Approximation: dp, sf, bounds, percentage errors, estimation practice questions.

Round 0.037249 to 2 significant figures.

• 0.04
• 0.037
• 0.0372
• 0.03

Answer: 0.037

Leading zeros are not significant, so the first two significant figures are 3 and 7; the next digit 2 means we round down to 0.037.

A length is recorded as 12.4 cm, correct to the nearest 0.1 cm. State the upper bound of the length.

• 12.45 cm
• 12.44 cm
• 12.5 cm
• 12.49 cm

Answer: 12.45 cm

The upper bound is the recorded value plus half the rounding unit: 12.4 + 0.05 = 12.45 cm. Values like 12.44 or 12.49 are not true bounds.

Using 3.14 as an approximation for π, the percentage error, correct to 3 s.f., is:

• 1.59%
• 0.507%
• 0.0507%
• 0.00051%

Answer: 0.0507%

Percentage error = |3.14 − π|/π × 100 = 0.0015927/3.14159 × 100 ≈ 0.0507%. Dividing by the exact value and multiplying by 100 are both essential.

Which of these numbers rounds to 6.0 when written correct to 2 significant figures?

• 6.05
• 5.94
• 6.06
• 5.96

Answer: 5.96

Numbers from 5.95 up to (but not including) 6.05 round to 6.0, so 5.96 works; 6.05 rounds up to 6.1 and 5.94 rounds to 5.9.

By rounding each number to 1 significant figure, estimate the value of (4.87 × 21.2) ÷ 0.492.

• 200
• 100
• 2000
• 20

Answer: 200

Rounding gives (5 × 20) ÷ 0.5 = 100 ÷ 0.5 = 200, close to the true value of about 210. Dividing by 0.5 doubles a number rather than halving it.

x = 4.3 and y = 2.6, each correct to 1 decimal place. The maximum possible value of x − y is:

• 1.70
• 1.60
• 1.80
• 1.75

Answer: 1.80

To maximise a difference, take the upper bound of x and the lower bound of y: 4.35 − 2.55 = 1.80. Using both upper bounds (4.35 − 2.65 = 1.70) is the common error.