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Number Bases · Form 4

Number Bases: Worked Examples (KBAT)

Higher-order thinking: real situations where the number base technique is hidden and must be spotted before solving over several steps. For students who can convert and compute and now need to decide when and how to apply it.

Written by the spmmath.com.my editorial team.

Worked example 1

A digital classroom lamp shows its brightness level as a row of 5 small lights, each either off (0) or on (1), read left to right as a base 2 number where the leftmost light has the largest value. On Monday the lights read 1 0 1 1 0.

A teacher raises the brightness by 5 steps. What will the row of lights read after the change?

  1. Recognise the row of lights is a base 2 number: 10110₂.
  2. Convert to base ten to add: (1×16)+(0×8)+(1×4)+(1×2)+(0×1) = 16 + 4 + 2 = 22.
  3. Raising brightness by 5 steps means add 5: 22 + 5 = 27.
  4. Convert 27 back to base 2 by repeated division: 27 ÷ 2 = 13 r 1; 13 ÷ 2 = 6 r 1; 6 ÷ 2 = 3 r 0; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1.
  5. Read remainders bottom to top: 1 1 0 1 1, which fills all 5 lights.
  6. Check: 11011₂ = 16 + 8 + 2 + 1 = 27. ✓

Worked example 2

A hostel warden counts stored rice sacks using a base 8 tally board because sacks are packed 8 to a pallet, 8 pallets to a bay. The board shows 246₈ sacks in the store.

During the week 57₈ sacks are used and a delivery of 30₈ sacks arrives. How many sacks are in the store now, written in base 8?

  1. Plan: work entirely in base 8. First subtract the used sacks, then add the delivery.
  2. 246₈ − 57₈. Units: 6 − 7 borrow: (6+8) − 7 = 7. Eights: 4 became 3; 3 − 5 borrow: (3+8) − 5 = 6. Sixty-fours: 2 became 1; 1 − 0 = 1. Result 167₈.
  3. Check the subtraction in base ten: 246₈ = 166, 57₈ = 47, 166 − 47 = 119, and 167₈ = 64+48+7 = 119. ✓
  4. Now 167₈ + 30₈. Units: 7 + 0 = 7. Eights: 6 + 3 = 9 = 8 + 1, write 1 carry 1. Sixty-fours: 1 + 0 + 1(carry) = 2. Result 217₈.
  5. Check in base ten: 119 + 24 = 143, and 217₈ = 128 + 8 + 7 = 143. ✓

Worked example 3

Two robotics teams write a shared code word by joining two numbers. Team A contributes 34₅ and Team B contributes 1101₂.

The rule is that the final code word is the sum of the two contributions, expressed in base 8. What is the code word?

  1. The two numbers are in different bases, so convert both to base ten first.
  2. Team A: 34₅ = (3×5) + (4×1) = 15 + 4 = 19.
  3. Team B: 1101₂ = (1×8)+(1×4)+(0×2)+(1×1) = 8 + 4 + 1 = 13.
  4. Add in base ten: 19 + 13 = 32.
  5. Convert 32 to base 8 by repeated division: 32 ÷ 8 = 4 r 0; 4 ÷ 8 = 0 r 4. Read remainders bottom to top: 4 0.
  6. Check: 40₈ = (4×8) + 0 = 32. ✓

Worked example 4

A car park machine prints ticket numbers in base 8. The last ticket printed was 377₈.

Find the ticket number, in base 8, of the car that arrives 5 cars later.

  1. Convert 377₈ to base ten: 3×64 + 7×8 + 7 = 192 + 56 + 7 = 255.
  2. The car 5 later has number 255 + 5 = 260 in base ten.
  3. Convert 260 to base 8: 260÷8 = 32 r 4, 32÷8 = 4 r 0, 4÷8 = 0 r 4.
  4. Read the remainders from last to first: 404.

Worked example 5

A control panel has 6 on/off switches showing a base-2 number, currently reading 100101₂. An operator changes the setting to add 14 (base ten).

Find the new reading in base 2.

  1. Convert 100101₂ to base ten: 32 + 0 + 0 + 4 + 0 + 1 = 37.
  2. Add 14: 37 + 14 = 51 in base ten.
  3. Convert 51 to base 2: 51 = 32 + 16 + 2 + 1.
  4. So the digits are 1 1 0 0 1 1, i.e. 110011.

Worked example 6

Two lockers use different code systems. Locker P opens with the code 1101₂ and locker Q with the code 23₅.

A student claims the two codes stand for the same number. Determine whether the student is correct, showing your working.

  1. Convert both codes to base ten to compare fairly.
  2. 1101₂ = 8 + 4 + 0 + 1 = 13.
  3. 23₅ = 2×5 + 3 = 13.
  4. Both codes equal 13 in base ten, so they represent the same number and the difference is 0.

Worked example 7

In base 8, the number 3d5₈ (where d is an unknown digit) is equal to 213 in base ten. Find the value of d.

  1. Expand 3d5₈ using place values: (3×8²) + (d×8¹) + (5×8⁰) = 213
  2. 192 + 8d + 5 = 213
  3. 197 + 8d = 213
  4. 8d = 16
  5. d = 2

Worked example 8

In a retro video game, the score display works in base 8 because the console's memory is organised in octal. After a level, Farah's score display reads 354₈ points.

She then loses 267₈ points as a penalty for missing a bonus item. Find the number of points remaining, giving your answer in base ten.

  1. Convert 354₈ to base ten: (3×64) + (5×8) + 4 = 192 + 40 + 4 = 236
  2. Convert 267₈ to base ten: (2×64) + (6×8) + 7 = 128 + 48 + 7 = 183
  3. Subtract: 236 − 183 = 53

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

How do number base questions get turned into KBAT-level problems?

They embed conversions inside a real context, like computer memory, coding, or pattern problems, requiring you to convert, compare, or solve an equation across bases before reaching the answer. Read carefully to identify which base each given number is in before starting.

What's the biggest pitfall in multi-step number base problems?

Losing track of which base you're working in partway through a long problem. Label every intermediate result with its base clearly, and convert back to base ten to check that your final answer is genuinely correct before writing it in the required base.

How should I approach a problem asking to solve for an unknown digit or base?

Set up an equation using place values with the unknown as a variable, then solve algebraically, and check that any digit found is valid for its base (a digit must always be less than the base). This validity check is often the deciding mark in this type of question.

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