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More questions on this topic: IGCSE Number: Upper & Lower Bounds (74)
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…)
19
\[R = \dfrac{P}{Q}\]\(P = 5.88 \times 10^8\) correct to 3 significant figures.
\(Q = 3.6 \times 10^5\) correct to 2 significant figures.
Work out the lower bound for \(R\).
Give your answer as an ordinary number correct to the nearest integer.
You must show all your working. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1610 | B1 | for \(5.875 \times 10^8\) or \(5.885 \times 10^8\) or \(3.55 \times 10^5\) or \(3.65 \times 10^5\) or the digits 5875 or 5885 or 355 or 365 |
| M1 | for method to find lower bound, \(\dfrac{[\text{LB}]}{[\text{UB}]}\) eg \((5.875 \times 10^8) \div (3.65 \times 10^5)\) oe | |
| A1 | for answer in range 1609 – 1610 from correct working |
Accept \(5.884\dot{9}\) for 5.885
Accept \(3.64\dot{9}\) for 3.65
\(5.875 \times 10^8 \leqslant [\text{LB}] \lt 5.88 \times 10^8\)
\(3.6 \times 10^5 \lt [\text{UB}] \leqslant 3.65 \times 10^5\)
If an answer is shown in the range in working and then incorrectly rounded award full marks
22 Ebony makes some bracelets to sell.
The materials to make all the bracelets cost £190, correct to the nearest £5
Ebony sells all the bracelets for a total of £875, correct to the nearest £5
The total time taken to make and sell all these bracelets was 72 hours, correct to the nearest hour.
Ebony uses this method to calculate her hourly rate of pay
\[\text{Hourly rate of pay} = \frac{\text{total selling price} - \text{total cost of materials}}{\text{total time taken}}\]The minimum hourly rate of pay for someone of Ebony’s age is £8.20
By considering bounds, determine if Ebony’s hourly rate of pay was definitely more than £8.20
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| (£)9.37(9…) and conclusion | B1 | for stating at least one bound 187.5 or 192.5 or 192.499… or 872.5 or 877.5 or 877.499… or 71.5 or 72.5 or 72.499… |
| P1 | for correct use of LB and UB find the lower bound for profit, eg [LB of selling price] – [UB of costs] \((= 872.5 - 192.5\ (= 680))\) | |
| P1 | (dep P1) for correct use of LB and UB to find the lower bound for hourly rate, eg [LB of profit] ÷ [UB of hours] | |
| C1 | for (£)9.37(9…) and conclusion that Ebony does earn more than minimum wage with value from working with correct bounds |
May be seen in calculations
Accept 192.499... or \(192.4\dot{9}\) or 877.499... \(877.4\dot{9}\) or 72.499… or \(72.4\dot{9}\)
\(872.5 \leqslant [\text{LB selling price}] \lt 875\)
\(190 \lt [\text{UB costs}] \leqslant 192.5\)
LB profit = "680"
\(72 \lt [\text{UB hours}] \leqslant 72.5\)
\(\dfrac{872.5 - 192.5}{72.5}\)
Accept bounds truncated or rounded to at least 2 dp
8 A number \(x\) is written correct to 2 significant figures.
The result is 1.9
Complete the error interval for \(x\). (2)
\(\ldots\ldots \leqslant x \lt \ldots\ldots\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(1.85 \leqslant x \lt 1.95\) | B1 | for 1.85 in correct position |
| B1 | for 1.95 in correct position |
Accept \(1.94\dot{9}\) or 1.9499(…)
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(…)
19 \(ABC\) is a right-angled triangle.
\(AB = 9.3\) cm correct to the nearest mm.
\(AC = 12.6\) cm correct to the nearest mm.
Calculate the lower bound for the size of the angle marked \(x\).
You must show all your working. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 46.989... | B1 | for stating at least one bound, 9.25 or 9.35 or 12.55 or 12.65 |
| P1 | for correct use of LB and UB, eg \(\sin x = \dfrac{[\text{LB of } AB]}{[\text{UB of } AC]}\) | |
| A1 | for answer in the range 46.98 to 47 from correct working |
Accept \(9.34\dot{9}\) or 9.3499(…) or \(12.64\dot{9}\) or 12.6499(…)
\(9.25 \leqslant \text{LB} \lt 9.3\)
\(12.6 \lt \text{UB} \leqslant 12.65\)
\(\sin(x) = 0.731(\ldots)\)
Answer only award no marks.
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(…)
25
\(x = 4700\) correct to 2 significant figures.
\(\ldots\ldots\ldots \leqslant x \lt \ldots\ldots\ldots\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.25 | B1 | for 1.25 or \(\dfrac{5}{4}\) or \(1\dfrac{1}{4}\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 4650 and 4750 | B1 | for 4650 in the correct position |
| B1 | for 4750 in the correct position |
Accept \(4749.\dot{9}\) or 4749.99(…)
23 A race is measured to have a distance of 10.6 km, correct to the nearest 0.1 km.
Sam runs the race in a time of 31 minutes 48 seconds, correct to the nearest second.
Sam’s average speed in this race is \(V\) km/hour.
By considering bounds, calculate the value of \(V\) to a suitable degree of accuracy.
You must show all your working and give a reason for your answer. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 20 | B1 | stating bound, eg 10.65 or 10.55 or 31 min 48.5sec or 31 min 47.5sec or 1908.5sec or 1907.5sec |
| P1 | (dep on B1) for a correct bound for time in hours, eg 0.5301(38…) or 0.5298(61…) Or a correct process to find one bound for speed in km per minute, eg [UB of S] = [UB of D]÷[LB of T] or [LB of S] = [LB of D]÷[UB of T] or a correct process to find one bound for speed in km per second, eg [UB of S] = [UB of D]÷[LB of T] or [LB of S] = [LB of D]÷[UB of T] | |
| P1 | (dep on P1) for correct process to find one bound for speed in km per hour, eg [UB of S] = [UB of D]÷ 0.5298(61…) or [LB of S] = [LB of D]÷ 0.5301(38…) OR Correct process to convert a bound for speed in km per minute to km per hour eg [UB of S] = [UB of D]÷[LB of T] × 60 or [LB of S] = [LB of D]÷[UB of T] × 60 OR Correct process to convert a bound for speed in km per second to km per hour eg [UB of S] = [UB of D]÷[LB of T] × 60 × 60 or [LB of S] = [LB of D]÷[UB of T] × 60 × 60 | |
| A1 | for both correct bounds from correct working, 20.099… and 19.900… | |
| C1 | for 20 correct to 2 significant figures as both bounds agree |
Bound rounded or truncated to 4 dp
Where \(10.6 \lt \) [UB of D] \(\leqslant 10.65\) and 31 min 47.5 sec \(\leqslant\) [LB of T] \(\lt \) 31 min 48 sec
Or \(10.55 \leqslant\) [LB of D] \(\lt 10.6\) and 31 min 48 sec \(\lt \) [UB of T] \(\leqslant\) 31 min 48.5 sec
Where \(10.6 \lt \) [UB of D] \(\leqslant 10.65\) and 1907.5 sec \(\leqslant\) [LB of T] \(\lt \) 1908 sec
Or \(10.55 \leqslant\) [LB of D] \(\lt 10.6\) and 1908 sec \(\lt \) [UB of T] \(\leqslant\) 1908.5 sec
Time used in hours
Where \(10.6 \lt \) [UB of D] \(\leqslant 10.65\) and 31 min 47.5 sec \(\leqslant\) [LB of T] \(\lt \) 31 min 48 sec
Or \(10.55 \leqslant\) [LB of D] \(\lt 10.6\) and 31 min 48 sec \(\lt \) [UB of T] \(\leqslant\) 31 min 48.5 sec
Where \(10.6 \lt \) [UB of D] \(\leqslant 10.65\) and 1907.5 sec \(\leqslant\) [LB of T] \(\lt \) 1908 sec
Or \(10.55 \leqslant\) [LB of D] \(\lt 10.6\) and 1908 sec \(\lt \) [UB of T] \(\leqslant\) 1908.5 sec
Figures rounded or truncated to 3 sf or better
16 \(p = \sqrt{\dfrac{2e}{f}}\)
\(e = 6.8\) correct to 1 decimal place.
\(f = 0.05\) correct to 1 significant figure.
Work out the upper bound for the value of \(p\).
Give your answer correct to 3 significant figures.
You must show all your working. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.4 | B1 | for stating any correct bound, eg 6.75 or 6.85 or 0.045 or 0.055 |
| M1 | using both UB of \(e\) and LB of \(f\) to work out value of \(2e \div f\), eg 2[UB of \(e\)] \(\div\) [LB of \(f\)] or \(\dfrac{2 \times 6.85}{0.045}\) | |
| A1 | for answer in the range 17.4 to 17.5 from correct working |
Accept \(6.84\dot{9}\) or 6.8499... for 6.85
and \(0.054\dot{9}\) or 0.05499.. for 0.055
\(6.8 \lt \text{UB}(e) \leqslant 6.85\)
\(0.045 \leqslant \text{LB}(f) \lt 0.05\)
If an answer is given in the range in working and then rounded incorrectly award full marks.
Award 0 marks for a correct answer with no (or incorrect) supportive working
21 The time period, \(T\) seconds, of a simple pendulum of length \(l\) cm is given by the formula
\[T = 2\pi\sqrt{\frac{l}{g}}\]
Katie uses a simple pendulum in an experiment to find an estimate for the value of \(g\).
Here are her results.
\(l = 52.0\) correct to 3 significant figures.
\(T = 1.45\) correct to 3 significant figures.
Work out the upper bound and the lower bound for the value of \(g\).
Use \(\pi = 3.142\)
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 984.(3677853) and 969.(0181643) | B1 | stating bound of 51.95 or 52.05 or 1.445 or 1.455 |
| P1 | for process to rearrange formula to give \(g\) as the subject, eg \(g = \dfrac{4\pi^2 l}{T^2}\) oe | |
| P1 | for process to use LB of \(l\) and UB of \(T\) in formula for \(g\) or \(T\) or process to use UB of \(l\) and LB of \(T\) in formula for \(g\) or \(T\) eg \(\dfrac{4\pi^2[\text{LB of } l]}{[\text{UB of } T]^2}\) or \(\dfrac{4\pi^2[\text{UB of } l]}{[\text{LB of } T]^2}\) | |
| A1 | for upper bound = 984.(3677853) or 984.(1125639..) and lower bound = 969.(0181643) or 968.(7669227..) |
Accept \(52.04\dot{9}\) or 52.0499… for 52.05
Accept \(1.454\dot{9}\) or 1.4549… for 1.455
Rearrangement may occur after substitution, in this case correct bounds are not needed for this mark
\(51.95 \leqslant [\text{LB of } l] \lt 52.0\)
\(1.45 \lt [\text{UB of } T] \leqslant 1.455\)
\(52.0 \lt [\text{UB of } l] \leqslant 52.05\)
\(1.445 \leqslant [\text{LB of } T] \lt 1.45\)
Rearrangement may not be correct
NB: correct answer without supportive working gets 0 marks
Accept answers rounded or truncated to 3sf or better
11 Freya writes down the value of \(x\), correct to 1 decimal place.
She writes \(x = 6.4\)
Complete the error interval for \(x\).
\(\ldots\ldots \leqslant x \lt \ldots\ldots\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6.35, 6.45 | B1 | for 6.35 in the correct position |
| B1 | for 6.45 in the correct position |
Accept \(6.44\dot{9}\) oe or 6.4499… oe
20 \(d = \dfrac{1}{8}c^3\)
\(c = 10.9\) correct to 3 significant figures.
By considering bounds, work out the value of \(d\) to a suitable degree of accuracy.
Give a reason for your answer. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 160 (supported) | B1 | stating bound of 10.85 or 10.95 |
| M1 | using both UB and LB to work out value of \(d\) eg [UB of \(c\)]\(^3 \div 8\) and [LB of \(c\)]\(^3 \div 8\) or gives a bound of 159.66… from correct working or gives a bound of 164.11… from correct working | |
| A1 | for 159.66… and 164.11… from correct working | |
| C1 | for 160 from 159.66… and 164.11… with a supporting reason eg “since both UB and LB round to 160” |
Accept 10.949 or 10.9499… for 10.95
\(10.9 \lt \text{UB} \leqslant 10.98\)
\(10.85 \leqslant \text{LB} \lt 10.9\)
Accept bounds rounded or truncated to at least 4 sig fig
25 Sally used her calculator to work out the value of a number \(y\).
The answer on her calculator display began
8.3
Complete the error interval for \(y\). (2)
\(\ldots\ldots\ldots \leqslant y \lt \ldots\ldots\ldots\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8.3 and 8.4 | B1 | for 8.3 in the correct position |
| B1 | for 8.4 in the correct position |
Accept \(8.3\dot{9}\) or 8.399...
19 \(D = \dfrac{u^2}{2a}\)
\(u = 26.2\) correct to 3 significant figures
\(a = 4.3\) correct to 2 significant figures
The lower bound for the value of \(D\) is 78.6003 correct to 6 significant figures.
| Answer | Mark | Mark scheme |
|---|---|---|
| 81.0662 | M1 | for one of 26.15 or 26.25 or 4.25 or 4.35 |
| M1 | for a correct process to find the upper bound for \(D\) [UB of \(u\)]\(^2 \div [2 \times\) LB of \(a\)] eg \(\dfrac{26.25^2}{2 \times 4.25}\) where \(26.2 \lt \) UB of \(u \leqslant 26.25\) and \(4.25 \leqslant\) LB of \(a \lt 4.3\) | |
| A1 | for answer given in the range 81.0661 to 81.0662 from correct working |
Accept \(26.24\dot{9}\) for 26.25 and \(4.34\dot{9}\) for 4.35
Award for \(\dfrac{26.25^2}{4.25}\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 80 explanation | B1 | for 80 ft answer to (a) with 78.6003 |
| C1 | for explanation relating to the upper bound found in (a) Acceptable examples bounds agree when rounded to 80 bounds agree to nearest 10 Not acceptable examples 80 79.83325 rounded to nearest tenth |
6 Sally used her calculator to work out the value of a number \(y\).
The answer on her calculator display began
8.3
Complete the error interval for \(y\).
……… \(\leqslant y \lt \) ……… (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8.3 and 8.4 | B1 | for 8.3 in the correct position |
| B1 | for 8.4 in the correct position |
Accept \(8.3\dot{9}\) or 8.399…
18 A high speed train travels a distance of 487 km in 3 hours.
The distance is measured correct to the nearest kilometre.
The time is measured correct to the nearest minute.
By considering bounds, work out the average speed, in km/minute, of the train to a suitable degree of accuracy.
You must show all your working and give a reason for your answer. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2.7 with statement | B1 | for 179.5 or 180.5 or 180.4999... |
| B1 | for 486.5 or 487.5 or 487.4999... | |
| P1 | for a correct process to find a bound for average speed, eg [upper bound of distance] ÷ [lower bound of time] where \(487 \lt\) [UB of distance] \(\leqslant 487.5\) and \(179.5 \leqslant\) [LB of time] \(\lt 180\) or for [lower bound of distance] ÷ [upper bound of time] where \(486.5 \leqslant\) [LB of distance] \(\lt 487\) and \(180 \lt\) [UB of time] \(\leqslant 180.5\) | |
| A1 | (dep on all previous marks) for 2.695(2…) and 2.715(8…) with both values clearly coming from working with correct values | |
| C1 | for 2.7 from 2.695… and 2.715… and statement that both LB and UB round to 2.7 |
Accept bounds truncated or rounded to at least 4 sig fig
21 Jackson is trying to find the density, in g/cm3, of a block of wood.
The block of wood is in the shape of a cuboid.
He measures
the length as 13.2 cm, correct to the nearest mm
the width as 16.0 cm, correct to the nearest mm
the height as 21.7 cm, correct to the nearest mm
He measures the mass as 1970 g, correct to the nearest 5 g.
By considering bounds, work out the density of the wood.
Give your answer 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 |
|---|---|---|
| 0.43 | B1 | for one correct bound for mass or length eg 1967.5 or 1972.5 or 13.15 or 15.95 or 21.65 or 13.25 or 16.05 or 21.75 |
| P1 | for a correct process to find a bound for the volume, eg \(13.15 \times 15.95 \times 21.65\) (=454(0.925125)) or \(13.25 \times 16.05 \times 21.75\) (=462(5.409375)) | |
| P1 | for a correct process to find a bound for density, eg [mass LB] \(\div\ \text{``}462(5.409375)\text{''}\) (=0.425(367755)) where \(1965 \leqslant\) mass LB \(\lt 1970\) or [mass UB] \(\div\ \text{``}454(0.925125)\text{''}\) (=0.434(3828506)) where \(1970 \lt\) mass UB \(\leqslant 1975\) | |
| A1 | for both correct bounds, 0.425(367755) and 0.434(3828506) | |
| C1 | (dep on A1) for a correct statement on degree of accuracy e.g. UB and LB both round to 0.43 to 2 decimal places or 2 significant figures |
B1: Can work in any units
P1 (volume): Accept volumes truncated or rounded to at least 3 sig fig
P1 (density): Accept densities truncated or rounded to at least 3 sig fig
A1: Accept bounds truncated or rounded to at least 3 sig fig
At this point correct units must be used
C1: Must be 0.43 not 0.4
23
Jess rounds a number, \(x\), to one decimal place.
The result is 9.8
| Answer | Mark | Notes |
|---|---|---|
| 0.625 | B1 | cao |
| Answer | Mark | Notes |
|---|---|---|
| \(9.75 \leqslant x \lt 9.85\) | B2 | \(9.75 \leqslant x \lt 9.85\) |
| [B1 | for 9.75 or 9.85 (or \(9.84\dot{9}\))] |
16 The petrol consumption of a car, in litres per 100 kilometres, is given by the formula
\(\text{Petrol consumption} = \dfrac{100 \times \text{Number of litres of petrol used}}{\text{Number of kilometres travelled}}\)
Nathan’s car travelled 148 kilometres, correct to 3 significant figures.
The car used 11.8 litres of petrol, correct to 3 significant figures.
Nathan says,
“My car used less than 8 litres of petrol per 100 kilometres.”
Could Nathan be wrong?
You must show how you get your answer. (3)
| Answer | Mark | Notes |
|---|---|---|
| Yes and correct working | B1 | for 147.5 or 148.5 or \(148.4999\ldots\) or 11.75 or 11.85 or \(11.84999\ldots\) |
| P1 | substitutes \(11.8 \lt \text{UB} \leqslant 11.85\) and \(147.5 \leqslant \text{LB} \lt 148\) in the formula to work out petrol consumption | |
| A1 | for ‘Yes’ and \(8.03(3898305\ldots)\) from correct working |
13 A factory makes 450 pies every day.
The pies are chicken pies or steak pies.
Each day Milo takes a sample of 15 pies to check.
The proportion of the pies in his sample that are chicken is the same as the proportion of the pies made that day that are chicken.
On Monday Milo calculated that he needed exactly 4 chicken pies in his sample.
On Tuesday, the number of steak pies Milo needs in his sample is 6 correct to the nearest whole number.
Milo takes at random a pie from the 450 pies made on Tuesday.
| Answer | Mark | Notes |
|---|---|---|
| 120 | P1 | for \(\dfrac{4 \times 450}{15}\) or \(\dfrac{4}{15} = \dfrac{x}{450}\) oe |
| A1 | cao |
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{165}{450}\) | P1 | 5.5 or 6.5 or 165 or \(\dfrac{5 \times 450}{15}\ (=150)\) and \(\dfrac{6 \times 450}{15}\ (=180)\) |
| A1 | for \(\dfrac{165}{450}\) oe |
5
Jess rounds a number, \(x\), to one decimal place.
The result is 9.8
| Answer | Mark | Notes |
|---|---|---|
| 0.625 | B1 | cao |
| Answer | Mark | Notes |
|---|---|---|
| \(9.75 \leqslant x \lt 9.85\) | B2 | for \(9.75 \leqslant x \lt 9.85\) |
| [B1 | for 9.75 or 9.85 (or \(9.84\dot{9}\))] |
23 A number, \(n\), is rounded to 2 decimal places.
The result is 4.76
Using inequalities, write down the error interval for \(n\). (2)
| Answer | Mark | Notes |
|---|---|---|
| \(4.755 \leqslant n \lt 4.765\) | B2 | for \(4.755 \leqslant n \lt 4.765\) |
| [B1 | for 4.755 or 4.765 or \(4.764\dot{9}\)] |
23 Harley’s house has a value of £160 000 correct to 2 significant figures.
The value of Rita’s house increased by 5%.
Her house then had a value of £210 000
| Answer | Mark | Notes |
|---|---|---|
| (i) 155 000 | B1 | cao |
| (ii) 165 000 or 164 999 or 164 999.99 | B1 | 165 000 or 164 999 or 164 999.99 |
| Answer | Mark | Notes |
|---|---|---|
| 200 000 | M1 | for recognising that 210 000 = 105% or a full method to find the original price eg \(210\,000 \div 1.05\) oe \((= 200\,000)\) |
| A1 | cao |
17 A train travelled along a track in 110 minutes, correct to the nearest 5 minutes.
Jake finds out that the track is 270 km long.
He assumes that the track has been measured correct to the nearest 10 km.
Jake’s assumption was wrong.
The track was measured correct to the nearest 5 km.
| Answer | Mark | Notes |
|---|---|---|
| No (supported) | P1 | for 265 or 275 or 274.999... or 107.5 or 112.5 or 112.4999... |
| P1 | process to find \(\dfrac{d}{t}\) where \(270 \lt d \leqslant 275\) and \(107.5 \leqslant t \lt 110\) oe | |
| P1 | for process to work in consistent units of time eg \(\dfrac{d}{t} \times 60\) or \(t \div 60\) where \(265 \leqslant d \leqslant 275\) and \(107.5 \leqslant t \lt 110\) oe or \(160 \div 60\ (= 2.666..)\) | |
| C1 | Conclusion supported with correct figure(s) given eg No and 153(.488..) or No and 2.66 to 2.7 and 2.5(581..) from correct working |
| Answer | Mark | Notes |
|---|---|---|
| Statement | C1 | e.g. Less distance in the same time so (max) speed would drop |
7 A number, \(n\), is rounded to 2 decimal places.
The result is 4.76
Using inequalities, write down the error interval for \(n\). (2)
| Answer | Mark | Notes |
|---|---|---|
| \(4.755 \leqslant n \lt 4.765\) | B2 | for \(4.755 \leqslant n \lt 4.765\) |
| [B1 | for 4.755 or 4.765 or \(4.764\dot{9}\)] |
22 \(a = \dfrac{v - u}{t}\)
\(v = 37.6\) correct to 3 significant figures.
\(u = 11.3\) correct to 3 significant figures.
\(t = 8.4\) correct to 2 significant figures.
Work out the upper bound for the value of \(a\).
Show your working clearly. (3)
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{528}{167}\) | 3 | B1 for 37.55 or 37.65 or 11.25 or 11.35 or 8.35 or 8.45 M1 for \(\dfrac{37.65 - 11.25}{8.35}\) for \(\dfrac{v_{ub} - u_{lb}}{t_{lb}}\) where \(37.6 \lt v_{ub} \leqslant 37.65\) and \(11.25 \leqslant u_{lb} \lt 11.3\) and \(8.35 \leqslant t_{lb} \lt 8.4\) A1 for answer in range 3.16 to 3.162 from correct working |
20 Jarek uses the formula \[\text{Area} = \frac{1}{2}ab\sin C\] to work out the area of a triangle.
For this triangle,
\(a = 7.8\) cm correct to the nearest mm.
\(b = 5.2\) cm correct to the nearest mm.
\(C = 63^\circ\) correct to the nearest degree.
Calculate the lower bound for the area of the triangle. (3)
| Answer | Mark | Notes |
|---|---|---|
| 17.7(014…) | 3 | B1 for 7.75 or 7.85 or 5.15 or 5.25 or 62.5 or 63.5 M1 for \(\dfrac{1}{2} \times 7.75 \times 5.15 \times \sin 62.5\) A1 for 17.7(0140994…) |
21 \[I = 5(v - u)\]
\(v = 14\) correct to 2 significant figures
\(u = 8.7\) correct to 2 significant figures
Work out the upper bound for the value of \(I\).
You must show your working. (3)
| Answer | Mark | Notes |
|---|---|---|
| 29.25 | 3 | B1 for one of 14.5, 13.5, 8.75, 8.65 M1 for \(\text{``}v_{\text{UB}}\text{''} - \text{``}u_{\text{LB}}\text{''}\) where \(14 \lt \text{``}v_{\text{UB}}\text{''} \leqslant 14.5\) and \(8.65 \leqslant \text{``}u_{\text{LB}}\text{''} \lt 8.7\) A1 for 29.25 from correct working |
19 The value of \(p\) is 4.3
The value of \(q\) is 0.4
Both \(p\) and \(q\) are given correct to the nearest 0.1
\(r = p + \dfrac{1}{q}\)
| Answer | Mark | Notes |
|---|---|---|
| 4.25 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7.20-7.21 | 3 | B1 4.35 or 0.35 M1 for \(4.35 + \dfrac{1}{0.35}\) A1 7.2(0)-7.21 or \(\dfrac{1009}{140}\) from a correct method seen |
24 Steve travelled from Ashton to Barnfield.
He travelled 235 miles, correct to the nearest 5 miles.
The journey took him 200 minutes, correct to the nearest 5 minutes.
Calculate the lower bound for the average speed of the journey.
Give your answer in miles per hour, correct to 3 significant figures.
You must show all your working. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{232.5}{202.5} \times 60\) | 68.9 | 4 | M1 for 232.5 or 237.5 or 197.5 (=3.29… hours) or 202.5 (= 3.375 hours) M1 for correct conversion of “upper bound of time” from minutes to hours, (202.5 to 205) ÷ 60 M1 for “lower bound of distance” ÷ “upper bound of time” (230 to 232.5) ÷ (3.375 to 3.41(6…)) A1 for 68.8 to 69 from correct working OR M1 for 232.5 or 237.5 or 197.5 (=3.29… hours) or 202.5 (= 3.375 hours) M1 for “lower bound of distance” ÷ “upper bound of time” (230 to 232.5) ÷ (202.5 to 205) M1 for correct conversion of “lower bound of speed” from miles per minute to miles per hour, ((1.12(1...) to 1.14(8...)) × 60 A1 for 68.8 to 69 from correct working |
*23 A road is 4530 m long, correct to the nearest 10 metres.
Kirsty drove along the road in 205 seconds, correct to the nearest 5 seconds.
The average speed limit for the road is 80 km/h.
Could Kirsty’s average speed have been greater than 80 km/h?
You must show your working. (5)
| Answer | Mark | Notes |
|---|---|---|
| Yes, average speed could have been as high as 80.622... | 5 | B1 for 4535 or 4534.999... or 202.5 M1 for 4535 (oe) ÷ 202.5 M1 for ×3600 and ÷1000 A1 for 80.622... C1 (dep on first M1) for correct conclusion from their calculations |
23 Dan does an experiment to find the value of \(\pi\).
He measures the circumference and the diameter of a circle.
He measures the circumference, \(C\), as 170 mm to the nearest millimetre.
He measures the diameter, \(d\), as 54 mm to the nearest millimetre.
Dan uses \(\pi = \dfrac{C}{d}\) to find the value of \(\pi\).
Calculate the upper bound and the lower bound for Dan’s value of \(\pi\). (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(d\): UB = 54.5 (or 54.499), LB = 53.5 \(C\): UB = 170.5 (or 170.499), LB = 169.5 \(170.5 \div 53.5\) \(169.5 \div 54.5\) | 3.19 3.11.. | 4 | B1 for any one correct bound quoted M1 for \(170.5 \div 53.5\) or \(169.5 \div 54.5\) A1 for UB = answer in range 3.18 to 3.19 from correct working A1 for LB = 3.11.. from correct working |
*24 \(m = \dfrac{\sqrt{s}}{t}\)
\(s = 3.47\) correct to 2 decimal places
\(t = 8.132\) correct to 3 decimal places
By considering bounds, work out the value of \(m\) to a suitable degree of accuracy.
You must show all your working and give a reason for your final answer. (5)
| Answer | Mark | Notes |
|---|---|---|
| 0.229 because the LB and UB agree to that number of figures | 5 | B1 for 3.465 or 3.475 or 3.474999… B1 for 8.1315 or 8.1325 or 8.132499… M1 for \(\dfrac{\sqrt{3.475}}{8.1315}\) as UB OR \(\dfrac{\sqrt{3.465}}{8.1325}\) as LB C1 (dep on all previous marks) for 0.2292… and 0.2288… both values must clearly come from working with correct values C1 for 0.229 from 0.2292… and 0.2288… and ‘both LB and UB round to 0.229’ |
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
The total mass of a plane is calculated by adding these three masses.
The maximum mass for the plane to take off safely is 1200 kg
Can this plane definitely take off safely?
Tick a box.
Show working to support your answer. [4 marks]
On the grid, draw a speed/time graph to represent this information. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: uses the upper bounds | ||
| 750 or 850 or 187.5 or 192.5 or 162.5 or 163.5 | B1 | condone 849.99… for 850 condone 192.49… for 192.5 condone 163.49… for 163.5 |
| 850 and 192.5 and 163.5 | B1 | |
| (800, 900] + (190, 195] + (163, 164] | M1 | |
| 1206 and No | A1 | oe 6 kg too heavy and No |
| Alternative method 2: uses other values | ||
| Correct sum of three possible values that exceeds 1200 and No | B4 | possible values must lie between the bounds, on the lower bound or on the upper bound eg \(849 + 192 + 163 = 1204\) and No |
| Units not required | |
| 850 and 195 and 163.5 with answer 1208.5 | B1B0M1A0 |
| 900 and 195 and 164 with answer 1259 | B0B0M1A0 |
| There are no part marks for Alt method 2 | |
| Mark to the Alt method which favours the student eg 849 and 192 and 163.5 with answer 1204.5 and ‘No’ could score B1B0M1A0 using Alt method 1, but scores B4 using Alt method 2 | |
| \(845 + 194 + 163 = 1202\) and No does not score using either Alt method |
| Answer | Mark | Comments |
|---|---|---|
| (Speed =) 500 | B1 | implied by 500 as highest point plotted or marked accept 8.33… with correct units miles per minute stated or implied |
| Vertical axis labelled ‘Speed’ with a linear scale to at least their 500 and Horizontal axis labelled ‘Time’ with a linear scale to at least 2 hours 30 minutes or 150 minutes | B1ft | ft their 500 which must not be 1250 accept their speed as the only value on the vertical axis and 2 hours 30 minutes as the only value on the horizontal axis ignore all units shown on their graph |
| Horizontal line from (0, their 500) stopping at (2 hours 30 minutes, their 500) or (150 minutes, their 500) | B1ft | ft their 500 which must not be 1250 ignore any vertical line at 2 hours 30 minutes |
| ‘Speed’ and ‘Time’ are the only acceptable labels | |
| Use of a distance/time graph is a maximum of B1 for 500 seen | |
| 1250/150 = 8.33… miles per minute and correctly plotted with correct labelling and scales | B1B1ftB1ft |
| \(1250 \div 2.3 = 543\ldots\), with horizontal line from (0, 543) to (2 hours 30 minutes, 543) with correct labelling and scales | B0B1ftB1ft |
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 |
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 |
16 When \(w\) is truncated the answer is 8
Sacha writes the error interval due to truncation as \(\quad 7.5 \leqslant w \lt 8.5\)
Give a reason why Sacha is wrong and state the correct error interval for \(w\). [2 marks]
Correct error interval \(\quad \ldots\ldots\ldots \leqslant w \lt \ldots\ldots\ldots\)
| Answer | Mark | Comments |
|---|---|---|
| Valid reason | B1 | see examples below |
| \(8 \leqslant w \lt 9\) | B1 | condone \(8.\dot{9}\) for 9 |
| Reasons | |
| These are the bounds for rounding to 8 | B1 |
| This is the error interval for the nearest whole number | B1 |
| The number must begin with 8 for it to truncate to 8 | B1 |
| The number cannot be less than 8 | B1 |
| When a number is truncated it gets smaller | B1 |
| \(w\) is being made lower, not higher | B1 |
| When a number is truncated the decimals are removed | B1 |
| The values have to be whole numbers/the bounds shouldn’t be decimals (with no decimals in their bounds) | B1 |
| 7.5 would truncate to 7 | B1 |
| 7.5 is wrong with 8 given as the correct lower bound or 8.5 is wrong with 9 given as the correct upper bound | B1 |
| 7.5 is wrong or 8.5 is wrong without the appropriate correct bound | B0 |
9 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
4 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 |
21 Jack is loading a van.
The van can safely carry 1375 kg of furniture.
Jack has already loaded 1200 kg of furniture to the nearest 50 kg
A table has mass 140 kg to the nearest 10 kg
Can the table safely be added to the furniture in the van?
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| 1225 or 1175 or 145 or 135 | M1 | |
| their 1225 + their 145 or 1370 and 1225 and 145 | M1 | must add two upper bounds their 1225 must be (1200, 1250] their 145 must be (140, 150] |
| 1370 and Yes and 1225 and 145 | A1 | |
| Alternative method 2 | ||
| 1225 or 1175 or 145 or 135 | M1 | |
| \(1375 -\) their 1225 or \(1375 -\) their 145 | M1 | their 1225 must be (1200, 1250] their 145 must be (140, 150] |
| 145 and 150 and Yes or 1225 and 1230 and Yes | A1 | |
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Note that M0M1A0 is possible eg \(1224 + 144\) | M0M1A0 |
| Accept correct use of decimals eg \(1224.\dot{9}\) for 1225 | |
| Yes may be implied eg The table can be added |
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
3 The length of a line is 8 cm to the nearest centimetre.
Complete the error interval. [2 marks]
______ cm \(\leqslant\) length \(\lt\) ______ cm
| Answer | Mark | Comments |
|---|---|---|
| 7.5 cm \(\leqslant\) length \(\lt\) 8.5 cm | B2 | B1 one length correct in correct position SC1 8.5 cm \(\leqslant\) length \(\lt\) 7.5 cm |
Accept \(8.4\dot{9}\) for 8.5
24 \(a = 65\) to the nearest integer
\(b = 30\) to 1 significant figure
Work out the upper bound for \(\quad 2a^2 - b^2\)
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 64.5 or 65.5 or 25 or 35 | M1 | allow \(65.4\dot{9}\) or \(34.\dot{9}\) implied by 4160.25 or 4290.25 or 8320.5 or 8580.5 or 625 or 1225 |
| \(2 \times\) their \(65.5^2 -\) their \(25^2\) or \(2 \times 4290.25 - 625\) or \(8580.5 - 625\) | M1 | their 65.5 must be (65, 66] their 25 must be [20, 30) |
| 65.5 and 25 and 7955.5 | A1 |
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| If multiple attempts are seen and one is fully correct, the correct one must be unambiguously selected (eg ticked or circled) to award A1 if the answer line is blank | |
| Note that M0M1A0 is possible eg \(2 \times 66^2 - 21^2\) | M0M1A0 |
| Condone eg 65.50 for 65.5 |
25 Two sides of a triangle are measured to 1 decimal place.
The angle between the sides is measured to the nearest degree.
Work out the upper bound for the area of the triangle.
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 7.15 or 7.25 or 13.55 or 13.65 or 109.5 or 110.5 | B1 | |
| 7.25 and 13.65 and 109.5 chosen | B1 | |
| \(0.5 \times\) their \(7.25 \times\) their \(13.65 \times \sin\) their 109.5 | M1 | their 7.25 must be [7.2, 7.25] their 13.65 must be [13.6, 13.65] their 109.5 must be [109.5, 110] or 110.5 |
| 46.6(4…) with correct bounds seen | A1ft | condone 47 with B1B1 scored ft their three bounds within M1 ranges which are not 7.2 or 13.6 or 110 |
| Accept \(7.24\dot{9}\) for 7.25 or \(13.64\dot{9}\) for 13.65 or \(110.4\dot{9}\) for 110.5 | |
| 7.25 and 13.65 and 110.5 used and answer 46.3… | B1B0M1A1ft |
| 7.25 and 13.65 and 110 used and answer 46.497… or 46.5 | B1B0M1A0ft |
| 7.2 and 13.6 and 110 used, with or without answer 46(.0…) | B0B0M1A0ft |
| 46.6(4…) or 47 with no working | B0B0M0A0 |
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 |
27 To be rented, a bedroom must have a floor area of at least 6.51 m2
A bedroom has a rectangular floor.
The floor measures 2.4 m by 2.9 m, each correct to 2 significant figures.
Show that the bedroom can be rented. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Sight of at least one of 2.35 or 2.45 or 2.85 or 2.95 | M1 | allow \(2.44\dot{9}\) for 2.45 and \(2.94\dot{9}\) for 2.95 |
| their \(2.35 \times\) their 2.85 | M1 | \(2.3 \leqslant\) their \(2.35 \lt 2.4\) \(2.8 \leqslant\) their \(2.85 \lt 2.9\) |
| \(2.35 \times 2.85\) selected and 6.6(975) | A1 | accept 6.7(0) or 6.698 with \(2.35 \times 2.85\) selected |
| Alternative method 2 | ||
| Sight of at least one of 2.35 or 2.45 or 2.85 or 2.95 | M1 | allow \(2.44\dot{9}\) for 2.45 and \(2.94\dot{9}\) for 2.95 |
| \(6.51 \div\) their 2.35 or \(6.51 \div\) their 2.85 | M1 | \(2.3 \leqslant\) their \(2.35 \lt 2.4\) \(2.8 \leqslant\) their \(2.85 \lt 2.9\) |
| \(6.51 \div 2.35\) and 2.7(7…) and 2.85 or \(6.51 \div 2.85\) and 2.2(8…) and 2.35 | A1 | |
| Alt 1 \(2.35 \times 2.85\) amongst other calculations eg \(2.45 \times 2.95\) and/or \(2.35 \times 2.95\) can still score the second M1 but it must be clear that they are considering \(2.35 \times 2.85 = 6.6(975)\) to show that the bedroom can be rented | |
| eg1 \(2.35 \times 2.85 = 6.6975 \qquad 2.45 \times 2.95 = 7.2275\) | M1M1A0 |
| eg2 \(2.35 \times 2.85 = 6.6975 \qquad 2.45 \times 2.95 = 7.2275\) \(2.35 \times 2.95 = 6.9325 \qquad\) The lower bounds show it can be rented | M1M1A1 |
| Ignore the calculation \(2.4 \times 2.9\) throughout | |
| Alt 1 6.6(975) or 6.7 or 6.698 without \(2.35 \times 2.85\) selected | A0 |
| 6.6975 only | M0M0A0 |
| Alt 2 2.7(7…) without \(6.51 \div 2.35\) and 2.85 seen | A0 |
| Alt 2 2.2(8…) without \(6.51 \div 2.85\) and 2.35 seen | A0 |
10 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 |
28 \(a = 4.72 \quad\) to 3 significant figures.
\(b = 158 \quad\) to 3 significant figures.
Work out the upper bound of \(\dfrac{a}{b}\)
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 4.715 or 4.725 or 157.5 or 158.5 | B1 | accept \(4.724\dot{9}\) or 4.7249… accept \(158.4\dot{9}\) or 158.49… |
| their \(4.725 \div\) their 157.5 | M1 | their 4.725 must be (4.72, 4.725] their 157.5 must be [157.5, 158) |
| 0.03 and correct working | A1 | oe eg \(\dfrac{3}{100}\) |
| Answer 0.03 with no correct working | B0M0A0 |
| 4.7249 exact value with no continuation dots seen | B0 |
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 |
26 Edith’s van can safely carry a maximum load of 920 kilograms.
She wants to use her van to carry
30 sacks of potatoes, each of mass 25 kilograms to the nearest kilogram
and
20 sacks of carrots, each of mass 7.5 kilograms to 1 decimal place.
Can she definitely use her van safely in one journey?
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 24.5 or 25.5 or 7.45 or 7.55 | B1 | accept \(25.4\dot{9}\) for 25.5 accept \(7.54\dot{9}\) for 7.55 |
| \(30 \times\) their 25.5 or 765 or \(20 \times\) their 7.55 or 151 | M1 | their 25.5 must be (25, 26] their 7.55 must be (7.5, 7.6] |
| \(30 \times\) their \(25.5 + 20 \times\) their 7.55 or \(765 + 151\) or 916 | M1dep | oe eg \(920 - 30 \times\) their \(25.5 - 20 \times\) their 7.55 their 25.5 must be (25, 26] their 7.55 must be (7.5, 7.6] |
| 25.5 and 7.55 and 916 and Yes | A1 | oe eg 25.5 and 7.55 and \(-4\) and Yes |
| Only using lower bounds can score a maximum of B1M0M0A0 | |
| Condone 25.50 for 25.5 etc | |
| 916 and Yes without both 25.5 and 7.55 is A0 but the B mark and M marks are possible eg \(30 \times 25.5 + 20 \times 7.54\ (= 915.8) = 916\) Yes | B1M1M1A0 |
| 916 and Yes with no working | Zero |
| Yes can be implied eg1 \(30 \times 25.5 + 20 \times 7.55 = 916\) which is less than 920 eg2 \(30 \times 25.5 + 20 \times 7.55 = 916\) so she can | B1M1M1A1 B1M1M1A1 |
20 To the nearest 1000, there are 18 000 people at a festival.
| Answer | Mark | Comments |
|---|---|---|
| 17 500 | B1 |
Accept response in words
| Answer | Mark | Comments |
|---|---|---|
| 18 499 | B1 |
| Accept response in words | |
| \(18\,499.\dot{9}\) or \(18\,49\dot{9}\) | B0 |
5 To the nearest 1000, there are 18 000 people at a festival.
| Answer | Mark | Comments |
|---|---|---|
| 17 500 | B1 |
Accept response in words
| Answer | Mark | Comments |
|---|---|---|
| 18 499 | B1 |
| Accept response in words | |
| \(18\,499.\dot{9}\) or \(18\,49\dot{9}\) | B0 |
4 When rounded to 3 significant figures, \(x = 6.37\)
Circle the correct error interval. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(6.365 \leqslant x < 6.375\) | B1 |
21 To the nearest pound, Jon has £9
To the nearest 50p, Ellie has £6.50
Work out the maximum possible total amount of money. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 8.5(0) or 9.49 or 9.5(0) or 6.25 or 6.74 or 6.75 | B1 | |
| 9.49 + 6.74 or (9, 9.5] + (6.5, 6.75] | M1 | |
| 16.23 | A1 | accept (£)16.23p SC2 16.25 or 16.24 |
| 9.5(0) and 6.55 with answer 16.05 | B1M1A0 |
| 9.4(0) and 6.25 with answer 15.65 | B1M0A0 |
| 9.4(0) and 6.55 with answer 15.95 | B0M1A0 |
17
\[m = \frac{p - 2b}{2}\]\(p = 68.3 \qquad\) correct to 1 decimal place.
\(b = 8.7 \qquad\) correct to 1 decimal place.
Work out the lower bound for \(m\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 68.3 – 0.05 or 68.25 or 68.3 + 0.05 or 68.35 or 8.7 – 0.05 or 8.65 or 8.7 + 0.05 or 8.75 | M1 | accept \(68.34\dot{9}\) for 68.35 accept \(8.74\dot{9}\) for 8.75 may be seen in an inequality eg \(68.25 \leqslant p < 68.35\) |
| \(\dfrac{[68.2, 68.3) - 2 \times (8.7, 8.8]}{2}\) | M1 | oe \(\dfrac{68.25 - 2 \times 8.75}{2}\) or \(\dfrac{68.25 - 17.5}{2}\) or \(\dfrac{50.75}{2}\) is M2 |
| 25.375 or \(\dfrac{203}{8}\) or \(25\dfrac{3}{8}\) | A1 | SC2 Answer 25.375 and 25.525 |
| 1st M1 If given as an inequality condone incorrect notation eg \(68.25 \leqslant p \leqslant 68.35\) | M1 |
| Ignore any subsequent rounding after 25.375 seen | |
| Condone eg 68.250 for 68.25 | M1 |
| Answer 25.3 or 25.4 with no correct working | M0M0A0 |
| Only working for upper bound eg \(\dfrac{68.35 - 2 \times 8.65}{2} = 25.525\) | M1M0A0 |
6 To the nearest pound, Jon has £9
To the nearest 50p, Ellie has £6.50
Work out the maximum possible total amount of money. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 8.5(0) or 9.49 or 9.5(0) or 6.25 or 6.74 or 6.75 | B1 | |
| 9.49 + 6.74 or (9, 9.5] + (6.5, 6.75] | M1 | |
| 16.23 | A1 | accept (£)16.23p SC2 16.25 or 16.24 |
| 9.5(0) and 6.55 with answer 16.05 | B1M1A0 |
| 9.4(0) and 6.25 with answer 15.65 | B1M0A0 |
| 9.4(0) and 6.55 with answer 15.95 | B0M1A0 |
21 The length of a table is 110 cm to the nearest cm
Complete the error interval. [2 marks]
______ cm \(\leqslant\) length \(<\) ______ cm
| Answer | Mark | Comments |
|---|---|---|
| 109.5 in the correct position | B1 | oe |
| 110.5 in the correct position | B1 | oe Allow \(110.4\dot{9}\) answers reversed score B0B1 |
| 110.4999… | B1 |
| 110.4999 | B0 |
19 The length of a roll of ribbon is 30 metres, correct to the nearest half-metre.
A piece of length 5.8 metres, correct to the nearest 10 centimetres, is cut from the roll.
Work out the maximum possible length of ribbon left on the roll. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 30.25 or 29.75 or 5.85 or 5.75 | B1 | |
| \(\text{their } 30.25 - \text{their } 5.75\) | M1 | Must be their max roll – their min cut their max must be (30, 30.5] their min must be [5.5, 5.8) |
| 24.5 | A1 |
| 30.5 – 5.75 = 24.75 | B1M1A0 |
5 The length of a table is 110 cm to the nearest cm
Complete the error interval. [2 marks]
.......... cm \(\leqslant\) length \(<\) .......... cm
| Answer | Mark | Comments |
|---|---|---|
| 109.5 in the correct position | B1 | oe |
| 110.5 in the correct position | B1 | oe Allow \(110.4\dot{9}\) answers reversed score B0B1 |
| 110.4999… | B1 |
| 110.4999 | B0 |
28 The length of each side of a regular pentagon is 8.4 cm to 1 decimal place.
\(\ldots\ldots\ldots\) cm \(\leqslant\) length \(<\) \(\ldots\ldots\ldots\) cm
\(\ldots\ldots\ldots\) cm \(\leqslant\) perimeter \(<\) \(\ldots\ldots\ldots\) cm
| Answer | Mark | Comments |
|---|---|---|
| 8.35 and 8.45 in the correct order | B2 | B1 8.35 on the left or 8.45 on the right or 8.45 and 8.35 in the wrong order accept \(8.44\dot{9}\) for 8.45 |
Do not accept 8.449… for \(8.44\dot{9}\)
| Answer | Mark | Comments |
|---|---|---|
| 41.75 and 42.25 | B1ft | correct or ft their two different values from (a) their 8.35 must be in the range (8.3, 8.4] their 8.45 must be in the range (8.4, 8.5] correct order or ft order accept \(42.24\dot{9}\) for 42.25 |
| (8.3, 8.4] does not include 8.3 but does include 8.4 (8.4, 8.5] does not include 8.4 but does include 8.5 | |
| Answer of 8.35 and 8.44 in part (a) leading to 41.75 and 42.2 | B1ft |
| Answer of 8 and 9 in part (a) leading to 40 and 45 | B0ft |
24 A tank is a cuboid measuring 50 cm by 35 cm by 20 cm
All lengths are to the nearest centimetre.
A container has a capacity of exactly 34 litres.
1 litre = 1000 cm\(^3\)
Which has the greater capacity?
Tick one box.
Show working to support your answer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Any product of three valid dimensions that would give a volume \(\lt\) 34 000 or any product of three valid dimensions that would give a volume \(\gt\) 34 000 | M1 | eg \(49.5 \times 34.5 \times 19.5\) or \(50.5 \times 35.5 \times 20.5\) or \(50 \times 35 \times 20\) ignore any evaluations of products |
| Any product of three valid dimensions that would give a volume \(\lt\) 34 000 and any product of three valid dimensions that would give a volume \(\gt\) 34 000 | M1dep | eg \(49.5 \times 34.5 \times 19.5\) and \(50.5 \times 35.5 \times 20.5\) ignore any evaluations of products |
| \(34 \times 1000\) or 34 000 | M1 | converts to cm\(^3\) |
| their volume \(\lt\) 34 000 and their volume \(\gt\) 34 000 and 34 000 and ticks Cannot tell | A1 | both volumes in cm\(^3\) must see working for M3 answers for their volumes must be seen and be correct or rounded or truncated to at least 2 sf (unless 34 000 to 2 sf when must be to at least 3 sf) |
| Alternative method 2 | ||
| Any product of three valid dimensions that would give a volume \(\lt\) 34 000 or any product of three valid dimensions that would give a volume \(\gt\) 34 000 | M1 | eg \(49.5 \times 34.5 \times 19.5\) or \(50.5 \times 35.5 \times 20.5\) or \(50 \times 35 \times 20\) ignore any evaluations of products |
| Any product of three valid dimensions that would give a volume \(\lt\) 34 000 and any product of three valid dimensions that would give a volume \(\gt\) 34 000 | M1dep | eg \(49.5 \times 34.5 \times 19.5\) and \(50.5 \times 35.5 \times 20.5\) ignore any evaluations of products |
| one of their volumes \(\div\) 1000 | M1dep | dep on first M1 converts to litres |
| their volume \(\lt\) 34 and their volume \(\gt\) 34 and ticks Cannot tell | A1 | both volumes in litres must see working for M3 answers for their volumes must be seen and be correct or rounded or truncated to at least 2 sf (unless 34 000 to 2 sf when must be to at least 3 sf) |
| There are an infinite number of sets of three valid dimensions Valid dimensions for 50 are [49.5, 50.5] for 35 are [34.5, 35.5] for 20 are [19.5, 20.5] | |
| \(49.5 \times 34.5 \times 19.5 = 33\,301.(\ldots)\) or 33 000 or 33 300 \(49.6 \times 34.6 \times 19.6 = [33\,636, 33\,637]\) or 33 000 or 33 600 or 33 630 or 33 640 \(49.7 \times 34.7 \times 19.7 = 33\,974.(\ldots)\) or 33 000 or 33 900 or 33 970 \(49.8 \times 34.8 \times 19.8 = 34\,314.(\ldots)\) or 34 300 or 34 310 \(49.9 \times 34.9 \times 19.9 = 34\,656.(\ldots)\) or 34 600 or 34 700 or 34 650 or 34 660 \(50 \times 35 \times 20 = 35\,000\) \(50.1 \times 35.1 \times 20.1 = 35\,346.(\ldots)\) or 35 000 or 35 300 or 35 340 or 35 350 \(50.2 \times 35.2 \times 20.2 = 35\,694.(\ldots)\) or 35 000 or 36 000 or 35 600 or 35 700 or 35 690 \(50.3 \times 35.3 \times 20.3 = 36\,044.(\ldots)\) or 36 000 or 36 040 \(50.4 \times 35.4 \times 20.4 = [36\,396, 36\,397]\) or 36 000 or 36 300 or 36 400 or 36 390 \(50.5 \times 35.5 \times 20.5 = 36\,751.(\ldots)\) or 36 000 or 37 000 or 36 700 or 36 800 or 36 750 | |
| Three valid dimensions do not have to follow a pattern eg \(49.6 \times 35 \times 20.4\) (\(= 35\,414.(\ldots)\) or 35 000 or 35 400 or 35 410) | M1 |
| 49.5 34.5 19.5 and 33 301 (answer implies multiplication signs) | M1 |
| 49.5 34.5 19.5 (no answer so multiplication signs not implied) | M0 |
| 33 301 but 49.5 34.5 19.5 not seen | M0 |
| Units do not have to be seen |
9 The length of each side of a regular pentagon is 8.4 cm to 1 decimal place.
| Answer | Mark | Comments |
|---|---|---|
| 8.35 and 8.45 in the correct order | B2 | B1 8.35 on the left or 8.45 on the right or 8.45 and 8.35 in the wrong order accept \(8.44\dot{9}\) for 8.45 |
Do not accept 8.449… for \(8.44\dot{9}\)
| Answer | Mark | Comments |
|---|---|---|
| 41.75 and 42.25 | B1ft | correct or ft their two different values from (a) their 8.35 must be in the range (8.3, 8.4] their 8.45 must be in the range (8.4, 8.5] correct order or ft order accept \(42.24\dot{9}\) for 42.25 |
| (8.3, 8.4] does not include 8.3 but does include 8.4 (8.4, 8.5] does not include 8.4 but does include 8.5 | |
| Answer of 8.35 and 8.44 in part (a) leading to 41.75 and 42.2 | B1ft |
| Answer of 8 and 9 in part (a) leading to 40 and 45 | B0ft |
25 The dimensions of a rectangular floor are to the nearest 0.1 metres.
Not drawn accurately
A force of 345 Newtons is applied to the floor.
The force is to the nearest 5 Newtons.
\(\text{pressure} = \dfrac{\text{force}}{\text{area}}\)
Work out the upper bound of the pressure.
Give your answer to 4 significant figures.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| 342.5 or 347.5 | B1 | Allow \(347.4\dot{9}\) for 347.5 |
| 6.35 or 6.45 or 2.55 or 2.65 | B1 | Allow \(6.44\dot{9}\) for 6.45 Allow \(2.64\dot{9}\) for 2.65 |
| their \(6.35 \times\) their 2.55 or 16.1925 | M1 | Must use their lower bounds for lengths their 6.35 must be [6.3, 6.4] their 2.55 must be [2.5, 2.6] |
| their \(347.5 \div\) their 16.1925 | M1dep | Must use their upper bound for force their 347.5 bound must be (345, 350] |
| 21.46 | A1 | Must come from \(347.5 \div (6.35 \times 2.55)\) or \(347.4\dot{9} \div (6.35 \times 2.55)\) |
| \(347.49 \div (6.35 \times 2.55) = 21.46\) | B0B1M1M1A0 |
| 21.4… or 21.5 does not score any marks if no working is seen |
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 |
8 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 | |
| \(39 + 33 + 13 = 85 \qquad (40 + 30 + 10 = 80)\) | M0 |
| Beware 82 from incorrect values, eg \(39 + 24 + 19 = 82\) | M0A0 |
| Ignore incorrectly evaluated trials that do not solely lead to the answer |
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 |
24 A kitchen has space for a fridge.
Complete the error interval for the width, \(w\).
\[\ldots\ldots\ldots\ldots \leqslant w < \ldots\ldots\ldots\ldots\][2]
Show that this fridge may not fit in the space. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 69.5 [\(\leqslant w <\)] 70.5 | 2 | B1 for each If 0 scored SC1 for
| |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair of values in which space < fridge from these ranges \(69.5 \leqslant\) space \(< 69.55\) and \(69.5 <\) fridge \(< 69.55\) e.g. s = 69.5 and f = 69.54 | 2 | B1 for one value for space or fridge where \(69.5 \leqslant\) space \(< 69.55\) e.g s = 69.5 or \(69.5 <\) fridge \(< 69.55\) e.g. f = 69.51 | Error interval (e.g. ...\(\leqslant\) space < ...) alone scores 0 Only mark candidate’s chosen value(s) Measurements used must be clearly associated with space and/or fridge Space < fridge does not need to be stated For B1 if one or multiple values for fridge/space seen and no clear choice all must be in correct ranges |
6 A kitchen has space for a fridge.
Complete the error interval for the width, \(w\).
\(\ldots\ldots \leqslant w \lt \ldots\ldots\) [2]
Show that this fridge may not fit in the space. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 69.5 [\(\leqslant w \lt\)] 70.5 | 2 | B1 for each If 0 scored SC1 for
| |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair of values in which space < fridge from these ranges 69.5 ≤ space < 69.55 e.g. s = 69.5 and 69.5 < fridge < 69.55 e.g. f = 69.54 | 2 | B1 for one value for space or fridge where 69.5 ≤ space < 69.55 e.g s = 69.5 or 69.5 < fridge < 69.55 e.g. f = 69.51 | Error interval (e.g. ...≤ space < ...) alone scores 0 Only mark candidate’s chosen value(s) Measurements used must be clearly associated with space and/or fridge Space < fridge does not need to be stated For B1 if one or multiple values for fridge/space seen and no clear choice all must be in correct ranges |
22 The width, \(w\), of a car parking space is 2.8 metres, correct to 1 decimal place.
Complete the error interval for the width, \(w\).
\[\ldots\ldots\ldots\ldots \leqslant w < \ldots\ldots\ldots\ldots\][2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.75 and 2.85 | 2 | B1 for each If 0 scored SC1 for both correct but reversed | |
13 A water company is laying pipes to cover a distance of 37 metres, correct to the nearest metre.
Each pipe has a length of 2.3 metres, correct to 1 decimal place.
Assume the pipes are laid end to end with no gaps or overlaps.
Work out the minimum number of pipes the water company needs to be sure of covering that distance.
You must show your working. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17 with correct working | 4 | B1 for 37.5 selected or 37.49[9…] B1 for 2.25 selected M1 for (36.5 to 37.5) ÷ (2.25 to 2.35) If 0 scored, instead award SC1 for an answer of 17 with no working or insufficient working or SC1 for either 36.5 and 37.49[9…]/37.5 or 2.25 and 2.349[9…]/2.35 | “Correct working” requires evidence of at least B1B1 alt. method B1 for 2.25 selected B1 for 37.5 selected M1 for 2.25 × 16 = 36 M1 for 2.25 × 17 = 38.25 Max. 3 marks unless the answer is 17 Note : 37 ÷ 2.3 = 16.08… with answer 17 scores M1 |
23 A garage is 5 metres long, correct to the nearest metre.
A car is 4.5 metres long, correct to 1 decimal place.
Show that the car may not fit in the garage. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair where car > garage matched pair of values or non-overlapping ranges and the values quoted are \(4.5 \leqslant \text{garage} \lt 4.55\) \(4.5 \lt \text{car} \lt 4.55\) | 3 | B1 for a value in \(4.5 \leqslant \text{garage} \lt 5\) B1 for a value in \(4.5 \lt \text{car} \lt 4.55\) | Ranges must not overlap for 3 marks Values must be clearly associated with garage (or 5) or car (or 4.5) as appropriate. For B1 if choice of values given all must be in range, unless acceptable value(s) indicated |
16 A lift can travel at a maximum speed of 9.83 m / s, correct to 3 significant figures.
The lift travels a distance of 174 m, correct to the nearest metre, between the ground floor and the top floor of a building.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17.6[4…] with correct working | 4 | B2 for 173.5 and 9.835 or 9.8349[9..] selected or B1 for one of these or M1 for 173.5 to 174.5/174.49[9..] and 9.825 to 9.835/9.8349[9..] M1 for (173.5 to 174.5) ÷ (9.825 to 9.835) If 0 or 1 scored, instead award SC2 for answer 17.6[4…] with no working or insufficient working | “Correct working” requires evidence of the selection of 173.5 and 9.835 or 9.8349[9..] |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Any viable reason e.g. lift has to speed up and slow down | 1 | see appendix | |
| Response | Mark |
|---|---|
| lift has to speed up and slow down | 1 |
| it will take time to reach this speed | 1 |
| It will not travel at the maximum speed all the way | 1 |
| the distance could be greater than 173.5 (e.g. 174.3) and the speed may be less than 9.835 (e.g. 9.826) therefore the time will be greater than the shortest time oe | 1 |
| We are using the bounds so it may not be accurate/achievable | 1 |
| The figure for the height has been rounded down which may not be the height of the building | 1 |
| The figure for the speed has been rounded up which may not be the actual speed | 1 |
| It makes other stops on the way | 1 |
| Extra weight may slow it down | 1 |
| Technical fault may occur | 1 |
| May not be a constant speed | 1 |
| Lift gets stuck | 1 |
| Lift may not be able to reach that speed | 1 |
| Weight on the lift is a factor [so may slow it down] | 1(BOD) |
| It does not make any other stops | 0 |
14 A number, \(n\), is given as 2.6, truncated to 1 decimal place.
Complete the error interval for \(n\) using a number and a symbol.
\(\ldots\ldots\ldots \leqslant n \ldots\ldots\ldots 2.7\) [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(2.6 \leqslant n \lt 2.7\) | 2 | B1 for each (2.6 and \(\lt\)) | Do not accept other figures or symbols on line with correct answer Accept trailing zeros |
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 | ||
14 Here is a right-angled triangle.
Not to scale
The area of the triangle is 100 cm2, correct to the nearest 10 cm2.
The length of the base of the triangle is 8 cm, correct to the nearest cm.
Calculate the largest possible height of the triangle. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 28 nfww | 4 | B1 for 105 B1 for 7.5 M1 for \(\dfrac{\textit{their}\ (95 \text{ to } 105)}{\frac{1}{2} \times \textit{their}\ (7.5 \text{ to } 8.5)}\) but not \(\dfrac{100}{\frac{1}{2} \times 8}\) | In part marks condone 104.99 or better for 105 |
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 |
12 A box contains 200 matches, correct to the nearest ten matches.
Show that the smallest possible height of the box is 6 cm. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 195 [\(\leqslant n \leqslant\)] 204 | 2 | B1 for 195 or 204 in correct position or for both 195 and 204 but in wrong position | Do not accept < 205 where \(\leqslant\) has been crossed out and replaced |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{247.5}{7.5 \times 5.5} = 6\) | 3 | Max 2 marks if answer is incorrect B1 for 247.5 B1 for 7.5 and 5.5 M1 for \(\frac{their\ 247.5}{their\ 7.5 \times their\ 5.5}\) where \(247.5 \leqslant their\ 247.5 \leqslant 248.5\), \(6.5 \leqslant their\ 7.5 \leqslant 7.5\) and \(4.5 \leqslant their\ 5.5 \leqslant 5.5\) | May be done in stages. 6 must not come from a rounded answer or other wrong working. If choice, mark the bounds used in the calculation. If no calculation, then 0 each time upper and lower bounds are both given 7.5 × 5.5 × 6 = 247.5 scores just the second mark. If they then explicitly state that 247.5 is the lower bound of 248 they also get the first mark but not for “247.5 rounds to 248” |
16 A plane flies from London to Tokyo.
The distance is 9600 km, correct to the nearest 100 km.
The plane travels at an average speed of 820 km/h, correct to the nearest 10 km/h.
Calculate the shortest possible flight time of the plane.
Give your answer in hours and minutes, correct to the nearest minute.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 11 [hours] 35 [mins] with correct working | 5 | B4 for answers 11 [hours] 34 [mins] or 11 [hours] 34.5[…] to 34.8 [mins], with correct working OR B1 for 9550 B1 for 825 M1 for (9550 to 9650) ÷ (815 to 825) M1 for their 0.57… × 60 oe If 0, 1 or 2 scored, instead award SC3 for answer 11 [hours] 35 [mins] with no working or insufficient working If 0 or 1 scored, instead award SC2 for answer 11 [hours] 34 [mins] or 11 [hours] 34.5[…] to 34.8 [mins] with no working or insufficient working If 0 scored, instead award SC1 for 11.57[…] or 11.58 with no working or insufficient working | “Correct working” requires evidence of at least B1B1M1 Allow 824.9[9…] for 825 With no contradictory choice With no contradictory choice Do not imply this first M1 mark. their 0.57 following from distance ÷ speed implied by 11.57 to 11.85 |
23 Four friends are going on holiday together.
They each take a suitcase.
The weight of each suitcase is 25 kg, correct to nearest kilogram.
[2]
Can the friends be certain that they will not have to pay extra?
Show how you decide. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 24.5 25.5 cao | 2 | B1 for each If 0 scored SC1 for correct but reversed or for 24.50[0] and 25.50[0] | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(4 \times 25.5\) or \(4 \times\) their upper bound in (a) | M1 | M0 if \(w = 25\) used | |
| 102 or correct answer to \(4 \times\) their upper bound | A1 | ||
| Yes and their 102 \(\lt / \leqslant\) 102.4 in symbols or words | 1 dep | Dep on previous 2 marks Conclusion and clear comparison FT their upper bound in (a) | Yes/ no conclusions must be consistent with their working |
| Alternative method 1 | |||
| \(102.4 \div 4\) | M1 | ||
| 25.6 | A1 | ||
| Yes and their 25.5 \(\lt / \leqslant\) 25.6 in symbols or words | 1 dep | Dep on previous 2 marks Conclusion and clear comparison FT their upper bound in (a) | |
| Alternative method 2 | |||
| \(102.4 \div 25.5\) or \(102.4 \div\) their upper bound | M1 | ||
| 4.01(…) or correct answer to \(102.4 \div\) their upper bound | A1 | ||
| Yes and \(4 \lt / \leqslant\) their 4.01(…) | 1 dep | Dep on previous 2 marks Conclusion and clear comparison FT their upper bound in (a) | |
11 Jamie buys fence panels that fit tightly together.
Not to scale
Each panel has a length of 1.8 m, correct to 1 decimal place.
Jamie measures the length of a garden as 42 m, correct to the nearest metre.
Work out the minimum number of panels Jamie should buy in order to be certain that there are enough panels for the length of the garden.
Show how you decide. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 42.5 or 42.49[9..] ÷ 1.75 | M3 | B1 for 1.75 B1 for 42.5 or 42.49[9…] | B1 seen anywhere and mark figures crossed out if you think they have been “rejected” for use |
| 25 | A1 | If 0 scored award B3 for 24.29 or 24.28[5…] or if 0 or 1 scored award SC1 as well for their 24.29 seen and correctly rounded up | |
18 A carpenter measures the length, \(k\) metres, of a piece of wood.
They write
[2]
\(3.35 \leqslant k \lt 3.45\) means that the length of the piece of wood is 3.4 metres correct to the nearest centimetre.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.39 and 3.44 only | 2 | B1 for one only or for two correct and one extra | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 10 cm [between 3.35 and 3.45] oe or [If to nearest cm it should be between] 3.395 and 3.405 | 1 | Mention of 10 cm (range or difference) oe | |
| (ii) 3.405 | 1 | ||
9 The width, \(w\), of a kitchen cupboard is 60 cm, correct to the nearest centimetre.
[2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 59.5 60.5 | 2 | B1 for either one correct or both correct but reversed | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched wall and 6 × cupboard where values quoted satisfy: wall < 6 × cupboard where 362.5 ⩽ wall < 363 and 362.5 < 6 × cupboard ⩽ 363 OR wall ÷ 6 < cupboard where 362.5 ⩽ wall < 363 and \(60.41\dot{6}\) to 60.42 < cupboard ⩽ 60.5 OR wall ÷ cupboard < 6 where 362.5 ⩽ wall < 363 and \(60.41\dot{6}\) to 60.42 < cupboard ⩽ 60.5 | 3 | B1 for 362.5 ⩽ wall value < 363 B1 for 362.5 < 6 × cupboard value ⩽ 363 or 60.416 to 60.42 < cupboard value ⩽ 60.5 | Example: [lower bound of] wall is 362.5 [upper bound of] cupboard is 60.5 362.5 ÷ 60.5 = 5.9[9…] < 6 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6.5 | 3 | M2 for 363.5 – 6 × their lower bound of cupboard or 363.5 – 357 or M1 for [upper bound of wall =] 363.5 or for 6 × their lower bound of cupboard or [6 cupboards =] 357 | |
11
Complete the error interval for the length, \(d\).
.................... \(\leqslant d \lt\) .................... [2]
Calculate the shortest possible time for Jamal’s journey.
Give your answer correct to the nearest minute. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4.715 4.725 | 2 | B1 for each or both correct and reversed | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 25 | 5 | B4 for 7.41[0...] or 7.411 OR B1 for 425 used B1 for 57.35 or 57.349[9...] used If B0 then B1 for 425 and 57.35 seen AND M2 for 425 ÷ 57.35 or M1 for use of distance ÷ speed e.g 430 ÷ 57.3 If 4 or 5 not scored SC1 for correctly changing their part hour to minutes | Condone use of 57.349[9...] for M2 and for M1 420 ≤ distance ≤ 440 and 56 ≤ speed ≤ 58 |
15 Andrea is 165 cm tall, correct to the nearest cm.
Joel is 170 cm tall, correct to the nearest 10 cm.
Show that Andrea could be taller than Joel. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correctly matched pair where Andrea > Joel and values quoted are \(165 \lt \text{Andrea} \lt 165.5\) \(165 \leqslant \text{Joel} \lt 165.5\) | 3 | B1 for value \(165 \lt \text{Andrea} \lt 165.5\) B1 for value \(165 \leqslant \text{Joel} \lt 170\) | For B1 if choice of values given all must be in range, unless acceptable value(s) indicated Values must be clearly associated with either Andrea or Joel as appropriate. |
12
How long did Arron run for?
Give your answer in minutes and seconds, to the nearest second. [4]
The distance is measured correct to the nearest km.
The time is given correct to 1 decimal place.
Calculate the lower and upper bounds of her average speed.
Give your answers correct to 2 decimal places. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 37 minutes 52 seconds to 37 minutes 53 seconds | 4 | B1 for 5000 or 0.0022 seen and M1 for figs 5 ÷ figs 22 oe soi by figs 2272 to 2273 and M1 for figs (2272 to 2273) ÷ 60 soi by figs 37[….] | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 19.09 20.19 | 6 | B5 for 19.09 to 19.1 and 20.18 to 20.2 as final answers OR B2 for 53.5, 52.5, 2.65 and 2.75 all seen or B1 for two of 53.5, 52.5, 2.65 or 2.75 seen and M2 for both 53.5 ÷ 2.65 and 52.5 ÷ 2.75 or M1 for (52.5 to 53.5) ÷ (2.65 to 2.75) and A1dep for 19.09 to 19.1 and 20.18 to 20.2 If 0 scored, allow SC3 for one answer either 19.09 to 19.1 or 20.18 to 20.2 | Allow 2.749[9…] for 2.75 or 53.49[9…] for 53.5 For M2 ignore other unnecessary divisions e.g. 53.5 ÷ 2.75 and 52.5 ÷ 2.65 Dep on M2 |
11 The area of a rectangle is 56 m\(^2\), correct to the nearest m\(^2\).
The length of the rectangle is 9.2 m, correct to the nearest 0.1 m.
Calculate the smallest possible width of the rectangle. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [55.5 ÷ 9.25 or \(9.24\dot{9}\) =] 6 | 4 | 6 must not come from a rounded answer B1 for 55.5 B1 for 9.25 condone 9.2499 or better M1 for their 55.5 ÷ their 9.25 \((55.5 \leqslant\) their \(55.5 \leqslant 56.5\) and \(9.15 \leqslant\) their \(9.25 \leqslant 9.25)\) | |
16 A £1 coin weighs 8.75 g, correct to the nearest 0.01 g.
Mitul weighs the contents of a large bag of £1 coins.
The coins weigh 2.63 kg, correct to the nearest 10 g.
Mitul says
I am sure that the bag contains exactly £300 because, using bounds,
2625 ÷ 8.755 = 299.8 to 1 decimal place.
Show that Mitul may not be correct. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| There could be £301 e.g. because 2635 ÷ 8.745 = 301[.3…] | 3 | M2 for a calculation of (2625 to 2635) ÷ (8.745 to 8.755) oe correctly evaluated to an answer of 301(….) or for a calculation of (2632.245 to 2635) ÷ 301 oe correctly evaluated to an answer of 8.745 to 8.755 or for a calculation of 301 × (8.745 to 8.7541528…) oe correctly evaluated to an answer of 2625 to 2635 or M1 for any further calculation of (2625 to 2635) ÷ (8.745 to 8.755) or (2625 to 2635) ÷ 301 or 301 × (8.745 to 8.755) but not 2625 ÷ 8.755 or 2630 ÷ 8.75 or B1 for 2635, 2.635, 8.745 or 8745 seen | For full marks, their conclusion must follow from a relevant calculation which shows that 301 is a possible answer (either use of 301 and two weights in range, or an answer of more than 301 rounded down, and not an answer of less than 301 rounded up) Calculations may be done in grams as shown, or converted to kg. Common calculations for at least M2 include: 2635 ÷ 8.75 = 301.1(…) 2635 ÷ 8.745 = 301.3(…) Common calculations scoring only M1 include: 2625 ÷ 8.75 (= 300) 2630 ÷ 8.745 = 300.7(…) |
4 Use the symbols \(\lt\), \(\leqslant\), \(=\), \(\gt\), or \(\geqslant\) to complete this statement.
If \(x = 4.7\), truncated to 1 decimal place, then
4.7 ..... \(x\) ..... 4.8
[2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (4.7) \(\leqslant\) (\(x\)) \(\lt\) (4.8) | 2 | B1 for each symbol | |
14
A house is built using 4663 standard bricks.
Joslin says
Placed end to end, the bricks from the house would definitely reach over 1 km.
Show that Joslin’s statement is correct. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 21.45 × 4663 ÷ 100 000 = 1.000 2[1..] (km) or 21.45 × 4663 = 100 020 to 100 021.4 > 100 000 (cm) or. 100 000 ÷ 21.45 = 4662[.0..] < 4663 or 100 000 ÷ 4663 = 21.44[5..] < 21.45 Note the first method does not require a comparison against 1 (km) | 4 | B1 for (minimum length =) 21.45 seen | Allow access to all marks if brick and 1 km are in consistent units. |
| B1 for 1 km = 100 000 cm soi oe such as ÷ 100 then ÷ 1000 or use of 1m = 100cm and 1km = 1000m if working in metres. | Allow these conversions even with their volume or surface area. eg 21.5 × 10.3 × 6.5 = 1439.425 cm/cm\(^2\)/cm\(^3\) = 0.014 394 25 km | ||
| M1 for their 21.45 × 4663 (÷ 100 000) or 100 000 ÷ their 21.45 or 100 000 ÷ 4663 | their 21.45 must be in the range 21.45 to 21.55 but accept equivalent if attempting the unit conversion first eg B0B0M1 for 21.5 cm = 0.0215 km followed by 0.0215 × 4663 | ||
| If M0 scored, allow SC1 for \(k\) × 4663 (÷ 100 000) or 100 000 ÷ \(k\) with \(k\) in the range 10.25 to 10.35 or 6.45 to 6.55 | Thus, use of width or height of the brick may score B0,B1,SC1 whereas use of volume may score B0/1,B1,SC0 Accept equivalent if working in m or km | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 7017 to 7020 | 3 | B1 for 20 000 or 2.849[…] or 2.85 or 0.0028[…] seen M1 for their 20 000 ÷ their 2.85 or 20 ÷ their 0.00285 | Ignore other bound ie a division after an attempt to reach consistent units their 2.85 must be in the range 2.75 to 2.85 inc.; their 0.00285 must be in the range 0.00275 to 0.00285. B0M0 for 20 ÷ 2.8 as no attempt to reach consistent units |
| (ii) The truck may not have enough room oe Safety regulations may not allow it | 1 | Mark their best reason. 0 for we do not know the exact weight of the bricks oe 0 for because the truck may need to carry other loads 0 there may not be enough bricks available | |
12 A log is 18 m long, correct to the nearest metre.
It is to be cut into fence posts which must be 80 cm long, correct to the nearest 10 centimetres.
What is the largest number of fence posts that can possibly be cut from this log? [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 24 | 4 | B3 for 24.7 or 24.6[6…] OR B1 for 18.5 or 1850 B1 for 0.75 or 75 M1 for their1850 ÷ their75 oe soi by 24.7 or 24.6[6…] | condone 18.49 or 1849 in this question 1750 ≤ their1850 ≤ 1850 and 70 ≤ their75 ≤ 90 allow work in metres e.g. use of 1.75, 1.85, .7, .9 |
4
Write down the shortest possible length of the line. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(-12\) | 1 | ||
| (ii) 256 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10.35 cao | 1 | ||
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 Sunil makes 7.5 litres of soup, correct to the nearest 0.5 litre.
He serves the soup in 300 ml portions, correct to the nearest 10 ml.
24 people order this soup.
Does Sunil definitely have enough soup to serve the 24 people?
Show how you decide. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| No, with correct calculation leading to 23.77 to 23.8 identified or with 7.32 compared with 7.25 oe or 302 compared with 305 oe | 4 | B1 for 7250 or 7.25 seen B1 for 305 or 0.305 seen M1 for their 7.25 ÷ their 0.305 with consistent units and at least one attempted bound or for their 0.305 × 24 oe or their 7250 ÷ 24 oe | Ignore upper bound Ignore lower bound Their 7.25 in range 7 to 8, their 0.305 in range 0.29 to 0.31 or equivs. Ignore other divisions or products M0 for 7500 ÷ 300 or 7.5 ÷ 0.3 |
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 | |
No questions match these filters.