Recurring decimals as fractions

(a) $\frac{7}{9}$

1. Let $x = 0.777\ldots$
   $$10x = 7.777\ldots$$
2. Subtract.
   $$9x = 7 \Rightarrow x = \frac{7}{9}$$

• M1 Writing $10x = 7.777\ldots$ and subtracting.
• A1 The correct answer, $\frac{7}{9}$.

(b) $\frac{3}{11}$

1. The repeating block has two digits, so multiply by 100.
   $$100x = 27.2727\ldots$$
2. Subtract.
   $$99x = 27 \Rightarrow x = \frac{27}{99} = \frac{3}{11}$$

• M1 Using $100x$ and subtracting to get $99x = 27$.
• A1 The correct answer, $\frac{3}{11}$.

(a) $99x = 45$, $x = \frac{45}{99} = \frac{5}{11}$

1. $$x = 0.4545\ldots, \qquad 100x = 45.4545\ldots$$
2. $$99x = 45 \Rightarrow x = \frac{45}{99} = \frac{5}{11}$$

• M1 Forming $100x - x = 45$.
• A1 Simplifying $\frac{45}{99}$ to $\frac{5}{11}$.

(b) $2\frac{5}{11}$

1. $2.\dot{4}\dot{5} = 2 + 0.\dot{4}\dot{5} = 2\frac{5}{11}$.

• B1 $2\frac{5}{11}$.

(a) $\frac{1}{6}$

1. $$100x-10x=15$$
2. Let x = 0.166666… . Then 10x = 1.666666… and 100x = 16.666666… .
3. Subtract: 100x $-$ 10x = 15, so 90x = 15.
4. x = 15/90 = 1/6.

• M1 Establishing $100x-10x=15$ or an equivalent valid method.
• A1 Correct answer: $\frac{1}{6}$

(a) $990x = 135$, so $x = \frac{135}{990} = \frac{3}{22}$

1. Line up the repeating blocks.
   $$10x = 1.363636\ldots, \qquad 1000x = 136.363636\ldots$$
2. Subtract.
   $$990x = 135$$
3. Simplify.
   $$x = \frac{135}{990} = \frac{27}{198} = \frac{3}{22}$$

• M1 Two multiples of $x$ with the same recurring part, such as $10x$ and $1000x$.
• M1 Subtracting to get $990x = 135$ (or $99x = 13.5$).
• A1 Simplifying $\frac{135}{990}$ to $\frac{3}{22}$ with the working shown.

(a) $\frac{1}{45}$

1. $0.0\dot{2}$ is $0.\dot{2} \div 10$.
2. $$\frac{2}{9} \div 10 = \frac{2}{90} = \frac{1}{45}$$

• M1 Dividing $\frac{2}{9}$ by 10.
• A1 The correct answer, $\frac{1}{45}$.

(b) $\frac{16}{45}$

1. $0.\dot{3} = \frac{1}{3}$.
2. $$\frac{1}{3} + \frac{1}{45} = \frac{15}{45} + \frac{1}{45} = \frac{16}{45}$$
3. Check: $0.333\ldots + 0.0222\ldots = 0.3555\ldots$ and $\frac{16}{45} = 0.3\dot{5}$.

• P1 Writing $0.\dot{3} = \frac{1}{3}$.
• P1 Adding with a common denominator of 45.
• A1 $\frac{16}{45}$.

(a) $\frac{9}{22}$

1. $$1000x-10x=135$$
2. $$3/22+3/11$$
3. For x, use 1000x $-$ 10x = 136.3636… $-$ 1.3636… = 135. Thus x = 135/990 = 3/22.
4. For y, 100y $-$ y = 27, giving y = 27/99 = 3/11.
5. Add using denominator 22: 3/22 + 6/22 = 9/22.

• P1 Establishing $1000x-10x=135$ or an equivalent valid method.
• P1 Establishing $3/22+3/11$ or an equivalent valid method.
• A1 Correct answer: $\frac{9}{22}$

(a) $\frac{19}{55}$

1. Multiply by 10 and by 1000 so the repeating tails align.
   $$1000x-10x=342$$
2. Divide by the resulting coefficient of x.
   $$x=342/990$$
3. Therefore $\frac{19}{55}$.

• M1 Multiply by 10 and by 1000 so the repeating tails align.
• M1 Divide by the resulting coefficient of x.
• A1 Correct answer: $\frac{19}{55}$

(b) $\frac{5}{11}$

1. $10x = 3.454545\ldots$, so subtracting 3 leaves the recurring block 45.
   $$10\times \frac{19}{55}-3=\frac{38}{11}-\frac{33}{11}$$
2. So $10x-3=\frac{5}{11}$.

• M1 Substituting $x=\frac{19}{55}$ into $10x-3$.
• A1 Correct answer: $\frac{5}{11}$

(a) $63$

1. $$(10a+b)/99+(10b+a)/99=1$$
2. $$a-b=3$$
3. A two-digit recurring block gives x = (10a + b)/99 and y = (10b + a)/99.
4. From x + y = 1, 11(a + b)/99 = 1, so a + b = 9.
5. From x $-$ y = 3/11, 9(a $-$ b)/99 = 3/11, so a $-$ b = 3.
6. Adding the equations gives 2a = 12, so a = 6 and b = 3. The integer ab is 63.

• P1 Establishing $(10a+b)/99+(10b+a)/99=1$ or an equivalent valid method.
• P1 Establishing $a-b=3$ or an equivalent valid method.
• A1 Correct answer: $63$