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Chapter 3 Practice Paper: Approximation and Estimation — mark scheme

QuizQuipPractice Series
QQ/S1MATH/CH3

Secondary 1 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Approximation and Estimation

Chapter 3 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer
1CThe digit after the first decimal place is 6 (at least 5), so 27.463 rounds up to 27.5.1
2AThe first two significant figures are 3 and 8. The next digit is 4, so the 8 stays: 3 800.1
3BZeros before the 7 are placeholders. The first significant figure is 7, and the next digit is 1, so it stays: 0.07.1
4DThe decimal part 0.5 rounds up, so 99.5 becomes 100.1
5BRound each number to 1 significant figure: 40 × 6 = 240.1
6AThe second decimal digit is 6, so the first decimal digit rounds up from 8 to 9.1
7DThe first three significant figures are 1, 9, 9 and the next digit is 9, so the third figure rounds up and carrying gives 20 000.1
8CNumbers from 335 up to but not including 345 round to 340 at 2 significant figures. The value 344 lies in that range; 334, 345 and 349.6 do not (345 itself rounds up to 350).1
9BTo 1 significant figure, 79 rounds to 80 and 1.9 rounds to 2, and 80 ÷ 2 = 40.1
10AThe first significant figure is 1, and the next digit is 4, so the 1 stays: 100 000.1

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
1(a)

2.36 [1]

1
1(b)

2.36 [1]

The answers to (a) and (b) agree here; the reasoning in the mark scheme differs, and either statement of method is accepted.

1
1(c)

46 000 [1]

1
1(d)

The whole-number part is 2, and rounding depends on the tenths digit. The tenths digit is 3, which is less than 5, so the number stays at 2. Only numbers from 2.5 upwards round to 3.

  • Identifies the deciding digit (the 3 in the tenths place) [1]
  • States 3 is less than 5, so no rounding up [1]
2
Total for Question 15
2(a)

50 × 20 = 1000

  • Rounds to 50 and 20 [1]
  • 1000 [1]
2
2(b)

(600 × 9) ÷ 30 = 5400 ÷ 30 = 180

  • Rounds 597, 9.1 and 31 to 600, 9 and 30 [1]
  • Numerator 5400 seen [1]
  • 180 [1]
3
Total for Question 25
3(a)

5000 ÷ 20 = 250 boxes

  • Rounds to 5000 and 20 [1]
  • 250 [1]
2
3(b)

4870 ÷ 24 = 202 remainder 22, so 202 boxes hold 4848 pens and 22 pens are left over. Every pen must be packed, so one more box is needed: 203 boxes. The answer must be rounded up, because rounding down would leave pens unpacked.

  • Division with remainder seen, or 202.9 noted [1]
  • Identifies that a partial box still needs a full box [1]
  • 203 boxes [1]

Do not accept 203 without the rounding-up reason; "round up in context" or equivalent is required.

3
Total for Question 35
4(a)

Yes. 12.64 has 6 in the hundredths place, and 12.64 rounds to 12.6 to 1 decimal place.

  • Yes, because 12.64 rounds to 12.6 [1]
1
4(b)

No. 12.71 rounds to 12.7, not 12.6.

  • No, because 12.71 rounds to 12.7 [1]
1
4(c)

Smallest 12.55 kg; largest: any mass up to but not including 12.65 kg. So the actual mass is at least 12.55 kg and less than 12.65 kg.

  • Lower value 12.55 [1]
  • Upper value 12.65, stated as not included or with the < symbol [1]
2
4(d)

12.6 kg

  • 12.6 kg (rounds half up) [1]
1
Total for Question 45

— End of mark scheme —