Jackson is trying to find the density, in g/cm³, of a block of wood. The block of wood is in the shape of a cuboid. He measures the length as 13.2 cm, correct to the nearest mm the width as 16.0 cm, correct to the nearest mm the height as 21.7 cm, correct to the nearest mm He measures the mass as 1970 g, correct to the nearest 5 g. By considering bounds, work out the density of the wood. Give your answer to a suitable degree of accuracy. You must show all your working and give a reason for your final answer.

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Jackson is trying to find the density, in g/cm³, of a block of wood - Edexcel - GCSE Maths - Question 1 - 2018 - Paper 3

Question 1

Jackson is trying to find the density, in g/cm³, of a block of wood. The block of wood is in the shape of a cuboid. He measures the length as 13.2 cm, correct to t... show full transcript

Worked Solution & Example Answer:Jackson is trying to find the density, in g/cm³, of a block of wood - Edexcel - GCSE Maths - Question 1 - 2018 - Paper 3

Step 1

Find bounds for the dimensions

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Answer

For the length, 13.2 cm, the bounds are:

Lower bound: 13.2 - 0.05 = 13.15 cm
Upper bound: 13.2 + 0.05 = 13.25 cm

For the width, 16.0 cm, the bounds are:

Lower bound: 16.0 - 0.05 = 15.95 cm
Upper bound: 16.0 + 0.05 = 16.05 cm

For the height, 21.7 cm, the bounds are:

Lower bound: 21.7 - 0.05 = 21.65 cm
Upper bound: 21.7 + 0.05 = 21.75 cm

Step 2

Find the volume using the bounds

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Answer

The volume of a cuboid is given by the formula: $V = ext{length} imes ext{width} imes ext{height}$ Using the upper and lower bounds:

Lower bound volume: $V_{min} = 13.15 imes 15.95 imes 21.65 = 4252.339875$ \
Upper bound volume: $V_{max} = 13.25 imes 16.05 imes 21.75 = 4270.303125$

Therefore, the volume bounds are from 4252.34 cm³ to 4270.30 cm³.

Step 3

Find bounds for the mass

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Answer

The mass is measured as 1970 g, correct to the nearest 5 g:

Lower bound: 1970 - 2.5 = 1967.5 g
Upper bound: 1970 + 2.5 = 1972.5 g

Step 4

Calculate density using bounds

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Answer

Density is given by the formula:

ho = rac{m}{V}$$ For the minimum density:

ho_{max} = rac{1970}{4270.30} = 0.461 $$ g/cm³

For the maximum density:

ho_{min} = rac{1967.5}{4252.34} = 0.462 $$ g/cm³ Thus, the density bounds are 0.461 g/cm³ to 0.462 g/cm³.

Step 5

Present density with suitable accuracy

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Answer

Given the bounds established, the density of the wood can be expressed to two significant figures as:

Density = 0.46 g/cm³

This reflects the uncertainty in the measurements and presents an appropriate degree of accuracy.

Jackson is trying to find the density, in g/cm³, of a block of wood - Edexcel - GCSE Maths - Question 1 - 2018 - Paper 3

Worked Solution & Example Answer:Jackson is trying to find the density, in g/cm³, of a block of wood - Edexcel - GCSE Maths - Question 1 - 2018 - Paper 3

Find bounds for the dimensions

Find the volume using the bounds

Find bounds for the mass

Calculate density using bounds

Present density with suitable accuracy

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