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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
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It is important to know that a number given in one base other than base ten can be converted to another base via base ten.

Example 1

Evaluate the following and express the answer in the bases indicated.

(i) 10401.117 to base eight

(ii) 2013 – 101012 + 2536 to base 6

Solution:

(i) Convert 10401.117 to base ten

i.e \( \scriptsize 10401.11_7 \\ \scriptsize = 1 \: \times \: 7^4 \: + \: 0 \: \times \: 7^3 \: + \: 4 \: \times \: 7^2 \\ \scriptsize \:+\: 0 \: \times \: 7^1 \: + \: 1 \: \times \: 7^0 \: + \: 1 \: \times \: 7^{-1} \: + \: 1 \: \times \: 7^{-2} \)

= \( \scriptsize 20141 \: + \: 0 \: + \: 196 \: + \: 0 \: + \: 1 \: + \: 0.142 \: + \: 0.0204 \)

= \( \scriptsize 2598.1624 \)

Convert to base 8

Whole number part

Screen Shot 2021 10 23 at 7.12.21 PM

Decimal part

Screen Shot 2021 10 23 at 7.25.09 PM

i.e \( \scriptsize 0.1624_8 = 0.123_8 \)

Aswer: \( \scriptsize 10401.11_7 = 5046.123_{8} \)

Solution:

(ii) 2013 – 101012 + 2536 to base 6

Convert all numbers to base ten

\(\scriptsize 201_3 \\ \scriptsize = 2 \: \times \: 3^2 \: + \: 0 \: \times \: 3^1 \: + \: 1 \: \times \: 3^0 \\ \scriptsize = 18 \: + \: 0 \: + \: 1 \\ \scriptsize = 19 \)

\( \scriptsize 10101_2 \\ \scriptsize = (1 \: \times \: 2^4 )\: + \: (0 \: \times \: 2^3) \: + \: (1 \: \times \: 2^2) \\ \scriptsize \: + \: (0 \: \times \: 2^1) \: + \: (1 \: \times \: 2^0) \\ \scriptsize = 16 \: + \: 0 \: + \: 4 \: + \: 0 \: + \: 1 \\ \scriptsize = 21 \)

\(\scriptsize 253_6 \\ \scriptsize = 2 \: \times \: 6^2 \: + \: 5 \: \times \: 6^1 \: + \: 3 \: \times \: 6^0 \\ \scriptsize = 72 \: + \: 30 \: + \: 3 \\ \scriptsize = 105 \)

i.e \( \scriptsize 201_3 \: – \: 10101_2 \: + \: 253_6 \\ \scriptsize = 19 \: – \: 21 \: + \: 105 \\ \scriptsize = 124 \: – \: 21 \\ \scriptsize = 103 \)

Screen Shot 2021 10 23 at 7.50.30 PM

Answer: \( \scriptsize 201_3 \: – \: 10101_2 \: + \: 253_6 = 251_6 \)

Example 2

Find the value of x in the following:

(i) \( \scriptsize 315_x \: -\: 223_x = 72_x \)

(ii) \( \scriptsize (251.25)_{10} = x_7\)

Solution (i):

\( \scriptsize 315_x = 3 \: \times \: x^2 \: + \: 1 \: \times \: x^1 \: + \: 5 \: \times \: x^0 \\ \scriptsize = 3x^2 \: + \: x \: + \: 5 \)

\( \scriptsize 223_x = 2 \: \times \: x^2 \: + \: 2 \: \times \: x^1 \: + \: 3 \: \times \: x^0 \\ \scriptsize = 2x^2 \: + \: 2x \: + \: 3 \)

\( \scriptsize 72_x = 7 \: \times \: x^1 \: + \: 2 \: \times \: x^0 \\ \scriptsize = 7x \: + \: 2 \)

i.e \(\scriptsize 3x^2 \: + \: x \: + \: 5 \: -\: (2x^2 \: + \: 2x \: + \: 3) = 7x \: + \: 2 \\ \scriptsize 3x^2 \: + \: x \: + \: 5 \: – \: 2x^2 \: – \: 2x \: – \: 3 = 7x \: + \: 2 \\ \scriptsize x^2 \: -\: x \: + \: 2 = 7x \: + \: 2\\ \scriptsize x^2 \: -x \: – \: 7x = 2 \: – \: 2 \\ \scriptsize x^2 \: – \: 8x = 0 \)

x = 0 or x = 8

But x cannot be base 0.

\( \scriptsize \therefore x = 8 \)

Solution (ii):

Convert \( \scriptsize (251.25)_{10} \) to base 7

i.e. whole number part

Screen Shot 2021 10 23 at 8.09.41 PM

i.e \( \scriptsize 251_{10} = 506_7 \)

Decimal part

Screen Shot 2021 10 23 at 8.13.36 PM

i.e \( \scriptsize 0.25_{10} = 151_7 \)

\( \scriptsize \therefore 251.25_{10} = \underset{whole\: number \: part}{506}.\underset{decimal\: part}{151_7} \)

\( \scriptsize 506.151_7 = x \)

\( \scriptsize x = 506.151_7 \)

Example 3

Find the number bases p and q  in the following simultaneous equations

(i)

\(\scriptsize 26p \: -\: 34q = 1000_2 \)

\(\scriptsize 38p \: -\: 21q = 111_5 \)

(ii)

\( \scriptsize A2p \: -\: 80q = 90_{10} \)

\( \scriptsize B4p \: +\: 77q = 250_{10} \)

Solution (i)

\(\scriptsize 26p \: -\: 34q = 1000_2 \)



\(\scriptsize 38p \: -\: 21q = 111_5 \)

Convert to base 10

\(\scriptsize 26p \: -\: 34q = 1000_2 \)

= \(\scriptsize 2 \: \times \: p^1 \: + \: 6 \: \times \: p^0 \: – \: (3 \: \times \: q^1 \: + \: 4 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 2^3 \: + \: 0 \: \times \: 2^2 \: + \: 0 \: \times \: 2^1 \: + \: 0 \: \times \: 2^0 \\ \scriptsize \therefore 2p \: + \: 6 \: – \: 3q \: -\: 4 = 8 \: + \: 0 \\ \scriptsize = 2p\: – \: 3q = 8 \: – \: 2 \\ \scriptsize 2p \: -\: 3q = 6 \: …..(1) \)

AlsoConvert to base 10, \(\scriptsize 38p \: -\: 21q = 111_5 \)

= \(\scriptsize 3 \: \times \: p^1 \: + \: 8 \: \times \: p^0 \: – \: (2 \: \times \: q^1 \: + \: 1 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 5^2 \: + \: 1 \: \times \: 5^1 \: + \: 1 \: \times \: 5^0 \\ \scriptsize \therefore 3p \: + \: 8 \: – \: 2q \: -\: 1 = 25 \: + \: 5 \: + \: 1 \\ \scriptsize = 3p\: – \: 2q = 31 \: – \: 7 \\ \scriptsize 3p \: -\: 2q = 24 \: …..(2) \)

Solving simultaneously and eliminate p by multiplying equation (1) by 3 and equation (2) by 2, we have:

\( \scriptsize 2p \: -\:3q = 6 \) ………….(1) x 3

\( \scriptsize 3p \: – \: 2q = 24 \) ……………(2) x 2

\( \scriptsize 6p \: -\:9q = 18 \) …………(3)

\( \scriptsize 6p \: – \: 4q = 48 \) ………….(4)

Subtract equation (3) from (4)

\( \scriptsize -\: 4q \: – \: (- \: 9q) = 48 \: -\: 18 \)

\( \scriptsize -\: 4q \: + \: 9q = 48 \: -\: 18 \)

\( \scriptsize 9q \: – \: 4q = 48 \: -\: 18 \)

\( \scriptsize 5q = 30 \)

divide both sides by 5

\( \frac{5q}{5} = \frac{30}{5} \)

\( \frac{\not{5}q}{\not{5}} = \frac{30}{5} \)

\( \scriptsize q = 6 \)

Substitute for q in equation (1) to obtain p 

\( \scriptsize 2p \: -\:3q = 6 \) ………….(1)

\( \scriptsize 2p \: -\:3 \: \times \: 6 = 6 \)

\( \scriptsize 2p \: -\:18 = 6 \)

\( \scriptsize 2p = 6 \: + \: 18 \)

\( \scriptsize 2p = 24 \)

divide both sides by 2

\( \frac{2p}{2} = \frac{24}{2} \)

\( \frac{\not{2}p}{\not{2}} = \frac{24}{2} \)

\( \scriptsize p = 12 \)

Solution (ii)

\( \scriptsize A2p \: -\: 80q = 90_{10} \)

\( \scriptsize B4p \: +\: 77q = 250_{10} \)

Convert to base 10

\( \scriptsize A2p \: -\: 80q = 90_{10} \)

A = 10

= \( \scriptsize 10 \: \times \: p^1 \: + \: 2 \: \times \: p^0 \: -\: 8 \: \times \: q^1 \: + \: 0 \: \times \: q^0 = 90 \)

\( \scriptsize \therefore 10p \: + \: 2 \: – \: 8q = 90 \)

\( \scriptsize = 10p \: – \: 8q = 90 \: -\: 2\)

\( \scriptsize = 10p \: – \: 8q = 88\)

simplify by dividing through by 2

\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1)\)

Also convert to base 10, \( \scriptsize B4p \: +\: 77q = 250_{10} \)

B = 11

= \( \scriptsize 11 \: \times \: p^1 \: + \: 4 \: \times \: p^0 \: +\: 7 \: \times \: q^1 \: + \: 7 \: \times \: q^0 = 250 \)

\( \scriptsize \therefore 11p \: + \: 4 \: + \: 7q \: + \: 7 = 250 \)

\( \scriptsize = 11p \: + \: 7q \: + \: 11 = 250\)

\( \scriptsize = 11p \: + \: 7q = 250 \: – \: 11\)

\( \scriptsize = 11p \: + \: 7q = 239\:…………..(2)\)

Eliminating q, by multiplying equation (1) by 7 and equation (2) by 4, we have:

\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1) \: \: \times \: 7\)

\( \scriptsize = 11p \: + \: 7q = 239\:…………..(2)\: \: \times \: 4\)

\( \scriptsize = 35p \: – \: 28q = 308\: ……………(3) \)

\( \scriptsize = 44p \: + \: 28q = 956\:…………..(4)\)

Add equation (4) to (3)

\( \scriptsize (44p \: + \: 25p) + (28 \: + (- 28) = (956 \: + \: 308) \)

\( \scriptsize (79p) + (0) = (1264) \)

\( \scriptsize 79p = 1264 \)

divide both sides by 79

\( \frac{79p}{79} = \frac{1264}{79} \)

\( \scriptsize p = 16 \)

Substitute for the value of p in equation (i) to obtain q

\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1) \)

\( \scriptsize = 5 \: \times \: 16 \: – \: 4q = 44 \)

\( \scriptsize 80 \: – \: 4q = 44 \)

\( \scriptsize \: – \: 4q = 44 \:-\:80 \)

\( \scriptsize \: – \: 4q =\: -36 \)

\( \scriptsize q = \normalsize \frac{-36}{-4} \)

\( \scriptsize q = 9 \)

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