IB MYP Standard Math Mock Tests 2 – 2026 Edition

To find the prime factors of 18, we factorize it:

18 ÷ 2 = 9

9 ÷ 3 = 3

3 ÷ 3 = 1

So, 18 = 2 × 3². The distinct prime factors are 2 and 3.

In set notation, the prime factors of 18 are represented as:

{2, 3}

Alternatively, using set-builder notation: {p: p is a prime and p divides 18} = {2, 3}

Set \(S\): Contains all factors of 12 and 18.

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 18: 1, 2, 3, 6, 9, 18

Union of factors: {1, 2, 3, 4, 6, 9, 12, 18}

Given \(S = \{1, 2, 3, a, b, 9, 12, 18\}\), compare with the full set:

Missing factor is 4 and 6 (since 1, 2, 3, 9, 12, 18 are present). However, the set has 8 elements, suggesting \(a\) and \(b\) replace duplicates or missing values. Testing: \(a = 4\), \(b = 6\) fits all factors except order.

Correct \(S = \{1, 2, 3, 4, 6, 9, 12, 18\}\), so \(a = 4\), \(b = 6\).

Set \(T\): Primes greater than 10 and less than 15.

Primes between 10 and 15: 11, 13

So, \(T = \{11, 13\}\), thus \(c = 11\), \(d = 13\).

Final Values: \(a = 4\), \(b = 6\), \(c = 11\), \(d = 13\).

Using \(S = \{1, 2, 3, 4, 6, 9, 12, 18\}\) and \(T = \{11, 13\}\) from 1b:

Step 1: Identify elements

Total elements in \(S\): 8

Two-digit numbers in \(S\): 12, 18 (2 numbers)

Total elements in \(T\): 2

One-digit numbers in \(T\): None (11 and 13 are two-digit)

Step 2: Calculate probabilities

Probability of a two-digit number from \(S\): \( \frac{\text{Number of two-digit numbers}}{\text{Total in } S} = \frac{2}{8} = \frac{1}{4} \)

Probability of a one-digit number from \(T\): \( \frac{0}{2} = 0 \)

Step 3: Interpret ‘or’

Since one number is chosen from \(S\) and one from \(T\), we consider the union of events (mutually exclusive here):

P(two-digit from \(S\)) or P(one-digit from \(T\)) = \( \frac{1}{4} + 0 = \frac{1}{4} \)

Correction Note: The original answer of \(\frac{1}{2}\) assumed 4 two-digit numbers, but only 12 and 18 are two-digit in the corrected \(S\). Thus, the correct probability is \(\frac{1}{4}\).