Past Paper Example 1
AQA Nov 2022 3H Q14
5 marks
Here are two solid cubes, X and Y.
The mass of X is 10.976 kg
The area of **each face** of X is 784 cm2
The mass of X is 10.976 kg
The area of **each face** of X is 784 cm2
(a) Zayan wants to know the density of Y.
He assumes that Y is identical to X.
What density should he get for Y?
Give your answer in **grams per cubic centimetre**. [4 marks]
He assumes that Y is identical to X.
What density should he get for Y?
Give your answer in **grams per cubic centimetre**. [4 marks]
(b) In fact,
the mass of Y is less than the mass of X
the area of each face of Y is greater than the area of each face of X.
What does this mean about the actual density of Y?
Tick **one** box.
[ ] It is less than the answer to part (a)
[ ] It is equal to the answer to part (a)
[ ] It is greater than the answer to part (a)
[ ] It is not possible to tell [1 mark]
the mass of Y is less than the mass of X
the area of each face of Y is greater than the area of each face of X.
What does this mean about the actual density of Y?
Tick **one** box.
[ ] It is less than the answer to part (a)
[ ] It is equal to the answer to part (a)
[ ] It is greater than the answer to part (a)
[ ] It is not possible to tell [1 mark]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
(Density = mass ÷ volume)
Side length of X = √784 = 28 cm M1
Volume of X = 28³ = 21 952 cm³ M1
Mass in grams = 10.976 × 1000 = 10 976 g M1
Density = 10 976 ÷ 21 952 = 0.5 g/cm³ A1
Side length of X = √784 = 28 cm M1
Volume of X = 28³ = 21 952 cm³ M1
Mass in grams = 10.976 × 1000 = 10 976 g M1
Density = 10 976 ÷ 21 952 = 0.5 g/cm³ A1
Answer (a): Answer 0.5 g/cm³
Part (b) Solution:
(Density = mass ÷ volume)
Smaller mass divided by a larger volume gives a smaller density B1
Smaller mass divided by a larger volume gives a smaller density B1
Answer (b): It is less than the answer to part (a)