Decimals

Q1: Express each of the following as a recurring decimal:

i. \( \frac{20}{3} \)

       
3) 20.000 (6.666...
  -18
  ----
    20
   -18
   ----
     20
    -18
    ----
      0
    ----
     ...

Answer: \(6.\overline{6}\)

ii. \( \frac{3}{11} \)

       
11) 3.000000 (0.272727...
  - 2.2
   ------
      80
     -77
    ----
      30
     -22
    ----
       80 (Cycle repeats)
      ...

Answer: \(0.\overline{27}\)

iii. \( \frac{5}{6} \)

       
6 ) 5.000 (0.8333...
  - 4.8
  ------
    20
   -18
   ----
     20
    -18
    ----
      20 (Cycle repeats)

Answer: \(0.8\overline{3}\)

iv. \( \frac{17}{90} \)

       
90 ) 17.000 (0.1888...
   -  0.0
    ------
     170.0
    - 90
      ----
      800
    - 720
   ------
      800 (Cycle repeats)
     

Answer: \(0.1\overline{8}\)

v. \( \frac{1}{37} \)

       
37) 1.000000 (0.027027...
  - 0
   ------
    100
   - 74
   ------
     260
    -259
    -----
       10
       -0
      -----
       100
     (repeats after full cycle every 3 digits)

Answer: \(0.\overline{027}\)

vi. \( \frac{22}{7} \)

       
7) 22.000000000000 (3.142857142857...
  -21
  ----
    10
   - 7
   ----
     30
    -28
    ----
      20
    - 14
    ----
       60
     - 56
     ----
        40
      - 35
      ----
         50
       - 49
       ----
          10 (Cycle repeats)

Answer: \(3.\overline{142857}\)

vii. \( \frac{2}{13} \)

       
13 ) 2.000000000 (0.153846153846...
   - 1.95
   -------
     50
   - 39
   -----
     110
   - 104
   -----
       60
     - 52
     -----
        80
      - 78
      -----
         20 (Cycle repeats)

Answer: \(0.\overline{153846}\)

Q2: Convert each of the following into a vulgar fraction:

i. \( 0.\overline{6} \)
The repeated digit is 6.
Thus, we write it as:
\(\frac{6}{9}\)
Simplify:
\(\frac{6}{9} = \frac{2}{3}\)
Answer: \( \frac{2}{3} \)

ii. \( 0.\overline{8} \)
The repeated digit is 8.
Thus, we write it as:
\(\frac{8}{9}\)
Answer: \( \frac{8}{9} \)

iii. \( 0.\overline{34} \)
The repeated digits are 34.
Thus, we write it as:
\(\frac{34}{99}\)
Answer: \( \frac{34}{99} \)

iv. \( 2.\overline{13} \)
The repeated digits are 13.
Thus, we write it as:
\(\frac{213}{99}\)
Simplify:
\(\frac{213}{99} = 2\frac{13}{99}\)
Answer: \( 2\frac{13}{99} \)

v. \( 1.\overline{243} \)
The repeated digits are 243.
Thus, we write it as:
\(\frac{1243}{999}\)
Simplify:
\(\frac{1243}{999} = 1\frac{243}{999}\)
Answer: \( 1\frac{243}{999} \)

Q3: Convert each of the following into a vulgar fraction:

i. \( 0.1\overline{6} \)
The number formed by the repeated digits is 16.
The number formed by the non-repeated digit is 1.
The difference is \(16 – 1 = 15\).
The denominator will have 9 for the repeating digit and 10 for the non-repeating digit: \(90\).
\(\frac{15}{90} = \frac{1}{6}\)
Answer: \( \frac{1}{6} \)

ii. \( 0.1\overline{43} \)
The number formed by the repeated digits is 143.
The number formed by the non-repeated digit is 1.
The difference is \(143 – 1 = 142\).
The denominator will have 99 for the repeating digits and 10 for the non-repeating digit: \(990\).
\(\frac{142}{990} = \frac{71}{495}\)
Answer: \( \frac{71}{495} \)

iii. \( 0.57\overline{4} \)
The number formed by the repeated digits is 574.
The number formed by the non-repeated digits is 57.
The difference is \(574 – 57 = 517\).
The denominator will have 9 for the repeating digit and 100 for the non-repeating digits: \(900\).
\(\frac{517}{900} = \frac{517}{900}\)
Answer: \( \frac{517}{900} \)

iv. \( 0.12\overline{34} \)
The number formed by the repeated digits is 1234.
The number formed by the non-repeated digits is 12.
The difference is \(1234 – 12 = 1222\).
The denominator will have 99 for the repeating digits and 100 for the non-repeating digits: \(9900\).
\(\frac{1222}{9900} = \frac{611}{4950}\)
Answer: \( \frac{611}{4950} \)