In order to know at what price the two services offered would be the same, we can actually use the process of trial and error starting from 1 onwards. The first one is 50.45 per month plus a standard fee of 60 for installation. So this will be 60+50.45x (number of days). After that. there is the other one that has free installation but charges 57.95. So it can be expressed as 57.95x. Now after trial and error with numbers as x, I came upon 8. If x is 8, then the first one will be 60+50.45 (8) = 463.6. For the second, 57.95(8) will also equal to 463.6. So the day in which the two services will charge the same is during the 8th day.
It will be 3:45. You can find this by going ahead 3 hours, then going ahead a half an hour.
Step-by-step explanation:
Given \:number (m) = £ 490Givennumber(m)=£490
Increase (g) = 11\%Increase(g)=11%
\begin{gathered} \red{ Required \:number } \\= m\Big( \frac{(100+g)}{100} \Big)\\= 490\Big(\frac{100+11}{100}\Big)\\= 490 \times \frac{111}{100} \\= 490 \times 1.11 \\= £ 543.9 \end{gathered}
Requirednumber
=m(
100
(100+g)
)
=490(
100
100+11
)
=490×
100
111
=490×1.11
=£543.9
Therefore.,
\red{ Required \:number }\green { = £ 543.9}Requirednumber=£543.9
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