IB DP Math AI HL Prediction Paper 1 for 2025 Exams
IB DP Math AI HL Prediction Paper 1 – April/May 2025 Exam
IB DP Math AI HL Prediction Paper 1: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
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Question 1: Financial Applications of Geometric Sequences and Series [15 marks]
Imani invests $3000 in a bank that pays a nominal annual interest rate of 1.25% compounded monthly.
a Question 1a [4 marks] – Future Value of Investment
Calculate the amount of money Imani will have in the bank at the end of 6 years:
Show Solution
Using the compound interest formula
\( FV = PV \left(1 + \frac{r}{n} \right)^{nt} \)
Substituting \( PV = 3000 \), \( r = 0.0125 \), \( n = 12 \), \( t = 6 \)
\( FV = 3233.53 \)
Detailed Solution:
- Formula: \( FV = PV \left(1 + \frac{r}{n} \right)^{nt} \).
- Given:
- \( PV = 3000 \) (initial investment).
- \( r = 1.25\% = 0.0125 \) (annual interest rate).
- \( n = 12 \) (compounded monthly).
- \( t = 6 \) years.
- Calculate:
- \( FV = 3000 \left(1 + \frac{0.0125}{12} \right)^{12 \times 6} \).
- \( FV = 3000 \left(1.00104167 \right)^{72} \).
- \( FV = 3233.53 \).
- Result: The amount is $3233.53.
b Question 1b [4 marks] – Time to Reach $3550
Calculate the number of months it takes until Imani has at least $3550 in the bank:
Show Solution
Using the compound interest formula
\( 3550 = 3000 \left(1 + \frac{0.0125}{12} \right)^{12t} \)
Solving for \( t \)
\( t \approx 162 \) months
Detailed Solution:
- Formula: \( FV = PV \left(1 + \frac{r}{n} \right)^{nt} \).
- Given: \( FV = 3550 \), \( PV = 3000 \), \( r = 0.0125 \), \( n = 12 \).
- Setup: \( 3550 = 3000 \left(1 + \frac{0.0125}{12} \right)^{12t} \).
- Simplify: \( \frac{3550}{3000} = \left(1.00104167 \right)^{12t} \).
- Logarithms:
- \( \log \left( \frac{3550}{3000} \right) = 12t \log \left(1.00104167 \right) \).
- \( t = \frac{\log \left( \frac{3550}{3000} \right)}{12 \log \left(1.00104167 \right)} \).
- \( t \approx 13.4738 \) years.
- Convert to months: \( 13.4738 \times 12 \approx 161.686 \).
- Result: Since it’s “at least” $3550, round up to 162 months.
c Question 1c [2 marks] – Loan Amount
Imani uses the \$3550 as a partial payment for a used car costing \$22000. Write down the amount of money that Imani takes out as a loan:
Show Solution
Subtracting partial payment from car cost
\( \text{Loan} = 22000 – 3550 \)
\( \text{Loan} = 18450 \)
Detailed Solution:
- Car cost: $22000.
- Partial payment: $3550.
- Calculate: \( 22000 – 3550 = 18450 \).
- Result: The loan amount is $18450.
d Question 1d [5 marks] – Monthly Loan Payment
The loan is for 8 years and the nominal annual interest rate is 12.6% compounded monthly. Imani will pay the loan in fixed monthly installments at the end of each month. Calculate the amount, correct to the nearest dollar, that Imani will have to pay the bank each month:
Show Solution
Using the loan payment formula
\( PMT = \frac{PV \times \frac{r}{n}}{1 – (1 + \frac{r}{n})^{-nt}} \)
Substituting \( PV = 18450 \), \( r = 0.126 \), \( n = 12 \), \( t = 8 \)
\( PMT = 306 \)
Detailed Solution:
- Formula: \( PMT = \frac{PV \times \frac{r}{n}}{1 – (1 + \frac{r}{n})^{-nt}} \).
- Given:
- \( PV = 18450 \) (loan amount).
- \( r = 12.6\% = 0.126 \) (annual interest rate).
- \( n = 12 \) (monthly compounding).
- \( t = 8 \) years.
- Calculate:
- Monthly rate: \( \frac{0.126}{12} = 0.0105 \).
- Number of payments: \( 12 \times 8 = 96 \).
- Numerator: \( 18450 \times 0.0105 \).
- Denominator: \( 1 – (1 + 0.0105)^{-96} \).
- \( PMT = \frac{18450 \times 0.0105}{1 – (1.0105)^{-96}} \).
- \( PMT \approx 306 \) (rounded to the nearest dollar).
- Result: The monthly payment is $306.
Syllabus Reference
Syllabus: Mathematics: Applications and Interpretation
Unit 1: Number and Algebra
- Geometric sequences and series
- Financial applications
Assessment Criteria: D (Applying mathematics in real-life contexts)
Question 2: Test for Proportion Using Binomial Distribution [10 marks]
Jerry makes handcrafted chocolates. On average, 1 in 25 of the chocolates that Jerry makes is flawed. Whether or not a chocolate is flawed is independent of all other chocolates.
a Question 2a [6 marks] – Probability of Flawed Chocolates
In a batch of 20 chocolates, chosen at random, find the probability that:
(i) Two are flawed:
(ii) More than two are flawed:
Show Solution
Recognizing binomial distribution \( M \sim B(20, 0.04) \)
(i) \( P(M=2) = \binom{20}{2} (0.04)^2 (0.96)^{18} = 0.146 \)
(ii) \( P(M > 2) = 1 – P(M \leq 2) = 0.0439 \)
Detailed Solution:
- Distribution: Number of flawed chocolates follows \( M \sim B(20, 0.04) \), where:
- \( n = 20 \) (total chocolates).
- \( p = \frac{1}{25} = 0.04 \) (probability of a flawed chocolate).
- (i) Exactly two flawed:
- Formula: \( P(M=k) = \binom{n}{k} p^k (1-p)^{n-k} \).
- Calculate: \( P(M=2) = \binom{20}{2} (0.04)^2 (0.96)^{18} \).
- \( \binom{20}{2} = 190 \), \( (0.04)^2 = 0.0016 \), \( (0.96)^{18} \approx 0.4798 \).
- \( P(M=2) = 190 \times 0.0016 \times 0.4798 \approx 0.145799 \).
- Result: \( 0.146 \) (3 significant figures).
- (ii) More than two flawed:
- Calculate: \( P(M > 2) = P(M \geq 3) = 1 – P(M \leq 2) \).
- \( P(M \leq 2) = P(M=0) + P(M=1) + P(M=2) \).
- Using cumulative probability: \( P(M \leq 2) = \text{bincdf}(20, 0.04, 2) \approx 0.956137 \).
- \( P(M \geq 3) = 1 – 0.956137 = 0.0438627 \).
- Result: \( 0.0439 \) (3 significant figures).
b Question 2b [4 marks] – Expected Revenue
Jerry sells the perfect chocolates for 50 pesos each and the flawed ones for 15 pesos each. Calculate the expected number of pesos Jerry makes from selling a batch of 20 randomly selected chocolates:
Show Solution
Expected flawed chocolates: \( E(M) = 20 \times 0.04 = 0.8 \)
Expected perfect chocolates: \( 20 – 0.8 = 19.2 \)
Expected revenue: \( 50 \times 19.2 + 15 \times 0.8 = 972 \) pesos
Detailed Solution:
- Expected flawed chocolates:
- For \( M \sim B(20, 0.04) \), \( E(M) = n \cdot p = 20 \times 0.04 = 0.8 \).
- Expected perfect chocolates:
- Total chocolates = 20, so \( E(P) = 20 – 0.8 = 19.2 \).
- Expected revenue:
- Perfect chocolates: 50 pesos each, so \( 50 \times 19.2 = 960 \).
- Flawed chocolates: 15 pesos each, so \( 15 \times 0.8 = 12 \).
- Total: \( 960 + 12 = 972 \) pesos.
- Result: Expected revenue is 972 pesos.
Syllabus Reference
Syllabus: Mathematics: Applications and Interpretation
Unit 4: Probability and Statistics
- Binomial distribution
- Expected value
Assessment Criteria: D (Applying mathematics in real-life contexts)
Question 3: Numerical Evaluation of Logarithms Using Technology [9 marks]
The pH scale is a measure of the acidity of a solution. Its value is given by the formula:
\( pH = -\log_{10}[H^+] \)
where \([H^+]\) is the concentration of hydrogen ions in the solution (measured in moles per litre).
a Question 3a [3 marks] – pH Value Calculation
Calculate the pH value if the concentration of hydrogen ions is 0.0003:
Show Solution
Using the formula \( pH = -\log_{10}[H^+] \)
Substituting \([H^+] = 0.0003\)
\( pH = 3.52 \)
Detailed Solution:
- Formula: \( pH = -\log_{10}[H^+] \).
- Given: \([H^+] = 0.0003\) moles per litre.
- Calculate:
- \( pH = -\log_{10}(0.0003) \).
- \( \log_{10}(0.0003) = \log_{10}(3 \times 10^{-4}) = \log_{10}(3) – 4 \approx 0.4771 – 4 = -3.5229 \).
- \( pH = -(-3.5229) = 3.5229 \).
- Rounded to 2 decimal places: \( 3.52 \).
- Result: The pH value is 3.52.
b Question 3b [3 marks] – Hydrogen Ion Concentration in Milk
The pH of milk is 6.6. Calculate the concentration of hydrogen ions in milk:
Show Solution
Using the formula \( pH = -\log_{10}[H^+] \)
Solving for \([H^+] = 10^{-pH}\)
Substituting \( pH = 6.6 \)
\( [H^+] = 2.51 \times 10^{-7} \)
Detailed Solution:
- Formula: \( pH = -\log_{10}[H^+] \), so \([H^+] = 10^{-pH} \).
- Given: \( pH = 6.6 \).
- Calculate:
- \([H^+] = 10^{-6.6} \).
- \( 10^{-6.6} = 10^{-6} \times 10^{-0.6} \approx 10^{-6} \times 0.25119 \).
- \([H^+] \approx 2.5119 \times 10^{-7} \).
- Rounded to 3 significant figures: \( 2.51 \times 10^{-7} \) moles per litre.
- Result: The hydrogen ion concentration is \( 2.51 \times 10^{-7} \) moles per litre.
c Question 3c [3 marks] – Acid Strength Comparison
The strength of an acid is measured by its concentration of hydrogen ions. A lemon has a pH value of 2 and a tomato has a pH value of 4.5. Calculate how many times stronger the acid in a lemon is when compared to the acid in a tomato:
Show Solution
Calculate \([H^+]\) for lemon: \( 10^{-2} = 0.01 \)
Calculate \([H^+]\) for tomato: \( 10^{-4.5} \approx 0.0000316227 \)
Ratio: \( \frac{0.01}{0.0000316227} \approx 316 \)
Detailed Solution:
- Formula: \([H^+] = 10^{-pH} \).
- Given:
- Lemon: \( pH = 2 \).
- Tomato: \( pH = 4.5 \).
- Calculate \([H^+]\):
- Lemon: \([H^+] = 10^{-2} = 0.01\) moles per litre.
- Tomato: \([H^+] = 10^{-4.5} \approx 10^{-4} \times 10^{-0.5} \approx 10^{-4} \times 0.316227 \approx 0.0000316227\).
- Ratio:
- Strength ratio = \( \frac{[H^+]_{\text{lemon}}}{[H^+]_{\text{tomato}}} = \frac{10^{-2}}{10^{-4.5}} \).
- \( = 10^{4.5 – 2} = 10^{2.5} \approx 316.227 \).
- Rounded to 3 significant figures: 316.
- Result: The acid in a lemon is approximately 316 times stronger than the acid in a tomato.
Syllabus Reference
Syllabus: Mathematics: Applications and Interpretation
Unit 1: Number and Algebra
- Logarithms and their properties
- Numerical evaluation using technology
Assessment Criteria: D (Applying mathematics in real-life contexts)