1. 1
   Three algebraic expressions are being simplified: $5x+3y-2x+7y$, then $4a-a+6b-2b$, then $3p^2+2p-p^2+5p$.

   (a)

   Simplify $5x+3y-2x+7y$.

   [1 mark]

   • A$7x+4y$
   • B$3x+10y$
   • C$3x+4y$
   • D$7x+10y$

   (b)

   Simplify $4a-a+6b-2b$.

   [1 mark]

   • A$3a+4b$
   • B$5a+4b$
   • C$3a+8b$
   • D$4a+4b$

   (c)

   Simplify $3p^2+2p-p^2+5p$.

   [1 mark]

   • A$4p^2+7p$
   • B$2p^2+3p$
   • C$3p^2+7p$
   • D$2p^2+7p$

   Total for question 1: 3 marks

2. 2
   An expression is given as $12a^5\div3a^{-2}$, and a second expression is given as $(2x^3)^4$.

   (a)

   Simplify $12a^5\div3a^{-2}$.

   [2 marks]

   (b)

   Simplify $(2x^3)^4$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   An expression is given as $(x+1)(x-4)(3x+2)$, and a separate expression $12x^2+18xy$ is to be factorised fully.

   (a)

   Expand and simplify $(x+1)(x-4)(3x+2)$ fully.

   [3 marks]

   (b)

   Factorise $12x^2+18xy$ fully.

   [3 marks]

   Total for question 3: 6 marks

4. 4
   A rectangular allotment has length $(2x+3)$ metres and width $(x-1)$ metres. Its perimeter is 46 metres.

   (a)

   Form an equation in $x$ using the perimeter, and solve it to find the value of $x$.

   [4 marks]

   (b)

   Hence find the length and width of the allotment. Given the width must be at least 5 metres, write down an inequality in $x$ for this condition, and state whether the current width satisfies it.

   [4 marks]

   (c)

   The allotment's length and width are each to be increased by $y$ metres, so the new area is 180 m$^2$. Form and solve a quadratic equation to find the positive value of $y$.

   [5 marks]

   Total for question 4: 13 marks

5. 5
   A formula is given as $v^2=u^2+2as$, used to model an object's motion.

   (a)

   Which of these correctly makes $s$ the subject of $v^2=u^2+2as$?

   [1 mark]

   • A$s=\frac{v^2+u^2}{2a}$
   • B$s=2a(v^2-u^2)$
   • C$s=\frac{v^2-u^2}{2a}$
   • D$s=\frac{v^2-u^2}{2}-a$

   (b)

   Which of these correctly makes $u$ the subject of $v^2=u^2+2as$ (where $u>0$)?

   [1 mark]

   • A$u=v^2-2as$
   • B$u=\sqrt{v^2-2as}$
   • C$u=\sqrt{v^2+2as}$
   • D$u=\sqrt{v^2}-2as$

   (c)

   Which of these correctly makes $a$ the subject of $v^2=u^2+2as$?

   [1 mark]

   • A$a=\frac{v^2-u^2}{2}$
   • B$a=2s(v^2-u^2)$
   • C$a=\frac{v^2+u^2}{2s}$
   • D$a=\frac{v^2-u^2}{2s}$

   Total for question 5: 3 marks

6. 6
   At a farm shop, 3 punnets of strawberries and 2 punnets of blueberries cost 13.50 dollars. 5 punnets of strawberries and 2 punnets of blueberries cost 19.50 dollars.

   (a)

   Form a pair of simultaneous equations, using $x$ for the price of one punnet of strawberries and $y$ for the price of one punnet of blueberries.

   [2 marks]

   (b)

   Solve your equations to find the price of one punnet of strawberries.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   Two expressions are given: $\frac{x}{4}+\frac{x-3}{3}$, and $\frac{2a}{5}\times\frac{15}{4a^2}$.

   (a)

   Simplify $\frac{x}{4}+\frac{x-3}{3}$, giving your answer as a single fraction.

   [3 marks]

   (b)

   Simplify $\frac{2a}{5}\times\frac{15}{4a^2}$ fully.

   [3 marks]

   Total for question 7: 6 marks

8. 8
   A sequence has $n$th term $T_n=n^2+3n$. A function is defined as $f(x)=2x-5$. A quantity $y$ is directly proportional to $x^2$, with $y=45$ when $x=3$.

   (a)

   Find the 6th term of the sequence, and find the positive value of $n$ for which $T_n=88$.

   [4 marks]

   (b)

   Find $f(4)$ and the inverse function $f^{-1}(x)$.

   [4 marks]

   (c)

   Given $y$ is directly proportional to $x^2$, and $y=45$ when $x=3$, find the constant of proportionality $k$, write the formula connecting $y$ and $x$, and hence find $y$ when $x=5$.

   [5 marks]

   Total for question 8: 13 marks

End of questions