Input/output and inverse number machines

(a) 7

1. $$21+7$$
2. Undo subtraction by adding 7: 21 + 7 = 28.
3. Undo multiplication by dividing by 4: 28 $\div$ 4 = 7.

• P1 Establishing $21+7$ or an equivalent valid method.
• A1 Correct answer: 7

(a) $13$

1. $5 \times 4 = 20$, then $20 - 7 = 13$.

• B1 The correct answer, $13$.

(b) $9$

1. Work backwards with the inverse operations: $29 + 7 = 36$, then $36 \div 4 = 9$.

• M1 Adding 7 to 29 first: $36$.
• A1 The correct answer, $9$.

(c) $4n - 7$

1. Multiply $n$ by 4 to get $4n$, then subtract 7: $4n - 7$.

• B1 The correct answer, $4n - 7$.

(a) $8$

1. Write each output with the input $x$.
   $$\text{A: } 3x + 2, \qquad \text{B: } 2(x + 5) = 2x + 10$$
2. Set them equal and solve.
   $$3x + 2 = 2x + 10 \Rightarrow x = 8$$
3. Check: A gives $26$ and B gives $26$.

• P1 Writing the outputs as $3x + 2$ and $2(x + 5)$.
• P1 Forming the equation $3x + 2 = 2(x + 5)$.
• A1 The correct answer, $8$.

(a) $52$

1. $$9\times 5$$
2. Undo the operations in reverse order. Multiply the output by 5: 9 $\times$ 5 = 45.
3. Undo subtracting 7 by adding 7: 45 + 7 = 52.

• P1 Establishing $9\times 5$ or an equivalent valid method.
• A1 Correct answer: $52$

(a) $7$

1. Undo the subtraction before undoing multiplication.
   $$\frac{14+7}{3}$$
2. Therefore $7$.

• M1 Undo the subtraction before undoing multiplication.
• A1 Correct answer: $7$

(b) Add 7, then divide by 3.

1. The final forward operation is the first inverse operation.
   $$y+7$$
2. Add 7, then divide by 3.

• M1 The final forward operation is the first inverse operation.
• C1 Correct conclusion with supporting reasoning: Add 7, then divide by 3.

(a) $(24,25)$

1. Evaluate each ordered sequence separately.
   $$2\times (8+4)$$
2. Therefore $(24,25)$.

• M1 Evaluate each ordered sequence separately.
• A1 Correct answer: $(24,25)$

(b) No. Their outputs differ by $2(4)-9=-1$, which is non-zero for every input.

1. No. Their outputs differ by $2(4)-9=-1$, which is non-zero for every input.

• C1 Correct conclusion with supporting reasoning: No. Their outputs differ by $2(4)-9=-1$, which is non-zero for every input.

(a) $x/2-3$

1. Therefore $x/2-3$.

• B1 Correct answer: $x/2-3$

(b) $4$

1. Reverse the machine: add 3, then double.
   $$(-1+3)\times 2$$
2. Therefore $4$.

• M1 Reverse the machine: add 3, then double.
• A1 Correct answer: $4$

(c) $-6$

1. Form an equation with the output equal to the input.
   $$x/2-3=x$$
2. Therefore $-6$.

• M1 Form an equation with the output equal to the input.
• A1 Correct answer: $-6$

(a) $6$ and $-6$

1. Work backwards.
   $$64 + 8 = 72, \qquad 72 \div 2 = 36$$
2. Undo the square: both $6^2$ and $(-6)^2$ are 36.

• P1 Reaching 36 by undoing the subtraction and the multiplication in reverse order.
• A1 Both $6$ and $-6$.

(b) The square is never negative, so $2x^2 - 8 \ge -8$.

1. The output is $2x^2 - 8$. Squaring never gives a negative number, so $2x^2 \ge 0$ and the output is at least $-8$.
2. $-10$ is less than $-8$, so it can never be an output.

• B1 Writing the output as $2x^2 - 8$ or saying the smallest output is $-8$.
• C1 Explaining that a square cannot be negative, so the output cannot be less than $-8$.