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Mathematical Modeling · Form 5

Mathematical Modeling: Worked Examples (KBAT)

This set gives full real situations where the model is hidden inside the story: profit built from income minus costs, a border around a photo, and a price that changes demand. Students must set up the model themselves, solve it, and check the answer fits the situation.

Written by the spmmath.com.my editorial team.

Worked example 1

A committee holds a fund-raising dinner. The hall rent is a fixed RM800, and each guest's meal costs RM25.

Tickets are sold at RM45 each. Form a model for the profit RM P from n tickets, then find (a) the number of tickets to break even, and (b) the number needed for a RM1000 profit.

  1. Income from tickets = 45n.
  2. Total cost = fixed rent + meals = 800 + 25n.
  3. Profit = income − cost: P = 45n − (800 + 25n) = 20n − 800.
  4. (a) Break even means P = 0: 20n = 800, so n = 40.
  5. (b) For P = 1000: 20n − 800 = 1000, so 20n = 1800, giving n = 90.

Worked example 2

A photo measuring 12 cm by 8 cm is mounted on a rectangular card so that there is a border of uniform width x cm all around it. The whole card has an area of 192 cm².

Form a quadratic model and find the width of the border.

  1. The border adds x to each side, so card length = 12 + 2x and card width = 8 + 2x.
  2. Area model: (12 + 2x)(8 + 2x) = 192.
  3. Expand: 96 + 40x + 4x² = 192.
  4. Rearrange: 4x² + 40x − 96 = 0, then divide by 4: x² + 10x − 24 = 0.
  5. Factorise: (x + 12)(x − 2) = 0, so x = −12 or x = 2; reject x = −12.
  6. Check: (12 + 4)(8 + 4) = 16 × 12 = 192. ✓

Worked example 3

A homestay charges RM100 a night and gets 20 bookings a month. The owner finds that every RM10 increase in the nightly price loses 1 booking.

Form a model for the monthly revenue RM R in terms of x, the number of RM10 increases, then find the price that gives a revenue of RM2250.

  1. New price = 100 + 10x, and new bookings = 20 − x.
  2. Revenue = price × bookings: R = (100 + 10x)(20 − x).
  3. Expand: R = 2000 + 100x − 10x².
  4. Set R = 2250: 2000 + 100x − 10x² = 2250.
  5. Rearrange: −10x² + 100x − 250 = 0, then divide by −10: x² − 10x + 25 = 0.
  6. Factorise: (x − 5)² = 0, so x = 5.
  7. Price = 100 + 10(5) = RM150, with 20 − 5 = 15 bookings; 150 × 15 = 2250. ✓

Worked example 4

A theatre currently sells tickets at RM30 each and sells 200 tickets per show. For every RM2 increase in the ticket price, 10 fewer tickets are sold.

Let x be the number of RM2 increases. (a) Form a model for the revenue R (in RM) in terms of x.

(b) Find the ticket price(s) that give a revenue of RM6080.

  1. (a) New price = (30 + 2x) and tickets sold = (200 − 10x), so R = (30 + 2x)(200 − 10x).
  2. Expand: R = 6000 − 300x + 400x − 20x² = 6000 + 100x − 20x².
  3. (b) Set R = 6080: 6000 + 100x − 20x² = 6080 → −20x² + 100x − 80 = 0 → x² − 5x + 4 = 0.
  4. Factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4; prices = 30 + 2(1) = RM32 or 30 + 2(4) = RM38.

Worked example 5

A farmer uses 60 m of fencing to enclose a rectangular pen against a long straight wall. The wall forms one long side, so fencing is only needed for the two widths and one length.

Let the width be x m. (a) Form a model for the area A (in m²) in terms of x.

(b) Find the value of x that gives the maximum area, and state that maximum area.

  1. (a) The fencing covers two widths and one length: 2x + length = 60, so length = 60 − 2x.
  2. Area A = x(60 − 2x) = 60x − 2x².
  3. (b) This is a quadratic (a = −2) with a maximum at x = −b/(2a) = −60/(2 × −2) = 15.
  4. Length = 60 − 2(15) = 30 m, so maximum area = 15 × 30 = 450 m².

Worked example 6

An open box is made from a rectangular card measuring 20 cm by 16 cm by cutting an equal square of side x cm from each corner and folding up the sides. (a) Form a model for the volume V (in cm³) in terms of x.

(b) Find the volume when x = 3 cm, and state the range of values of x for which the box can be made.

  1. (a) After cutting corners of side x, the base measures (20 − 2x) by (16 − 2x) and the height is x.
  2. Volume V = x(20 − 2x)(16 − 2x).
  3. (b) Substitute x = 3: V = 3(20 − 6)(16 − 6) = 3 × 14 × 10 = 420 cm³.
  4. The box exists only when every dimension is positive: x > 0 and 16 − 2x > 0 (the smaller side), so 0 < x < 8.

Worked example 7

A zoo builds two identical adjacent rectangular enclosures that share one common dividing fence, using a total of 120 m of fencing. Let y m be the height of each enclosure (the length of each of the three parallel fences) and let L m be the total length (the two outer fences).

(a) Form a model for the total enclosed area A (m²) in terms of y. (b) Find the value(s) of y for which the total area is 450 m².

  1. Total fencing: 3y + 2L = 120, so L = (120−3y)/2 = 60 − 1.5y.
  2. Area A = L × y = (60−1.5y)y = 60y − 1.5y².
  3. Set A = 450: 60y − 1.5y² = 450 → multiply by −2/3: y² − 40y + 300 = 0.
  4. Factorise: (y−10)(y−30) = 0, so y = 10 or y = 30.
  5. Check: y=10 gives L=45 m (a long, narrow layout); y=30 gives L=15 m (a wide, shallow layout). Both are geometrically valid.

Worked example 8

A ball is thrown upward from the top of a 20 m tower. If it had been thrown from ground level, the height gained above the throwing point would be modelled by 25t − 5t² (metres) after t seconds.

(a) Form a model for h(t), the ball's height above the ground when thrown from the tower. (b) Find the height of the ball after 2 seconds.

(c) Find the time when the ball hits the ground, correct to 2 decimal places.

  1. Height above ground = tower height + height gained from the throw: h(t) = 20 + 25t − 5t².
  2. At t = 2: h(2) = 20 + 25(2) − 5(2)² = 20+50−20 = 50 m.
  3. Ball hits the ground when h(t) = 0: 20+25t−5t² = 0 → divide by −5: t² − 5t − 4 = 0.
  4. Use the quadratic formula: t = [5 ± √(25+16)]/2 = [5 ± √41]/2 = [5 ± 6.403]/2.
  5. Taking the positive root: t ≈ 11.403/2 ≈ 5.70 s (rejecting the negative root).

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What makes KBAT mathematical-modeling questions the hardest in this chapter?

KBAT items rarely tell you which equation to form, you must decide what to model, choose your own variables, and often optimise or evaluate multiple possible models against a real constraint (cost, time, capacity). There is usually no single obvious first step, so planning before writing anything matters more than at any other level.

What does the examiner reward in a KBAT modeling response?

A justified choice of variables and model, working that shows how each equation was derived from the given information, and a final evaluation or conclusion that answers the actual question asked (not just a number). Marks reward the reasoning trail as much as any correct value produced.

What common pitfalls affect strong students on KBAT modeling questions?

Building a mathematically correct model that doesn't actually address what the question asked, forgetting to state assumptions the model depends on, and stopping once an equation is solved without interpreting the result back in the real-world context. Re-reading the question after solving, to check the answer actually responds to it, catches most of these.

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