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Either paper 2 marks
A bottling plant fills \(m\) crates of bottles per minute. Each crate holds 40 bottles.
Show that the plant fills \(\frac{ 2m }{ 3 }\) bottles per second.
Show answer
Per minute: \(40m\) bottles. There are 60 seconds in one minute, so per second it is \(\frac{ 40m }{ 60 }\), and dividing top and bottom by 20 gives \(\frac{ 2m }{ 3 }\).Each minute, the plant fills \(m\) crates with 40 bottles in each, so the number of bottles per minute is
\(40 \times m = 40m\)
There are 60 seconds in one minute, so divide by 60 to get the number per second:
\(\frac{ 40m }{ 60 }\)
Both 40 and 60 divide by 20. Divide the top and the bottom by 20:
\(\frac{ 40m \div 20 }{ 60 \div 20 } = \frac{ 2m }{ 3 }\)
- M1builds \(40m\) bottles per minute and divides by 60 to get \(\frac{ 40m }{ 60 }\)
- A1cancels the common factor 20 to reach \(\frac{ 2m }{ 3 }\)
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Either paper 3 marks
Reuben is asked to simplify \(\frac{ 6n^2 - 5n }{ n }\).
Reuben's method is to divide the second term of the numerator by \(n\), then add the first term.
(a)State the expression Reuben gets.
(b)Simplify \(\frac{ 6n^2 - 5n }{ n }\) correctly.
(c)Explain why Reuben's answer is wrong.
Show answer
(a) \(6n^2 - 5\)
(b) \(6n - 5\)
(c) Reuben divided only the second term by \(n\). The whole of the numerator is divided by \(n\), so the other term must be divided as well: \(\frac{ 6n^2 }{ n } - \frac{ 5n }{ n } = 6n - 5\), not \(6n^2 - 5\).Following Reuben's method, the second term is divided by \(n\) and the first term is added as it stands:
\(6n^2 - 5\)
The denominator \(n\) divides the whole of the numerator, so it divides every term of it:
\(\frac{ 6n^2 - 5n }{ n } = \frac{ 6n^2 }{ n } - \frac{ 5n }{ n }\)
Simplify each fraction:
\(6n - 5\)
Dividing a sum by a number means dividing every term of the sum. Reuben divided the second term and left the other one undivided, so the answer Reuben got, \(6n^2 - 5\), is not equal to \(\frac{ 6n^2 - 5n }{ n }\).
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Either paper 1 mark
Circle the expression that is equivalent to
\(\frac{ 15x^2 }{ 10x^2 + 15 }\)
A \(\frac{ 3x }{ 2x + 3 }\)
B \(\frac{ 3x^2 }{ 10x^2 + 3 }\)
C \(\frac{ 3x^2 }{ 2x^2 + 3 }\)
D \(\frac{ 3x^2 }{ 2x^2 + 15 }\)
Show answer
option C: \(\frac{ 3x^2 }{ 2x^2 + 3 }\)Take the common factor 5 out of the top and out of the bottom:
\(\frac{ 15x^2 }{ 10x^2 + 15 } = \frac{ 5 \times 3x^2 }{ 5(2x^2 + 3) }\)
5 multiplies the whole of the top and the whole of the bottom, so it cancels:
\(\frac{ 3x^2 }{ 2x^2 + 3 }\)
Only a factor of the whole top and the whole bottom can be cancelled, never a single term, so the answer is option C.
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Either paper 6 marks(a)
Write \(\frac{ 5x + 1 }{ 4 } + \frac{ 3x + 4 }{ 6 }\) as a single fraction.
Give your answer in its simplest form.
(b)Write \(\frac{ 5x - 1 }{ 4 } - \frac{ 4x + 1 }{ 7 }\) as a single fraction.
Give your answer in its simplest form.
Show answer
(a) \(\frac{ 21x + 11 }{ 12 }\)
(b) \(\frac{ 19x - 11 }{ 28 }\)The lowest common denominator of 4 and 6 is 12.
\(\frac{ 5x + 1 }{ 4 } = \frac{ 3\left(5x + 1\right)}{ 12 }\)
\(\frac{ 3x + 4 }{ 6 } = \frac{ 2\left(3x + 4\right)}{ 12 }\)
Now the bottoms match, so add the tops:
\(3\left(5x + 1\right) + 2\left(3x + 4\right) = 21x + 11\)
\(\frac{ 21x + 11 }{ 12 }\)
The lowest common denominator of 4 and 7 is 28.
\(\frac{ 5x - 1 }{ 4 } = \frac{ 7\left(5x - 1\right)}{ 28 }\)
\(\frac{ 4x + 1 }{ 7 } = \frac{ 4\left(4x + 1\right)}{ 28 }\)
The whole of the second top is being subtracted, so keep it in a bracket:
\(7\left(5x - 1\right) - 4\left(4x + 1\right) = 19x - 11\)
\(\frac{ 19x - 11 }{ 28 }\)
- (a) M1common denominator \(12\), tops multiplied by \(3\) and \(2\)
- (a) M1adds the two tops
- (a) A1\(\frac{ 21x + 11 }{ 12 }\)
- (b) M1common denominator \(28\), tops multiplied by \(7\) and \(4\)
- (b) M1subtracts the whole of the second top
- (b) A1\(\frac{ 19x - 11 }{ 28 }\)
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Either paper 2 marks
Write \(\frac{ 6 }{ x^3 } - \frac{ 1 }{ 2x }\) as a single fraction.
Give your answer in its simplest form.
Show answer
\(\frac{ 12 - x^2 }{ 2x^3 }\)Both bottoms must divide into the common denominator.
Numbers: the lowest common multiple of 1 and 2 is 2.
Letters: take the highest power of each letter that appears in a bottom, which gives \(x^3\).
So the common denominator is \(2x^3\).
Multiply the top and bottom of each fraction by whatever turns its bottom into \(2x^3\):
\(\frac{ 6 }{ x^3 }\) becomes \(\frac{ 12 }{ 2x^3 }\)
\(\frac{ 1 }{ 2x }\) becomes \(\frac{ x^2 }{ 2x^3 }\)
The bottoms now match, so combine the tops over the one bottom:
\(\frac{ 12 }{ 2x^3 } - \frac{ x^2 }{ 2x^3 } = \frac{ 12 - x^2 }{ 2x^3 }\)
Nothing divides into every term of the top and the bottom, so this is in its simplest form.
- M1common denominator \(2x^3\) and both fractions rewritten over it
- A1\(\frac{ 12 - x^2 }{ 2x^3 }\)
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Either paper 3 marks
Simplify fully \(\frac{ 2(3x + 1)^{ 3 } }{ (3x + 1) }\)
Your answer must be in the form \(ax^2 + bx + c\), where \(a\), \(b\) and \(c\) are integers.
Show answer
\(18x^2 + 12x + 2\)The bracket \((3x + 1)\) is on the top and the bottom, so cancel it. Subtract the powers: \(3 - 1 = 2\).
\(\frac{ 2(3x + 1)^{ 3 } }{ (3x + 1) } = 2(3x + 1)^2\)
Expand the squared bracket:
\((3x + 1)^2 = (3x + 1)(3x + 1) = 9x^2 + 6x + 1\)
Multiply every term by 2:
\(2(9x^2 + 6x + 1) = 18x^2 + 12x + 2\)
- M1cancels the repeated bracket to leave \(2(3x + 1)^2\)
- M1expands the bracket to \(9x^2 + 6x + 1\)
- A1multiplies through by 2 to reach \(18x^2 + 12x + 2\)
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Either paper 3 marks
\(a\) and \(b\) are positive numbers.
Show that \(\frac{ 21a + 6b - 3a }{ 14a + 9b + a - 4b }\) always simplifies to the same value.
Show answer
Top \(18a + 6b = 6(3a + b)\), bottom \(15a + 5b = 5(3a + b)\); the bracket cancels, leaving \(\frac{ 6 }{ 5 }\) whatever \(a\) and \(b\) are.Collect the like terms on the top:
\(21a + 6b - 3a = 18a + 6b\)
Collect the like terms on the bottom:
\(14a + 9b + a - 4b = 15a + 5b\)
Take out the largest common factor from each, which leaves the same bracket:
\(\frac{ 18a + 6b }{ 15a + 5b } = \frac{ 6(3a + b) }{ 5(3a + b) }\)
\((3a + b)\) is not zero because \(a\) and \(b\) are positive, so it cancels:
\(\frac{ 6 }{ 5 }\)
This is a number with no \(a\) or \(b\) in it, so the fraction always simplifies to the same value, \(\frac{ 6 }{ 5 }\).
- M1collects the top, \(18a + 6b\)
- M1collects the bottom, \(15a + 5b\)
- A1takes out the common bracket \((3a + b)\), cancels it, and states the value \(\frac{ 6 }{ 5 }\) does not depend on \(a\) or \(b\)
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Either paper 3 marks
Simplify fully \(\frac{x^2 - 4x + 3 }{x^2 - 3x + 2 }\)
Show answer
\(\frac{x - 3 }{x - 2 }\)Factorise the top and the bottom separately.
The top needs two numbers multiplying to 3 and adding to -4. Those are -1 and -3:
\(x^2 - 4x + 3 = (x - 1)(x - 3)\)
The bottom needs two numbers multiplying to 2 and adding to -3. Those are -1 and -2:
\(x^2 - 3x + 2 = (x - 1)(x - 2)\)
The bracket \((x - 1)\) appears on both, so it cancels:
\(\frac{(x - 1)(x - 3)}{(x - 1)(x - 2)} = \frac{x - 3 }{x - 2 }\)
Nothing is left that both share, so this is as far as it goes.
- M1top factorised: \((x - 1)(x - 3)\)
- M1bottom factorised: \((x - 1)(x - 2)\)
- A1common bracket cancelled: \(\frac{x - 3 }{x - 2 }\)
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Either paper 3 marks
Simplify \(\frac{x^2 - 16 }{x^2 - 2x - 24 }\) as far as possible.
Show answer
\(\frac{x - 4 }{x - 6 }\)There is no \(x\) term on the top, and 16 is a square number, so the top is a difference of two squares:
\(x^2 - 16 = (x + 4)(x - 4)\)
The bottom needs two numbers multiplying to -24 and adding to -2. Those are 4 and -6:
\(x^2 - 2x - 24 = (x + 4)(x - 6)\)
\((x + 4)\) cancels from the top and the bottom:
\(\frac{(x + 4)(x - 4)}{(x + 4)(x - 6)} = \frac{x - 4 }{x - 6 }\)
- M1top as a difference of two squares: \((x + 4)(x - 4)\)
- M1bottom factorised: \((x + 4)(x - 6)\)
- A1common bracket cancelled: \(\frac{x - 4 }{x - 6 }\)
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Either paper 6 marks(a)
Write \(\frac{ 3 }{x - 2 } + \frac{ 7 }{x + 1 }\) as a single fraction.
Give your answer in its simplest form.
(b)Write \(\frac{ 9 }{x - 1 } - \frac{ 8 }{x - 3 }\) as a single fraction.
Give your answer in its simplest form.
Show answer
(a) \(\frac{ 10x - 11 }{(x - 2)(x + 1)}\)
(b) \(\frac{ x - 19 }{(x - 1)(x - 3)}\)The two bottoms have no factor in common, so the lowest common denominator is their product, \((x - 2)(x + 1)\).
\(\frac{ 3 }{x - 2 } = \frac{ 3(x + 1)}{(x - 2)(x + 1)}\)
\(\frac{ 7 }{x + 1 } = \frac{ 7(x - 2)}{(x - 2)(x + 1)}\)
Add the tops:
\(3(x + 1) + 7(x - 2) = 3x + 3 + 7x - 14 = 10x - 11\)
\(\frac{ 10x - 11 }{(x - 2)(x + 1)}\)
The bottom is left in its factorised form, which is how an exam answer is normally written.
The common denominator is again the product \((x - 1)(x - 3)\).
\(\frac{ 9 }{x - 1 } - \frac{ 8 }{x - 3 } = \frac{ 9(x - 3) - 8(x - 1)}{(x - 1)(x - 3)}\)
Expand the top, taking care with the minus sign in front of the second bracket:
\(9(x - 3) - 8(x - 1) = 9x - 27 - 8x + 8 = x - 19\)
\(\frac{ x - 19 }{(x - 1)(x - 3)}\)
- (a) M1common denominator \((x - 2)(x + 1)\)
- (a) M1tops \(3(x + 1)\) and \(7(x - 2)\) added
- (a) A1numerator \(10x - 11\)
- (b) M1common denominator \((x - 1)(x - 3)\)
- (b) M1tops \(9(x - 3)\) and \(8(x - 1)\) subtracted
- (b) A1numerator \(x - 19\)