WMA11 June 2021

5.​​ WMA11/01​​ Edexcel IAL P1 June 2021 Q5​​ (Quadratics​​ &​​ Straight Line Graphs)

The share value of two companies, company A and company B, has been monitored over a 15-year period.​​ 

The share value PA​​ of company A, in millions of pounds, is modelled by the equation​​ 

PA = 53 – 0.4t – 82         t≥ 0 

where t is the number of years after monitoring began.​​ 

The share value PB​​ of company B, in millions of pounds, is modelled by the equation​​ 

PB = –1.6t + 44.2            t≥ 0 

where t is the number of years after monitoring began.​​ 

Figure 2 shows a graph of both models.​​ 

Use the equations of one or both models to answer parts (a) to (d).

(a) Find the difference between the share value of company A and the share value of company B at the point monitoring began.​​ 

(2)

(b) State the maximum share value of company A during the 15-year period.​​ 

(1)

(c) Find, using algebra and showing your working, the times during this 15-year period when the share value of company A was greater than the share value of company B.​​ 

(4)

(d) Explain why the model for company A should not be used to predict its share value when t = 20​​ 

(1)

SOLUTION​​ 

a-​​ 

At the time, when monitoring began,​​ t=0.​​ 

The share value of company A is​​ 

PA = 53 – 0.4t – 82

PA=53-0.4  -82

PA=  27.4

The share value of company B is​​ 

PB=-1.60+44.2

PB=44.2

Hence, the difference in the share value of company A and B when monitoring began​​ 

PB-PA=16.8 

Difference=16.8 million pounds

b-​​ 

Always remember, when the word ‘state’ is used in the question, it means that there isn’t a need of much working. You just have to answer it directly. And since this is of just 1 marks, we have to look for the answer very carefully.​​ 

Now, if we look at the modelled equation of company A, we can see that the maximum value of company A is when​​ 0.4(t–8)2​​ is equal to zero, so that there is nothing to be subtracted from 53.​​ 

Hence, the maximum value is 53 million pounds.

Max share value=53 million pounds

c-​​ 

It can be observed from the graph that after the point where the curves cut each other the share value of company A has remain greater than B till 15 years period. So first we need to find the point of intersection (year in which both companies had the same share value). ​​ 

PA=PB

53-0.4 t-82= -1.6t+44.2

53-0.4t2-16t+64=-1.6t+44.2

53-0.4t2+6.4t-25.6= -1.6t+44.2

-0.4t2+6.4t+27.4= -1.6t+44.2

0=0.4t2-8t+169

t2-20t+42=0

t2-20t=-42

t-102-100=-42

t-102=100-42

t-102=58

t-10= ±58

t=10±58

t1=10-58=2.384

t2=10+58=17.616

But since t2>15, we will reject it.​​ 

Hence, the times for which the share value of company A is greater than company B is

2.38 <t≤15

d-​​ 

The share value model for company A gives the negative value of time, the model can’t handle the values after 15 years of period.​​ 

When​​ t=20

PA=53-0.42-82

PA=-4.6