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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…)
19 There are some sheets of paper in a pile.
The height of the pile is 10.4 cm, correct to the nearest mm.
The thickness of each sheet of paper is 0.17 mm, correct to 2 significant figures.
Calculate the upper bound for the number of sheets of paper in the pile.
You must show all your working. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 633 | B1 | for stating any correct bound, eg 10.35 or 10.45 or 103.5 or 104.5 or 0.165 or 0.175 or 0.0165 or 0.0175 |
| P1 | for process to find the upper bound, eg [UB of height] \(\div\) [LB of thickness] or \(104.5 \div 0.165\) or \(10.45 \div 0.0165\) or eg [UB of height] \(\div\) [LB of thickness] \(\times\ 10\) or \(10.45 \div 0.165 \times 10\) | |
| A1 | (dep on all previous marks) for an answer of 633 or 633(.33…) clearly coming from working with correct values eg \(104.5 \div 0.165 = 633\) |
Accept \(10.44\dot{9}\) or 10.4499(9…) for 10.45
Accept \(0.174\dot{9}\) or 0.17499(9…) for 0.175
\(104 \lt [\text{UB of height}] \leqslant 104.5\) and \(0.165 \leqslant [\text{LB of thickness}] \lt 0.17\)
or
\(10.4 \lt [\text{UB of height}] \leqslant 10.45\) and \(0.0165 \leqslant [\text{LB of thickness}] \lt 0.017\)
Condone 630 from a correct calculation
If correct answer is seen and then incorrectly rounded award full marks
Award 0 marks for a correct answer with no (or incorrect) supportive working
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…)
23 Jana used her calculator to find the value of a number \(t\).
The answer on her calculator began 10.2
Complete the error interval for \(t\). (2)
………… \(\leqslant t \lt \) …………
| Answer | Mark | Mark scheme |
|---|---|---|
| 10.2 and 10.3 | B1 | for 10.2 in the correct place |
| B1 | for 10.3 in the correct place |
Accept \(10.2\dot{9}\) or 10.299(99…)
19 \(T = \dfrac{w}{a - c}\)
\(w = 435\) correct to the nearest 5
\(a = 9.8\) correct to 2 significant figures.
\(c = 2.5\) correct to 2 significant figures.
By considering bounds, calculate the value of \(T\) to a suitable degree of accuracy.
You must show all your working and give a reason for your final answer. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 60 and reason | B1 | for 432.5 or 437.5 or 9.75 or 9.85 or 2.45 or 2.55 |
| M1 | for a correct process to find a bound for \(T\) eg [LB of \(w\)] \(\div\) [UB of \(a\) – LB of \(c\)] where \(432.5 \leqslant\) [LB of \(w\)] \(\lt 435\) and \(9.8 \lt \) [UB of \(a\)] \(\leqslant 9.85\) and \(2.45 \leqslant\) [LB of \(c\)] \(\lt 2.5\) or [UB of \(w\)] \(\div\) [LB of \(a\) – UB of \(c\)] where \(435 \lt \) [UB of \(w\)] \(\leqslant 437.5\) and \(9.75 \leqslant\) [LB of \(a\)] \(\lt 9.8\) and \(2.5 \lt \) [UB of \(c\)] \(\leqslant 2.55\) | |
| M1 | for a correct process to find both LB and UB bound for \(T\) eg [LB of \(w\)] \(\div\) [UB of \(a\) – LB of \(c\)] where \(432.5 \leqslant\) [LB of \(w\)] \(\lt 435\) and \(9.8 \lt \) [UB of \(a\)] \(\leqslant 9.85\) and \(2.45 \leqslant\) [LB of \(c\)] \(\lt 2.5\) and [UB of \(w\)] \(\div\) [LB of \(a\) – UB of \(c\)] where \(435 \lt \) [UB of \(w\)] \(\leqslant 437.5\) and \(9.75 \leqslant\) [LB of \(a\)] \(\lt 9.8\) and \(2.5 \lt \) [UB of \(c\)] \(\leqslant 2.55\) | |
| A1 | (dep on all previous marks) for 58.44(5...) and 60.76(3...) with both values clearly coming from working with correct values | |
| C1 | for 60 from 58.44... and 60.76... and statement that both LB and UB round to 60 |
| Letter | Given | LB | UB |
|---|---|---|---|
| \(w\) | 435 | 432.5 | 437.5 |
| \(a\) | 9.8 | 9.75 | 9.85 |
| \(c\) | 2.5 | 2.45 | 2.55 |
UB
| Letter | Given | LB | UB |
|---|---|---|---|
| \(w\) | 435 | 432.5 | 437.5 |
| \(a\) | 9.8 | 9.75 | 9.85 |
| \(c\) | 2.5 | 2.45 | 2.55 |
LB
| Letter | Given | LB | UB |
|---|---|---|---|
| \(w\) | 435 | 432.5 | 437.5 |
| \(a\) | 9.8 | 9.75 | 9.85 |
| \(c\) | 2.5 | 2.45 | 2.55 |
Accept bounds rounded or truncated to at least 4 sf
7 Jana used her calculator to find the value of a number \(t\).
The answer on her calculator began 10.2
Complete the error interval for \(t\). (2)
………… \(\leqslant t \lt \) …………
| Answer | Mark | Mark scheme |
|---|---|---|
| 10.2 and 10.3 | B1 | for 10.2 in the correct place |
| B1 | for 10.3 in the correct place |
Accept \(10.2\dot{9}\) or 10.299(99…)
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 |
23 The length of a wall is 9 metres to the nearest metre.
Complete the error interval for the length of the wall. [2 marks]
\(\ldots\ldots\ldots\ \text{m} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{m}\)
| Answer | Mark | Comments |
|---|---|---|
| \(8.5\ \text{m} \leqslant \text{length} \lt 9.5\ \text{m}\) | B2 | oe B1 8.5 or 9.5 in correct position SC1 \(9.5\ \text{m} \leqslant \text{length} \lt 8.5\ \text{m}\) |
Accept \(9.4\dot{9}\) for 9.5
Accept eg 8.50 for 8.5
22 To the nearest pound, Rosie has £12
She wants to buy 6 drinks.
Each drink costs £1.89
Show that Rosie definitely has enough money to buy the 6 drinks. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: Compares lower bound with cost of 6 drinks | ||
| \(12 - 0.5(0)\) or 11.5(0) | M1 | oe eg \(1200 - 50\) allow mixed units eg \(12 - 50\) |
| \(1.89 \times 6\) or 11.34 | M1 | oe eg \(189 \times 6\) |
| 11.5(0) and 11.34 | A1 | oe eg 1150 and 1134 units must be consistent |
| Alternative method 2: Compares rounded cost of drinks with 12 | ||
| \(1.89 \times 6\) or 11.34 | M1 | oe eg \(189 \times 6\) |
| 11.34 rounds to 11 | M1dep | oe eg \(11.34 \to 11\) |
| 11.34 rounds to 11 and 11 is less than 12 | A1 | oe units must be consistent |
| Alternative method 3: Uses lower bound to work out cost or number of drinks | ||
| \(12 - 0.5(0)\) or 11.5(0) | M1 | oe eg \(1200 - 50\) allow mixed units eg \(12 - 50\) |
| their \(11.5(0) \div 6\) or [1.91, 1.92] or their \(11.5(0) \div 1.89\) or [6.08, 6.1] | M1 | oe eg \(1150 \div 6\) or [191, 192] their 11.5(0) must be [11, 11.99] |
| \(11.5(0) \div 6\) and [1.91, 1.92] or \(11.5(0) \div 1.89\) and [6.08, 6.1] | A1 | oe eg \(1150 \div 6\) and [191, 192] units must be consistent |
| Ignore any use of upper bound | |
| Condone inconsistent notation eg 11.50 and £11.34p | M1M1A1 |
| 11.34 rounds to 11 not 12 so she will have enough | M1M1A1 |
| 11.34 rounds to 11 so she will have enough | M1M1A0 |
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
23
Complete the error interval for the length of the pipe. [2 marks]
Answer \(\underline{\qquad\qquad}\ \text{m} \leqslant \text{length} \lt \underline{\qquad\qquad}\ \text{m}\)
Olly says,
“The total length of the two pipes is 11 metres to the nearest metre.”
Give an example to show that he could be correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 5.5 in the correct position | B1 | oe |
| 6.5 in the correct position | B1 | oe |
| 5.50 or \(5\dfrac{1}{2}\) or \(\dfrac{11}{2}\) | B1 |
| 6.50 or \(6\dfrac{1}{2}\) or \(\dfrac{13}{2}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| One correctly evaluated trial using (6, 6.5] + (4, 4.5) or (6, 6.5) + (4, 4.5] or two values in the ranges given that work if correctly evaluated | M1 | eg \(6.3 + 4.1 = 10.4\) eg 6.4, 4.2 |
| One correctly evaluated trial using (6, 6.5) + (4, 4.5) with an answer that rounds to 11 | A1 | eg \(6.4 + 4.2 = 10.6\) Ignore fw |
| \(6.4 + 4.4 = 10.8\ (= 11)\) do not need to show 11 | M1A1 |
| \(6.4999 + 4.4999 = 10.9998\) | M1A1 |
| \(6.5 + 4.4 = 10.9\) | M1A0 |
| \(4.5 + 6.2 = 10.7\) | M1A0 |
| \(6 + 4 = 10\) | M0 |
| \(6.5 + 4.5 = 11\) | M0 |
| \(6.4\dot{9} + 4.4\dot{9} = 11\) | M0 |
20 This sign shows when a lift is safe to use.
Total mass of people must be 450 kg or less
Ben and some other people are in the lift.
Their total mass is 525 kg to the nearest 5 kg
Ben gets out.
He has a mass of 78 kg to the nearest kg
Is the lift now safe to use?
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 522.5 or 527.5 | B1 | oe Accept 527.499(999…) |
| 77.5 or 78.5 | B1 | oe Accept 78.499(999…) |
| \(527.5 - 77.5\) | M1 | their max total – their min Ben their max total must be (525, 530] their min Ben must be [77, 78) Accept \(527.4\dot{9}\) or 527.499(999…) for 527.5 |
| 450 and Yes with correct working seen | A1 | Accept [449.999, 450] |
| 525 – 78 = 447 and yes | B0B0M0A0 |
| 525 = 520 to 530 78 = 77.5 to 78.5 520 – 78.5 = 441.5 520 – 77.5 = 442.5 530 – 78.5 = 451.5 530 – 77.5 = 452.5 Answer No | B0 B1 M1 A0 |
7
Complete the error interval for the length of the pipe. [2 marks]
Answer \(\;\ldots\ldots\ldots\;\) m \(\;\leqslant\;\) length \(\;\lt\;\) \(\;\ldots\ldots\ldots\;\) m
Olly says,
“The total length of the two pipes is 11 metres to the nearest metre.”
Give an example to show that he could be correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 5.5 in the correct position | B1 | oe |
| 6.5 in the correct position | B1 | oe |
| 5.50 or \(5\dfrac{1}{2}\) or \(\dfrac{11}{2}\) | B1 |
| 6.50 or \(6\dfrac{1}{2}\) or \(\dfrac{13}{2}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| One correctly evaluated trial using (6, 6.5] + (4, 4.5) or (6, 6.5) + (4, 4.5] or two values in the ranges given that work if correctly evaluated | M1 | eg 6.3 + 4.1 = 10.4 eg 6.4, 4.2 |
| One correctly evaluated trial using (6, 6.5) + (4, 4.5) with an answer that rounds to 11 | A1 | eg 6.4 + 4.2 = 10.6 Ignore fw |
| 6.4 + 4.4 = 10.8 (= 11) do not need to show 11 | M1A1 |
| 6.4999 + 4.4999 = 10.9998 | M1A1 |
| 6.5 + 4.4 = 10.9 | M1A0 |
| 4.5 + 6.2 = 10.7 | M1A0 |
| 6 + 4 = 10 | M0 |
| 6.5 + 4.5 = 11 | M0 |
| \(6.4\dot{9} + 4.4\dot{9} = 11\) | M0 |
No questions match these filters.