| Answer Key at the end of the table | |
| 1 | $\mathrm{( Edexcel/MathB/2015/june/paper02/q6) }$ Figure 2 shows a cylindrical can, of diameter 40 cm, open at its top. Water is poured into the can until the depth of the water in the can is 10 cm. (a) Calculate as a multiple of $\pi$, the volume, in cm$3$, of the water in the can. (2) 30 solid glass spheres, each of radius r cm, are placed in the can. The water completely covers all the spheres and no water overflows from the can. The depth of the water in the can is now 16.4 cm. (b) Find the exact value of r. (6) Area of circle = $\pi r^2$ Volume of sphere = $\frac43 \pi r^3$ |
| 2 | $\mathrm{( Edexcel/MathB/2015/jan/paper01/q24) }$ A rectangle has length 2 m and width 50 cm. Inside the rectangle are 300 identical triangles. Each triangle is isosceles, with sides of length 5 cm, 5 cm and 6 cm. Each triangle is completely within the rectangle and no triangle overlaps any other triangle. Express the total area covered by these 300 triangles as a percentage of the area of the rectangle (5) |
| 3 | $\mathrm{( Edexcel/MathB/2015/june/paper01/q18) }$
A, B and C are points on a circle with diameter AC. AB = 8 cm BC = 10 cm Calculate the area, in cm2 to 3 significant figures, of the circle. (Total for Question 18 is 4 marks) |
| 4 | $\mathrm{( Edexcel/MathB/2016/june/paper01/q8) }$ |
| 5 | $\mathrm{( Edexcel/MathB/2014/jan/paper01/q14) }$ The diagram shows a triangular prism ABCDEF of length 5 cm. AB = FE = 2 cm, BC = ED = 3 cm and $\angle$ABC = $\angle$FED = 30° Calculate the volume, in cm$^3$, of the prism. (3) |
| 6 | $\mathrm{( Edexcel/MathB/2014/janR/paper01/q23) }$ A layer of a microchip has a length of 4 mm, a width of 5 mm and a thickness of 1.8x10$^–4$ mm. (a) Calculate the volume, in mm$^3$, of one layer of the microchip. Give your answer in standard form. (3) (b) Calculate the number of these microchip layers needed to give a total thickness of 0.9 mm. (2) |
| 7 | $\mathrm{( Edexcel/MathB/2014/janR/paper02/q1) }$ Figure 1 shows a ring. The inner radius of the ring is 24 mm and the outer radius is 26 mm. Given that the ring is 4 mm thick, calculate the volume, in mm$^3$, of the ring. Give your answer to 3 significant figures. [Area of a circle = $\pi r^2$] (4) |
| 8 | $\mathrm{( Edexcel/MathB/2014/june/paper01/q2) }$ The volume of a right circular cylinder of height 13 cm is 117ʌ cm3. Calculate, in cm, the radius of the cylinder. (2) |
| 9 | $\mathrm{( Edexcel/MathB/2014/june/paper01/q15) }$ The diagram shows an inverted hollow right circular cone of height h cm. The area of the open end of the cone is 280 cm$^2$. Water is poured into the cone to a height of 9 cm.The area of the surface of the water is 70 cm$^2$, as shown in the diagram. Calculate the value of h. (3) |
| 10 | $\mathrm{( Edexcel/MathB/2014/juneR/paper01/q12) }$ The height of a right circular cylinder is four times the radius of the base of the cylinder. The volume of the cylinder is 500 cm$^3$. Calculate, in cm, the radius of the base of the cylinder. (3) |
| 11 | $\mathrm{( Edexcel/MathB/2014/juneR/paper01/q22) }$ Diagram NOT accurately drawn $ABCD$ is a trapezium with $BC=11$ cm, $AD=15$ cm and $BC$ parallel to $AD.$ The point $E$ on $AD$ is such that $EC$ is parallel to $AB.$ The area of $\triangle ECD=7$ cm$^{2}.$ Find the area, in cm$^{2}$, of the trapezium $ABCD.$ cm$^{2}$ (Total for Question 22 is 4 marks) |
| 12 | $\mathrm{( Edexcel/MathB/2015/janR/paper02/q1) }$ Diagram NOT accurately drawn Figure l shows a solid right circular cone with base radius 9 cm and height 40 cm. Show that the total surface area of the cone is 450$\pi$ cm$^{2}.$ [Area of circle $=\pi r^2$, Curved surface area of right circular cone $=\pi rl.]$ Total for question l is 4 Marks |
| 13 | $\mathrm{( Edexcel/MathB/2016/jan/paper01/q21) }$ The length of the perimeter of the square base of a pyramid is 920 m. The height of the pyramid is 129 m. Calculate the volume, in m$^3$, of the pyramid. Give your answer in standard form to 3 significant figures. m$^{3}$ (Total for Question 21 is 4 marks) |
| 14 | $\mathrm{( Edexcel/MathB/2016/janR/paper01/q22) }$
Diagram NOT accurately drawn A toy is made by joining a solid hemisphere of radius r to a right circular cone of base radius $r$. The plane face of the cone coincides with the plane face of the hemisphere. The axis of symmetry of the toy is vertical and the hemisphere is on top of the cone, as shown in the diagram. Given that the volume of the cone is equal to the volume of the hemisphere, find the height of the toy, in terms of $r.$ (Total for Question 22 is 4 marks) |
| 15 | $\mathrm{( Edexcel/MathB/2016/juneR/paper02/q2a) }$
Diagram NOT accurately drawn A hollow right circular cone, $A$ has height 20cm and radius 6cm. The cone is held with its axis vertical and its vertex at the bottom. A funnel is formed by removing the right circular cone, $B$, of height 8 cm from the bottom of $A$, as shown in Figure l. (a) Calculate the radÃus, in cm, of cone $B.$ (2) (b) Calculate the volume, in cm$^{3}$ to 3 significant figures, of the funnel $\begin{pmatrix}2\end{pmatrix}$ (2) Diagram NOT accurately draw: Figure 2, below, shows the funnel placed on a sheet of metal. The funnel is completely filled with water and no water escapes from the bottonı of the funnel. The sheet of metal is suddenly removed. Given that water flows out of the funnel at a constant rate of 54 cm$^{3}/$s (c) calculate the time, to the nearest second, to completely empty the funnel of water. $\begin{pmatrix}2\end{pmatrix}$ (Total for question 2 is 6 marks) |
| 16 | $\mathrm{( Edexcel/MathB/2014/janR/paper01/q8) }$
Two models of elephants were bought by Shenaz. The models are mathematically similar, with elephant A twice the height of elephant B. Given that the volume of elephant A is
240 cm$^{3}$, calculate the volume, in cm$^3$, of elephant B.
cm$^{3}$
(Total for Question 8 is 2 marks) |
| 17 |
$\mathrm{( Edexcel/MathB/2014/jan/paper01/q29) }$
Diagram NOT accurately drawn $ABCD$ is a square of side l2 cm. The circle $PQRS$ touches the sides of the square at $P$, ${\mathcal{Q}}, R$ and S as shown in the diagram. (a) Write down the length of a radius of the circle $PQRS.$ cm (b) Find, in terms of $\pi$, the total area, in cm^2, of the shaded region in the diagram restisteresse $cm^{z}$ (c) Express your answer to part (b) as a percentage of the area of the square $ABCD.$ Give your answer to 3 significant figures. (Total for Question 29 is 4 marks) |
| 18 | $\mathrm{( Edexcel/MathB/2014/juneR/paper01/q14) }$ The angle of a sector of a circle of radius 12 cm is 85° Calculate, in cm^2 to 3 significant figures, the area of the sector. cm$^{2}$ (Total for Question 14 is 3 marks) |
| 19 | $( \mathrm{Edexcel/Math~B/2015/june/paper01/q26) }$ The circumference of a circle is 12 cm. A sector of this circle has an angle of 72° at the centre of the circle The area of this sector is $A$ cm$^{2}$ (a) Find an expression for $A$ in terms of $\pi$ Simplify your expression. $\begin{pmatrix}3\end{pmatrix}$ The perimeter of the sector is $P$cm. (b) Show that $P=\frac{12(\pi+5)}{5\pi}$ $\begin{pmatrix}3\end{pmatrix}$ (Total for Question 26 is 6 marks) |
| 20 | $\mathrm{( Edexcel/MathB/2015/juneR/paper01/q23) }$
$AOB$ is a sector of a circle, centre $O$, with $\angle AOB=75^{\circ}$ The area of the sector is $180 cm^2$
Find, to 3 significant figures,
(a) the radius, in cm, of the circle. cm $\begin{pmatrix}2\end{pmatrix}$ (b) the length, in cm, of the perimeter of the sector. cm $\begin{pmatrix}3\end{pmatrix}$ (Total for Question 23 is 5 marks) |
Mensuration (2014-2016) Short Answer
- (a) $4000\pi$ cm$^3$ (b) $40\pi r^3,r=4$
- Fraction $=35 \% \quad$
- $129 \mathrm{~cm}^2$
- $h=14$
- $7 .3 \mathrm{~cm}^3$
- (a) $3.6 \times 10^{-3} $ (b) $ 5000 \quad$
- $1260 \mathrm{~mm}^3 \quad $
- $ r=3 \mathrm{~cm} \quad$
- $ h=18 \mathrm{~cm}$
- $r=5$
- $45.5 \mathrm{~cm}^2 \quad$
- Slant height $=41,450 \pi \quad $
- Volume $=2.27 \times 10^6 \mathrm{~m}^3$.
- $h( cone )=2 r$. height $r+2 r=3 r$
- (a) $ r=2.4 \mathrm{~cm} \quad$ (b) $ 706 / 705\left(\mathrm{~cm}^3\right)$ (c) $\operatorname{time}(s)=13$
- $ 30 $
- (a) radius $=6 \mathrm{~cm} $ (b) $12^2-\pi 36 $ (c) $21.45>21.5$ OR. $21.43>21.4$
- $107\left(\mathrm{~cm}^2\right)$
- (a) Area $=\frac{36}{5 \pi} \mathrm{cm}^2$ (b) $\frac{12(\pi+5)}{5 \pi}$
- (a) $r=16.6 \mathrm{~cm}$ (b) $54.9 \mathrm{~cm}$
- (a) $29.3 \mathrm{~cm}$ (b) $4.66 / 4.67$ (c) $13.2 \mathrm{~cm} \quad$
- $5 .2 \mathrm{~cm}, \frac{5}{3} \pi$















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