Work out the value of \(xy\). Give your answer as a fraction in its simplest form. (5)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{5}{33}\)
M1
for \(10x = 2.\dot{2}\) or 2.22… or \((10x - x =)\ 2.\dot{2} - 0.\dot{2}\ (= 2)\) or \(2.22\ldots - 0.22\ldots\ (= 2)\) or \(\dfrac{2}{9}\) oe fraction
M1
for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(y\) eg \((1000y - 10y =)\ 681.\dot{8}\dot{1} - 6.\dot{8}\dot{1}\ (= 675)\) or \(681.81\ldots - 6.81\ldots\ (= 675)\) or \(\dfrac{675}{990}\) or \((100y - y) = 68.\dot{1}\dot{8} - 0.6\dot{8}\dot{1}\ (= 67.5)\) or \(68.181\ldots - 0.681\ldots\ (= 67.5)\) or \(\dfrac{67.5}{99}\)
A1
for \((x =)\ \dfrac{2}{9}\) oe and \((y =)\ \dfrac{675}{990}\) oe
M1
for \(\text{``}\dfrac{2}{9}\text{''} \times \text{``}\dfrac{675}{990}\text{''}\)
A1
cao
Additional guidance
eg \(\dfrac{20}{90}, \dfrac{22}{99}\)
Accept \((y =)\ \dfrac{67.5}{99}\)
Award 4 marks for an answer equivalent to \(\dfrac{5}{33}\), eg \(\dfrac{15}{99}, \dfrac{135}{891}, \dfrac{1350}{8910}\) unless from incorrect working
14 Prove algebraically that \(0.4\dot{6}\dot{2}\) can be written as \(\dfrac{229}{495}\) (3)
Mark scheme
Answer
Mark
Mark scheme
Shown
M1
for start to find multiples of \(x\) with the same recurring pattern eg for \((10x =)\ 4.626262\ldots\) or \((100x =)\ 46.262626\ldots\) or \((1000x =)\ 462.626262\ldots\)
M1
(dep on M1) for a correct subtraction that would lead to a terminating decimal, eg \((1000x - 10x) = 462.6262\ldots - 4.6262\ldots\ (= 458)\) or \((100x - x) = 46.2626\ldots - 0.4626\ldots\ (= 45.8)\)
C1
for correct working leading to the correct answer
Additional guidance
Any recurring notation acceptable throughout. Proofs with terminating decimals (at least 5 figures) score M1M1C0
18 Show that \(0.\dot{1}\dot{5} + 0.2\dot{2}\dot{7}\) can be written in the form \(\dfrac{m}{66}\) where \(m\) is an integer. (3)
Mark scheme
Answer
Mark
Mark scheme
Shown
M1
for \(0.1515\ldots + 0.22727\ldots\ (= 0.37878\ldots\) or \(0.3\dot{7}\dot{8})\)
M1
for finding two correct recurring decimals that when subtracted would result in a terminating decimal or integer, eg \((1000x - 10x =)\ 378.7878\ldots - 3.7878\ldots\ (= 375)\) or \(\dfrac{375}{990}\) or \((100x - x =)\ 37.8787\ldots - 0.37878\ldots\ (= 37.5)\) or \(\dfrac{37.5}{99}\)
C1
for correct working leading to \(\dfrac{25}{66}\)
OR
M1
for start of a method to convert \(0.1515\ldots\) or \(0.22727\ldots\) to a fraction, eg \(100x = 15.1515\ldots\) or \(\dfrac{15}{99}\) or \(\dfrac{5}{33}\) oe or \(10y = 2.2727\ldots\) or \(100y = 22.7272\ldots\) or \(1000y = 227.2727\ldots\) or \(\dfrac{225}{990}\) or \(\dfrac{22.5}{99}\) or \(\dfrac{5}{22}\) oe
M1
for a method to convert \(0.1515\ldots\) and \(0.22727\ldots\) to fractions, eg \((100x - x =)\ 15.1515\ldots - 0.1515\ldots\ (= 15)\) or \(\dfrac{15}{99}\) or \(\dfrac{5}{33}\) oe and \((1000y - 10y =)\ 227.2727\ldots - 2.2727\ldots\ (= 225)\) or \((100y - y =)\ 22.7272\ldots - 0.22727\ldots\ (= 22.5)\) or \(\dfrac{225}{990}\) or \(\dfrac{22.5}{99}\) or \(\dfrac{5}{22}\) oe
C1
for correct working leading to \(\dfrac{25}{66}\)
Additional guidance
Recurring decimal notation acceptable for this mark
Recurring decimal notation acceptable for both M marks
20 Prove algebraically that \(0.1\dot{2}\dot{3}\) can be written as \(\dfrac{61}{495}\) (3)
Mark scheme
Answer
Mark
Mark scheme
Proof
M1
for \((10x =)\ 1.2323\ldots\) or \((100x =)\ 12.323\ldots\) or \((1000x =)\ 123.23\ldots\)
M1
(dep M1) for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(x\) eg \((1000x - 10x =)\ 123.23\ldots - 1.23\ldots\ (= 122)\) or \(\dfrac{122}{990}\) or \((100x - x =)\ 12.323\ldots - 0.123\ldots\ (= 12.2)\) or \(\dfrac{12.2}{99}\)
C1
for completing algebra to \(\dfrac{61}{495}\)
Additional guidance
Any recurring notation acceptable throughout. Proofs with terminating decimals (at least 5 figures) score M1M1C0
14 Using algebra, prove that \(1.06\dot{2}\) can be written as \(1\dfrac{14}{225}\) (3)
Mark scheme
Answer
Mark
Mark scheme
Shown
M1
for \((x =)\ 1.0622\ldots\) or \((10x =)\ 10.622\ldots\) or \((100x =)\ 106.22\ldots\) or \((1000x =)\ 1062.2\ldots\)
OR
for \((x =)\ 0.0622\ldots\) or \((10x =)\ 0.622\ldots\) or \((100x =)\ 6.22\ldots\) or \((1000x =)\ 62.2\ldots\)
M1
(dep M1) for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(x\) eg \((1000x - 100x =)\ 1062.2\ldots - 106.22..\ (= 956)\) or \(\dfrac{956}{900}\)
OR
(dep M1) for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(x\) eg \((1000x - 100x =)\ 62.2\ldots - 6.22...\ (= 56)\) or \(\dfrac{56}{900}\)
12 Express \(0.1\dot{1}\dot{7}\) as a fraction. You must show all your working. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{116}{990}\)
M1
for \((x =)\ 0.11717\ldots\) or \((10x =)\ 1.\dot{1}\dot{7}\) or \(1.1717\ldots\) or \((100x =)\ 11.\dot{7}\dot{1}\) or \(11.7171\ldots\) or \((1000x =)\ 117.\dot{1}\dot{7}\) or \(117.1717\ldots\)
M1
(dep M1) for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(x\) eg \((1000x - 10x =)\ 117.\dot{1}\dot{7} - 1.\dot{1}\dot{7}\ (= 116)\) or \(117.1717\ldots - 1.1717\ldots\ (= 116)\)
A1
for \(\dfrac{116}{990}\) oe, eg \(\dfrac{58}{495}\)
Additional guidance
Accept \((100x - x =)\ 11.\dot{7}\dot{1} - 0.1\dot{1}\dot{7}\) or \(11.7171\ldots - 0.11717\ldots\ (= 11.6)\)
\(\dfrac{11.6}{99}\) must be written in the form \(\dfrac{a}{b}\) where \(a\) and \(b\) are integers to gain the accuracy mark
Acceptable examples he should have used \(100(x)\) rather than \(10(x)\) he should have used \(1000x\) and \(10x\) Ted’s working does not eliminate the recurring decimals the recurring numbers after the decimal point have to be in the same sequence he should have multiplied by 100 to subtract easier after the decimal point he should have multiplied by 100 because two numbers are recurring
Not acceptable examples it is not correct the method is not complete he should have used \(1000x\) he should have multiplied by 100 he should have multiplied by 100 and then done \(100x - 10x\) to give 43/90
15 Express \(0.4\dot{1}\dot{8}\) as a fraction. You must show all your working. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{414}{990}\)
M1
for (\(x =\)) 0.41818… or (\(10x =\)) \(4.\dot{1}\dot{8}\) or 4.1818… or (\(100x =\)) \(41.\dot{8}\dot{1}\) or 41.818… or (\(1000x =\)) \(418.\dot{1}\dot{8}\) or 418.18…
M1
for using two recurring decimals with a terminating decimal difference, eg. (\(1000x - 10x =\)) \(418.\dot{1}\dot{8} - 4.\dot{1}\dot{8}\) or \(418.18\ldots - 4.1818\ldots\ (= 414)\)
A1
for \(\dfrac{414}{990}\) oe, eg \(\dfrac{23}{55}\)
Additional guidance
Accept (\(100x - x =\)) \(41.\dot{8}\dot{1} - 0.4\dot{1}\dot{8}\) or \(41.818\ldots - 0.41818\ldots\ (= 41.4)\)
\(\dfrac{41.4}{99}\) must be simplified to gain the accuracy mark
16 Prove algebraically that \(0.2\dot{5}\dot{6}\) can be written as \(\dfrac{127}{495}\) (3)
Mark scheme
Answer
Mark
Mark scheme
Proof with \(\dfrac{127}{495}\)
M1
0.25656... or 0.2 + 0.05656.. or \((10 \times 0.2\dot{5}\dot{6} =)\ 2.\dot{5}\dot{6}\) or 2.5656… or \((100 \times 0.2\dot{5}\dot{6} =)\ 25.\dot{6}\dot{5}\) or 25.6565…or \((1000 \times 0.2\dot{5}\dot{6} =)\ 256.\dot{5}\dot{6}\) or 256.5656…
M1
for finding two correct recurring decimals that when subtracted would result in a terminating decimal or integer, eg. 256.5656….. – 2.5656….. or 25.6565….. – 0.25656….. or \(256.\dot{5}\dot{6} - 2.\dot{5}\dot{6}\) or \(25.\dot{6}\dot{5} - 0.2\dot{5}\dot{6}\) or for \(\dfrac{254}{990}\) or \(\dfrac{25.4}{99}\)
16 Using algebra, prove that \(0.1\dot{3}\dot{6} \times 0.\dot{2}\) is equal in value to \(\dfrac{1}{33}\) (3)
Mark scheme
Answer
Mark
Notes
M1
for the start of a method to convert \(0.22\ldots\) to a fraction, eg \(10y = 2.22\ldots\) or \((y =)\ \dfrac{2}{9}\)
M1
for the start of a method to convert \(0.13636\ldots\) to a fraction, \(10x = 1.3636\ldots\) or \(100x = 13.6363\ldots\) or \(1000x = 136.3636\ldots\) or \((x =)\ \dfrac{13.5}{99}\) or \((x =)\ \dfrac{135}{990}\)
C1
for correct arithmetic and concluding the proof
OR
M1
for \(0.1\dot{3}\dot{6} \times 0.\dot{2} = 0.\dot{0}\dot{3}\ (= z)\)
M1
for complete method to find two appropriate recurring decimals the difference of which is a rational number, eg. \(100z = 3.0303\ldots\), \((z =)\ 0.0303\ldots\) or \(\dfrac{3}{99}\)
Prove algebraically that \(x\) can be written as \(\dfrac{1}{22}\) (3)
Mark scheme
Answer
Mark
Notes
Proof
3
M1 for \((x =)\) 0.04545(...) or \(1000x = 45.4545(...)\), accept \(1000x = 45.\dot{4}\dot{5}\) or \(100x = 4.54545(...)\), accept \(100x = 4.\dot{5}\dot{4}\) or \(10x = 0.4545(...)\), accept \(10x = 0.\dot{4}\dot{5}\) M1 for finding the difference between two correct, relevant recurring decimals for which the answer is a terminating decimal A1 (dep on M2) for completing the proof by subtracting and cancelling to give a correct fraction eg \(\dfrac{45}{990} = \dfrac{1}{22}\) or \(\dfrac{4.5}{99} = \dfrac{1}{22}\)
M1 for 0.28181(...) or 0.2 + 0.08181(...) or evidence of correct recurring decimal eg. 281.81(...) M1 for two correct recurring decimals that, when subtracted, would result in a terminating decimal, and attempting the subtraction eg. \(100x = 28.1818\ldots\), \(x = 0.28181\ldots\) and subtracting eg. \(1000x = 281.8181\ldots\), \(10x = 2.8181\ldots\) and subtracting OR \(\dfrac{27.9}{99}\) or \(\dfrac{279}{990}\) oe A1 cao
If the answer is given as 2.4545… there must be at least four digits after the decimal point and at least two dots
eg 2.454545..
M1A1
eg 2.4545
M1A0
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1: using \(x = 0.155\ldots\)
Multiplication by a power of 10 with correct result
M1
eg \(10x = 1.55\ldots\) or \(100x = 15.55\ldots\) or \(1000x = 155.55\ldots\) any or no letter
Correct equation formed from subtraction of two equations to eliminate recurring digits or correct fraction with a decimal numerator
M1dep
eg \(9x = 1.4\) or \(90x = 14\) or \(99x = 15.4\) eg \(\dfrac{15.4}{99}\)
\(\dfrac{7}{45}\)
A1
oe fraction eg \(\dfrac{14}{90}\) or \(\dfrac{154}{990}\)
Alternative method 2: using \(x = 0.055\ldots\)
Correct equation formed from subtraction of two equations to eliminate recurring digits or correct fraction, which may include a decimal numerator
M1
eg \(9x = 0.5\) or \(90x = 5\) or \(99x = 5.5\) eg \(\dfrac{55}{990}\) or \(\dfrac{5.5}{99}\)
\(\dfrac{1}{10}\) + their correct fraction, which may include a decimal numerator
M1dep
eg \(\dfrac{1}{10} + \dfrac{5}{90}\)
\(\dfrac{7}{45}\)
A1
oe fraction eg \(\dfrac{14}{90}\) or \(\dfrac{154}{990}\)
Additional guidance
Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Ignore incorrect simplifying if correct fraction seen
Unless recovered, recurring decimals must be in correct notation or denoted by at least two repeated digits after the decimal point with at least two dots
21 Prove algebraically that \(\quad 1.0\dot{1}\dot{8} = \dfrac{56}{55}\) [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1: using \(x = 1.018\ldots\)
Multiplication by power of 10
M1
eg \(10x = 10.18\ldots\) or \(100x = 101.81\ldots\) or \(1000x = 1018.18\ldots\) any or no letter
Correct equation formed from subtraction of two equations to eliminate recurring digits
M1dep
eg \(99x = 100.8\) or \(990x = 1008\) or \(x = \dfrac{1008}{990}\)
\((x =)\ \dfrac{1008}{990}\) and \(\dfrac{56}{55}\) with no incorrect working
A1
oe from using different powers of 10
Alternative method 2: using \(x = 0.018\ldots\)
Multiplication by power of 10
M1
eg \(10x = 0.1818\ldots\) or \(100x = 1.818\ldots\) or \(1000x = 18.18\ldots\) any or no letter
Correct equation formed from subtraction of two equations to eliminate recurring digits
M1dep
eg \(99x = 1.8\) or \(990x = 18\) or \(x = \dfrac{1.8}{99}\)
\((x =)\ \dfrac{1.8}{99}\) or \(\dfrac{18}{990}\) and \((x =)\ \dfrac{1}{55}\) and \(\dfrac{56}{55}\) or \(\dfrac{100.8}{99}\) or \(\dfrac{1008}{990}\) and \(\dfrac{56}{55}\) with no incorrect working
A1
oe from using different powers of 10
Alternative method 3: using \(x = 1.018\ldots\) and addition
Multiplication by power of 10
M1
eg \(10x = 10.18\ldots\) or \(100x = 101.81\ldots\) or \(1000x = 1018.18\ldots\) any or no letter
Correct addition of two correct equations leading to 0.9 recurring
M1dep
eg \(110x = 111.99\ldots\) or \(1100x = 1119.99\ldots\)
\((x =)\ \dfrac{112}{110}\) and \(\dfrac{56}{55}\) with no incorrect working
A1
oe from using different powers of 10
Additional guidance
Up to M2 may be awarded for correct work with no answer or incorrect answer if this is seen amongst multiple attempts
For all marks, numbers must be correct
Working with 1.018018018… scores 0
Recurring decimals should be denoted by correct notation or at least two of the recurring digits followed by at least two dots
In alt1 and alt2 condone incorrect recurring notation if the result of the subtraction is a correct equation
19 When converted to a fraction \(\quad 0.\dot{7} = \dfrac{7}{9}\)
Work out \(\quad 0.\dot{4} + 0.0\dot{7}\)
Give your answer as a fraction. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\((0.\dot{4} =)\ \dfrac{4}{9}\) or \(10x - x = 4\) or \(9x = 4\)
M1
oe eg \(100x - x = 44\) or \(99x = 44\) or \(100x - 10x = 40\) or \(90x = 40\) any letter
\(\dfrac{7}{9} \div 10\) or \(\dfrac{7}{90}\) or \(10y - y = 0.7\) or \(9y = 0.7\)
M1
oe eg \(100y - y = 7.7\) or \(99y = 7.7\) or \(100y - 10y = 7\) or \(90y = 7\) any letter
\(\dfrac{47}{90}\)
A1
oe single fraction
Alternative method 2
\(0.5\dot{2}\)
M1
oe
\(10x - x = 4.7\) or \(9x = 4.7\)
M1dep
oe eg \(100x - x = 51.7\) or \(99x = 51.7\) or \(100x - 10x = 47\) or \(90x = 47\) any letter
\(\dfrac{47}{90}\)
A1
oe single fraction
Additional guidance
For M marks, allow fractions with decimal numerator or denominator eg in alt 1, \(\dfrac{0.7}{9}\) scores M1 and in alt 2, \(\dfrac{4.7}{9}\) scores M2
23 Write \(0.1\dot{3}\) as a fraction in its simplest form. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1 – subtracting powers of 10 algebraically
Denotes the given recurring decimal by a letter and multiplies by one of 10, 100, etc
M1
eg \(10x = 1.33\ldots\) or \(100x = 13.3\ldots\)
Denotes the given recurring decimal by a letter and multiplies by one or two of 10, 100, etc and subtracts accordingly
M1dep
eg \(10x - x = 1.333\ldots - 0.1333\ldots\) or \(9x = 1.2\) or \(\dfrac{1.2}{9}\) or \(100x - x = 13.333\ldots - 0.1333\ldots\) or \(99x = 13.2\) or \(\dfrac{13.2}{99}\) or \(100x - 10x = 13.333\ldots - 1.333\ldots\) or \(90x = 12\) or \(\dfrac{12}{90}\)
\(\dfrac{2}{15}\)
A1
Alternative method 2 – subtracting powers of 10 numerically
Multiplies the given decimal by one of 10, 100, etc
M1
eg \(0.1\dot{3} \times 10 = 1.\dot{3}\)
Multiplies the given decimal by one or two of 10, 100, etc and subtracts appropriately in fraction form
M1dep
eg \(0.1\dot{3} \times 100 = 13.\dot{3}\) and \(0.1\dot{3} \times 10 = 1.\dot{3}\) and \(\dfrac{13.3 - 1.3}{100 - 10}\) or \(\dfrac{12}{90}\)
\(\dfrac{2}{15}\)
A1
Alternative method 3 – splitting into a known fraction and a recurring decimal
Splits into 0.1 and \(0.0\dot{3}\) and uses a correct first step from alt 1 or alt 2 with \(0.0\dot{3}\)
M1
eg \(10x = 0.33\ldots\) or \(0.0\dot{3} \times 100 = 3.33\ldots\) 0.1 does not need to be seen separately at this stage
Correct method to evaluate \(0.0\dot{3}\) as a fraction and addition to \(\dfrac{1}{10}\) or \(\dfrac{1}{30} + \dfrac{1}{10}\) or \(\dfrac{4}{30}\)
M1dep
oe fraction
\(\dfrac{2}{15}\)
A1
Additional guidance
Condone decimals within fractions up to M2 eg \(\dfrac{1.2}{9}\)
M2
Equals signs may be implied throughout
Subtraction signs must be seen or the results correct
Recurring decimals should be denoted by correct notation or at least two of the recurring digits followed by at least two dots. However, condone missing dots if the result is, or would be, correct eg condone \(13.3 - 1.3 = 100x - 10x\)
21 Convert \(\quad 0.6\dot{1} \quad\) to a fraction. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\(10x = 6.11\ldots\) and \(x = 0.61\ldots\) or \(100x = 61.11\ldots\) and \(10x = 6.11\ldots\)
M1
oe two powers of 10
\(10x - x = 6.11\ldots - 0.61\ldots\) or \(9x = 5.5\)
M1dep
oe subtraction of powers of 10 eg \(100x - 10x = 61.1\ldots - 6.1\ldots\)
\(\dfrac{11}{18}\) or \(\dfrac{55}{90}\) or \(\dfrac{605}{990}\)
A1
oe fraction
Alternative method 2
\((0.6\dot{1} =)\ 0.6 + 0.0\dot{1}\) and \(10x = 0.11\ldots\) and \(x = 0.01\ldots\) or \(100x = 1.11\ldots\) and \(10x = 0.11\ldots\)
M1
oe two powers of 10
\(10x - x = 0.11\ldots - 0.01\ldots\) or \(9x = 0.1\) and \(\dfrac{6}{10} +\) their \(\dfrac{1}{90}\)
M1dep
oe subtraction of powers of 10, with \(x\) evaluated as a fraction and added to \(\dfrac{6}{10}\) eg \(1000x - 10x = 11.11\ldots - 0.11\ldots\) or \(990x = 11\) and \(\dfrac{3}{5} + \dfrac{11}{990}\) sum of correct fractions implies M1M1
\(\dfrac{11}{18}\) or \(\dfrac{55}{90}\) or \(\dfrac{605}{990}\)
A1
oe fraction
Additional guidance
Ignore incorrect simplification of a correct fraction eg \(\dfrac{605}{990}\) and \(\dfrac{121}{190}\)
M1M1A1
Otherwise correct fraction with fraction(s) or decimal(s) as the numerator and/or denominator, eg \(\dfrac{5.5}{9}\)
13 Convert \(\quad \dfrac{5}{6} \quad\) to a recurring decimal. [2 marks]
Mark scheme
Answer
Mark
Comments
\(5 \div 6\) attempted with at least 0.8 shown and a carry of any integer from 1 to 7 or \(0.1\dot{6} \times 5\) or \(1.\dot{6} \div 2\) or \(1 - 0.1\dot{6}\)
M1
oe calculation involving a correct recurring decimal which would give an outcome of \(0.8\dot{3}\)
\(0.8\dot{3}\)
A1
condone any number of 3s immediately before the recurring 3
Additional guidance
Condone other recurring symbols or repeated dots eg \(0.83^{\text{r}}\) or 0.83…
26 Prove algebraically that \(\quad 3.4\dot{7} = \dfrac{313}{90}\) [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1 Working with \(3.47\ldots\)
\(10x = 34.7\ldots\) or \(100x = 347.7\ldots\)
M1
oe multiplication by a power of 10 eg \(1000x = 3477.7\ldots\) any letter
\(10x - x = 34.7\ldots - 3.47\ldots\) or \(9x = 31.3\) with \(10x = 34.7\ldots\) seen or \(100x - 10x = 347.7\ldots - 34.7\ldots\) or \(90x = 313\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen or \(100x - x = 347.7\ldots - 3.47\ldots\) or \(99x = 344.3\) with \(100x = 347.7\ldots\) seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 3477.7\ldots - 34.7\ldots\) or \(990x = 3443\) with \(1000x = 3477.7\ldots\) and \(10x = 34.7\ldots\) seen numbers must all be correct
\(x = 3.47\ldots\) stated and M2 scored and \(9x = 31.3\) and \((x =)\ \dfrac{31.3}{9}\) and \(\dfrac{313}{90}\) or \(x = 3.47\ldots\) stated and M2 scored and \(90x = 313\) and \((x =)\ \dfrac{313}{90}\) or \(x = 3.47\ldots\) stated and M2 scored and \(99x = 344.3\) and \((x =)\ \dfrac{344.3}{99}\) and \(\dfrac{313}{90}\)
A1
oe eg \(x = 3.47\ldots\) stated and M2 scored and \(990x = 3443\) and \((x =)\ \dfrac{3443}{990}\) and \(\dfrac{313}{90}\)
Alternative method 2 Working with \(0.47\ldots\)
\(10x = 4.7\ldots\) or \(100x = 47.7\ldots\)
M1
oe multiplication by a power of 10 eg \(1000x = 477.7\ldots\) any letter
\(10x - x = 4.7\ldots - 0.47\ldots\) or \(9x = 4.3\) with \(10x = 4.7\ldots\) seen or \(100x - 10x = 47.7\ldots - 4.7\ldots\) or \(90x = 43\) with \(100x = 47.7\ldots\) and \(10x = 4.7\ldots\) seen or \(100x - x = 47.7\ldots - 0.47\ldots\) or \(99x = 47.3\) with \(100x = 47.7\ldots\) seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 477.7\ldots - 4.7\ldots\) or \(990x = 473\) with \(1000x = 477.7\ldots\) and \(10x = 4.7\ldots\) seen numbers must all be correct
\(x = 0.47\ldots\) stated and M2 scored and \(9x = 4.3\) and \((x =)\ \dfrac{4.3}{9}\) and \(3\dfrac{4.3}{9}\) and \(\dfrac{313}{90}\) or \(x = 0.47\ldots\) stated and M2 scored and \(90x = 43\) and \((x =)\ \dfrac{43}{90}\) and \(3\dfrac{43}{90}\) and \(\dfrac{313}{90}\) or \(x = 0.47\ldots\) stated and M2 scored and \(99x = 47.3\) and \((x =)\ \dfrac{47.3}{99}\) and \(3\dfrac{47.3}{99}\) and \(\dfrac{313}{90}\)
A1
oe eg \(x = 0.47\ldots\) stated and M2 scored and \(990x = 473\) and \((x =)\ \dfrac{473}{990}\) and \(3\dfrac{473}{990}\) and \(\dfrac{313}{90}\)
Alternative method 3 Working with \(0.07\ldots\)
\(10x = 0.7\ldots\) or \(100x = 7.7\ldots\)
M1
oe multiplication by a power of 10 eg \(1000x = 77.7\ldots\) any letter
\(10x - x = 0.7\ldots - 0.07\ldots\) or \(9x = 0.7\) with \(10x = 0.7\ldots\) seen or \(100x - 10x = 7.7\ldots - 0.7\ldots\) or \(90x = 7\) with \(100x = 7.7\ldots\) and \(10x = 0.7\ldots\) seen or \(100x - x = 7.7\ldots - 0.07\ldots\) or \(99x = 7.7\) with \(100x = 7.7\ldots\) seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 77.7\ldots - 0.7\ldots\) or \(990x = 77\) with \(1000x = 77.7\ldots\) and \(10x = 0.7\ldots\) seen numbers must all be correct
\(x = 0.07\ldots\) stated and M2 scored and \(9x = 0.7\) and \((x =)\ \dfrac{0.7}{9}\) and \(3.4 + \dfrac{0.7}{9}\) and \(\dfrac{313}{90}\) or \(x = 0.07\ldots\) stated and M2 scored and \(90x = 7\) and \((x =)\ \dfrac{7}{90}\) and \(3.4 + \dfrac{7}{90}\) and \(\dfrac{313}{90}\) or \(x = 0.07\ldots\) stated and M2 scored and \(99x = 7.7\) and \((x =)\ \dfrac{7.7}{99}\) and \(3.4 + \dfrac{7.7}{99}\) and \(\dfrac{313}{90}\)
A1
oe eg \(x = 0.07\ldots\) stated and M2 scored and \(990x = 77\) and \((x =)\ \dfrac{77}{990}\) and \(3.4 + \dfrac{77}{990}\) and \(\dfrac{313}{90}\)
Additional guidance
\(313 \div 90 = 3.47\ldots\)
M0M0A0
Alt 1 M1dep oe subtraction to eliminate recurring decimals includes \(100x - 10x = 313\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen or \(90x = 347.7\ldots - 34.7\ldots\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen (apply same principle in Alt 2 and Alt 3)
Alt 2 equivalents for final part of A1 eg For \(3\dfrac{43}{90}\) and \(\dfrac{313}{90}\) allow \(3 + \dfrac{43}{90}\) and \(\dfrac{313}{90}\)
Alt 3 equivalents for final part of A1 eg For \(3.4 + \dfrac{7}{90}\) and \(\dfrac{313}{90}\) allow \(3 + \dfrac{4}{10} + \dfrac{7}{90}\) and \(\dfrac{313}{90}\)
27 Prove algebraically that \(\; 2.7\dot{5} \;\) converts to the fraction \(\dfrac{124}{45}\) [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1 Working with 2.75…..
\(10x = 27.5\)… or \(100x = 275.5\)…
M1
oe multiplication by a power of 10 eg \(1000x = 2755.5\)… any letter
\(10x - x = 27.5\)… \(-\) 2.75… or \(9x = 24.8\) with \(10x = 27.5\)… seen or \(100x - 10x = 275.5\)… \(-\) 27.5… or \(90x = 248\) with \(100x = 275.5\)… and \(10x = 27.5\)… seen or \(100x - x = 275.5\)… \(-\) 2.75… or \(99x = 272.8\) with \(100x = 275.5\)… seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 2755.5\)… \(-\) 27.5… or \(990x = 2728\) with \(1000x = 2755.5\)… and \(10x = 27.5\)… seen numbers must all be correct
\(x = 2.75\)… stated and M2 scored and \(9x = 24.8\) and \(x = \dfrac{24.8}{9} = \dfrac{124}{45}\) or \(x = 2.75\)… stated and M2 scored and \(90x = 248\) and \(x = \dfrac{248}{90} = \dfrac{124}{45}\) or \(x = 2.75\)… stated and M2 scored and \(99x = 272.8\) and \(x = \dfrac{272.8}{99} = \dfrac{124}{45}\)
A1
oe eg \(x = 2.75\)… stated and M2 scored and \(990x = 2728\) and \(x = \dfrac{2728}{990} = \dfrac{124}{45}\)
Alternative method 2 Working with 0.75…..
\(10x = 7.5\)… or \(100x = 75.5\)…
M1
oe multiplication by a power of 10 eg \(1000x = 755.5\)… any letter
\(10x - x = 7.5\)… \(-\) 0.75… or \(9x = 6.8\) with \(10x = 7.5\)… seen or \(100x - 10x = 75.5\)… \(-\) 7.5… or \(90x = 68\) with \(100x = 75.5\)… and \(10x = 7.5\)… seen or \(100x - x = 75.5\)… \(-\) 0.75… or \(99x = 74.8\) with \(100x = 75.5\)… seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 755.5\)… \(-\) 7.5… or \(990x = 748\) with \(1000x = 755.5\)… and \(10x = 7.5\)… seen numbers must all be correct
\(x = 0.75\)… stated and M2 scored and \(9x = 6.8\) and \(x = \dfrac{6.8}{9}\) and \(2\dfrac{6.8}{9} = \dfrac{124}{45}\) or \(x = 0.75\)… stated and M2 scored and \(90x = 68\) and \(x = \dfrac{68}{90}\) and \(2\dfrac{68}{90} = \dfrac{124}{45}\) or \(x = 0.75\)… stated and M2 scored and \(99x = 74.8\) and \(x = \dfrac{74.8}{99}\) and \(2\dfrac{74.8}{99} = \dfrac{124}{45}\)
A1
oe eg \(x = 0.75\)… stated and M2 scored and \(990x = 748\) and \(x = \dfrac{748}{990}\) and \(2\dfrac{748}{990} = \dfrac{124}{45}\)
Alternative method 3 Working with 0.05…..
\(10x = 0.5\)… or \(100x = 5.5\)…
M1
oe multiplication by a power of 10 eg \(1000x = 55.55\)… any letter
\(10x - x = 0.5\)… \(-\) 0.05… or \(9x = 0.5\) with \(10x = 0.5\)… seen or \(100x - 10x = 5.5\)… \(-\) 0.5… or \(90x = 5\) with \(100x = 5.5\)… and \(10x = 0.5\)… seen or \(100x - x = 5.5\)… \(-\) 0.05… or \(99x = 5.5\) with \(100x = 5.5\)… seen
M1dep
oe subtraction to eliminate recurring digits eg \(1000x - 10x = 55.5\)… \(-\) 0.5… or \(990x = 55\) with \(1000x = 55.5\)… and \(10x = 0.5\)… seen numbers must all be correct
\(x = 0.05\)… stated and M2 scored and \(9x = 0.5\) and \(x = \dfrac{0.5}{9}\) and \(2.7 + \dfrac{0.5}{9} = \dfrac{124}{45}\) or \(x = 0.05\)… stated and M2 scored and \(90x = 5\) and \(x = \dfrac{5}{90}\) and \(2.7 + \dfrac{5}{90} = \dfrac{124}{45}\) or \(x = 0.05\)… stated and M2 scored and \(99x = 5.5\) and \(x = \dfrac{5.5}{99}\) and \(2.7 + \dfrac{5.5}{99} = \dfrac{124}{45}\)
A1
oe eg \(x = 0.05\)… stated and M2 scored and \(990x = 55\) and \(x = \dfrac{55}{990}\) and \(2.7 + \dfrac{55}{990} = \dfrac{124}{45}\)
Additional guidance
\(124 \div 45 = 2.75\)…
M0M0A0
Alt 1 M1dep oe subtraction to eliminate recurring decimals includes \(100x - 10x = 248\) with \(100x = 275.5\)… and \(10x = 27.5\)… seen or \(90x = 275.5\)… \(-\) 27.5… with \(100x = 275.5\)… and \(10x = 27.5\)… seen (apply same principle in Alts 2 and 3)
Alt 2 equivalents for final part of A1 eg For \(2\dfrac{68}{90} = \dfrac{124}{45}\) allow \(2 + \dfrac{68}{90} = \dfrac{124}{45}\)
Alt 3 equivalents for final part of A1 eg For \(2.7 + \dfrac{5}{90} = \dfrac{124}{45}\) allow \(2 + \dfrac{7}{10} + \dfrac{5}{90} = \dfrac{124}{45}\)
Use this fact to show that \(\quad 0.0\dot{7} = \dfrac{7}{90}\) [1 mark]
(b) Using part (a) or otherwise, convert \(\quad 0.2\dot{7} \quad\) to a fraction.
Give your answer in its simplest form. [3 marks]
Mark scheme (a)
Answer
Mark
Comments
\(0.\dot{7} \div 10 = 0.0\dot{7}\) and \(\dfrac{7}{9} \div 10 = \dfrac{7}{90}\) or \(0.0\dot{7} \times 10 = 0.\dot{7}\) and \(\dfrac{7}{90} \times 10 = \dfrac{7}{9}\) or \(0.\dot{7} \div 10 = 0.0\dot{7}\) and \(\dfrac{7}{90} \times 10 = \dfrac{7}{9}\) or because the decimal is divided by 10 the 9 has to be multiplied by 10
B1
oe
Additional guidance
Algebraic methods
B0
Division of 7 by 90
B0
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
\(0.2 + 0.0\dot{7}\) or \(\dfrac{2}{10} + \dfrac{7}{90}\)
M1
\(\dfrac{18}{90} + \dfrac{7}{90}\) or \(\dfrac{25}{90}\)
M1dep
\(\dfrac{5}{18}\)
A1
Alternative method 2
\(10x = 2.777\)… or \(100x = 27.777\)…
M1
Any letter
\(10x - x = 2.777\)… \(-\) 0.277… or \(9x = 2.5\) or \(\dfrac{2.5}{9}\) or \(100x - x = 27.777\)… \(-\) 0.277… or \(99x = 27.5\) or \(\dfrac{27.5}{99}\) or \(100x - 10x = 27.777\)… \(-\) 2.777… or \(90x = 25\) or \(\dfrac{25}{90}\)
Give your answer as a mixed number in its simplest form.
You must show your working. [5]
Mark scheme
Answer
Marks
Part marks and guidance
\(6\frac{2}{5}\) final answer with correct working
5
Correct working requires \(\frac{2}{9}\) oe fraction seen and \(\frac{128}{90}\) oe fraction seen or alternate convincing approach
B4 for \(\frac{64}{10}\) oe with correct working isw cancelling
e.g. B4 for \(6\frac{4}{10}\), \(\frac{32}{5}\), \(\frac{576}{90}\), \(\frac{1152}{180}\) with correct working
OR
B1 for \([0.\dot{2} =]\ \frac{2}{9}\) oe fraction
B2 for \(\frac{128}{90}\) oe fraction
For B2 condone e.g. \(\frac{12.8}{9}\)
or M1 for [\(n\) =1.422….] and [100\(n\) =] 142.2… oe or better or [\(n\) =] 1.422…. and [10\(n\) =] 14.22… or better
oe any correct approach that allows elimination of the recurring decimal Implied by e.g. 90\(n\) = 128 , 9\(n\) = 12.8
or shows e.g. [\(n\) = 0.422….] and [100\(n\) =] 42.2… oe or better
implied by [1]\(\frac{38}{90}\) oe seen
If 0, 1 or 2 scored, instead award SC3 for \(6\frac{2}{5}\) with no or insufficient working If 0 or 1 scored, instead award SC2 for answer \(\frac{64}{10}\) oe with no or insufficient working
Find the value of R. Give your answer as a mixed number in its simplest form. You must show your working. [5]
Mark scheme
Answer
Marks
Part marks and guidance
\(2\frac{1}{9}\) final answer with correct working
5
Correct working requires evidence of at least M2 M1 or other alternate correct approach leading to \(2\frac{1}{9}\), correct working may be shown on a diagram
M2 for \(0.\dot{4}\) or 0.44[4…] oe or M1 for \(1.\dot{6} - 1.\dot{2}\) oe seen
For M1 accept 1.66[6…] – 1.22[2…] or better
M2 for \(2.\dot{1}\) oe or M1 for \(1.\dot{6} + 0.\dot{4}\) oe
If 0 or 1 scored, instead award SC2 for answer \(2\frac{1}{9}\)
For M2, accept any clear indication that the decimal figure 1 recurs e.g. 2.1r, 2.111… For M1 accept 1.66[6…] + 0.44[4…] or better Accept oe e.g. \(1.\dot{2} + 2(0.\dot{4})\)
Alt method M2 for \(\frac{11}{9}\) oe fraction or M1 for \(1.\dot{2}\) and \(12.\dot{2}\) oe seen
Allow oe for any pair allowing ‘recurrence’ to be removed, accept e.g. 1.22… and 12.22…
M2 for \(\frac{15}{9} + \left(\frac{15}{9} - \textit{their } \frac{11}{9}\right)\) oe
For M2 and M1 must be using a common denominator for FT and could work with mixed numbers . The method must be shown for FT e.g. M2 for their \(\frac{11}{9} + 2\left(\frac{15}{9} - \textit{their } \frac{11}{9}\right)\) oe
or M1 for \(\left(\frac{15}{9} - \textit{their } \frac{11}{9}\right)\) oe
If 0 or 1 scored, instead award SC2 for answer \(2\frac{1}{9}\)
M1 for \(100x - x = 18.1818\ldots - 0.1818\ldots\) or better
For full marks, clear step by step process must be evident \(x\) and \(100x\) to same number of dp (min of 2dp) or both to min of 4dp or accept \(x = 0.\dot{1}\dot{8}\) and \(100x = 18.\dot{1}\dot{8}\)
Apply marks in a similar way to other methods eg M1 and M1 for \(10000x - 100x\) etc
If attempting non-calculator 2÷11: For full marks, need to see all “carry-overs” of 9 and 2 and answer to division of 0.181… M2 for “carry-over” of 9 and 2 and answer to division of 0.18 or M1 for “carry-over” of 9 and answer 0.1…
Mark scheme (b)
Answer
Marks
Part marks and guidance
\(0.\dot{1}\dot{8} \times 5\) or \(\frac{2}{11} \times 5\) or ‘multiply by 5’ oe
1
Accept equivalent calculations or words e.g. \(0.\dot{1}\dot{8} \div 2 = 0.\dot{0}\dot{9}\) and \(0.\dot{0}\dot{9} \times 10\) ‘divide by 2 and then multiply by 10’ Working may be to min of 4dp but answer must use recurring notation
where \(\frac{p}{q}\) is a fraction in its lowest terms.
Find the value of \(p\) and the value of \(q\). [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
16
2
M1 for \(\left[\sqrt[3]{64} =\right]\) 4 or [\(64^2\) =] 4096
M1 implied by answer \(\frac{1}{16}\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
[\(p\) = ] 14 [\(q\) = ] 33
4
B3 for 42 and 99 OR B2 for \(\frac{13}{99}\) or 0.4242…. or M1 for [\(100x\) = ] 13.13….. or 0.555…
M1 for \(\frac{55}{99} - \frac{13}{99}\) FT their fraction converted to common denominator (dep on a conversion needed of their fraction or \(\frac{5}{9}\))
B3 for \(\frac{42}{99}\)
or any pair that allow elimination of recurring decimal
M1 may be implied from the value of their p and their q
16 Work out \(0.\dot{6} \times 0.\dot{5}\dot{4}\) giving your answer as a fraction in its simplest form. You must show your working. [5]
Mark scheme
Answer
Marks
Part marks and guidance
\(\frac{4}{11}\) final answer with correct working
5
“Correct working” requires \(\frac{6}{9}\) oe fraction seen and \(\frac{54}{99}\) oe fraction seen or B1 and B2 if alt method
B4 for \(\frac{324}{891}\) oe with correct working isw cancelling or B1 for [\(0.\dot{6}\) =] \(\frac{6}{9}\) oe fraction B2 for \(\frac{54}{99}\) oe fraction or M1 for [100\(n\) =] 54.54… oe or better
e.g. \(\frac{36}{99}\) with correct working gets B4
oe any correct method for conversion
Alternative method: B4 for \(\frac{324}{891}\) oe with correct working or B1 for [\(0.\dot{6}\) =] \(\frac{6}{9}\) oe fraction B2 for [100\(n\) =] 36.36… or better or M1 for [\(\frac{2}{3} \times 0.\dot{5}\dot{4}\) =] \(0.\dot{3}\dot{6}\)
e.g. \(\frac{36}{99}\) with correct working gets B4
If 0, 1 or 2 scored, instead award SC3 for \(\frac{4}{11}\) with no or insufficient working If 0 or 1 scored, instead award SC2 for answer \(\frac{324}{891}\) oe with no or insufficient working
18 Write \(0.4\dot{1}\dot{6}\) as a fraction in its simplest form. You must show full working in support of your answer. [3]
Mark scheme
Answer
Marks
Part marks and guidance
\(\frac{412}{990} = \frac{206}{495}\) with correct working
or
\(\frac{41.2}{99} = \frac{206}{495}\) with correct working
3
M2 for \(1000x\) [–] \(10x\) = 416.16[1...] [–] 4.16[1...] leading to \(990x = 412\) or for \(100x\) [–] \(x\) = 41.616[1...] [–] 0. 416[1...] leading to \(99x = 41.2\)
or M1 for \(10x\) = 4.16[1...] or \(100x\) = 41.616[1...] or \(1000x\) = 416.16[1...] seen
(a) Write \(\frac{1}{6}\) as a recurring decimal. [2]
(b) Elsa divides a two-digit number by another two-digit number. She gets the answer \(0.1\dot{5}\).
She says that there is only one possible pair of numbers that will give this answer. Is she correct? Show how you decide. [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(0.1\dot{6}\) final answer
2
B1 for 0.16….
Accept unambiguous alternate notation for the recurring decimal e.g. B1 for 0.166, 0.168, \(0.\dot{1}\dot{6}\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
She is correct oe OR She is not correct oe AND 7 [and] 45 and 14 [and] 90 or 14 [and] 90 and \(14k\) [and] \(90k\) shown
4
B3 for \(\frac{7}{45}\) and \(\frac{14}{90}\) or for \(\frac{14}{90}\) and \(\frac{14k}{90k}\) or B2 for \(\frac{14}{90}\) oe fraction or M1 for 1.5... [× 10n] and 15.5… [× 10n] seen
For 4 marks there must be no incorrect fractions shown Accept yes for she is correct and no for she is not correct
Where \(k\) is a positive integer For B2 and B3 accept as pairs of values instead of fractions
Where \(n\) is an integer e.g. allow M1 for 0.15… and 1.5… or for 15.5.. and 155.5…