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Either paper 3 marks
\(y = 7x + 6\)
(a)Work out the value of \(y\) when \(x = 12\).
(b)Work out the value of \(y\) when \(x = -9\).
Show answer
(a) \(90\)
(b) \(-57\)\(y = 7 \times 12 + 6 = 84 + 6 = 90\)
A negative number multiplied by a positive number is negative:
\(y = 7 \times (-9) + 6 = -63 + 6 = -57\)
- (b) M1substitutes \(x = -9\): \(7 \times (-9) = -63\)
- (b) A1adds the constant: \(y = -57\)
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Either paper 1 mark
Circle the expression that has the smallest value when \(x = 3\).
\(10 - x\) \(\frac{ 11 }{ x }\) \(x + 3\) \(x + 4\)
Show answer
the second expression: \(10 - 3 = 7\), \(\frac{ 11 }{ 3 } = 3\frac{2}{3}\), \(3 + 3 = 6\), \(3 + 4 = 7\), so the smallest value is \(3\frac{2}{3}\)Substitute \(x = 3\) into each expression.
\(10 - 3 = 7\)
\(\frac{ 11 }{ 3 } = 3\frac{2}{3}\)
\(3 + 3 = 6\)
\(3 + 4 = 7\)
The smallest of these values is \(3\frac{2}{3}\), so the answer is the second expression.
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Either paper 2 marks
Work out the value of \(6p - 2q\) when \(p = 6\) and \(q = -2\).
Show answer
\(40\)Substitute both values in, keeping the brackets round the negative one:
\(6 \times 6 + (-2) \times (-2) = 36 + 4 = 40\)
- M1substitutes both values: \(36\) and \(4\)
- A1total \(40\)
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Either paper 4 marks
A taxi firm works out the cost of a journey using the formula
\(C = 3m + 5\)
where \(C\) is the cost in pounds and \(m\) is the number of miles travelled.
(a)Ben travels 5 miles. Work out the cost of his journey.
(b)Calculate how much more a journey of 11 miles costs than a journey of 5 miles.
Show answer
(a) \(\text{£}20\)
(b) \(\text{£}18\)\(C = 3 \times 5 + 5 = 15 + 5 = 20\)
So Ben pays £20.
The extra distance is \(11 - 5 = 6\) miles.
The £5 is paid once whatever the distance, so only the miles matter:
\(6 \times 3 = 18\)
- (a) M1substitutes \(m = 5\): \(3 \times 5 + 5\)
- (a) A1cost \(20\)
- (b) M1extra distance \(6\) miles multiplied by \(3\), or both costs found
- (b) A1difference \(18\)
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Either paper 1 mark
Circle the point that does not lie on the curve \(y = x^{2} + x - 1\)
\((-2, 1)\) \((-9, 71)\) \((-5, 29)\) \((-3, 5)\)
Show answer
\((-5, 29)\)Put each \(x\)-value into \(x^{2} + x - 1\) and compare with the \(y\)-value given. Keep a negative \(x\)-value in brackets so that both the square and the \(x\)-term keep its sign.
\((-2)^{2} + (-2) - 1 = 1\), given \(y = 1\): the point \((-2, 1)\) is on the curve.
\((-9)^{2} + (-9) - 1 = 71\), given \(y = 71\): the point \((-9, 71)\) is on the curve.
\((-5)^{2} + (-5) - 1 = 19\), given \(y = 29\): the point \((-5, 29)\) is not on the curve.
\((-3)^{2} + (-3) - 1 = 5\), given \(y = 5\): the point \((-3, 5)\) is on the curve.
The point that does not lie on the curve is \((-5, 29)\).
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Either paper 3 marks
Find the value of \(5x^2 + 2x + 2\) when \(x = -4\).
Show answer
\(74\)Square first — squaring a negative gives a positive:
\((-4)^2 = 16\), so \(5x^2 = 5 \times 16 = 80\)
The \(x\) term gives \(2 \times (-4) = -8\).
\(80 - 8 + 2 = 74\)
- M1squares the negative value: \((-4)^2 = 16\), so \(5x^2 = 80\)
- M1linear term \(2 \times (-4) = -8\)
- A1total \(74\)
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Either paper 3 marks
\(y = \frac{ 3x + 30 }{ 6 }\)
Work out the value of \(y\) when \(x = -4\).
Show answer
\(3\)Work out the numerator first:
\(3 \times (-4) + 30 = -12 + 30 = 18\)
Then divide by 6:
\(y = \frac{ 18 }{ 6 } = 3\)
- M1substitutes \(x = -4\) into the numerator: \(3 \times (-4) + 30\)
- M1numerator \(18\) divided by \(6\)
- A1\(y = 3\)
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Either paper 3 marks
A formula connecting four quantities is
\(v = u + at\)
\(v = 50\), \(u = 8\) and \(t = 6\).
Work out the value of \(a\).
Show answer
\(7\)Substitute the three values you are given and solve the equation that is left.
\(50 = 8 + a \times 6\)
Subtract 8 from both sides.
\(42 = 6a\)
\(a = 42 \div 6 = 7\)
- M1substitutes the given values: \(50 = 8 + a \times 6\)
- M1subtracts \(8\) from both sides: \(42 = 6a\)
- A1\(a = 7\)
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Either paper 3 marks
Here is a formula.
\(t = 2n^2 - 3n\)
(a)Find \(t\) when \(n = 5\).
(b)Explain why \(t\) is always positive when \(n\) is negative.
Show answer
(a) \(35\)
(b) When \(n\) is negative, \(n^2\) is positive (a negative times a negative), and multiplying by the positive coefficient keeps it positive. Also \(3n\) is negative, so \(-3n\) is positive. Both terms are positive, so their sum \(t\) is positive.Substitute \(n = 5\).
\(t = 2(5)^2 - 3 \times 5\)
\(t = 50 - 15 = 35\)
Take any negative value of \(n\).
\(n^2\) is a negative number multiplied by itself, so it is positive, and multiplying by the positive coefficient keeps it positive.
\(3n\) is a positive number times a negative number, so it is negative. Subtracting a negative number gives a positive, so \(-3n\) is positive.
\(t\) is a positive number plus a positive number, so \(t\) is always positive.
- (a) M1substitutes \(n = 5\): \(50\) and \(15\)
- (a) A1subtracts to get \(t = 35\)
- (b) B1both terms positive: \(n^2\) is positive and \(-3n\) is positive when \(n\) is negative
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Either paper 4 marks
This formula converts a temperature in degrees Celsius (°C) to degrees Fahrenheit (°F).
\(F = 1.8C + 32\)
\(F\) is the number of degrees Fahrenheit and \(C\) is the number of degrees Celsius.
(a)Show that 5°C is the same temperature as 41°F.
(b)The temperature is -2°C.
Nick says, "Because the temperature is negative in Celsius, it must be negative in Fahrenheit."
Is Nick correct? You must show your working.
Show answer
(a) Substituting \(C = 5\) gives \(F = 1.8 \times 5 + 32 = 9 + 32 = 41\)
(b) No. \(F = 1.8 \times (-2) + 32 = -3.6 + 32 = 28.4\), which is positive.Substitute \(C = 5\) into the formula.
\(F = 1.8 \times 5 + 32\)
\(1.8 \times 5 = 9\), so
\(F = 9 + 32 = 41\)
So 5°C is 41°F.
Substitute \(C = -2\) into the formula to find the temperature in Fahrenheit.
\(F = 1.8 \times (-2) + 32\)
\(1.8 \times (-2) = -3.6\), so
\(F = -3.6 + 32 = 28.4\)
28.4°F is positive, so Nick is not correct: a temperature that is negative in Celsius can be positive in Fahrenheit.
- (a) M1substitutes \(C = 5\) and finds \(1.8 \times 5 = 9\)
- (a) A1adds 32 to reach \(F = 41\), matching the stated value
- (b) M1substitutes \(C = -2\) and adds 32 to reach \(F = 28.4\)
- (b) A1No, Nick is not correct, because 28.4°F is positive