Question
Scott purchases food for his dog in large bags and feeds the dog the same amount of
dog food each day. The amount of dog food left in the bag at the end of each day can be
modelled by an arithmetic sequence.
On a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the
third day there were 115.5 cups of dog food remaining in the bag and at the end of the eighth
day there were 108 cups of dog food remaining in the bag.
(a) Find the number of cups of dog food
(i) fed to the dog per day;
(ii) remaining in the bag at the end of the first day. [4]
(b) Calculate the number of days that Scott can feed his dog with one bag of food. [2]
In 2021, Scott spent $625 on dog food. Scott expects that the amount he spends on dog food
will increase at an annual rate of 6.4%.
(c) Determine the amount that Scott expects to spend on dog food in 2025. Round your
answer to the nearest dollar. [3]
(d) (i) Calculate the value of \(\sum_{n=1}^{10}(625\times 1.064^{(n-1)})\)
(ii) Describe what the value in part (d)(i) represents in this context. [3]
(e) Comment on the appropriateness of modelling this scenario with a geometric sequence. [1]
▶️Answer/Explanation
Ans
1. (a) (i) EITHER
115.5 = u1 + (3 − 1) × d (115.5 = u1 + 2d )
108 = u1 + (8 − 1) × d (108 = u1 + 7d) (M1)(A1)
Note: Award M1 for attempting to use the arithmetic sequence term formula, A1 for
both equations correct. Working for M1 and A1 can be found in parts (i) or (ii).
d = − ( 1.5)
1.5 (cups/day) A1
Note: Answer must be written as a positive value to award A1.
OR
\((d=)\frac{115.5-108}{5}\) M1A1
Note: Award M1 for attempting a calculation using the difference between term 3
and term 8; A1 for a correct substitution.
(d=) 1.5 (cups/day) A1
(ii) ( u1 =) 118.5 (cups) A1
[4 marks]
(b) attempting to substitute their values into the term formula for arithmetic
sequence equated to zero (M1)
0 = 118.5 + ( n –1) × 1.(−1. 5)
( n ) = 80 days A1
Note: Follow through from part (a) only if their answer is positive.
[2 marks]
(c) (t5 =) 625 × 1.064(5-1) (M1)(A1)
Note: Award M1 for attempting to use the geometric sequence term formula;
A1 for a correct substitution.
$ 801 A1
Note: The answer must be rounded to a whole number to award the final A1.
[3 marks]
continued…
Question 1 continued
(d) (i) (S10) ($) 8390 (8394.39…) A1
(ii) EITHER
the total cost (of dog food) R1
for 10 years beginning in 2021 OR 10 years before 2031 R1
OR
the total cost (of dog food) R1
from 2021 to 2030 (inclusive) OR from 2021 to (the start of) 2031 R1
[3 marks]
(e) EITHER
According to the model, the cost of dog food per year will eventually
be too high to keep a dog.
OR
The model does not necessarily consider changes in inflation rate.
OR
The model is appropriate as long as inflation increases at a similar rate.
OR
The model does not account for changes in the amount of food the dog
eats as it ages/becomes ill/stops growing.
OR
The model is appropriate since dog food bags can only be bought in
discrete quantities. R1
Note: Accept reasonable answers commenting on the appropriateness of the model for the
specific scenario. There should be a reference to the given context. A reference to
the geometric model must be clear: either “model” is mentioned specifically, or other
mathematical terms such as “increasing” or “discrete quantities” are seen. Do not
accept a contextual argument in isolation, e.g. “The dog will eventually die”.
[1 mark]
Total [13 marks]
Question
A large underground tank is constructed at Mills Airport to store fuel.
The tank is in the shape of an isosceles trapezoidal prism, ABCDEFGH .
AB = 70 m , AF = 200 m , AD = 40 m , BC = 40 m and CD = 110 m . Angle ADC = 60° and angle BCD = 60° . The tank is illustrated below.
Find h , the height of the tank. [2]
Show that the volume of the tank is 624 000 m3, correct to three significant figures. [3]
Once construction was complete, a fuel pump was used to pump fuel into the empty tank. The amount of fuel pumped into the tank by this pump each hour decreases as an arithmetic sequence with terms u1 , u2 , u3 , … , un .
Part of this sequence is shown in the table.
Hour (n)
1st
2nd
3rd
…
Amount of fuel pumped into the tank in each hour, in m3 (un )
45 000
43 200
41 400
…
Write down the common difference, d . [1]
Find the amount of fuel pumped into the tank in the 13th hour. [2]
(i) Find the value of n such that un = 0 .
(ii) Write down the number of hours that the pump was pumping fuel into the tank. [3]
At the end of the 2nd hour, the total volume of fuel in the tank was 88 200 m3.
Find the total amount of fuel pumped into the tank in the first 8 hours. [2]
Show that the tank will never be completely filled using this pump. [3]
▶️Answer/Explanation
Ans:
(a)
sin 60° = \(\frac{h}{40}\)
OR tan60 = \(\frac{h}{20} \)
\(OR 20^{2}+ h^{2}= 40 ^{2}(\sqrt{40^{2}-20^{2}})\)
(h=)34.6(m)\((\sqrt{1200,})20\sqrt{3}\),34.6410…
(b)
\(\frac{1}{2}(70+110)\)(34.6410…)\(\times 200\)
OR
\((2 \times \frac{1}{2}\times 20\times 34.6410…+70 \times 34.6410….)\)\(\times 200\)
OR
\(70\times 34.6410…\times 200+2\times \frac{1}{2}\times 34.6410…\times 20\times 200\)
623538…
624000 (cm3)
(c)
(d=)-1800
(d)
\((u_{13}=) 45000+ (13-1(-1800))\)
\(23400(m^{3})\)
(e)
(i)
0= 45000+ ((n-1)-1800))
(n=26)
(ii)25
(f)
\((S_{8})\frac{8}{2}(2\times 45000+(8-1)\times (-1800))\)
\(310000(m^{3})(3096000)\)
(g)
\((S_{25}=)\frac{25}{2}(2\times 45000+(25-)\times (-1800)),\)
\((S_{25=})\frac{25}{2}(45000+1800)\)
\((S_{25})= 585000 (m^{3})\)<624000 m3
OR
\(S_{n}= \frac{n}{2}(2\times 45000+(n-1)\times (-1800))\)
Maximum of this function 585225 (m3 )
585225(m3 ) < 624000(m3 ) hence it will never be filled
OR
sketch with concave down curve and labelled 624000 horizontal line
curve explicitly labelled as
\(s_{n}= \frac{n}{2}(2\times 450000+ (n-1)\times (-1800))\)or equivalent
the line and the curve do not intersect hence it will never be filled
OR
624000=\(\frac{n}{2}(2\times 45000+(n-1)\times (-1800))\)
Demonstrates there is no solution
There is no (real) solution (to this equation) hence it will never be filled
Question
The natural numbers: 1, 2, 3, 4, 5… form an arithmetic sequence.
A geometric progression \(G_1\) has 1 as its first term and 3 as its common ratio.
i.a.State the values of u1 and d for this sequence.[2]
i.b.Use an appropriate formula to show that the sum of the natural numbers from 1 to n is given by \(\frac{1}{2}n (n +1)\).[2]
i.c.Calculate the sum of the natural numbers from 1 to 200.[2]
ii.a.The sum of the first n terms of G1 is 29 524. Find n.[3]
ii.b.A second geometric progression G2 has the form \(1,\frac{1}{3},\frac{1}{9},\frac{1}{{27}}…\)[1]
ii.c.Calculate the sum of the first 10 terms of G2.[2]
ii.d.Explain why the sum of the first 1000 terms of G2 will give the same answer as the sum of the first 10 terms, when corrected to three significant figures.[1]
ii.e.Using your results from parts (a) to (c), or otherwise, calculate the sum of the first 10 terms of the sequence \(2,3\frac{1}{3},9\frac{1}{9},27\frac{1}{{27}}…\)
Give your answer correct to one decimal place.[3]
▶️Answer/Explanation
Markscheme
\(u_1 = d = 1\). (A1)(A1)[2 marks]
Sum is \(\frac{1}{2}n(2{u_1} + d(n – 1))\) or \(\frac{1}{2}n({u_1} + {u_n})\) (M1)
Award (M1) for either sum formula seen, even without substitution.
So sum is \(\frac{1}{2}n(2 + (n – 1)) = \frac{1}{2}n(n + 1)\) (A1)(AG)
Award (A1) for substitution of \({u_1} = 1 = d\) or \({u_1} = 1\) and \({u_n} = n\) with simplification where appropriate. \(\frac{1}{2}n(n + 1)\) must be seen to award this (A1).[2 marks]
\(\frac{1}{2}(200)(201) = 20 100\) (M1)(A1)(G2)
(M1) is for correct formula with correct numerical input. Original sum formula with u, d and n can be used.[2 marks]
\(\frac{{1 – {3^n}}}{{1 – 3}} = 29524\) (M1)(A1)
(M1) for correctly substituted formula on one side, (A1) for = 29524 on the other side.
n = 10. (A1)(G2)
Trial and error is a valid method. Award (M1) for at least \(\frac{{1 – {3^{10}}}}{{1 – 3}}\) seen and then (A1) for = 29524, (A1) for \(n = 10\). For only unproductive trials with \(n \ne 10\), award (M1) and then (A1) if the evaluation is correct.[3 marks]
Common ratio is \(\frac{1}{3}\), (0.333 (3sf) or 0.3) (A1)
Accept ‘divide by 3’.[1 mark]
\(\frac{{1 – {{\left( {\frac{1}{3}} \right)}^{10}}}}{{1 – \frac{1}{3}}}\) (M1)
= 1.50 (3sf) (A1)(ft)(G1)
1.5 and \(\frac{3}{2}\) receive (A0)(AP) if AP not yet used Incorrect formula seen in (a) or incorrect value in (b) can follow through to (c). Can award (M1) for \(1 + \left( {\frac{1}{3}} \right) + \left( {\frac{1}{9}} \right) + ……\)[2 marks]
Both \({\left( {\frac{1}{3}} \right)^{10}}\) and \({\left( {\frac{1}{3}} \right)^{1000}}\) (or those numbers divided by 2/3) are 0 when corrected to 3sf, so they make no difference to the final answer. (R1)
Accept any valid explanation but please note: statements which only convey the idea of convergence are not enough for (R1). The reason must show recognition that the convergence is adequately fast (though this might be expressed in a much less technical manner).[1 mark]
The sequence given is \(G_1 + G_2\) (M1)
The sum is 29 524 + 1.50 (A1)(ft)
= 29 525.5 (A1)(ft)(G2)
The (M1) is implied if the sum of the two numbers is seen. Award (G1) for 29 500 with no working. (M1) can be awarded for
\(2 + 3\frac{1}{3} + …\) Award final (A1) only for answer given correct to 1dp.[3 marks]
Question
Give all answers in this question correct to the nearest dollar.
Clara wants to buy some land. She can choose between two different payment options. Both options require her to pay for the land in 20 monthly installments.
Option 1: The first installment is \(\$ 2500\). Each installment is \(\$ 200\) more than the one before.
Option 2: The first installment is \(\$ 2000\). Each installment is \(8\% \) more than the one before.
If Clara chooses option 1,
a.(i) write down the values of the second and third installments;
(ii) calculate the value of the final installment;
(iii) show that the total amount that Clara would pay for the land is \(\$ 88000\).[7]
b.If Clara chooses option 2,
(i) find the value of the second installment;
(ii) show that the value of the fifth installment is \(\$ 2721\).[4]
c.The price of the land is \(\$ 80000\). In option 1 her total repayments are \(\$ 88000\) over the 20 months. Find the annual rate of simple interest which gives this total.[4]
d.Clara knows that the total amount she would pay for the land is not the same for both options. She wants to spend the least amount of money. Find how much she will save by choosing the cheaper option.[4]
▶️Answer/Explanation
Markscheme
(i) Second installment \( = \$ 2700\) (A1)
Third installment \( = \$ 2900\) (A1)
(ii) Final installment \( = 2500 + 200 \times 19\) (M1)(A1)
Note: (M1) for substituting in correct formula or listing, (A1) for correct substitutions.
\( = \$ 6300\) (A1)(G2)
(iii) Total amount \( = \frac{{20}}{2}(2500 + 6300)\)
OR
\( \frac{{20}}{2}(5000 + 19 \times 200)\) (M1)(A1)
Note: (M1) for substituting in correct formula or listing, (A1) for correct substitution.
\( = \$ 88000\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[7 marks]
(i) Second installment \(2000 \times 1.08 = \$ 2160\) (M1)(A1)(G2)
Note: (M1) for multiplying by \(1.08\) or equivalent, (A1) for correct answer.
(ii) Fifth installment \( = 2000 \times {1.08^4} = 2720.98 = \$ 2721\) (M1)(A1)(AG)
Notes: (M1) for correct formula used with numbers from the problem. (A1) for correct substitution. The \(2720.9 \ldots \) must be seen for the (A1) mark to be awarded. Accept list of 5 correct values. If values are rounded prematurely award (M1)(A0)(AG).[4 marks]
Interest is \( = \$ 8000\) (A1)
\(80000 \times \frac{r}{{100}} \times \frac{{20}}{{12}} = 8000\) (M1)(A1)
Note: (M1) for attempting to substitute in simple interest formula, (A1) for correct substitution.
Simple Interest Rate \( = 6\% \) (A1)(G3)
Note: Award (G3) for answer of \(6\% \) with no working present if interest is also seen award (A1) for interest and (G2) for correct answer.[4 marks]
Financial accuracy penalty (FP) is applicable where indicated in the left hand column.
(FP) Total amount for option 2 \( = 2000\frac{{(1 – {{1.08}^{20}})}}{{(1 – 1.08)}}\) (M1)(A1)
Note: (M1) for substituting in correct formula, (A1) for correct substitution.
\( = \$ 91523.93\) (\( = \$ 91524\)) (A1)
\(91523.93 – 88000 = \$ 3523.93 = \$ 3524\) to the nearest dollar (A1)(ft)(G3)
Note: Award (G3) for an answer of \(\$ 3524\) with no working. The difference follows through from the sum, if reasonable. Award a maximum of (M1)(A0)(A0)(A1)(ft) if candidate has treated option 2 as an arithmetic sequence and has followed through into their common difference. Award a maximum of (M1)(A1)(A0)(ft)(A0) if candidate has consistently used \(0.08\) in (b) and (d).[4 marks]