Home / IB Mathematics SL 1.8 Systems of linear equations AI HL Paper 1- Exam Style Questions

IB Mathematics SL 1.8 Systems of linear equations AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI SL Paper 1 - All Topics

Question

A hat manufacturer sells custom hats to businesses. The pricing policy is as follows:
  • For orders of 15 hats or fewer, each hat costs \( p \) euros (EUR).
  • For larger orders, the price per hat after the first 15 is reduced by 5 EUR.
The total cost of ordering \( n \) hats is denoted by \( C(n) \).
(a) Write an expression, in terms of \( p \), for the total cost of ordering exactly 15 hats.
(b) Write an equation for \( C(n) \), in terms of \( p \) and \( n \), for the total cost when \( n > 15 \).
A company places an order for 100 hats. The average price per hat for this order is 25.73 EUR.
(c) (i) Calculate the exact total cost of the order.
  (ii) Hence, determine the value of \( p \), correct to two decimal places.

Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL 2025):

• AHL 2.9: Piecewise models — parts (a), (b)
• SL 1.8: Use technology to solve polynomial equations — part (c)
▶️ Answer/Explanation

(a)
For \( n \leq 15 \), each hat costs \( p \) euros.
Total cost = \( 15p \).
\( \boxed{15p} \)

(b)
For \( n > 15 \):
– First 15 hats cost \( 15p \).
– The remaining \( n-15 \) hats each cost \( p-5 \) euros.
Thus \( C(n) = 15p + (p-5)(n-15) \).
(Alternative equivalent forms are acceptable.)
\( \boxed{C(n) = 15p + (p-5)(n-15)} \)

(c)

(i) Average price = total cost ÷ number of hats.
\( 25.73 = \frac{C(100)}{100} \) ⇒ \( C(100) = 100 \times 25.73 = 2573 \) EUR.
\( \boxed{2573 \, \text{EUR}} \)

(ii) Substitute \( n = 100 \) and total cost = 2573 into the formula from (b):
\( 2573 = 15p + (p-5)(100-15) \)
\( 2573 = 15p + (p-5)(85) \)
\( 2573 = 15p + 85p – 425 \)
\( 2573 = 100p – 425 \)
\( 100p = 2998 \)
\( p = 29.98 \)
\( \boxed{29.98 \, \text{EUR}} \) (to 2 d.p.)

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