(i) Write \(\ 5.3 \times 10^4\ \) as an ordinary number. (1)
(ii) Write \(\ 7.4 \times 10^{-5}\ \) as an ordinary number. (1)
(b) Calculate the value of \(\quad 9.7 \times 10^6 + 2.45 \times 10^7\) Give your answer in standard form. (2)
Mark scheme (a)(i)
Answer
Mark
Mark scheme
53 000
B1
cao
Mark scheme (a)(ii)
Answer
Mark
Mark scheme
0.000 074
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
\(3.42 \times 10^7\)
M1
for \(9\,700\,000 + 24\,500\,000\ (= 34\,200\,000)\) or \(3.42 \times 10^n\ (n \ne 7)\) oe or \(3.4 \times 10^7\) or correct answer in incorrect form eg \(34.2 \times 10^6\) or both in a form ready for addition, eg \(9.7 \times 10^6 + 24.5 \times 10^6\)
(i) Write \(5.3 \times 10^4\) as an ordinary number. (1)
(ii) Write \(7.4 \times 10^{-5}\) as an ordinary number. (1)
(b) Calculate the value of \(9.7 \times 10^6 + 2.45 \times 10^7\) Give your answer in standard form. (2)
Mark scheme (a)
Answer
Mark
Mark scheme
(i) 53 000
B1
cao
(ii) 0.000 074
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
\(3.42 \times 10^7\)
M1
for \(9\,700\,000 + 24\,500\,000\ (= 34\,200\,000)\) or \(3.42 \times 10^n\ (n \neq 7)\) oe or \(3.4 \times 10^7\) or correct answer in incorrect form eg \(34.2 \times 10^6\) or both in a form ready for addition, eg \(9.7 \times 10^6 + 24.5 \times 10^6\)
(a) Write \(\quad (9 \times 10^4) : (4.5 \times 10^6) \quad\) in the form \(1 : n\) where \(n\) is an integer. (2)
(b) Write the following numbers in order of size. Start with the smallest number.\[5.625 \times 10^4 \qquad 5625 \qquad 56\,250 \times 10^{-3} \qquad 0.005\,625 \times 10^5\] (2)
Mark scheme (a)
Answer
Mark
Mark scheme
1 : 50
M1
for an equivalent ratio eg 9 : 450 or \(9 : 4.5 \times 10^2\) or 90000 : 4500000 oe or for \(4500000 \div 90000\) or \(\dfrac{4500000}{90000}\ (=50)\)
(a) Write \(\;6.75 \times 10^{-4}\;\) as an ordinary number. (1)
(b) Work out \(\quad \dfrac{2.56 \times 10^{6} \times 4.12 \times 10^{-3}}{1.6 \times 10^{-2}}\) Give your answer in standard form. (2)
Mark scheme (a)
Answer
Mark
Mark scheme
0.000675
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
\(6.592 \times 10^5\)
M1
for \(10.5472 \times 10^3\) oe or \(1.6 \times 10^8\) oe or \(2.575 \times 10^{-1}\) oe or for \(6.592 \times 10^n\) where \(n \neq 5\) or for \(6.59 \times 10^5\) or for \(6.6 \times 10^5\) or for 659200 oe
A1
cao
Additional guidance
If the answer (for 2 marks) is seen in working and then rounded or truncated, award full marks.
“6.212 is bigger than 4.73 so \(A\) is bigger than \(B\).”
(c) Is Asma correct? You must give a reason for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
\(3.246 \times 10^7\)
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
0.00496
B1
cao
Mark scheme (c)
Answer
Mark
Mark scheme
No with explanation
C1
No and explanation that B is bigger as the power of 10 is bigger. Acceptable examples She is incorrect as \(10^8\) is smaller than \(10^9\) No, because B has more digits than A No, A is millions but B is billions No, if you subtract A from B the answer is positive (but if you subtract B from A the answer is negative) A= 621200000, B=4730000000, B is bigger No because she did not take into account standard form No as when you find the ordinary number B is greater than A
Not acceptable examples Yes... A = 5 zeros after the number where as B = 7 zeros after the number No as \(4.73 \times 10^9\) is one more than \(6.212 \times 10^8\) 6.212 is to the power of 8 and 4.73 is to the power of 9 so there is an extra digit Asma is wrong because she has more numbers behind the decimal point which means that it will be bigger than A No B has more zeros
Additional guidance
Decision eg “No” may be seen by the question. “She is incorrect” is equivalent to “no”
27 Work out \((3.42 \times 10^{-7}) \div (7.5 \times 10^{-6})\) Give your answer in standard form. (2)
Mark scheme
Answer
Mark
Mark scheme
\(4.56 \times 10^{-2}\)
M1
for \(0.000000342 \div 0.0000075\) OR for 0.0456 oe eg \(0.456 \times 10^{-1}\) or \(45.6 \times 10^{-3}\) or \(\dfrac{57}{1250}\) OR for an answer of \(4.56 \times 10^n\)
7 Work out \((3.42 \times 10^{-7}) \div (7.5 \times 10^{-6})\) Give your answer in standard form. (2)
Mark scheme
Answer
Mark
Mark scheme
\(4.56 \times 10^{-2}\)
M1
for \(0.000000342 \div 0.0000075\) OR for 0.0456 oe eg \(0.456 \times 10^{-1}\) or \(45.6 \times 10^{-3}\) or \(\dfrac{57}{1250}\) OR for an answer of \(4.56 \times 10^n\)
“6.212 is bigger than 4.73 so A is bigger than B.”
(c) Is Asma correct? You must give a reason for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
\(3.246 \times 10^7\)
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
0.00496
B1
cao
Mark scheme (c)
Answer
Mark
Mark scheme
No with explanation
C1
No and explanation that B is bigger as the power of 10 is bigger.
Additional guidance
Acceptable examples She is incorrect as \(10^8\) is smaller than \(10^9\) No, because B has more digits than A No, A is millions but B is billions No, if you subtract A from B the answer is positive (but if you subtract B from A the answer is negative) A = 621200000, B = 4730000000, B is bigger No because she did not take into account standard form No as when you find the ordinary number B is greater than A
Not acceptable examples Yes… A = 5 zeros after the number where as B = 7 zeros after the number No as \(4.73 \times 10^9\) is one more than \(6.212 \times 10^8\) 6.212 is to the power of 8 and 4.73 is to the power of 9 so there is an extra digit Asma is wrong because she has more numbers behind the decimal point which means that it will be bigger than A No B has more zeros
Decision eg “No” may be seen by the question. “She is incorrect” is equivalent to “no”
11 In May 2019, the distance between Earth and Mars was \(3.9 \times 10^7\) km.
In May 2019, a signal was sent from Earth to Mars. Assuming that the signal sent from Earth to Mars travelled at a speed of \(3 \times 10^5\) km per second,
(a) how long did the signal take to get to Mars? (2)
The speed of the signal sent from Earth to Mars in May 2019 was actually less than \(3 \times 10^5\) km per second.
(b) How will this affect your answer to part (a)? (1)
Mark scheme (a)
Answer
Mark
Mark scheme
130
P1
for process to divide eg \((3.9 \times 10^7) \div (3 \times 10^5)\)
A1
cao
Additional guidance
Condone missing brackets
Accept \(1.3 \times 10^2\)
Mark scheme (b)
Answer
Mark
Mark scheme
Explanation
C1
Explanation referring to the time Acceptable examples The time will be more It will take longer The answer will be bigger Not acceptable examples The answer will be wrong The answer will be different
(a) Write the number 0.000 086 23 in standard form. (1)
(b) Work out \(\dfrac{3.2 \times 10^3 + 5.1 \times 10^{-2}}{4.3 \times 10^{-4}}\) Give your answer in standard form, correct to 3 significant figures. (2)
Mark scheme (a)
Answer
Mark
Mark scheme
\(8.623 \times 10^{-5}\)
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
\(7.44 \times 10^6\)
M1
for \(\dfrac{3200 + 0.051}{0.00043}\) or \(\dfrac{3200.051}{0.00043}\) or performs an operation eg shows 163.2, 7441860.5, 118.6(…) or an answer or \(7.44(\ldots.) \times 10^n\) where \(n \neq 6\) or 7441979(….) or an answer of \(7.4 \times 10^6\)
A1
for \(7.44(1979\ldots) \times 10^6\)
Additional guidance
7441979.0689…
If a correct answer is shown in working and then rounded incorrectly, award full marks. Answer need only be given correctly to 3 sig fig; if following digits are incorrect ignore them.
(a) Work out the value of \(T\). Give your answer in standard form correct to 3 significant figures. (2)
\(w\) is increased by 10% \(d\) is increased by 5%
Lottie says,
“The value of \(T\) will increase because both \(w\) and \(d\) are increased.”
(b) Lottie is wrong. Explain why. (2)
Mark scheme (a)
Answer
Mark
Mark scheme
\(4.52 \times 10^3\)
M1
for \(2.04.... \times 10^7\) oe eg \(2.04.... \times 10^{-5} \div 10^{-12}\) or \(20.4... \times 10^6\) or 204(08163.27) or for correct value of \(T\), 4517.(53....), not written in standard form, eg 4520
A1
for answer in the range \(4.51 \times 10^3\) to \(4.52 \times 10^3\)
(SC B1 for \(6.32... \times 10^{-1}\))
Additional guidance
May be given correct to 3 sig figs or more
Mark scheme (b)
Answer
Mark
Mark scheme
Explanation
M1
for method to find the scale factor or decreased value in \(T\), eg \(\sqrt{\dfrac{1.1}{1.05^3}}\) (= 0.97......) oe or \(\sqrt{\dfrac{5.6 \times 10^{-5} \times 1.1}{(1.4 \times 10^{-4} \times 1.05)^3}}\) (\(= 4.40.... \times 10^3\)) oe
C1
(dep M1) for explanation eg value of scale factor less than 1, so a decrease in \(T\) OR compares \(4.40... \times 10^3\) with their value of \(T\) from (a) provided answer to (a) is greater
Additional guidance
Award mark for a correct method to calculate the scale factor or the percentage increases in \(w\) and \(d^3\) or the decreased value of \(T\)
This mark may only be awarded if supported by numerical evidence
for \(\dfrac{6 \times 10^{-2} \times 3 \times 10^{-4}}{1 \times 10^{-2}}\) or \(18 \times 10^{-4}\) or 0.0018 as the answer
(M1
for \(6 \times 0.0003\) or \(0.06 \times 0.03\) or \(1.8 \times 10^{n}\ (n \neq -3)\) or \(0.000018 \div 0.01\) or rewriting one number in standard form)
(a) Write \(7.97 \times 10^{-6}\) as an ordinary number. (1)
(b) Work out the value of \((2.52 \times 10^{5}) \div (4 \times 10^{-3})\) Give your answer in standard form. (2)
Mark scheme (a)
Answer
Mark
Notes
0.00000797
B1
cao
Mark scheme (b)
Answer
Mark
Notes
\(6.3 \times 10^{7}\)
M1
for partial calculation involving powers of 10 e.g. \(0.63 \times 10^{5 - -3}\) or \(6.3 \times 10^{n}\) where \(n \ne 7\) or for \(n \times 10^{8}\) or for 63000000
20 William is building a planetary path for people to walk along. The planetary path will have a model of the Sun and models of the planets.
William uses two different scales. He uses
1 cm to 1000 km for the diameter of each planet 1 m to 1 000 000 km for the distance from the Sun to each planet
William makes a model of the planet Venus. The model has a diameter of 12.1 cm.
(a) Work out the real diameter of the planet Venus. Give your answer in standard form. (2)
William works out the distance from the model of the Sun to the model of the planet Neptune.
The real distance from the Sun to the planet Neptune is \(4.503 \times 10^9\) km.
(b) Work out the distance from the model of the Sun to the model of the planet Neptune. Give your answer in km, correct to 1 decimal place. (3)
Mark scheme (a)
Answer
Mark
Notes
\(1.21 \times 10^4\)
2
M1 for \(12.1 \times 1000\) (= 12100) oe A1 cao
Mark scheme (b)
Answer
Mark
Notes
4.5
3
M1 for \(4.503 \times 10^9 \div 10^6\) (= 4503) oe or for \(4.503 \times 10^9 \div 10^3\) (= 4503000) oe M1 for \(4.503 \times 10^9 \div 10^6 \div 10^3\) (=4.503 with no other digits) A1 for cao
M1 for converting all numbers to same form with at least one conversion correct A1 for fully correct order with correct numbers in any correct form (SC B1 if one number incorrectly placed or all 4 numbers listed in reverse order)
*15 One sheet of paper is \(9 \times 10^{-3}\) cm thick.
Mark wants to put 500 sheets of paper into the paper tray of his printer. The paper tray is 4 cm deep.
Is the paper tray deep enough for 500 sheets of paper? You must explain your answer. (3)
Mark scheme
Answer
Mark
Notes
No + explanation
3
M1 for \(500 \times 9 \times 10^{-3}\) oe A1 for 4.5 C1 (dep M1) for correct decision based on comparison of their paper height with 4
OR
M1 for \(4 \div 500\) oe A1 for 0.008 C1 (dep M1) for correct decision based on comparison of their paper thickness with 0.009
OR
M1 for \(4 \div (9 \times 10^{-3})\) oe A1 for 444(.4...) C1 (dep M1) for correct decision based on comparison of their number of sheets of paper with 500
Rob makes a model of the Sun. He also makes a model of the planet Jupiter.
Rob is going to hang the two models in the school hall. Rob wants a distance of 16 m between the two models.
The real distance between the planet Jupiter and the Sun is \(8 \times 10^8\) km.
Work out the scale Rob should use. Give your answer in the form \(1 : n\) (3)
Mark scheme
Working
Answer
Mark
Notes
16 metres : \(8 \times 10^8\) km. 16 : \(8 \times 10^8 \times 1000\) 16 : \(8 \times 10^{11}\) 1 : \(5 \times 10^{10}\) OR 2 m to \(10^8\) km 2m to 100 000 000 000m 1m to 50 000 000 000m
1 : \(5 \times 10^{10}\)
3
M1 (indep) correct method to convert to consistent units M1 \(\dfrac{\text{‘}8 \times 10^8\text{’}}{\text{‘16’}}\) (units may not be consistent) or \(5 \times 10^{10}\) oe or \(5 \times 10^7\) oe A1 1: \(5 \times 10^{10}\) or 1: 50 000 000 000 OR M1 (indep) correct method to convert to consistent units M1 \(\dfrac{\text{‘16’}}{8}\) to ‘\(10^8\)’ A1 1: \(5 \times 10^{10}\) or 1: 50 000 000 000
B3 for \(4.88 \times 10^{-5}\) to \(4.9 \times 10^{-5}\) (B2 for digits 238(23529) or 24 or 488(09353) or 49) (B1 for digits 81 or 34) OR B3 for \(4.88 \times 10^{-5}\) to \(4.9 \times 10^{-5}\) (B2 for digits 238(23529) or 24 or 488(09353) or 49) (B1 for digits 25 or 117(647))
(b) Write \(6.7 \times 10^{-5}\) as an ordinary number. (1)
(c) Work out the value of \((3 \times 10^7) \times (9 \times 10^6)\) Give your answer in standard form. (2)
Mark scheme (a)
Answer
Mark
Notes
1
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
0.000067
1
B1 cao
Mark scheme (c)
Answer
Mark
Notes
\(2.7 \times 10^{14}\)
2
M1 for \(27 \times 10^{7 + 6}\) or \(27 \times 10^{13}\) oe or an answer of \(2.7 \times 10^n\) where \(n\) is an integer or an answer of \(a \times 10^{14}\) where \(1 \leqslant a \lt 10\) A1 cao
condone extra zeros which do not affect the value eg \(6.0 \times 10^{-3}\) B1 correct value not in standard form (allow value written as \(a \times 10^n\)) eg 0.006 or \(0.6 \times 10^{-2}\) or given calculation with 0.6 written as \(6 \times 10^{-1}\) eg \(6 \times 10^{-1} \div 100\) or value whose only non-zero digit is 6 correctly converted to standard form answer eg \(0.0006 = 6 \times 10^{-4}\)
Additional guidance
\(6 \times 10^{-1} \div 10^2\)
B1
Mark scheme (b)
Answer
Mark
Comments
\(1.2 \times 10^8\)
B2
condone extra zeros which do not affect the value eg \(1.20 \times 10^8\) B1 120 000 000 or 120 million or \(12 \times 10^7\) or \(120 \times 10^6\) or \(1.2 \times 10^a \times 10^b\) where \(a + b = 8\) or value whose only non-zero digits are 12 correctly converted to standard form answer eg \(12\,000\,000 = 1.2 \times 10^7\)
condone extra zeros which do not affect the value eg \(6.0 \times 10^{-3}\) B1 correct value not in standard form (allow value written as \(a \times 10^n\)) eg 0.006 or \(0.6 \times 10^{-2}\) or given calculation with 0.6 written as \(6 \times 10^{-1}\) eg \(6 \times 10^{-1} \div 100\) or value whose only non-zero digit is 6 correctly converted to standard form answer eg \(0.0006 = 6 \times 10^{-4}\)
Additional guidance
\(6 \times 10^{-1} \div 10^2\)
B1
Mark scheme (b)
Answer
Mark
Comments
\(1.2 \times 10^8\)
B2
condone extra zeros which do not affect the value eg \(1.20 \times 10^8\) B1 120 000 000 or 120 million or \(12 \times 10^7\) or \(120 \times 10^6\) or \(1.2 \times 10^a \times 10^b\) where \(a + b = 8\) or value whose only non-zero digits are 12 correctly converted to standard form answer eg \(12\,000\,000 = 1.2 \times 10^7\)
B1 400 000 oe correct answer not in standard form eg \(40 \times 10^4\) or \(8 \times 10^7\) or \(2 \times 10^2\) or \(8 \times 10^5 \div 2\) or \(4 \times 10^7 \div 100\) or any value seen and then correctly converted to standard form eg 4 000 000 and \(4 \times 10^6\) 40 000 and \(4 \times 10^4\)
Additional guidance
Ignore incorrect position of commas or spacing in long numbers
Condone 400 000 and \(4 \times 10^5\) on the answer line, in either order Condone 40 000 and \(4 \times 10^4\) on the answer line, in either order
B2 B1
400 000 only on the answer line
B1
Do not award both marks for the correct answer from incorrect working but B1 can be awarded for one or both numbers incorrectly converted to standard form and the result of their division given correctly in standard form
B1 400 000 oe correct answer not in standard form eg \(40 \times 10^4\) or \(8 \times 10^7\) or \(2 \times 10^2\) or \(8 \times 10^5 \div 2\) or \(4 \times 10^7 \div 100\) or any value seen and then correctly converted to standard form eg 4 000 000 and \(4 \times 10^6\) 40 000 and \(4 \times 10^4\)
Additional guidance
Ignore incorrect position of commas or spacing in long numbers
Condone 400 000 and \(4 \times 10^5\) on the answer line, in either order Condone 40 000 and \(4 \times 10^4\) on the answer line, in either order
B2 B1
400 000 only on the answer line
B1
Do not award both marks for the correct answer from incorrect working but B1 can be awarded for one or both numbers incorrectly converted to standard form and the result of their division given correctly in standard form
(b) Work out \(\quad \dfrac{1.8 \times 10^2}{3 \times 10^{-1}}\)
Give your answer as an ordinary number. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
\(2 \times 10^3\) or \(7 \times 10^4\) or 140 000 000
M1
oe correct value not in standard form eg \(14 \times 10^7\)
\(1.4 \times 10^8\)
A1
SC1 Correctly converts an ordinary number with at least four digits to standard form
Additional guidance
Condone extra zeros on 1.4 eg \(1.40\,000\,000 \times 10^8\)
M1A1
\(1.4 \times 10^8\) from 1 400 000 000
M0A0
\(2 \times 10^3\) is implied by \((2 \times 7) \times (10^3 \times 10^a)\) \(7 \times 10^4\) is implied by \((2 \times 7) \times (10^b \times 10^4)\)
M1
1 400 000 000 converted to \(1.4 \times 10^9\)
SC1
Mark scheme (b)
Answer
Mark
Comments
180 or 0.3 or \((1.8 \div 3 =)\) 0.6 or \((10^2 \div 10^{-1} =)\ 10^3\) or calculation which would have the outcome 600 or correct value not given as an ordinary number
M1
eg \(1800 \div 3\) eg \(6 \times 10^2\)
600
A1
Additional guidance
\(1800 \div 0.3 = 600\) scores M1 only, as 600 comes from incorrect working
M1A0
\(1800 \div 30 = 600\) scores zero, as 600 comes from incorrect working
(b) Work out \(\quad \dfrac{1.8 \times 10^2}{3 \times 10^{-1}}\)
Give your answer as an ordinary number. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
\(2 \times 10^3\) or \(7 \times 10^4\) or 140 000 000
M1
oe correct value not in standard form eg \(14 \times 10^7\)
\(1.4 \times 10^8\)
A1
SC1 Correctly converts an ordinary number with at least four digits to standard form
Additional guidance
Condone extra zeros on 1.4 eg \(1.40\,000\,000 \times 10^8\)
M1A1
\(1.4 \times 10^8\) from 1 400 000 000
M0A0
\(2 \times 10^3\) is implied by \((2 \times 7) \times (10^3 \times 10^a)\) \(7 \times 10^4\) is implied by \((2 \times 7) \times (10^b \times 10^4)\)
M1
1 400 000 000 converted to \(1.4 \times 10^9\)
SC1
Mark scheme (b)
Answer
Mark
Comments
180 or 0.3 or (\(1.8 \div 3\) =) 0.6 or (\(10^2 \div 10^{-1}\) =) \(10^3\) or calculation which would have the outcome 600 or correct value not given as an ordinary number
M1
eg \(1800 \div 3\) eg \(6 \times 10^2\)
600
A1
Additional guidance
\(1800 \div 0.3 = 600\) scores M1 only, as 600 comes from incorrect working
M1A0
\(1800 \div 30 = 600\) scores zero, as 600 comes from incorrect working
(a) Write \(\quad 0.000\,97 \quad\) in standard form. [1 mark]
(b) Work out \(\quad \dfrac{3 \times 10^5}{4 \times 10^3}\)
Give your answer as an ordinary number. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
\(9.7 \times 10^{-4}\)
B1
Additional guidance
Condone \(9.7\,.\,10^{-4}\) or \(9.7 \cdot 10^{-4}\)
B1
Ignore zeroes before the ‘9’ eg \(00009.7 \times 10^{-4}\)
B1
\(9.7 \times 10^{4-}\)
B0
Mark scheme (b)
Answer
Mark
Comments
300 000 and 4000 or \((10^5 \div 10^3 =)\ 10^2\) or \((10^5 \div 10^3 =)\ 100\) or \(7.5 \times 10^{(1)}\) or \(75 \times 10^0\) or \(\dfrac{3 \times 10^2}{4}\) or \(\dfrac{300}{4}\)
M1
75
A1
Additional guidance
If the answer is given in standard form and as 75 the student must indicate that 75 is their chosen answer or it must be the final answer given eg1 \(7.5 \times 10^{(1)} = 75\) on the answer line eg2 \(75 = 7.5 \times 10^{(1)}\) on the answer line
M1A1 M1A0
\(\dfrac{300}{4}\) or 75 from incorrect working scores zero eg1 \(3 \times 10^5 = 30\,000\) and \(4 \times 10^3 = 400\) and \(30\,000 \div 400 = \dfrac{300}{4} = 75\) eg2 \(\dfrac{30\,000}{400} = 75\)
M0A0 M0A0
For the method mark, ignore incorrect work from a correct expression eg \(0.75 \times 10^2 = 7.5 \times 10^3\)
M1A0
If the student attempts two methods (simplifying the powers and attempting to convert to ordinary numbers) mark both methods and award the higher mark
(a) Write \(\quad 0.000\,97 \quad\) in standard form. [1 mark]
(b) Work out \(\quad \dfrac{3 \times 10^5}{4 \times 10^3}\)
Give your answer as an ordinary number. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
\(9.7 \times 10^{-4}\)
B1
Additional guidance
Condone \(9.7\,.\,10^{-4}\) or \(9.7 \cdot 10^{-4}\)
B1
Ignore zeroes before the ‘9’ eg \(00009.7 \times 10^{-4}\)
B1
\(9.7 \times 10^{4-}\)
B0
Mark scheme (b)
Answer
Mark
Comments
300 000 and 4000 or \((10^5 \div 10^3 =)\ 10^2\) or \((10^5 \div 10^3 =)\ 100\) or \(7.5 \times 10^{(1)}\) or \(75 \times 10^0\) or \(\dfrac{3 \times 10^2}{4}\) or \(\dfrac{300}{4}\)
M1
75
A1
Additional guidance
If the answer is given in standard form and as 75 the student must indicate that 75 is their chosen answer or it must be the final answer given eg1 \(7.5 \times 10^{(1)} = 75\) on the answer line eg2 \(75 = 7.5 \times 10^{(1)}\) on the answer line
M1A1 M1A0
\(\dfrac{300}{4}\) or 75 from incorrect working scores zero eg1 \(3 \times 10^5 = 30\,000\) and \(4 \times 10^3 = 400\) and \(30\,000 \div 400 = \dfrac{300}{4} = 75\) eg2 \(\dfrac{30\,000}{400} = 75\)
M0A0 M0A0
For the method mark, ignore incorrect work from a correct expression eg \(0.75 \times 10^2 = 7.5 \times 10^3\)
M1A0
If the student attempts two methods (simplifying the powers and attempting to convert to ordinary numbers) mark both methods and award the higher mark
One correct conversion to a comparable form \(0.08 \times 10^{-2}\) or 0.0008 \(400 \times 10^{-4}\) or 0.04 \(0.06 \times 10^{-2}\) or 0.0006 \(7 \times 10^{-2}\) or \(700 \times 10^{-4}\)
M1
\(6 \times 10^{-4}\) \(8 \times 10^{-4}\) \(4 \times 10^{-2}\) 0.07 with no clearly incorrect working
A1
oe accept in converted form
Additional guidance
Correct answer from clearly incorrect working
A0
Accept numbers with two decimal points if it is clear that the point has been moved to the correct place eg 0.0008.0 with curved lines between each place value between the decimal points
If the numbers are converted into fractions, at least two must be given correctly with common denominators to score the first mark
eg \(\dfrac{4}{100}\) and \(\dfrac{7}{100}\)
M1
eg \(\dfrac{6}{1000}\) and \(\dfrac{8}{1000}\) only
M0
eg \(\dfrac{6}{10\,000}\) and \(\dfrac{7}{100}\) only
One correct conversion to a comparable form \(0.08 \times 10^{-2}\) or 0.0008 \(400 \times 10^{-4}\) or 0.04 \(0.06 \times 10^{-2}\) or 0.0006 \(7 \times 10^{-2}\) or \(700 \times 10^{-4}\)
M1
\(6 \times 10^{-4}\) \(8 \times 10^{-4}\) \(4 \times 10^{-2}\) 0.07 with no clearly incorrect working
A1
oe accept in converted form
Additional guidance
Correct answer from clearly incorrect working
A0
Accept numbers with two decimal points if it is clear that the point has been moved to the correct place eg 0.0008.0 with curved lines between each place value between the decimal points
If the numbers are converted into fractions, at least two must be given correctly with common denominators to score the first mark
eg \(\dfrac{4}{100}\) and \(\dfrac{7}{100}\)
M1
eg \(\dfrac{6}{1000}\) and \(\dfrac{8}{1000}\) only
M0
eg \(\dfrac{6}{10\,000}\) and \(\dfrac{7}{100}\) only
\(9.56 \times 3^{10} \quad 9563 \quad 9.56 \times 10^3\) or 564 508 (.44) \(\quad\) 9563 \(\quad\) 9560 with no incorrect evaluations seen
B2
B1 \(9.563 \times 10^3\) or 9560 or 564 508 (.44) or \(5.6(450844) \times 10^5\) SC1 \(9.56 \times 10^3 \quad 9563 \quad 9.56 \times 3^{10}\) with no incorrect evaluations seen
Additional guidance
Allow numbers to be written in original or converted form or as a mixture for B2 or SC1
(b) A country consists of a south island and a north island. The table below shows the population and area of each island in June 2023.
Population
Area (km\(^2\))
South island
\(1.2 \times 10^6\)
\(1.5 \times 10^5\)
North island
\(4.0 \times 10^6\)
\(1.1 \times 10^5\)
Work out the overall population density of the country in June 2023. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(3.7 \times 10^{-3}\)
1
Condone trailing zeros
Mark scheme (b)
Answer
Marks
Part marks and guidance
20 or \(2 \times 10^1\) or \(2 \times 10\) nfww
4
M1 for \(1.2 \times 10^6 + 4[.0] \times 10^6\) or \(5.2 \times 10^6\) oe M1 for \(1.5 \times 10^5 + 1.1 \times 10^5\) or \(2.6 \times 10^5\) oe
M1 for \(\dfrac{\textit{a population value}}{\textit{an area value}}\)
may be implied by figs 52 may be implied by figs 26
Award M marks even if not used to find final answer as long as it is not contradicted.
Population and area values must be consistent eg both south island, both north island or their summed population and area e.g. M1 implied by 8 or 36.36 to 36.4 or 44.36 or 44.4
(b) A country consists of a south island and a north island. The table below shows the population and area of each island in June 2023.
Population
Area (km\(^2\))
South island
\(1.2 \times 10^6\)
\(1.5 \times 10^5\)
North island
\(4.0 \times 10^6\)
\(1.1 \times 10^5\)
Work out the overall population density of the country in June 2023. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(3.7 \times 10^{-3}\)
1
Condone trailing zeros
Mark scheme (b)
Answer
Marks
Part marks and guidance
20 or \(2 \times 10^1\) or \(2 \times 10\) nfww
4
M1 for \(1.2 \times 10^6 + 4[.0] \times 10^6\) or \(5.2 \times 10^6\) oe M1 for \(1.5 \times 10^5 + 1.1 \times 10^5\) or \(2.6 \times 10^5\) oe
may be implied by figs 52 may be implied by figs 26
M1 for \(\dfrac{\textit{a population value}}{\textit{an area value}}\)
Award M marks even if not used to find final answer as long as it is not contradicted.
Population and area values must be consistent eg both south island, both north island or their summed population and area e.g. M1 implied by 8 or 36.36 to 36.4 or 44.36 or 44.4
\(1.655 \times 10^{12}\) raindrops fall in a storm
the mass of each raindrop is \(6 \times 10^{-5}\) grams.
Calculate the total mass of all of the raindrops that fall in the storm. Give your answer in standard form in kilograms. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(1.025 \times 10^{-3}\)
1
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(9.93 \times 10^4\)
4
B3 for 99 300 or B2 for figs 993 or M2 for \(\dfrac{1.655 \times 10^{12} \times 6 \times 10^{-5}}{1000}\) oe or M1 for \(1.655 \times 10^{12} \times 6 \times 10^{-5}\) oe
May be in stages
May be seen as product of two st form numbers converted
B3 for all 4 expressed in an equivalent comparable form
or
B2 for 3 expressed in an equivalent comparable form
or
B1 for 2 expressed in an equivalent comparable form
Equivalent comparable form is either decimal, percentage or standard index form. Fractions only acceptable as comparable form with common denominators.
For final answer, accept some or all given as equivalent form on answer line. eg. Full marks for 0.003[003…], 0.0031, 0.0036, 0.03 or \(\frac{1}{333}\), \(\frac{31}{10\,000}\), \(\frac{9}{2500}\), \(\frac{3}{100}\)
Where choice of comparable form, mark to candidate’s advantage.
Likely comparable form (award best row):
0.36%
\(\frac{1}{333}\)
0.03
\(3.1 \times 10^{-3}\)
0.36[%]
0.3[003…][%]
3[%]
0.31[%]
0.0036
0.003[003…]
0.03
0.0031
\(3.6 \times 10^{-3}\)
\(3[.003\ldots] \times 10^{-3}\)
\(3 \times 10^{-2}\)
\(3.1 \times 10^{-3}\)
\(\frac{11\,988}{3\,330\,000}\)
\(\frac{10\,000}{3\,330\,000}\)
\(\frac{99\,900}{3\,330\,000}\)
\(\frac{10\,323}{3\,330\,000}\)
Notes
In the percentage row of the table, \(\frac{1}{333}\) is printed in the mark scheme as 0.00003[003…][%]; it has been corrected here to 0.3[003…][%].
Mark the answer line and look for order of indices –1, 1, 2, 3 Condone TE if unambiguous If answer given as ordinary numbers must be correct 0.49, 95, 240, 1300
(b) The speed of sound is \(3.43 \times 10^{-1}\) km/s. An object is travelling at the speed of sound.
Work out how far the object travels in one day. [2]
(c) In a science fiction story, a spacecraft travelling faster than the speed of light is said to be travelling at ‘warp \(n\)’ where \(n\) is an integer.
Warp \(n\) is defined as \(n^3 \times\) the speed of light.
In the story, a spacecraft needs to travel from Earth to Neptune in less than 2 minutes.
The speed of light is \(3.00 \times 10^5\) km/s.
The distance from Earth to Neptune is \(4.41 \times 10^9\) km.
Find the smallest possible warp \(n\) at which the spacecraft can travel. You must show your working. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(3.86 \times 10^{-3}\)
1
Condone trailing zeros
Mark scheme (b)
Answer
Marks
Part marks and guidance
29 635[.2] or 29 640 or 29 600
2
M1 for \(3.43 \times 10^{-1} \times 60^2 \times 24\) oe If 0 scored SC1 for 14 817.6, 14 818, 14 820 or 14 800
Note: \(60^2 \times 24 = 86\,400\)
Mark scheme (c)
Answer
Marks
Part marks and guidance
5 with correct working
3
M2 for \(\sqrt[3]{\frac{4.41 \times 10^9}{120 \times 3.00 \times 10^5}}\) oe
OR
M1 for \(\frac{4.41 \times 10^9}{3.00 \times 10^5}\) oe or \(\frac{4.41 \times 10^9}{120}\) oe or \(120 \times 3.00 \times 10^5\) oe
If 0 scored SC1 for answer 5 with no or insufficient working
“Correct working” requires evidence of at least M1 or M2 if trials are used M2 implied by 4.966… or 4.97 or \(\sqrt[3]{122.5}\) or \(n^3 = 122.5\) condone use of a value \(119 \leqslant\) value \(\leqslant 120\) instead of 120 which should lead to 4.98
M1 may be seen within a larger calculation e.g. \(\frac{4.41 \times 10^9}{120 \times 3.00 \times 10^5}\) or \(\sqrt[3]{\frac{4.41 \times 10^9}{3.00 \times 10^5}}\) M1 implied by 14 700 or \(3.675 \times 10^7\) oe or \(3.6 \times 10^7\) oe or 122.5
use of trials M1 for each correct trial using integer values of \(n\) up to a maximum of M2 e.g. \(\frac{their\ 14700}{5^3}\) oe and 117 to 118 or 1.95 to 1.97 \(\frac{their\ 14700}{4^3}\) oe and 229[. ..] or 3.8[…] their 14 700 comes from \(3 \times 10^5\) and \(4.41 \times 10^9\) (see appendix for more examples)
Appendix: Question 5(c)
Other trials, condone rot : \(\frac{their\ 14700}{2^3}\) oe and 1837[.5…] or 30[.6…] \(\frac{their\ 14700}{3^3}\) oe and 544[.4….] or 9[.07…] \(\frac{their\ 14700}{6^3}\) oe and 68[.05…] or 1[.1…]
Or they might try to see how far they go at each warp speed in 2 minutes (rot at least 2 figures):
21 The table below shows the approximate population of three countries in 2020.
Country
Population
China
\(1.44 \times 10^9\)
Kiribati
\(1.19 \times 10^5\)
Tuvalu
\(1.18 \times 10^4\)
(a) Calculate the approximate total population of Kiribati and Tuvalu in 2020.
Give your answer in standard form, correct to 3 significant figures. [4]
(b) Show that in 2020 the population of China was approximately 120 000 times the population of Tuvalu. [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(1.31 \times 10^5\)
4
B3 for 131000 or \(131 \times 10^3\) oe or \(1.3 \times 10^5\) or \(1.30[8] \times 10^5\) or B2 for 130800 or \(1308 \times 10^2\) oe or M1 for figs 119 + figs 118
If 0 scored SC1 for their value correctly rounded to 3 sf
Addition may be implied by figs 131 or 1308, Must clearly be their answer from attempt at addition
The unrounded value must be seen
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(\dfrac{1.44 \times 10^9}{1.18 \times 10^4}\)
M1
Accept in ordinary numbers e.g. \(\dfrac{1440000000}{11800} =\)
12 Dinosaurs first appeared on Earth \(2.4 \times 10^8\) years ago. Dinosaurs became extinct on Earth \(7 \times 10^7\) years ago.
(a) Explain why it is appropriate to use standard form for these numbers. [1]
(b) Use the given information to work out how long dinosaurs existed on Earth. Give your answer in standard form. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
Correct comment about large numbers or too many zeros oe
1
e.g. Standard form is used for [very] large/long numbers [or very small numbers] Writing out all the zeros may lead to errors
See Appendix
Appendix
Exemplar responses for Question 12a
Response
Mark
1
Writing all the 0’s out is very long/difficult to read/may lead to errors
1
2
As they contain a lot of 0s
1
3
Because there are too many zeros
1
4
Because the numbers are extremely long/high/large, shortening the number.
1
5
Because of how long/high/large the number will be
1
6
The numbers are too long to write down e.g. \(7 \times 10^7 = 70\,000\,000\)
1
7
It shortens the numbers due to them having so many 0s
1
8
Because they’re too big (Given the wording of the demand – refers to numbers)
1
9
It takes too long to write it all out
1
10
Because the digits will be too long (BOD infers numbers are too long)
BOD 1
11
Wastes time writing out the digits, this is simple shorthand (similar to above)
BOD 1
12
To makes it easier to read (no reason given as to why)
0
13
To make the numbers more simpler (no reason given e.g. numbers too large, too many zeros)
0
14
Its easier to understand (no reason given as to why)
0
15
Because otherwise you need to write 2.4 billion years ago so \(2.4 \times 10^8\) is easier to say in standard form (Contains error in number, if 240 million years this would be OK)
0
16
Because the numbers are billions so it’s easier to read them in standard form (see above)
0
17
It shortens the numbers (no reason given why e.g. too large, too many zeros)
0
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(1.7 \times 10^8\)
3
M2 for figs 17 in final answer or B1 for 240 000 000 or 70 000 000 or \(24 \times 10^7\) or \(0.7 \times 10^8\)
or M1 for \(2.4 \times 10^8 - 7 \times 10^7\) oe
Accept both converted to same power of 10 e.g. \(240 \times 10^6\) and \(70 \times 10^6\)
For M1, if standard form subtraction written correctly, isw incorrect conversion to values
3 Dinosaurs first appeared on Earth \(2.4 \times 10^8\) years ago. Dinosaurs became extinct on Earth \(7 \times 10^7\) years ago.
(a) Explain why it is appropriate to use standard form for these numbers. [1]
(b) Use the given information to work out how long dinosaurs existed on Earth. Give your answer in standard form. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
Correct comment about large numbers or too many zeros oe
1
e.g. Standard form is used for [very] large/long numbers [or very small numbers] Writing out all the zeros may lead to errors See Appendix
Appendix: additional guidance for Q3(a)
Response
Mark
1
Writing all the 0’s out is very long/difficult to read/may lead to errors
1
2
As they contain a lot of 0s
1
3
Because there are too many zeros
1
4
Because the numbers are extremely long/high/large, shortening the number.
1
5
Because of how long/high/large the number will be
1
6
The numbers are too long to write down eg \(7 \times 10^7\) = 70 000 000
1
7
It shortens the numbers due to them having so many 0s
1
8
Because they’re too big (Given the wording of the demand – refers to numbers)
1
9
It takes too long to write it all out
1
10
Because the digits will be too long (BOD infers numbers are too long)
BOD 1
11
Wastes time writing out the digits, this is simple shorthand (similar to above)
BOD 1
12
To makes it easier to read (no reason given as to why)
0
13
To make the numbers more simpler (no reason given e.g. numbers too large, too many zeros)
0
14
Its easier to understand (no reason given as to why)
0
15
Because otherwise you need to write 2.4 billion years ago so \(2.4 \times 10^8\) is easier to say in standard form (Contains error in number, if 240 million years this would be OK)
0
16
Because the numbers are billions so it’s easier to read them in standard form (see above)
0
17
It shortens the numbers (no reason given why e.g. too large, too many zeros)
0
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(1.7 \times 10^8\)
3
M2 for figs 17 in final answer or B1 for 240 000 000 or 70 000 000 or \(24 \times 10^7\) or \(0.7 \times 10^8\)
Accept both converted to same power of 10 e.g. \(240 \times 10^6\) and \(70 \times 10^6\)
or M1 for \(2.4 \times 10^8 - 7 \times 10^7\) oe
For M1, if standard form subtraction written correctly, isw incorrect conversion to values
17 The table below shows the number of barrels of oil produced per day by some countries.
Country
Barrels of oil produced per day
USA
\(1.17 \times 10^{7}\)
China
\(3.98 \times 10^{6}\)
UK
\(9.39 \times 10^{5}\)
Cameroon
\(9.32 \times 10^{4}\)
Japan
\(3.92 \times 10^{3}\)
(a) Write the number of barrels of oil produced per day by Cameroon as an ordinary number. [1]
(b) How many more barrels of oil per day did China produce than the UK? Give your answer in standard form, correct to 3 significant figures. [4]
(c) Jamal says the USA produced approximately three times more barrels of oil than Japan.
Is he correct? Show how you decide.
Jamal is .................. because .................. [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
93 200
1
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(3.04 \times 10^{6}\)
4
B3 for 3 040 000 or \(3.041[0..] \times 10^6\) or \(30.4 \times 10^5\) oe rounded to 3sf OR B2 for 3 041 000 or \(30.41[0..] \times 10^5\) oe index form OR M1 for \((3.98 \times 10^6) - (9.39 \times 10^5)\) or 3 980 000 – 939 000 and M1 for their final value correctly rounded to 3sf
M1 may be implied by figs 3041….
The unrounded value must be seen
Mark scheme (c)
Answer
Marks
Part marks and guidance
Wrong/Incorrect it is 3 000 or 2 984 to 2 985 times bigger
or No, difference is [order of] \(3 \times 10^3\) which is 3 000
or Incorrect 11 760 is 3 times bigger than 3 920 or 3 900 000 is 3 times smaller than 11 700 000
or Incorrect and evaluates USA’s production ÷ 3 or Japan’s production × 3 with comment comparing the values
2
M1 for difference is [order of] \(10^3\)
or \(\dfrac{1.17 \times 10^7}{3.92 \times 10^3}\)
or \((1.17 \times 10^7) \div 3 = 3.9 \times 10^6\)
or 11 700 000 ÷ 3 = 3 900 000
or \((3.92 \times 10^3) \times 3 = 1.176 \times 10^4\) or \(1.18 \times 10^4\)
or 3 920 × 3 = 11 760
Wrong/Incorrect and a comment for 2 marks to answer the question Condone No
Give your answer in standard form, correct to 2 significant figures. [4]
Mark scheme
Answer
Marks
Part marks and guidance
\(1.5 \times 10^3\)
4
B3 for 1500 or 1.50[0] × 103 or 1.48[9..] × 103 or 1.49 × 103 or B2 for 1489.1… or M1 for \(\frac{5.8 \times 10^6}{\sqrt{4.1 \times 10^8 \times 3.7 \times 10^{-2}}}\) and M1 for writing their answer correct to 2s.f. and in standard form
implied by 3894.8… or 3894.9
need to see where their answer comes from e.g. longer figure or a simple fraction
One sesame seed weighs \(3.64 \times 10^{-6}\) kilograms.
Write the three types of seed in order according to the weight of one seed. Write the lightest type of seed first. You must show how you decide.[4]
Mark scheme
Answer
Marks
Part marks and guidance
Poppy, Sesame, Pumpkin with correct comparable values shown
4
B3 for all 3 quantities seen correct in comparable form or B2 for \(8.4 \times 10^{-5}\) or \(8.4 \times 10^{-2}\) seen or seen correct in comparable form: pumpkin with poppy, e.g. implied by [250 poppy =] 0.075 or pumpkin with sesame, e.g. implied by [250 sesame =] 0.91 or B1 poppy and sesame seen correct in comparable form or [pumpkin =] 0.084 or 0.000 084 seen or [250 poppy =] 0.000 075 oe seen or [250 sesame =] 0.000 91 oe seen
Condone weights as answer Quantities given in the question need not be rewritten
Comparable forms include:
In kilograms
In grams
Poppy
0.000 000 3 \(3 \times 10^{-7}\)
0.000 3 \(3 \times 10^{-4}\)
Pumpkin
0.000 084 \(8.4 \times 10^{-5}\)
0.084 \(8.4 \times 10^{-2}\)
Sesame
0.000 003 64 \(3.64 \times 10^{-6}\)
0.003 64 \(3.64 \times 10^{-3}\)
Must not be a mix of standard and ordinary form
Accept consistent multiples for full marks, e.g. 250 poppy = 0.075 and 250 sesame seeds = 0.91
22 A newborn baby has an approximate mass of 3.5 kilograms.
A human cell has an approximate mass of \(2.7 \times 10^{-11}\) grams.
Use these values to estimate the number of human cells in a newborn baby. Give your answer in standard form, correct to 2 significant figures. [5]
Mark scheme
Answer
Marks
Part marks and guidance
\(1.3 \times 10^{14}\)
5
B4 for \(1.30 \times 10^{14}\) or \(1.29[6\ldots] \times 10^{14}\) or 130 000 000 000 000 as final answers
or B3 for \(1.3 \times 10^n\) (\(n \ne 0\)) or \(1.29[6\ldots] \times 10^{14}\) written in full or M3 for 3500 \(\div\) \((2.7 \times 10^{-11})\) oe or B2 for \(1.29[6\ldots] \times 10^n\) (\(n \ne 0\)) or figs 13
OR
M1 for figs 35 \(\div\) figs 27 soi by figs 129[6…]
B1 for 3500 or \(2.7 \times 10^{-14}\) oe or \(3.5 \times 10^3\) seen
For 5 marks and M marks, condone use of correctly rounded values in correct calculations
3 A newborn baby has an approximate mass of 3.5 kilograms. A human cell has an approximate mass of \(2.7 \times 10^{-11}\) grams.
Use these values to estimate the number of human cells in a newborn baby. Give your answer in standard form, correct to 2 significant figures. [5]
Mark scheme
Answer
Marks
Part marks and guidance
\(1.3 \times 10^{14}\)
5
B4 for \(1.30 \times 10^{14}\) or \(1.29[6\ldots] \times 10^{14}\) or 130 000 000 000 000 as final answers or B3 for \(1.3 \times 10^{n}\) (\(n \ne 0\)) or \(1.29[6\ldots] \times 10^{14}\) written in full or M3 for \(3500 \div (2.7 \times 10^{-11})\) oe or B2 for \(1.29[6\ldots] \times 10^{n}\) (\(n \ne 0\)) or figs 13 OR M1 for figs 35 ÷ figs 27 soi by figs 129[6…] B1 for 3500 or \(2.7 \times 10^{-14}\) oe or \(3.5 \times 10^{3}\) seen
For 5 marks and M marks, condone use of correctly rounded values in correct calculations
(b) \(4.3 \times 10^5 + 3.8 \times 10^4\) Give your answer in standard form. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
2
2
B1 for [\(\sqrt[3]{64} =\)] 4 or for ½ oe
Accept \(4^3 = 64\) for B1
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(4.68 \times 10^5\) final answer
3
B2 for 468 000 or answer \(4.7 \times 10^5\) or B1 for 430 000 or \(43 \times 10^4\) or 38 000 or \(0.38 \times 10^5\) seen M1 for figs 468 oe in final answer
13 A company makes sweets. The sweets are put into packets.
Here are some facts.
\(1.47 \times 10^7\) sweets are made every day
\(3.5 \times 10^5\) packets of sweets are produced every day
(a) Calculate the mean number of sweets in one packet. [2]
(b) Sweets are made on 288 days each year.
Calculate the number of sweets made each year. Give your answer in standard form. [3]
(c) The company has 152 machines making the sweets. Each machine operates for 15 hours each day.
(i) Calculate the number of sweets made by one machine each hour. Give your answer as an ordinary number correct to the nearest 10. [3]
(ii) State one assumption you have made in part (c)(i). [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
42
2
M1 for \(\dfrac{1.47 \times 10^7}{3.5 \times 10^5}\) oe
If 0 scored SC1 for figs 42 in answer
Eg. \(\dfrac{14\,700\,000}{350\,000}\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(4.2[3\ldots] \times 10^9\)
3
B2 for 4 233 600 000 oe as answer or M1 for their \(1.47 \times 10^7 \times 288\)
If 0 scored SC1 for figs 423[…] in answer
Eg. \(423.[36] \times 10^7\)
their \(1.47 \times 10^7\) converted from info in (a)
Mark scheme (c)
Answer
Marks
Part marks and guidance
(i) 6450
3
B2 for 6447 to 6448 or M1 for \(\dfrac{1.47 \times 10^7}{(152 \times 15)}\) oe or figs 6447 in answer
May be in stages. NB: 152 \(\times\) 15 = 2280
(ii) Each machine makes the same amount of sweets. or There are no breakdowns oe or Machines running at same rate oe or All machines run for the same time oe
5 A company makes sweets. The sweets are put into packets.
Here are some facts.
(a) Calculate the mean number of sweets in one packet. [2]
(b) Sweets are made on 288 days each year.
Calculate the number of sweets made each year. Give your answer in standard form. [3]
(c) The company has 152 machines making the sweets. Each machine operates for 15 hours each day.
(i) Calculate the number of sweets made by one machine each hour. Give your answer as an ordinary number correct to the nearest 10. [3]
(ii) State one assumption you have made in part (c)(i). [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
42
2
M1 for \(\dfrac{1.47 \times 10^7}{3.5 \times 10^5}\) oe
If 0 scored SC1 for figs 42 in answer
Eg. \(\dfrac{14\,700\,000}{350\,000}\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(4.2[3\ldots] \times 10^9\)
3
B2 for 4 233 600 000 oe as answer or M1 for their \(1.47 \times 10^7 \times 288\)
If 0 scored SC1 for figs 423[...] in answer
Eg. \(423.[36] \times 10^7\)
their \(1.47 \times 10^7\) converted from info in (a)
Mark scheme (c)
Answer
Marks
Part marks and guidance
(i) 6450
3
B2 for 6447 to 6448 or M1 for \(\dfrac{1.47 \times 10^7}{(152 \times 15)}\) oe or figs 6447 in answer
May be in stages. NB: 152 × 15 = 2280
(ii) Each machine makes the same amount of sweets or There are no breakdowns oe or Machines running at same rate oe or All machines run for the same time oe
1 This table shows the populations of the four countries of the UK in 2012. All values are given correct to 3 significant figures.
Country
Population
England
\(5.35 \times 10^7\)
Wales
\(3.07 \times 10^6\)
Scotland
\(5.31 \times 10^6\)
Northern Ireland
\(1.82 \times 10^6\)
(a) Write the population of England as an ordinary number. [1]
(b) Work out the total population of Wales, Scotland and Northern Ireland. Give your answer in standard form. [2]
(c) The total population of the UK is predicted to reach 73.3 million in 2037.
Calculate the predicted percentage increase in the UK population from 2012 to 2037. Give your answer correct to 2 significant figures. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
53 500 000
1
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(1.02 \times 10^7\)
2
B1 for answer figs102
Mark scheme (c)
Answer
Marks
Part marks and guidance
15 nfww
4
B3 for 15.07 to 15.1 nfww
OR
B1 for figs 637 and M1 for \(\dfrac{73.3 \times 10^6 - \mathit{their}\,63.7 \times 10^6}{\mathit{their}\,63.7 \times 10^6}\) oe or \(\dfrac{73.3 \times 10^6}{\mathit{their}\,63.7 \times 10^6}\) oe and M1dep interpretation of answer to division as a percentage increase soi by answer
their \(63.7 \times 10^6\) can be an error from their sum, their (b) or one of the country populations