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Question
Ramiro walks to work each morning. During the first minute he walks \(80\) metres. In each subsequent minute he walks \(90\% \) of the distance walked during the previous minute.
The distance between his house and work is \(660\) metres. Ramiro leaves his house at 08:00 and has to be at work by 08:15.
Explain why he will not be at work on time.
Answer/Explanation
Markscheme
METHOD 1
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
correct substitution into the sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{1 – {{0.9}^n}}}{{1 – 0.9}}} \right)\)
any correct equation with sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{{{0.9}^n} – 1}}{{0.9 – 1}}} \right) = 660,{\text{ }}1 – {0.9^n} = \frac{{66}}{{80}}\)
attempt to solve their equation involving the sum of a GP (M1)
eg\(\;\;\;\)graph, algebraic approach
\(n = 16.54290788\) A1
since \(n > 15\) R1
he will be late AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 2
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
correct substitution into the sum of a geometric sequence (A1)
eg\(\;\;\;80\left( {\frac{{1 – {{0.9}^n}}}{{1 – 0.9}}} \right)\)
attempt to substitute \(n = 15\) into sum of a geometric sequence (M1)
eg\(\;\;\;{S_{15}}\)
correct substitution (A1)
eg\(\;\;\;80\left( {\frac{{{{0.9}^{15}} – 1}}{{0.9 – 1}}} \right)\)
\({S_{15}} = 635.287\) A1
since \(S < 660\) R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
METHOD 3
recognize that the distance walked each minute is a geometric sequence (M1)
eg\(\;\;\;r = 0.9\), valid use of \(0.9\)
recognize that total distance walked is the sum of a geometric sequence (M1)
eg\(\;\;\;{S_n},{\text{ }}a\left( {\frac{{1 – {r^n}}}{{1 – r}}} \right)\)
listing at least 5 correct terms of the GP (M1)
15 correct terms A1
\(80,{\rm{ }}72,{\rm{ }}64.8,{\rm{ }}58.32,{\rm{ }}52.488,{\rm{ }}47.2392,{\rm{ }}42.5152,{\rm{ }}38.2637,{\rm{ }}34.4373,{\rm{ }}30.9936,{\rm{ }}27.8942,{\rm{ }}25.1048,{\rm{ }}22.59436,{\rm{ }}20.3349,{\rm{ }}18.3014\)
attempt to find the sum of the terms (M1)
eg\(\;\;\;{S_{15}},{\text{ }}80 + 72 + 64.8 + 58.32 + 52.488 + \ldots + 18.301433\)
\({S_{15}} = 635.287\) A1
since \(S < 660\) R1
he will not be there on time AG N0
Note: Do not award the R mark without the preceding A mark.
[7 marks]
Question
A population of rare birds, \({P_t}\), can be modelled by the equation \({P_t} = {P_0}{{\text{e}}^{kt}}\), where \({P_0}\) is the initial population, and \(t\) is measured in decades. After one decade, it is estimated that \(\frac{{{P_1}}}{{{P_0}}} = 0.9\).
(i) Find the value of \(k\).
(ii) Interpret the meaning of the value of \(k\).
Find the least number of whole years for which \(\frac{{{P_t}}}{{{P_0}}} < 0.75\).
Answer/Explanation
Markscheme
(i) valid approach (M1)
eg\(\,\,\,\,\,\)\(0.9 = {{\text{e}}^{k(1)}}\)
\(k = – 0.105360\)
\(k = \ln 0.9{\text{ (exact), }} – 0.105\) A1 N2
(ii) correct interpretation R1 N1
eg\(\,\,\,\,\,\)population is decreasing, growth rate is negative
[3 marks]
METHOD 1
valid approach (accept an equality, but do not accept 0.74) (M1)
eg\(\,\,\,\,\,\)\(P < 0.75{P_0},{\text{ }}{P_0}{{\text{e}}^{kt}} < 0.75{P_0},{\text{ }}0.75 = {{\text{e}}^{t\ln 0.9}}\)
valid approach to solve their inequality (M1)
eg\(\,\,\,\,\,\)logs, graph
\(t > 2.73045{\text{ }}({\text{accept }}t = 2.73045){\text{ }}(2.73982{\text{ from }} – 0.105)\) A1
28 years A2 N2
METHOD 2
valid approach which gives both crossover values accurate to at least 2 sf A2
eg\(\,\,\,\,\,\)\(\frac{{{P_{2.7}}}}{{{P_0}}} = 0.75241 \ldots ,{\text{ }}\frac{{{P_{2.8}}}}{{{P_0}}} = 0.74452 \ldots \)
\(t = 2.8\) (A1)
28 years A2 N2
[5 marks]