Algebra: expressions and factorising – Year 8 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Only like terms can be combined. The a terms give \(5a-2a=3a\) and the b terms give \(3b+4b=7b\). The answer is \(3a+7b\). (1 mark for each correct term)
   2. Multiply each term inside the bracket by 3: \(3\times2x=6x\) and \(3\times(-7)=-21\). The answer is \(6x-21\). (1 mark)
   3. The highest common factor of 8 and 12 is 4. Write 4 outside the bracket and divide each term by 4: \(8p\div4=2p\) and \(12\div4=3\). The answer is \(4(2p+3)\). (1 mark)

   Check (c) by expanding: \(4\times2p+4\times3=8p+12\). An answer such as \(2(4p+6)\) is correct but not fully factorised.

2. Answer to Question 2 (4 marks)

   1. Expand both brackets: \(4(x+3)=4x+12\) and \(2(x-1)=2x-2\) (1 mark). Collect the x terms, \(4x+2x=6x\), and the numbers, \(12-2=10\): \(6x+10\) (1 mark).
   2. The minus sign in front of the second bracket multiplies both of its terms: \(-3\times y=-3y\) and \(-3\times(-4)=+12\). So \(5(2y-1)-3(y-4)=10y-5-3y+12\) (1 mark for expanding with correct signs). Collect: \(10y-3y=7y\) and \(-5+12=7\), giving \(7y+7\) (1 mark).

   Check (b) with \(y=1\): the original gives \(5\times1-3\times(-3)=5+9=14\), and \(7\times1+7=14\). The answer can also be written \(7(y+1)\).

3. Answer to Question 3 (4 marks)

   1. Substitute, keeping the negative value in brackets: \(2\times4-3\times(-3)=8-(-9)=8+9=\)17. (1 mark)
   2. \(a^{2}=16\) and \(b^{2}=(-3)^{2}=9\), because a negative number squared is positive. So \(16+9=\)25. (1 mark)
   3. Look for the highest common factor of each part. The numbers 12 and 18 share 6; both terms contain at least one x and at least one y. The HCF is \(6xy\). Dividing: \(12x^{2}y\div6xy=2x\) and \(18xy^{2}\div6xy=3y\). The answer is \(6xy(2x+3y)\). (1 mark for a correct partial factor, 1 mark for the full HCF)

   Check (c) by expanding: \(6xy\times2x+6xy\times3y=12x^{2}y+18xy^{2}\).

4. Answer to Question 4 (4 marks)

   1. The bus costs £60 plus \(5n\) for n pupils, and the snacks cost \(3n\) (1 mark). Total cost \(=60+5n+3n=\)\(60+8n\) pounds (1 mark).
   2. Substitute \(n=24\): \(60+8\times24=60+192=\)£252. (1 mark)
   3. The cost must not be more than £400: \(60+8n\le400\). Subtract 60: \(8n\le340\). Divide by 8: \(n\le42.5\). Pupils come in whole numbers, so round down to 42 pupils. (1 mark)

   42 pupils cost \(60+8\times42=£396\), while 43 would cost £404, which is over the budget. Rounding up here would break the budget, so this is a case where 42.5 must be rounded down.

5. Answer to Question 5 (4 marks)

   1. Multiplying n by 3 gives \(3n\); adding 12 gives \(3n+12\); dividing by 3 gives \(\dfrac{3n+12}{3}\) (1 mark). Divide each term on top by 3: \(\dfrac{3n}{3}+\dfrac{12}{3}=\)\(n+4\) (1 mark).
   2. The simplified expression \(n+4\) means "the starting number plus 4" for every value of n, so Hana is correct (1 mark). An example illustrates it: with \(n=10\), \(3\times10+12=42\) and \(42\div3=14=10+4\) (1 mark).

   One example on its own is not a proof, since it shows only a single case. The algebra in part (a) is what proves the claim for every number, including negatives and decimals.

Total : 20 marks