Question 14:
Table 2 shows the information related to four ingredients used to produce a type of biscuit.
(a)(i) Given that the price index of ingredient P for the year 2021 based on the year 2019 is 130, find the value of z.
The price index of ingredient Q for the year 2021 based on the year 2019 is 140. The price per kg of ingredient Q in the year 2019 is RM6 less than its corresponding price in the year 2021.
Calculate the value of x and of y.
[4 marks]
(b) Given that the composite index to produce the biscuit for the year 2021 based on the year 2019 is 132, express p in terms of q. [3 marks]
(c) The price of the biscuit in the year 2019 is RM40. The prices of all the ingredients have increased by 20% from the year 2021 to the year 2022.
Find the selling price of the biscuit in the year 2022, if the company is targeting a profit of 15%.
[3 marks]
Answer:
(a)(i) $$ \begin{aligned} \frac{z}{15}(100) & =130 \\ z & =\mathrm{RM} 19.50 \end{aligned} $$
(a)(ii) $$ \begin{aligned} &\begin{aligned} \frac{y}{x}(100) & =140, x+6=y \\ 140 x & =(x+6) 100 \\ x & =\text { RM15 } \end{aligned}\\ &\begin{aligned} 15+6 & =y \\ y & =\text { RM } 21 \end{aligned} \end{aligned} $$
(b) $$ \begin{aligned} \frac{130(5)+140 p+160 q+110(3)}{(5+p+q+3)} & =132 \\ 980+140 p+160 q & =1056+132 p+132 q \\ 8 p & =1056-980-28 q \\ p & =\frac{19}{2}-\frac{7}{2} q \end{aligned} $$
(c) $$ \begin{aligned} \bar{I} & =\frac{132 \times 120}{100} \\ & =158.4 \end{aligned} $$
$$ \begin{aligned} \frac{x}{\text { RM40 }} \times 100 & =158.4 \\ x & =\text { RM63.36 } \end{aligned} $$
$$ \begin{aligned} & \text { Biscuit’s price in } 2022\\ &\begin{aligned} & =\frac{115}{100} \times \text { RM } 63.36 \\ & =\text { RM72.86 } \end{aligned} \end{aligned} $$
Table 2 shows the information related to four ingredients used to produce a type of biscuit.
(a)(i) Given that the price index of ingredient P for the year 2021 based on the year 2019 is 130, find the value of z.
The price index of ingredient Q for the year 2021 based on the year 2019 is 140. The price per kg of ingredient Q in the year 2019 is RM6 less than its corresponding price in the year 2021.
Calculate the value of x and of y.
[4 marks]
(b) Given that the composite index to produce the biscuit for the year 2021 based on the year 2019 is 132, express p in terms of q. [3 marks]
(c) The price of the biscuit in the year 2019 is RM40. The prices of all the ingredients have increased by 20% from the year 2021 to the year 2022.
Find the selling price of the biscuit in the year 2022, if the company is targeting a profit of 15%.
[3 marks]
Answer:
(a)(i) $$ \begin{aligned} \frac{z}{15}(100) & =130 \\ z & =\mathrm{RM} 19.50 \end{aligned} $$
(a)(ii) $$ \begin{aligned} &\begin{aligned} \frac{y}{x}(100) & =140, x+6=y \\ 140 x & =(x+6) 100 \\ x & =\text { RM15 } \end{aligned}\\ &\begin{aligned} 15+6 & =y \\ y & =\text { RM } 21 \end{aligned} \end{aligned} $$
(b) $$ \begin{aligned} \frac{130(5)+140 p+160 q+110(3)}{(5+p+q+3)} & =132 \\ 980+140 p+160 q & =1056+132 p+132 q \\ 8 p & =1056-980-28 q \\ p & =\frac{19}{2}-\frac{7}{2} q \end{aligned} $$
(c) $$ \begin{aligned} \bar{I} & =\frac{132 \times 120}{100} \\ & =158.4 \end{aligned} $$
$$ \begin{aligned} \frac{x}{\text { RM40 }} \times 100 & =158.4 \\ x & =\text { RM63.36 } \end{aligned} $$
$$ \begin{aligned} & \text { Biscuit’s price in } 2022\\ &\begin{aligned} & =\frac{115}{100} \times \text { RM } 63.36 \\ & =\text { RM72.86 } \end{aligned} \end{aligned} $$
Question 15:
Solution by graph sketching is not accepted.
Diagram 7 shows two particles, A and B, move along a straight line.
At instant t = 0, particle A and particle B move off from point M and point N respectively. The displacement, s m, from point M for particle A and particle B is given by sA = t3 + 4t2 + 5t and sB = t3 – t2 + 4t + 18 respectively, such that t is time, in seconds.
[3 marks]
(a) Find the initial velocity, in m s-1, of the particle A and the particle B. [3 marks]
(b) Find the distance, in m , travelled by the particle A and the particle B at the instant they collide. [4 marks]
(c) Calculate the acceleration, in m s-2, for the particle A and the particle B immediately before the collision occurs. [3 marks]
Answer:
(a) $$ \begin{aligned} &\operatorname{Particle} A \\ &\begin{aligned} V_0 & =3(0)^2+8(0)+5 \\ & =5 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Particle B }\\ &\begin{aligned} V_0 & =3(0)^2-2(0)+4 \\ & =4 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(b) $$ \begin{aligned} S_A & =S_B \\ t^3+4 t^2+5 t & =t^3-t^2+4 t+18 \\ 5 t^2+t-18 & =0 \\ (5 t-9)(t+2) & =0 \\ t=\frac{9}{5} \text { and } t & =-2( \text { Ignore }) \end{aligned} $$
$$ \begin{aligned} & \text { Distance travelled by } A \text { when colliding }\\ &\begin{aligned} & =\left(\frac{9}{5}\right)^3+4\left(\frac{9}{5}\right)^2+5\left(\frac{9}{5}\right) \\ & =27.79 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Distance travelled by } B \text { when colliding }\\ &\begin{aligned} & =27.79-18 \\ & =9.79 \end{aligned} \end{aligned} $$
(c) $$ \begin{aligned} & t=\frac{9}{5} \\ & \quad a_A=6 t+8 \\ & 6\left(\frac{9}{5}\right)+8=18.8 \end{aligned} $$
$$ \begin{aligned} a_B & =6 t-2 \\ 6\left(\frac{9}{5}\right)-2 & =8.8 \end{aligned} $$
Solution by graph sketching is not accepted.
Diagram 7 shows two particles, A and B, move along a straight line.
At instant t = 0, particle A and particle B move off from point M and point N respectively. The displacement, s m, from point M for particle A and particle B is given by sA = t3 + 4t2 + 5t and sB = t3 – t2 + 4t + 18 respectively, such that t is time, in seconds.
[3 marks]
(a) Find the initial velocity, in m s-1, of the particle A and the particle B. [3 marks]
(b) Find the distance, in m , travelled by the particle A and the particle B at the instant they collide. [4 marks]
(c) Calculate the acceleration, in m s-2, for the particle A and the particle B immediately before the collision occurs. [3 marks]
Answer:
(a) $$ \begin{aligned} &\operatorname{Particle} A \\ &\begin{aligned} V_0 & =3(0)^2+8(0)+5 \\ & =5 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Particle B }\\ &\begin{aligned} V_0 & =3(0)^2-2(0)+4 \\ & =4 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(b) $$ \begin{aligned} S_A & =S_B \\ t^3+4 t^2+5 t & =t^3-t^2+4 t+18 \\ 5 t^2+t-18 & =0 \\ (5 t-9)(t+2) & =0 \\ t=\frac{9}{5} \text { and } t & =-2( \text { Ignore }) \end{aligned} $$
$$ \begin{aligned} & \text { Distance travelled by } A \text { when colliding }\\ &\begin{aligned} & =\left(\frac{9}{5}\right)^3+4\left(\frac{9}{5}\right)^2+5\left(\frac{9}{5}\right) \\ & =27.79 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Distance travelled by } B \text { when colliding }\\ &\begin{aligned} & =27.79-18 \\ & =9.79 \end{aligned} \end{aligned} $$
(c) $$ \begin{aligned} & t=\frac{9}{5} \\ & \quad a_A=6 t+8 \\ & 6\left(\frac{9}{5}\right)+8=18.8 \end{aligned} $$
$$ \begin{aligned} a_B & =6 t-2 \\ 6\left(\frac{9}{5}\right)-2 & =8.8 \end{aligned} $$