Function Machines: Practise Online

Either paper 2 marks

Here is a function machine.

Input \(\times 5\) \(- 8\) Output

\(3\) → → ?

Work out the output when the input is \(3\).
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\(7\)
Work through the boxes from left to right.

Multiply by 5: \(3 \times 5 = 15\)

Then subtract 8: \(15 - 8 = 7\)
- M1multiplying by \(5\) to get \(15\)
- A1output \(7\)
Either paper 3 marks

Here is a function machine.

Input \(+ 2\) \(\times 6\) \(- 2\) Output

\(3\) → → → ?

The input is \(3\).

Work out the output.
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\(28\)
\(3 + 2 = 5\)

\(5 \times 6 = 30\)

\(30 - 2 = 28\)
- M1add 2 to the input: \(5\)
- M1multiply by 6: \(30\)
- A1subtract 2 to get the output \(28\)
Either paper 3 marks

Here is a function machine.

Input \(\times 4\) \(- 7\) Output

? → → \(9\)

The machine gives an output of \(9\).

Find the input.
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\(4\)
Run the machine backwards, inverting each operation in reverse order.

Add 7: \(9 + 7 = 16\)

Divide by 4: \(16 \div 4 = 4\)
- M1reverse the order: add 7 to the output giving \(16\)
- M1divide \(16\) by 4
- A1input \(4\)
Either paper 1 mark

Anya wants to make a number machine so that \(y = 8x + 5\)

Here is her attempt.

Input \(x\) × 8 − 5 Output \(y\)

\(x\) → → \(y\)

What mistake has she made?
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Multiplied by 8 but subtracted 5 instead of adding it.
Test the machine with an input of 2.

The target \(y = 8x + 5\) gives \(2 \times 8 + 5 = 21\)

The attempt gives 2 × 8 − 5 = 11

The outputs are different, so the attempt is wrong. Comparing the boxes with the target: Multiplied by 8 but subtracted 5 instead of adding it.
Either paper 4 marks

Here is a function machine.

Input \(- 4\) \(\times 6\) Output

? → → ?

(a)
Calculate the output when the input is \(9\).

(b)
Calculate the input when the output is \(24\).
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(a) \(30\)
(b) \(8\)
\(9 - 4 = 5\)

\(5 \times 6 = 30\)
Reverse the machine: divide by 6 first, then add 4.

\(24 \div 6 = 4\)

\(4 + 4 = 8\)
- (a) M1subtract 4 from the input: \(5\)
- (a) A1multiply by 6 to get the output \(30\)
- (b) M1divide the output \(24\) by 6: \(4\)
- (b) A1add 4 to get the input \(8\)
Either paper 2 marks

Here is a function machine. The first operation is missing.

Input \(\times \, ?\) \(- 7\) Output

\(3\) → → \(14\)

Work out the missing number.
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\(7\)
Undo the last box first. Add 7: \(14 + 7 = 21\)

So 3 was multiplied to give 21.

\(21 \div 3 = 7\)
- M1undoing the last box to get \(21\)
- A1missing number \(7\)
Either paper 4 marks

Here are two number machines.

Machine 1

Input \(\times 7\) \(+ 5\) Output

\(a\) → → ?

Machine 2

Input \(\times 5\) \(+ 21\) Output

\(a\) → → ?

The outputs are the same.

Work out the value of \(a\).
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\(8\)
Machine 1 gives \(7a + 5\).

Machine 2 gives \(5a + 21\).

The outputs are the same:

\(7a + 5 = 5a + 21\)

Take \(5a\) from both sides so the \(a\) terms are together:

\(2a + 5 = 21\)

Subtract 5 from both sides:

\(2a = 16\)

Divide both sides by 2:

\(a = 8\)

Check: machine 1 gives \(7 \times 8 + 5 = 61\) and machine 2 gives \(5 \times 8 + 21 = 61\).
- M1machine 1 output \(7a + 5\)
- M1machine 2 output \(5a + 21\)
- M1equating and collecting to \(2a = 16\)
- A1\(a = 8\)
Either paper 3 marks

Each number machine changes the input \(x\) into the output \(y\).

(a)
Complete this number machine so that \(y = 9x + 16\)

Input \(?\) \(?\) Output

\(x\) → → \(y\)

(b)
Complete this number machine so that \(y = 2x - 26\)

Input \(?\) \(?\) Output

\(x\) → → \(y\)

(c)
Complete this number machine so that \(y = x\)

Input \(\div 9\) \( ?\) Output

\(x\) → → \(y\)

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(a) The first box is \(\times 9\) and the second box is \(+ 16\)
(b) The first box is \(\times 2\) and the second box is \(- 26\)
(c) The second box is \(\times 9\)
In \(y = 9x + 16\), \(x\) is multiplied by \(9\) first and then \(16\) is added.

So the first box is \(\times 9\) and the second box is \(+ 16\).
In \(y = 2x - 26\), \(x\) is multiplied by \(2\) first and then \(26\) is subtracted.

So the first box is \(\times 2\) and the second box is \(- 26\).
For \(y = x\) the output must be the same as the input, so the two boxes undo each other.

The first box divides by 9, so the second box must multiply by 9.
Either paper 3 marks

Each number machine changes an input into an output.

(a)
Here is a number machine.

Input \(+ 10\) \(\times 6\) Output

\(n\) → → \(d\)

Write down the output \(d\) in terms of \(n\).

(b)
Here is another number machine, with both boxes left blank.

Input \(?\) \(?\) Output

\(u\) → → \(k\)

Complete the number machine so that \(k = \frac{ u + 8 }{ 2 }\)
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(a) \(d = 6(n + 10)\)
(b) The first box is \(+ 8\) and the second box is \(÷ 2\)
The first box adds 10, so after it the value is \(n + 10\).

The second box multiplies that whole value by 6, so it needs brackets.

\(d = 6(n + 10)\)
In \(k = \frac{ u + 8 }{ 2 }\), the top of the fraction, u + 8, is worked out first and the result is divided by 2.

So the first box is \(+ 8\) and the second box is \(÷ 2\).
- (a) B1the first box gives \(n + 10\)
- (a) B1the second box multiplies all of that by 6: \(6(n + 10)\)
Either paper 4 marks

Reuben is investigating number machines.

(a)
Here is a number machine.

Input Output

\(x\) × 9 + 3 \(y\)

Show that when the input increases by 2, the output increases by 18.

(b)
Now \(f(x) = x^{ 2 }\)

Reuben says that \(\frac{ f(6) }{ f(2) }\) is equal to \(f(3)\) because \(6 \div 2 = 3\).

Is Reuben correct?

Show working to support your answer.
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(a) Input \(x\) gives \(y = 9x + 3\). Input \(x + 2\) gives \(9(x + 2) + 3 = 9x + 18 + 3\). The difference is \((9x + 18 + 3) - (9x + 3) = 18\).
(b) Yes. \(f(6) = 36\) and \(f(2) = 4\), so \(\frac{ f(6) }{ f(2) } = 9\), while \(f(3) = 9\). The ratio equals \(f(3)\).
Call the original input \(x\). The output is

\(y = 9x + 3\)

When the input increases by 2 it becomes \(x + 2\), and the output is

\(9(x + 2) + 3 = 9x + 18 + 3\)

The increase in the output is the new output minus the old output:

\((9x + 18 + 3) - (9x + 3) = 18\)

The terms in \(x\) and the constant cancel, so the output increases by \(9 \times 2 = 18\) whatever \(x\) is.
Work out each side of the claim.

\(f(6) = 1 \times 6^{ 2 } = 36\)

\(f(2) = 1 \times 2^{ 2 } = 4\)

\(\frac{ f(6) }{ f(2) } = \frac{ 36 }{ 4 } = 9\)

\(f(3) = 1 \times 3^{ 2 } = 9\)

The constant multiplier cancels in the ratio, which is \(\left( \frac{ 6 }{ 2 } \right)^{ 2 } = 3^{ 2 }\), whereas \(f(3) = 1 \times 3^{ 2 }\). The two are equal only when the constant is 1.

The ratio equals \(f(3)\), so Reuben is correct.
- (a) M1output for input \(x + 2\) written as \(9(x + 2) + 3\)
- (a) A1subtracts the outputs so that \(x\) and 3 cancel, leaving \(9 \times 2 = 18\)
- (b) M1evaluates \(\frac{ f(6) }{ f(2) } = \frac{ 36 }{ 4 } = 9\) and \(f(3) = 9\)
- (b) A1Yes: Reuben is correct, since 9 equals 9