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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3Topics
    |
    1 Quiz
  3. Number Base System III | Week 3
    2Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2Topics
  5. Modular Arithmetic II | Week 5
    2Topics
  6. Modular Arithmetic III | Week 6
    3Topics
    |
    1 Quiz
  7. Indices I | Week 7
    2Topics
  8. Indices II | Week 8
    1Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3Topics
  10. Logarithms II
    1Topic
    |
    1 Quiz
Lesson Progress
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Here, we will learn the methods to convert from other bases to Base 10

Example 1

Convert

(i) 7603eight to base 10

(ii) 1101twoto base 10

(iii) 5B9Twelve to base 10

Solution (i)7603eight to base 10

a. First Method: (Power Expansion)

Power Expansion entails multiplying out the number base using their base values.

  • First, write the place values starting from the right-hand side.
  • Write each digit under its place value.
  • Multiply each digit by its corresponding place value.
  • Add up the products. The answer will be the decimal number in base ten.
place value Binary

Students often find it difficult to locate the place values. The easiest way is to start from the last digit to the first digit. The last Digit always starts with a 0, the digit that follows a 1, and so on. (0, 1 2, 3, etc)

83828180
7603

Multiply each digit by its place value

7603eight = (7 x 83) + (6 x 82) + (0 x 81) + (3 x 80)

              = 3584 + 384 + 0 + 3

              = 3,971ten

               ∴    7603eight = 3,971ten

b. Alternate Method – (Successive Multiplication)

7 x 8 = 56 + 6 = 62 x 8  = 496 + 0 = 496 x 8 = 3968 + 3 = 3,971ten  

Solution (ii)1101twoto base 10

a. First Method: (Power Expansion)

23222120
1101

Multiply each digit by its place value

1101two = (1 x 23) + (1 x 22) + (0 x 21) + (1 x 20)

= 8 + 4 + 0 + 1

= 13ten  

Alternate Method

1 x 2 = 2 + 1 = 3 x 2 = 6 + 0 = 6 x 2 = 12 + 1 =13ten  

Solution (iii)5B9Twelve to base 10

a. First Method: (Power Expansion)

122121120
5B9

Multiply each digit by its place value (Note: B =11)

5B9Twelve = (5 x 122) + (11 x 121) + (9 x 120)

= 720 + 132 + 9

= 861ten  

Alternate Method

5 x 12 = 60 + 11 = 71 x 12  = 852 + 9 = 861ten  

Example 2

Convert 1101.101two to base ten

\( \scriptsize 2^{3} \)\( \scriptsize 2^{2} \)\( \scriptsize 2^{1} \)\( \scriptsize 2^{0} \)\( \scriptsize 2^{-1} \)\( \scriptsize 2^{-2} \)\( \scriptsize 2^{-3} \)
1101.101
decimal number e1601570808972

In the conversion of a decimal number to base 10, the whole number is treated as the examples above (i.e from the last digit the numbering is 0, 1, 2, 3, etc up to the first digit).

The decimal part on the other hand is numbered -1, -2, -3, etc immediately after the decimal point. The first number after the decimal point will be -1, the second number will be -2, etc.

Hence. We multiply the first number by the base raised to the power minus one (x-1) and so on.

Solution

1101.101two = (1 x 23) + (1 × 22) + (0 × 21) + (1 × 20) + (1 × 2-1) + (0 × 2-2) + (1 × 2-3)

= 8 + 4 + 0 + 1 + \( \frac {1}{2} \) + 0 + \( \frac {1}{8} \)

= 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125

=  13.625

1101.101two= 13.625ten

Alternate Method

Whole number part = 1 x  2 = 2 + 1 = 3 x 2 = 6 + 0 = 6 x 2 = 12 + 1 = 13                                      

Fractional part = 1 x 2 = 2 + 0 = 2 x 2 = 4 + 1 = 5 

The result for the whole number part is: 1101two = 13ten

The place value of the last column of the fractional part is 2-3

  ∴   0.101two= 5 × 2-3 = 5 × \( \frac {1}{8} \) = 0.625

Summing up the results, then

1101.101two = 13 + 0.625

1101.101two = 13.625ten

Example 3

Convert

(i) 4A3D.216 to base 10

(ii) 4B3.A612to base 10

Solution (i)4A3D.216 to base 10

16316216116016-1
4A3D.2

Remember D = 13 from our Number Bases table

4A3D.216 = (4 × 163) + (A × 162) + (3 × 161) + (D × 160) + (2 × 16-1)

= (16,384) + (10 × 256) + (48) + (13 × 1) + (2 × 0.0625)

= 16,384 + 2,560 + 48 + 13 + 0.125

= 19,005.125

Solution (ii)4B3.A612to base 10

12212112012-112-2
4B3.A6

A = 10, B = 11

4B3.A612 = (4 × 122) + (B × 121) + (3 × 120) + (A × 12-1) + (6 × 12-2)

= (576) + (11 × 12) + (3) + (11 × 0.0833) + (6 × 0.00694)

= 576 + 132 + 3 + 0.9163 + 0.04164

= 711.96

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