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J277 1.2.4 Binary shifts and consolidation
Part 5 of 5 · J277 1.2.4 · Number systems
Binary shifts, then a mixed checkpoint over the whole OCR number-systems topic.
Students will:
- carry out left and right binary shifts
- connect the number of places with a power of two
- identify direction and places from a before-and-after pair
- explain when a fixed-width shift loses information
- apply conversion, addition and overflow independently
Inside: 5 explanation cells, 5 number grids, 3 multiple-choice questions and 3 written answers. 45 marks, about 75 minutes.
Series: J277 1.2.4 · Number systems, part 5 of 5.
Shared by Coding PathwayVerified teacher
- 16 cells
- About 75 minutes
- CC BY-SA 4.0
- Shared 17 Aug 2026
- Updated 9 Sept 2026
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Number systems 5: binary shifts and consolidation
A binary shift moves every bit by the same number of places. You need to track the direction, the number of places, the zeros inserted and any bits lost at the edge. The second half of this worksheet then brings the whole OCR number-systems topic together.
By the end of this worksheet, you should be able to:
- carry out left and right binary shifts
- connect the number of shift places with a power of two
- identify both the direction and number of places from a before-and-after pair
- explain when a fixed-width shift loses information
- apply conversion, addition and overflow knowledge independently
Use the early tasks to identify gaps, then complete the application and final challenge independently.
1. Carry out a shift
These rules apply to the values from 0 to 255 used in OCR GCSE questions.
Left shift
- move every bit left by the stated number of places
- insert zeros into the empty positions on the right
- a shift of n places has a multiplication factor of
2^nwhen the result fits
Example: 00010110 shifted left by 2 becomes 01011000. The denary value changes from 22 to 88, so it has multiplied by 4.
Right shift
- move every bit right by the stated number of places
- insert zeros into the empty positions on the left
- a shift of n places performs whole-number division by
2^n; any remainder represented by discarded low bits is lost
Example: 10110000 shifted right by 3 becomes 00010110. The denary value changes from 176 to 22, so it has divided by 8.
The platform may call these logical shifts. For the values used here, that means zeros are inserted into the empty positions.
Which statement correctly describes what these shifts do when no 1 bit is lost at the edge?
- ALeft by 3 multiplies by 3; right by 3 divides by 3
- BLeft by 2 divides by 4; right by 2 multiplies by 4
- CLeft by 2 multiplies by 4; right by 3 divides by 8
- DLeft by 8 multiplies by 8; right by 8 divides by 8
Apply a 2-place left shift, then state the denary value of the shifted result.
Move every bit two places left and insert two zeros on the right. These examples fit, so the new value is four times the original.
a)Apply a logical shift left of 2 places to 00010110.
b)Apply a logical shift left of 2 places to 00100011.
c)Apply a logical shift left of 2 places to 00001101.
Apply a 3-place right shift, then state the denary value of the shifted result.
Move every bit three places right and insert three zeros on the left. These examples divide exactly by eight.
a)Apply a logical shift right of 3 places to 10110000.
b)Apply a logical shift right of 3 places to 01101000.
c)Apply a logical shift right of 3 places to 11100000.
2. Bits can leave the 8-bit row
A shift keeps the same fixed width. A bit moved beyond an edge is discarded.
Example: shift 11110000 left by 1.
- move left: the first 1 leaves the 8-bit row
- insert 0 on the right
- result:
11100000
The simple statement “left shift multiplies by two” is exact only when no significant 1 bit is discarded. This does not change the movement rule: move every bit, discard anything beyond the edge and fill the empty position with 0.
For a right shift, discarded 1 bits at the right represent a lost remainder. This is why we describe the result as whole-number division.
What is the 8-bit result of shifting 11110000 left by 1 place?
- A01111000
- B11100000
- C11110001
- D111100000
Identify the direction and number of places used in each shift.
Compare the before and after positions of the 1 bits. Both direction and number of places are required.
a)00110101 became 00001101. Which shift was applied, and by how many places?
b)11001011 became 10010110. Which shift was applied, and by how many places?
c)01010010 became 00001010. Which shift was applied, and by how many places?
d)10000100 became 00010000. Which shift was applied, and by how many places?
Describe the binary shift used to multiply a value by 8 when the result fits in the available bits.
OCR awards separate points for the direction and number of places. Use 8 = 2^3.
Students type their answer here.
3. Mixed OCR-style consolidation
Complete the next three tasks without returning to the worked examples.
Conversion table
Use nibbles for binary and hexadecimal, and place values for binary and denary. Include 8 bits and 2 hex digits.
Binary addition
Work from right to left and record every carry.
Overflow
Decide whether the complete answer fits within 8 bits. Explain the decision using the available width.
Complete the three tasks independently, then use the feedback to identify the first method that needs another example.
Complete each row with the two missing number representations.
Check the requested widths before moving to the next row.
a)Hexadecimal FE. Fill in the denary and binary.
b)Binary 01001100. Fill in the denary and hexadecimal.
c)Hexadecimal 8A. Fill in the denary and binary.
Add each pair of 8-bit binary numbers and show every carry.
Use binary column addition directly. The answer and carry row are both marked.
a)Add these 8-bit binary numbers: 00001011 + 01010011.
b)Add these 8-bit binary numbers: 01011110 + 01100010.
Add 11110000 and 01010101. Show your binary working, write the complete answer and explain whether an overflow error occurs.
Continue beyond eight places if the final carry creates a ninth bit. Overflow occurs when the complete answer needs more than 8 bits.
Students type their answer here.
Which four statements are accurate? Select four.
- AAny carry inside an 8-bit addition proves that overflow has occurred
- BThe maximum denary value represented by 8 bits is 255
- CMultiplication by 8 requires an 8-place left shift
- DThe right-hand zero in hexadecimal B0 must be kept because removing it changes the value
- ECarries within a binary addition are normal and do not automatically mean overflow
- FAn 8-bit answer may contain any number of digits as long as its value is correct
- GA 3-place right shift divides the value by 8 when no remainder is lost
Challenge: explain why an 8-bit left shift does not always produce the exact multiplication result predicted by 2^n. Use a specific binary example in your answer.
Discuss fixed width, a 1 bit leaving beyond the MSB and the resulting loss of information.
Students type their answer here.
Final confidence check
You should now be able to:
- convert between denary 0 to 255, binary 00000000 to 11111111 and hexadecimal 00 to FF
- identify the MSB and LSB
- explain why leading zeros may be required
- add two binary numbers and show every carry
- explain an 8-bit overflow
- carry out left and right shifts
- state both the direction and number of shift places
Choose one statement you cannot yet demonstrate without notes. Return to that task, complete one fresh example and explain the method aloud.