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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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Example 1

11.44eight – 3.65eight + 2.61eight

Apply BODMAS

Rewrite 11.448 – 3.658 + 2.618 as 11.448 + 2.618 – 3.658

add sub e1601669418412

Answer = 10.48

Example 2

Work out the base in the following and use it to find the missing figures

add questions e1601741096429

The highest number here is 5, therefore the bases can be 6, 7, e.t.c …….

Lets try Base 6

1st Column (from right to left) = 2 + 1 = 3. 3 is less than 6 so we write down 3

2nd Column = x + 5 = 3. 3, in this case, is the remainder. Through trial and error we know that 4 + 5 = 9 9 ÷ 6 = 1 r 3. So 4 in this case is the missing number and we carry 1 to the 3rd column.

3rd Column = The remainder here is 0. Since we are trying base 6 and 6 ÷ 6 = 1 r 0….. 1 + x + 3 = 6. (remember we are carrying 1 from the 2nd column). ∴ x = 6 – 4 = 2. Carry 1 to the 4th column

4th column = We are carrying 1 from the 3rd column so 1 + 3 + 2 = 6. Since we are trying base 6, 6 ÷ 6 = 1 r 0. Write 0 and carry 1 to the 5th column

5th column = 1 + 0 = 1

From our analysis we can see that the required base is 6 and the missing numbers are 2, 4

∴ 32426 – 23516 = 100336

Class Activity

Confirm the previous example by solving 100336 – 23516

Example 3

sub3 e1601757485624

Explanation

The highest number is 4 so we can try 5, 6, 7 etc

1st Column (from right to left) = 1 – 3. We have to borrow. If we borrow 5 from 2nd column we get 5 + 1 – 3 = 1, so we know it can’t be 5 because the result is 2, not 1. Next let’s borrow 6 ∴ 6 – 1 + 3 = 2. So we can keep trying 6. Remember we borrowed 1 from 2nd column.

2nd Column = 0 – 1 = -1. – 1 is less than the result 3 so we have to borrow from the 3rd column∴ 6 + (- 1) – x = 3. We now have 5 – x = 3. The missing number in this column is 2. Remember we borrowed 1 from 2nd column.

3rd Column = 2 -1 = 1. 1 is less than the result 4 so we have to borrow from the 4th column ∴ 6 + 1 – x = 4. We now have 7 – x = 4. The missing number in this column is 3.

4th column = 4 – 1 = 3, ∴ 3 – 3 = 0

We can see that from the equation the numbers are in base 6 and the missing numbers are 3, 2.

∴ 42016 – 33226 = 4326

Example 4

Find the missing number in the addition of the following numbers, in base seven

sub 5 e1601760163152

Solution: 1st step

Solve 43217 + 12347

The resulting numbers are all less than seven so this addition can be done easily.

1st Column (from right to left) = 1 + 4 = 5

2nd Column = 2 + 3 = 5

3rd Column = 3 + 2 = 5

4th column = 4 + 1 = 5

sub6 e1601760521766

Subtract the result from the 1st step from the final result to get the missing numbers.

sub 6 e1601761739666

Explanation

The highest number is 4 so we can try 5, 6, 7 etc

1st Column (from right to left) = 7 + 1 – 5 = 3

2nd Column = 4 reduces to 3. 7 + 3 – 5 = 5

3rd Column = 3 reduces to 2 = 7 + 2 – 5 = 4

4th column = 3 reduces to 1. 7 + 1 – 5 = 3

5th column = 1 – 1 = 0

∴ Answer = 34537

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