Algebraic notation and simplifying expressions – Year 7 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. In algebra we leave out the multiplication signs and write the number first. The two letters a multiply to give \(a^{2}\), and b stays as it is, so the term is \(5a^{2}b\). (1 mark)
   2. Group the like terms: the x terms give \(7x-2x=5x\) and the y terms give \(3y+y=4y\). Remember that y on its own means \(1y\). The simplified expression is \(5x+4y\). (1 mark for each correct term)
   3. Write the division as a fraction \(\dfrac{12ab}{4a}\). Divide the numbers, \(12\div4=3\), and cancel the a on the top with the a underneath. Only b is left: \(3b\). (1 mark)

   Check (c): \(4a\times3b=12ab\), so the division is correct.

2. Answer to Question 2 (4 marks)

   1. Each notebook costs £n, so four of them cost \(4\times n=4n\). Three pens at £p each cost \(3p\). Add the two amounts: \(4n+3p\) pounds. These are unlike terms, so nothing more can be collected. (1 mark)
   2. Multiply every term inside each bracket: \(2(x+3)=2x+6\) and \(3(x-1)=3x-3\) (1 mark). Now collect: \(2x+3x=5x\) and \(6-3=3\), giving \(5x+3\) (1 mark).
   3. Only x is squared, not the 3. Square first: \((-2)^{2}=4\), because a negative times a negative is positive. Then multiply by 3: \(3\times4=\)12. (1 mark)

   A common error in (c) is working out \((3\times-2)^{2}=36\); the order of operations puts the power before the multiplication.

3. Answer to Question 3 (4 marks)

   1. Collect the a terms: \(4a-a=3a\) (the lonely a counts as \(1a\)). Collect the b terms: \(5b+2b=7b\). The simplified expression is \(3a+7b\). Award 1 mark for each correct term.
   2. Replace a by 3 and b by \(-1\), keeping the negative number in brackets: \(3\times3+7\times(-1)\) (1 mark). This gives \(9+(-7)=9-7=\)2 (1 mark).

   Check by substituting into the original expression instead: \(4\times3+5\times(-1)-3+2\times(-1)=12-5-3-2=2\). Both routes agree, which confirms that the simplification in part (a) was correct. Simplifying first made the substitution much quicker.

4. Answer to Question 4 (4 marks)

   1. In the morning she sells 12 cups at £c, worth \(12c\), and 8 slices at £d, worth \(8d\). The morning money is \(12c+8d\) pounds. (1 mark)
   2. The afternoon brings \(15c+10d\). Add the two sessions: \((12c+8d)+(15c+10d)\) (1 mark). Collect the c terms, \(12c+15c=27c\), and the d terms, \(8d+10d=18d\): \(27c+18d\) (1 mark).
   3. Work in pounds, so 50p becomes \(c=0.5\) and £1.20 becomes \(d=1.2\). Then \(27\times0.5=13.5\) and \(18\times1.2=21.6\). The total takings are \(13.5+21.6=\)£35.10. (1 mark)

   Writing 50 instead of 0.5 would mix pence with pounds and give a silly answer.

5. Answer to Question 5 (4 marks)

   1. A rectangle has two lengths and two widths, so the perimeter is \(2(3x+2)+2(x+1)\). Expanding gives \(6x+4+2x+2\) (1 mark). Collecting like terms: \(6x+2x=8x\) and \(4+2=6\), so the perimeter is \(8x+6\) metres (1 mark).
   2. Set the expression equal to the given perimeter: \(8x+6=36\) (1 mark). Subtract 6 from both sides to get \(8x=30\), then divide both sides by 8: \(x=\)3.75 (1 mark).

   Check: the length is \(3\times3.75+2=13.25\) m and the width is \(3.75+1=4.75\) m. Then \(2(13.25+4.75)=2\times18=36\) m, exactly the perimeter given in the question.

Total : 20 marks