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IB Maths AI SL – Topic 1: Number and Algebra

Approximation, financial maths, sequences, exponents and logarithms.

12 questions  ·  66 marks  ·  answers included  ·  IB AI SL
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This worksheet covers:

  • Approximation and percentage error
  • Arithmetic and geometric sequences
  • Financial mathematics using the TVM solver
  • Exponents
  • Logarithms

A GDC is allowed throughout; give answers to 3 significant figures unless stated. Allow about 70 minutes.

Questions

Section A – Skills

GDC allowed. Short answers.
1

(a) Write $0.0034567$ correct to 3 significant figures.(b) Write your answer to (a) in the form $a \times 10^k$ where $1 \le a < 10$ and $k \in \mathbb{Z}$.(c) A student measures a mass as $12.5$ g. The exact mass is $12.32$ g. Calculate the percentage error in the student’s measurement.

Show answer
(a) $0.00346$(b) $3.46 \times 10^{-3}$(c) $\left|\frac{12.5-12.32}{12.32}\right| \times 100 = 1.46\%$
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2

An arithmetic sequence has first term $7$ and common difference $4$.(a) Find the 20th term.(b) Find the sum of the first 20 terms.(c) Find the smallest value of $n$ for which $u_n > 200$.

Show answer
(a) $u_{20} = 7 + 19(4) = 83$(b) $S_{20} = \frac{20}{2}(14 + 76) = 900$(c) $7 + 4(n-1) > 200 \Rightarrow n > 49.25$, so $n = 50$
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3

The first three terms of a geometric sequence are $3, 6, 12, \ldots$(a) Write down the common ratio.(b) Find the 10th term.(c) Find the sum of the first 10 terms.

Show answer
(a) $r = 2$(b) $u_{10} = 3 \times 2^9 = 1536$(c) $S_{10} = \frac{3(2^{10}-1)}{2-1} = 3069$
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4

Evaluate $\sum_{k=1}^{15} (3k + 2)$.

Show answer
$3 \times 120 + 2 \times 15 = 390$ (arithmetic series, $u_1 = 5$, $u_{15} = 47$, $S_{15} = \frac{15}{2}(5+47) = 390$)
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5

(a) Solve $2^{x+1} = 50$, giving your answer to 3 significant figures.(b) Without a calculator, simplify $\log_{10} 8 + \log_{10} 5 – \log_{10} 4$.(c) Write $\frac{x^3 \sqrt{x}}{x^{-2}}$ in the form $x^n$.

Show answer
(a) $x + 1 = \log_2 50 \Rightarrow x = 4.64$(b) $\log_{10}\frac{8 \times 5}{4} = \log_{10} 10 = 1$(c) $x^{3 + 0.5 + 2} = x^{5.5}$
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Section B – Exam-style

GDC allowed. Show your method, including any TVM parameters used.
6

Maya invests $5000$ USD in an account paying $4.2\%$ per annum, compounded quarterly.(a) Find the value of the investment after 6 years.(b) Find the total interest earned.(c) Find how many years it takes for the investment to double in value (compounded quarterly).

Show answer
(a) $5000\left(1 + \frac{0.042}{4}\right)^{24} = 6424.53$ USD (accept $6420$)(b) $1424.53$ USD(c) $N$ from TVM: $n = 66.4$ quarters, so about $16.6$ years (accept 17 years)
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7

Ben takes out a loan of $15000$ EUR at $6\%$ per annum, compounded monthly, to be repaid in equal monthly payments over 5 years.(a) Find the monthly payment.(b) Find the total amount of interest Ben pays over the 5 years.

Show answer
(a) TVM: $N = 60$, $I = 6$, $PV = 15000$, $FV = 0$, $P/Y = C/Y = 12 \Rightarrow PMT = 289.99$ EUR (accept $290$)(b) $60 \times 289.99 – 15000 = 2399.52$ EUR (accept $2400$)
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8

A car is bought for $24000$ GBP and depreciates by $15\%$ each year.(a) Write down an expression for the value $V$ of the car after $n$ years.(b) Find the value of the car after 4 years.(c) Find the number of complete years after which the value first falls below $10000$ GBP.

Show answer
(a) $V = 24000(0.85)^n$(b) $12528$ GBP (accept $12500$)(c) $0.85^n < \frac{10000}{24000} \Rightarrow n > 5.39$, so 6 years
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9

Aisha saves $200$ AUD at the end of every month into an account paying $3\%$ per annum, compounded monthly.(a) Find the value of her savings after 10 years.(b) State the total amount Aisha deposited and hence find the interest earned.

Show answer
(a) TVM: $N = 120$, $I = 3$, $PV = 0$, $PMT = -200$, $P/Y = C/Y = 12 \Rightarrow FV = 27948.28$ AUD (accept $27900$)(b) Deposited $24000$ AUD; interest $= 3948.28$ AUD (accept $3950$)
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10

A savings account pays a nominal rate of $4.5\%$ per annum. Inflation is $2.1\%$ per annum.(a) Explain what is meant by the real rate of return.(b) Calculate the real rate of return, giving your answer to 2 decimal places.

Show answer
(a) The rate of growth of the purchasing power of the money, i.e. the interest rate after adjusting for inflation.(b) $\frac{1.045}{1.021} – 1 = 2.35\%$ (accept the approximation $4.5 – 2.1 = 2.4\%$)
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Section C – Challenge

Extended response.
11

The side of a square tile is measured as $12$ cm, correct to the nearest cm.(a) Write down the lower and upper bounds of the side length.(b) Find the lower and upper bounds of the area of the tile.(c) A rectangular floor is measured as $8.3$ m by $5.1$ m, each correct to 2 significant figures. The exact dimensions are $8.34$ m by $5.08$ m. Calculate the percentage error in the area found using the rounded measurements.

Show answer
(a) $11.5 \le L < 12.5$ cm(b) $132.25 \le A < 156.25$ cm$^2$(c) Rounded area $42.33$; exact $42.3672$; percentage error $= 0.0878\%$
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12

A population of bacteria grows so that its size after $t$ hours is $N = 500 \times 1.8^t$.(a) Find the population after 5 hours.(b) Find the time, in hours, for the population to reach $100000$. Give your answer to 3 significant figures.(c) A second colony has population $M = 20000 \times 0.7^t$. Use logarithms or your GDC to find the time when the two populations are equal.

Show answer
(a) $500 \times 1.8^5 = 9447.84 \approx 9450$(b) $1.8^t = 200 \Rightarrow t = \frac{\ln 200}{\ln 1.8} = 9.01$ hours(c) $500 \times 1.8^t = 20000 \times 0.7^t \Rightarrow \left(\frac{1.8}{0.7}\right)^t = 40 \Rightarrow t = \frac{\ln 40}{\ln(18/7)} = 3.91$ hours
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