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Consumer Mathematics: Insurance · Form 5

Consumer Mathematics: Insurance: Worked Examples (KBAT)

Real-situation, multi-step problems where the technique is hidden: an under-insured shop claim that needs the average clause then an excess, comparing two policies by total out-of-pocket cost, and a home policy combining building and contents premiums with a discount and SST before a storm payout. General information only, not financial or insurance advice.

Written by the spmmath.com.my editorial team.

Worked example 1

A shop building is valued at RM500,000 but the owner insured it for RM400,000. A fire causes a loss of RM120,000.

The policy applies the average clause and also carries an excess of RM2,000. How much is the final payout?

  1. Value RM500,000 but insured RM400,000 → under-insured, so the average clause applies.
  2. Step 1, apply average: (400,000 ÷ 500,000) × 120,000 = 0.8 × 120,000 = RM96,000.
  3. Step 2, subtract the excess: 96,000 − 2,000 = RM94,000.

Worked example 2

For a sum insured of RM100,000, Policy A charges RM3.00 per RM1,000 with an RM300 excess, while Policy B charges RM2.50 per RM1,000 with an RM800 excess. Over one year with a single claim of RM5,000, which policy gives the lower total out-of-pocket cost (premium plus the excess borne), and by how much?

  1. Premium A = (100,000 ÷ 1,000) × 3.00 = 100 × 3.00 = RM300; Premium B = 100 × 2.50 = RM250.
  2. With one claim, out-of-pocket = premium + the excess the owner bears on that claim.
  3. Policy A total = 300 + 300 = RM600; Policy B total = 250 + 800 = RM1,050.
  4. Compare: RM600 vs RM1,050, so Policy A is lower by 1,050 − 600 = RM450.

Worked example 3

A family takes a home policy: building sum insured RM180,000 at RM2.20 per RM1,000, and contents sum insured RM40,000 at RM4.00 per RM1,000. The insurer gives a 15% discount on the combined premium, then 8% SST is added.

(a) Find the total premium payable, rounded to the nearest sen. (b) Later a storm causes RM25,000 damage to the fully-insured building, with an excess of RM1,000.

Find the payout.

  1. Building premium = (180,000 ÷ 1,000) × 2.20 = 180 × 2.20 = RM396.
  2. Contents premium = (40,000 ÷ 1,000) × 4.00 = 40 × 4.00 = RM160.
  3. Subtotal = 396 + 160 = RM556.
  4. Apply 15% discount: 556 × (1 − 0.15) = 556 × 0.85 = RM472.60.
  5. Add 8% SST: 472.60 × 1.08 = RM510.408 ≈ RM510.41.
  6. (b) Building is fully insured, so no average clause; payout = 25,000 − 1,000 = RM24,000.

Worked example 4

Mr Lim's building is worth RM400000. Option A: insure it fully for RM400000.

Option B: insure it for only RM300000. Both are at RM1.80 per RM1000 and have an average clause.

If a fire causes RM80000 of damage, compare the total cost (premium + uninsured loss) of each option and state which is cheaper.

  1. Option A premium = (400000 ÷ 1000) × 1.80 = 400 × 1.80 = RM720
  2. Option A: fully insured, so the RM80000 loss is paid in full; out-of-pocket loss = RM0; total cost = RM720
  3. Option B premium = (300000 ÷ 1000) × 1.80 = 300 × 1.80 = RM540
  4. Option B payout (average clause) = (300000 ÷ 400000) × 80000 = 0.75 × 80000 = RM60000
  5. Option B out-of-pocket loss = 80000 − 60000 = RM20000; total cost = 540 + 20000 = RM20540
  6. Difference = 20540 − 720 = RM19820

Worked example 5

A shop building worth RM360000 is insured for RM270000, and its stock worth RM80000 is insured fully for RM80000. A fire damages the building by RM90000 and the stock by RM30000.

The average clause applies to each item separately, and a single excess of RM2000 applies to the whole claim. Calculate the total insurer payout.

  1. Building payout (average clause) = (270000 ÷ 360000) × 90000 = 0.75 × 90000 = RM67500
  2. Stock is fully insured, so its payout = RM30000 (no averaging)
  3. Subtotal = 67500 + 30000 = RM97500
  4. Less single excess RM2000: 97500 − 2000 = RM95500

Worked example 6

A medical card charges an annual deductible of RM500 (once a year), then 10% co-insurance on the balance, with an annual payout limit of RM50000. In one year the member is admitted twice, with bills of RM12000 and RM6000.

Find the total the insurer pays and the total the member pays.

  1. Total bills = 12000 + 6000 = RM18000
  2. Annual deductible applied once = RM500; balance = 18000 − 500 = RM17500
  3. Member co-insurance 10% = 0.10 × 17500 = RM1750
  4. Insurer pays 90% = 0.90 × 17500 = RM15750 (below the RM50000 limit)
  5. Member total = 500 + 1750 = RM2250

Worked example 7

Encik Chong's car is insured for a sum insured of RM60,000. The basic premium rate is 5% of the sum insured.

He qualifies for a No-Claim Discount (NCD) of 25% because he made no claims last year. This year he is involved in an accident; repair costs come to RM4,500, and his policy has an excess of RM300.

(a) Find the premium he pays after the NCD discount. (b) Find how much the insurer pays for the repair.

  1. Basic premium = 5% × RM60,000 = RM3,000
  2. NCD discount = 25% × RM3,000 = RM750
  3. Premium payable = RM3,000 − RM750 = RM2,250
  4. Insurer's payout = RM4,500 − RM300 (excess) = RM4,200

Worked example 8

A retailer's warehouse stock is worth RM240,000. He insures it for only RM180,000, at a premium rate of RM2.50 per RM1,000, with an excess of RM1,500 on each claim.

During the year there are two separate claims: a break-in causing RM30,000 of loss, and a fire causing RM50,000 of loss. The average clause applies to each claim separately.

(a) Find the annual premium. (b) Find the total amount the insurer pays for the two claims combined.

  1. Annual premium = (RM180,000 ÷ RM1,000) × RM2.50 = 180 × RM2.50 = RM450
  2. Fraction insured = RM180,000 ÷ RM240,000 = 0.75
  3. Break-in claim: 0.75 × RM30,000 = RM22,500, minus excess RM1,500 = RM21,000
  4. Fire claim: 0.75 × RM50,000 = RM37,500, minus excess RM1,500 = RM36,000
  5. Total insurer payout = RM21,000 + RM36,000 = RM57,000

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

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Frequently asked questions

What kind of scenario appears in insurance KBAT questions?

KBAT insurance questions typically present a family or individual's real situation, multiple assets, a budget limit, or changing needs, and ask you to decide on suitable coverage, calculate costs under given constraints, and justify the decision with figures. You must extract only the relevant information from a longer scenario.

What is the biggest challenge in solving these questions?

The hardest part is usually filtering out irrelevant details in the scenario and identifying exactly which values (premium rates, budget, sum insured needed) apply to the calculation. Students who rush past the wording often use the wrong figures, even if their formula and arithmetic are otherwise correct.

What separates a full-mark answer from a partial one at this level?

Full marks come from a complete chain: correct extraction of data, accurate calculation, and a clear final decision or recommendation that directly answers what the question asked (not just a number). Leaving out the concluding justification, even with correct working, is one of the most common reasons for lost marks.

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