to one significant figure, estimate the cost of 3.9 kg of apples at 87p per kg. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [£]3.6[0] | 2 | B1 for 4 or 90 or 0.9 or M1 for 4 lots of 87 or 4 lots of their attempt to round 87 | Condone 3.6[0] rounded to give an answer of 4 for 2 marks M1 implied by answer 360 or 3.48 |
1 Calculate.
\[\sqrt[3]{\frac{210}{10^2 + 5^2}}\]Give your answer correct to 3 significant figures. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.19 | 3 | B2 for 1.18[8…] or 1.189 or B1 for 125 or 1.68 and B1 for their answer seen to more than 3sf correctly rounded to 3sf (both rounded and unrounded values must be seen) | |
12
[2]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 20 | 2 | M1 for 80 \(\div\) 4 oe | NOT 8 \(\div\) 0.4 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{20 \times 40}{2} = 400\) | 3 | M2 for \(\dfrac{20 \times 40}{2}\) or M1 for 20 or 40 or 2 | For 3 marks all 3 values must be rounded to 1sf |
7
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 81.5 or \(8.15 \times 10^1\) cao | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 0.005 69 or \(5.69 \times 10^{-3}\) cao | 1 | ||
2 By writing each number correct to 1 significant figure, find an estimate for this calculation.
\[\frac{606.3 \times 0.312}{19.93}\][3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 9 with correct rounded values shown nfww | 3 | M1 for two of 600, 0.3 and 20 M1 for a first correct calculation with one pair of their rounded values or M1 FT for second correct calculation | Condone e.g. 0.300 M1 can be implied by 180, 30, 0.015, 9.09, 9.15 |
19 Asha worked out \(\dfrac{326.8 \times (6.94 - 3.4)}{59.4}\).
She got an answer of 19.5, correct to 3 significant figures.
Write each number correct to 1 significant figure to decide if Asha’s answer is reasonable. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{300 \times (7 - 3)}{60} = 20\) AND it is close to 19.5 oe or 19.5 rounds to 20 oe or [Asha’s estimate] is reasonable | 3 | B2 for 300, 7, 3 and 60 seen or B1 for two of 300, 7, 3 and 60 seen or 300, 4 and 60 seen or 300.0, 7.0, 3.0. 60.0 AND B1dep for result 20 and correct conclusion following B1 or B2 | Actual answer 19.475959...(may be rounded) scores 0 Accept “Yes” or “She’s right” or “It is” or equivalent comment |
9 The length, \(a\), of a pencil is 15.3 cm, correct to 1 decimal place.
Complete the error interval for the length of the pencil.
.................. \(\leqslant a \lt\) .................. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 15.25 15.35 | 2 | B1 for one correct or both correct but reversed | |
3 Calculate.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 42.9 cao | 2 | B1 for 42.8 or 42.87[5] or 42.88 or 43 seen | |
3 Asha worked out \(\dfrac{326.8 \times (6.94 - 3.4)}{59.4}\).
She got an answer of 19.5, correct to 3 significant figures.
Write each number correct to 1 significant figure to decide if Asha’s answer is reasonable. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{300 \times (7 - 3)}{60} = 20\) AND it is close to 19.5 oe or 19.5 rounds to 20 oe or [Asha’s estimate] is reasonable | 3 | B2 for 300, 7, 3 and 60 seen or B1 for two of 300, 7, 3 and 60 seen or 300, 4 and 60 seen or 300.0, 7.0, 3.0. 60.0 AND B1dep for result 20 and correct conclusion following B1 or B2 | Actual answer 19.475959...(may be rounded) scores 0 Accept “Yes” or “She’s right” or “It is” or equivalent comment |
7
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 3 | 1 | ||
| (ii) 22 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 32 | 1 | ||
| (ii) 20 | 1 | Accept \(\pm\) 20 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 | 3 | M2 for two values from 20, 4 and 8 used correctly in calculation or M1 for 20 or 4 or 8 | eg \(\dfrac{23 \times 4}{8}\) or (24 \(\div\) 8) \(\times\) 4 |
3
[1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 7900 | 1 | ||
| (ii) 8000 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 1 | Do not allow \(3^7\) | |
16 Last year, Katie earned £16 200.
Her total loan repayments were £6400.
Katie estimates that the ratio of her loan repayments to her earnings is approximately 3 : 8.
Is she correct?
Show your reasoning. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| A - Yes with appropriate reasoning involving rounding and correct simplification to 3 : 8 or 3 : 11 or 8 : 11 or ratios reversed OR B - Yes it is approximately correct oe and simplification of 6400 : 16200 to 32 : 81 OR C - Yes with a correctly evaluated calculation using e.g. ratio 3 : 8 with a comparison comment OR D - Yes and e.g. \(16200 \div 8\) and \(6400 \div 3\) correctly evaluated | 3 | M2 eg for showing 6000 : 16000 and reducing to 3 : 8 or for appropriate rounding at some stage in correctly simplifying ratio leading to 3 : 8 isw or reduces 6400 : 16200 to 32 : 81 isw or reduces 6400 : 22600 to 32 : 113 isw or for ratio calculation leading to one of the following values seen 6075, 6163 to 6165, 16436 to 16440, 17066 to 17067 or 22 275 or 23 463 to 23 467 seen isw or for 2025 and 2133 to 2134 seen isw or 2025 and 2054 to 2055 seen isw or 2133 to 2134 and 2054 to 2055 seen isw Or M1 for 6000 or 16 000 or 20 000 or 22 000 or 23 000 seen or for appropriate rounding of one number at some stage in simplifying ratio or for intention to find \(\dfrac{3}{8}\) of 16 200 or for \(\dfrac{8}{3}\) of 6400 or \(\dfrac{3}{11}\) of (16 200 + 6400) or \(\dfrac{8}{11}\) of (16 200 + 6400) isw or for \(6400 \div 3\) and one of \(16\,200 \div 8\) or \((6400 + 16\,200) \div 11\) seen isw or \(16\,200 \div 8\) and \((6400 + 16\,200) \div 11\) seen isw | For all marks accept method with equivalent fractions or decimals [3sf or better] Allow equivalent methods working with the totals e.g. 3 : 11, condone 22600 rounded to 22000 For 3 or M2, allow clear ‘reverse’ methods working from e.g. 3 : 8 to 6000 and 16000 Accept clear working if not in ratio form e.g. 3.2 and 8.1 shown not in ratio The figures in the part marks column are guidance on accuracy required for 3 marks or M2 SEE APPENDIX B |
Exemplar responses for Q16
| Response | Mark |
|---|---|
| 6400 : 16200 = 64 : 162 = 32 : 81 which is roughly 30 : 80 = 3 : 8 so yes | 3B |
| 6000 + 16000 = 22000, 22000 ÷ (3 + 8) = 2000 3 × 2000 = 6000 , 8 × 2000 = 16 000 so yes she is correct Method C but better | 3C |
| 3.2 and 8.1 in working. Close as he can round the decimals to nearest whole number. Decision not clear | M2B |
| 3 : 8 = 6 : 16 = 6000 : 16000 yes Katie is correct if she rounds to the nearest 1000 Reverse method | 3A |
| 16 200 ÷ 8 = 2040, 2040 × 3 = 6120 she is not correct Error made in calculation M2 not available | M1C |
| 16 200 ÷ 8 = 2025, 2025 × 3 = 6075 which is close to 6400 so yes she is approximately correct | 3C |
| 16200 + 6400 = 22600, 22600 ÷ 11 =2540, 2540 × 3 Intention to find 3/11 with errors | M1C |
| 16200 + 6400 = 22600 = 22000, 22000 ÷ 11 = 2000 M1 for one correct rounding | M1D |
| 16200 ÷ 8 × 3 = 6075 No it is not correct Correctly evaluated calculation with ratio 3 : 8 | M2C |
| 16200 : 6400 = 81 : 32 = 8.1 : 3.2 which is approximately 8 : 3 so she is correct | 3B |
| 6000 ÷ 3 = 2000 and 16000 ÷ 8 = 2000 so yes Equivalent to 4th the method but better | 3D |
| 16000÷8 = 2000, 6400÷3 = 2138.3×11 = 23466. Approximately 400 off so No. Error in calculation | M1C/D |
| 16200 × 3/8 =6075. No not correct as for ratio to be correct her loan would have to be £6075. | M2C |
| 6400÷3 = 2133.33, 16200÷8 = 2025. Not correct as ratio parts are not equal. | M2D |
| 16200 – 6400 = 9800, 9800 ÷ 5 x 3 = 5880 No | M2C |
13
Complete the error interval for the mass, \(m\).
\[\ldots\ldots\ldots\ldots \leqslant m \lt \ldots\ldots\ldots\ldots\][2]
Show that the piece of wood could be longer than the metal rod. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 12.65 12.75 | 2 | B1 for one correct or both correct and reversed | For B1, correct value must be in correct place |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair in which wood > metal from \(8.45 \lt \text{wood} \leqslant 8.49[9\ldots]\) and \(8.45 \leqslant \text{metal} \leqslant 8.49[9\ldots]\) | 2 | Wood > metal does not need to be stated but it must be clear which measurement refers to wood and which to metal B1 for one measurement in range \(8.45 \lt \text{wood} \leqslant 8.49[9\ldots]\) seen or \(8.45 \leqslant \text{metal} \leqslant 8.49[9\ldots]\) seen | Eg “Wood 8.46 [m] but Metal 8.45 [m]” scores 2 marks Wood or 8 Metal or 8.5 For B1, measurement must be correctly associated with either wood or metal, as appropriate 8.45 seen as end of range Eg \(8.45 \leqslant 8.5 \lt 8.55\) scores 1 mark |
11
Calculate the cost of 380 g of grapes. [2]
Estimate the cost of one ticket.
Show your working. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [0].76 cao | 2 | M1 for \(\dfrac{380}{1000} \times 2\) oe or B1 for figs 76 as answer | Condone 0.76p for 2 marks Eg \(0.38 \times 2\) If using a method of scaling, must be complete method eg 1000g = £2 100g = 20p 20g = 4p Their 20p \(\times\) 3 + their 4p \(\times\) 4 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{300}{20}\) or \(\dfrac{280}{20}\) | M1 | Answers from \(280.25 \div 19\), and rounded, score 0 | |
| 15 or 14 | A1 | 15 dep on \(\dfrac{300}{20}\) and 14 dep on \(\dfrac{280}{20}\) If 0 scored SC1 for one of 300 or 280 or 20 | Accept 15 (14) and 15.00 (14.00) but not 15.0 (14.0) or 15.00p (14.00p) |
11 320 people go on a coach trip.
Each coach holds 53 people.
Gary says 6 coaches are needed.
Is Gary correct?
You must show your working. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Identifying there are not enough coaches or too many people with correct justification | 2 | M1 for \(320 \div 53\) soi by 6.03[..] or \(53 \times 6\) soi by 318 or 2 or \(320 \div 6\) soi by 53.3 | No, he needs 7 coaches alone scores 0 See appendix |
Exemplar responses for Q11
| Response | Mark |
|---|---|
| No – 6 coaches only hold 318 people which is 2 short | 2 |
| Gary needs an extra coach for the other 2 people | 2 |
| \(320 \div 6 = 53.3333\).. no you can’t split a person up into 3 parts, you will need 7 coaches | 2 |
| 320/53 = 6.03 No because it’s a decimal you need to round up not down | 2 |
| 53x6=318 Gary is not correct as 6 coaches will mean 2 people will not be aloud on the coaches | 2 |
| Gary is incorrect 6 coaches only hold 318 people so he needs 7 | 2 |
| 6 coaches hold 318 so there will be 2 people left | 2 bod |
| 320/53 = 6.03…. No he is wrong he will need more | 2 |
| Gary is correct 320/53 = 6.0 | 1 |
| 320/6=53.3 Not enough room one person will be left behind (in fact two people are left behind!) | 1 |
9
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.31 cao | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 408 000 cao | 1 | ||
| (ii) [0].006 14 cao | 1 | ||
6 Work out 17% of 54.
Give your answer correct to 1 decimal place. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 9.2 | 3 | M1 for \(0.17 \times 54\) oe A1 for 9.18 If 0 scored SC1 for their answer rounded to 1dp, if two dp or more seen. | Allow fully correct non calculator method for 1 mark allow 1 error in addition |
3 Last year, Katie earned £16 200.
Her total loan repayments were £6400.
Katie estimates that the ratio of her loan repayments to her earnings is approximately 3 : 8.
Is she correct?
Show your reasoning. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| A - Yes with appropriate reasoning involving rounding and correct simplification to 3 : 8 or 3 :11 or 8 : 11 or ratios reversed OR B - Yes it is approximately correct oe and simplification of 6400 : 16200 to 32 : 81 OR C - Yes with a correctly evaluated calculation using e.g. ratio 3 : 8 with a comparison comment OR D - Yes and e.g. 16200 ÷ 8 and 6400 ÷ 3 correctly evaluated | 3 | M2 eg for showing 6000 : 16000 and reducing to 3 : 8 or for appropriate rounding at some stage in correctly simplifying ratio leading to 3 : 8 isw or reduces 6400 : 16200 to 32 : 81 isw or reduces 6400 : 22600 to 32 : 113 isw or for ratio calculation leading to one of the following values seen 6075, 6163 to 6165, 16436 to16440, 17066 to 17067 or 22 275 or 23 463 to 23 467 seen isw or for 2025 and 2133 to 2134 seen isw or 2025 and 2054 to 2055 seen isw or 2133 to 2134 and 2054 to 2055 seen isw Or M1 for 6000 or 16 000 or 20 000 or 22 000 or 23 000 seen or for appropriate rounding of one number at some stage in simplifying ratio or for intention to find \(\frac{3}{8}\) of 16 200 or for \(\frac{8}{3}\) of 6400 or \(\frac{3}{11}\) of (16 200 + 6400) or \(\frac{8}{11}\) of (16 200 + 6400) isw or for 6400 ÷ 3 and one of 16 200 ÷ 8 or (6400 + 16 200) ÷ 11 seen isw or 16 200 ÷ 8 and (6400 + 16 200)÷ 11 seen isw | For all marks accept method with equivalent fractions or decimals [3sf or better] Allow equivalent methods working with the totals e.g. 3 : 11, condone 22600 rounded to 22000 For 3 or M2, allow clear ‘reverse’ methods working from e.g. 3 : 8 to 6000 and 16000 Accept clear working if not in ratio form e.g. 3.2 and 8.1 shown not in ratio The figures in the part marks column are guidance on accuracy required for 3 marks or M2 SEE APPENDIX |
| Response | Mark | ||
|---|---|---|---|
| 1 | 6400 : 16200 = 64 : 162 = 32 : 81 which is roughly 30 : 80 = 3 : 8 so yes | 3B | |
| 2 | 6000 + 16000 = 22000, 22000 ÷ (3 + 8) = 2000 3 × 2000 = 6000 , 8 × 2000 = 16 000 so yes she is correct | Method C but better | 3C |
| 3 | 3.2 and 8.1 in working. Close as he can round the decimals to nearest whole number. | Decision not clear | M2B |
| 4 | 3 : 8 = 6 : 16 = 6000 : 16000 yes Katie is correct if she rounds to the nearest 1000 | Reverse method | 3A |
| 5 | 16 200 ÷ 8 = 2040, 2040 × 3 = 6120 she is not correct | Error made in calculation M2 not available | M1C |
| 6 | 16 200 ÷ 8 = 2025, 2025 × 3 = 6075 which is close to 6400 so yes she is approximately correct | 3C | |
| 7 | 16200 + 6400 = 22600, 22600 ÷ 11 =2540, 2540 × 3 | Intention to find 3/11 with errors | M1C |
| 8 | 16200 + 6400 = 22600 = 22000, 22000 ÷ 11 = 2000 | M1 for one correct rounding | M1D |
| 9 | 16200 ÷ 8 × 3 = 6075 No it is not correct | Correctly evaluated calculation with ratio 3 : 8 | M2C |
| 10 | 16200 : 6400 = 81 : 32 = 8.1 : 3.2 which is approximately 8 : 3 so she is correct | 3B | |
| 11 | 6000 ÷ 3 = 2000 and 16000 ÷ 8 = 2000 so yes | Equivalent to 4th the method but better | 3D |
| 12 | 16000÷8 = 2000, 6400÷3 = 2138.3×11 = 23466. Approximately 400 off so No. | Error in calculation | M1C/D |
| 13 | 16200 × 3/8 =6075. No not correct as for ratio to be correct her loan would have to be £6075. | M2C | |
| 14 | 6400÷3 = 2133.33, 16200÷8 = 2025. Not correct as ratio parts are not equal. | M2D | |
| 15 | 16200 – 6400 = 9800, 9800 ÷ 5 × 3 = 5880 No | M2C | |
| 16 | 6400 × 8 = 51200, 16200 x 3 = 48600 No they are different | M2C |
2 The length, \(L\), of a steel rod is 8.3 m, correct to 1 decimal place.
Complete the error interval for length \(L\).
\(\ldots\ldots\ldots \leqslant L \lt \ldots\ldots\ldots\) [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 8.25 8.35 | 2 | B1 for either one correct or for both correct but reversed | |
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