Practice Test Geeks(NBT) National Benchmark Test Practice Test
← All NBT Flashcard Decks

Quantitative Literacy Flashcards

6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Quantitative Literacy flashcards as text

  1. Thabo has R10 000 to invest for exactly 3 years. Bank A offers 9% simple interest per year. Bank B offers 8% interest per year, compounded annually. Which option gives the larger final amount, and by how much?

    Answer: Bank A, by R102.88

    Bank A (simple): interest = 10000 × 0.09 × 3 = R2 700, so final = R12 700. Bank B (compound): 10000 × 1.08^3 = 10000 × 1.259712 = R12 597.12. Bank A is larger by 12 700 − 12 597.12 = R102.88. The lower compound rate does not overtake the higher simple rate over just 3 years.

  2. A retailer increases the price of a jacket by 20%, then later advertises '20% OFF' on the increased price. Compared to the original price, the final price is:

    Answer: 4% lower

    Apply the factors in sequence to an original price P: after +20% it is 1.20P; after −20% it is 1.20P × 0.80 = 0.96P. That is a 4% decrease. Successive percentage changes multiply rather than cancel, so a 20% rise followed by a 20% fall does not return to the start.

  3. A shop buys a kettle for R400. It applies a 25% markup on cost, then offers a 10% discount off that marked-up price. Finally, 15% VAT is added to the discounted price. What does the customer pay?

    Answer: R517.50

    Markup: 400 × 1.25 = R500. Discount: 500 × 0.90 = R450. VAT: 450 × 1.15 = R517.50. Each step multiplies the running total, and VAT is applied last to the already-discounted price.

  4. A survey of 200 learners recorded gender and whether they passed a test. Results: 60 males passed, 90 females passed, 30 males failed, 20 females failed. Given that a randomly chosen learner FAILED, what is the probability that the learner is female?

    Answer: 40%

    The condition 'failed' restricts the sample space to failers only: 30 + 20 = 50 learners. Of these, 20 are female. So P(female | failed) = 20/50 = 0.40 = 40%. Common errors: dividing by 110 (total females), by 200 (grand total), or by 90 (total who passed).

  5. A machine lays fibre-optic cable at a steady speed of 54 km/h. The cable costs R2 per metre laid. Assuming continuous operation, what is the total cost of cable laid in exactly one minute?

    Answer: R1 800

    Convert speed: 54 km/h = 54 × 1000 ÷ 3600 = 15 m/s. In one minute (60 s) the machine lays 15 × 60 = 900 m. Cost = 900 × R2 = R1 800. R900 forgets the R2/m rate; R3 600 doubles the speed in conversion; R108 mishandles unit conversion entirely.

  6. A course mark uses Test 1 (weight 40%), Test 2 (weight 40%), and a Project (weight 30%) — note the weights sum to 110%. The school normalises by dividing the weighted total by the sum of weights. A learner scores 60, 70, and 80 respectively. What is her final course mark (to one decimal place)?

    Answer: 69.1%

    Weighted total = 0.40×60 + 0.40×70 + 0.30×80 = 24 + 28 + 24 = 76. Since the weights sum to 1.10, normalise: 76 ÷ 1.10 ≈ 69.1%. Choosing 76.0% forgets to divide by the total weight; 70.0% is the simple (unweighted) average; 63.3% divides by the wrong figure.