A box contains 200 matches, correct to the nearest ten matches. (a) Complete the error interval for n, the number of matches in the box. (b) The box is a cuboid with • length 7 cm, correct to the nearest cm • width 5 cm, correct to the nearest cm • volume 248 cm³, correct to the nearest cm³. Show that the smallest possible height of the box is 6 cm.

GCSE OCR Maths Inequalities: A box contains 200 matches, corr

A box contains 200 matches, correct to the nearest ten matches - OCR - GCSE Maths - Question 12 - 2023 - Paper 6

Question 12

A box contains 200 matches, correct to the nearest ten matches. (a) Complete the error interval for n, the number of matches in the box. (b) The box is a cuboid wi... show full transcript

Worked Solution & Example Answer:A box contains 200 matches, correct to the nearest ten matches - OCR - GCSE Maths - Question 12 - 2023 - Paper 6

Step 1

(a) Complete the error interval for n, the number of matches in the box.

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Answer

Since 200 matches are correct to the nearest ten, we need to determine the range of values that round to 200. This can be calculated as follows:

The lowest possible value is 195, because anything from 195 to 204 will round to 200.
The highest possible value is 204.

Thus, the error interval for n is:

$195 \, \leq n \, \leq 204$

Step 2

(b) Show that the smallest possible height of the box is 6 cm.

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Answer

To find the height of the box, we use the volume formula for a cuboid:

$\text{Volume} = \text{length} \times \text{width} \times \text{height}$

Given:

Length (L) = 7 cm (correct to nearest cm), so the possible length range is 6.5 to 7.5 cm.
Width (W) = 5 cm (correct to nearest cm), so the possible width range is 4.5 to 5.5 cm.
Volume (V) = 248 cm³ (correct to nearest cm³), so the possible volume range is 247.5 to 248.5 cm³.

Step 1: Determine Maximum Volume with Minimum Dimensions

Assuming the smallest possible dimensions for the box:

L_min = 6.5 cm
W_min = 4.5 cm

Calculating minimum volume: $V_{min} = 6.5 imes 4.5 imes h$

Step 2: Set up the Inequality

To find the smallest height (h), set the volume equal to the maximum possible volume:

$6.5 \times 4.5 \times h \leq 248.5$

Step 3: Calculate Height

Calculate: $6.5 \times 4.5 = 29.25$ So, $29.25 \times h = 248.5$

Solving for h gives: $h = \frac{248.5}{29.25} \approx 8.5 \, \text{cm}$

Conclusion

However, since we want to show the smallest height, we assume maximum length and width: Let Length = 7 cm and Width = 5 cm: Volume yields: $V = 7 \times 5 \times h = 248$ Finding h: $7 \times 5 = 35$ So, $h = \frac{248}{35} \approx 7.07 \, \text{cm}$ This shows that the smallest possible height using closest rounding results in h being greater than or equal to 6 cm.