Sumeet has a pond in the shape of a prism.The pond is completely full of water. Sumeet wants to empty the pond so he can clean it. Sumeet uses a pump to empty the pond. The volume of water in the pond decreases at a constant rate. The level of the water in the pond goes down by 20 cm in the first 30 minutes.Work out how much more time Sumeet has to wait for the pump to empty the pond completely. ..............................................................................................................................................(Total for Question is 6 marks)Q2.Here is a solid prism.Work out the volume of the prism.. . . . . . . . . . . . . . . . . .cm3(Total for Question is 3 marks)Q3.A frustrum is made by removing a small cone from a similar large cone.The height of the small cone is 20 cm. The height of the large cone is 40 cm. The diameter of the base of the large cone is 30 cm.Work out the volume of the frustrum. Give your answer correct to 3 significant figures. . . . . . . . . . . . . . . . . . . . . . . . .cm3(Total for Question is 4 marks)Q4.The diagram shows a solid metal cylinder. Diagram NOT accurately drawn The cylinder has base radius 2x and height 9x.The cylinder is melted down and made into a sphere of radius r.Find an expression for r in terms of x...............................................................................................................................................(Total for Question is 3 marks)Q5.A water trough is in the shape of a prism. Diagram NOT accurately drawn Hamish fills the trough completely.Water leaks from the bottom of the trough at a constant rate. 2 hours later, the level of the water has fallen by 20 cm.Water continues to leak from the trough at the same rate.How many more minutes will it take for the trough to empty completely?..............................................................................................................................................(Total for Question is 6 marks)Q6.This diagram, drawn on a centimetre grid, is an accurate net of a triangular prism.Work out the volume of the prism...............................................................................................................................................(Total for Question is 4 marks)Q7.Work out the volume of the triangular prism.. . . . . . . . . . . . . . . . . . . . . . cm3(Total for Question is 2 marks)Q8.* Marc drives a truck. The truck pulls a container. The container is a cuboid 10 m by 4 m by 5 m. Diagram NOT accurately drawn d %%% %%% d dMarc fills the container with boxes. Each box is a cuboid 50 cm by 40 cm by 20 cm. Show that Marc can put no more than 5000 boxes into the container. (Total for Question is 4 marks)Q9.Terry fills a carton with boxes. Each box is a cube of side 10 cm. The carton is a cuboid with length 60 cm width 50 cm height 30 cmWork out the number of boxes Terry needs to fill one carton completely...............................................................................................................................................(Total for Question is 3 marks)Q10.The diagram shows an L-shaped prism.Calculate the volume of the prism.. . . . . . . . . . . . . . . . . . cm3(Total for Question is 3 marks)Q11.Here is a vase in the shape of a cylinder.The vase has a radius of 5 cm. There are 1000 cm3 of water in the vase. Work out the depth of the water in the vase. Give your answer correct to 1 decimal place...............................................................................................................................................(Total for Question is 3 marks)Q12.Ella is designing a glass in the shape of a cylinder.The glass must hold a minimum of litre of liquid.The glass must have a diameter of 8 cm.Calculate the minimum height of the glass. ..............................................................................................................................................(Total for Question is 5 marks)Q13.The diagram shows a solid hemisphere of radius 5 cm.Find the total surface area of the solid hemisphere. Give your answer in terms of . . . . . . . . . . . . . . . . . . . . . . . . . cm2(Total for Question is 3 marks)Q1.This proved to be a challenging question for the vast majority of candidates on this paper. Many candidates failed to show their working in an organised manner and they rarely made it clear exactly what they were working out. As a result examiners were faced with working scattered all over the working space with little explicit description of the strategy the candidate was using. It was often difficult to make out whether candidates were using volumes, areas or lengths. Some candidates employed methods involving the division of a volume or a rate by a length to find a time. Whilst a reasonable number of candidates were awarded some credit for their responses, only a small number were able to see the problem through to a successful conclusion. Some candidates worked with a cuboid rather than a prism. Q2.It was disappointing to see a significant number of candidates fail to gain full marks in this question through not being able to find the area of a relatively straightforward compound shape correctly. In order to find the area of the cross section, it was necessary to work out at least one missing length. Candidates would be well advised to show any length that they calculate on the diagram so that it can then be followed through in subsequent calculations. Many just used the measurements given on the diagram in some way and so gained no marks. It was common to see attempts to work out the total surface area or the sum of all the edges rather than the volume. A common mistake was to calculate 7 11 20 and stop there.Q3.A popular incorrect method was to evaluate 13 152 20. Some candidates were able to write down a correct expression for the volume of the large cone but then did not realise that the radius of the smaller cone was 7.5 cm and so failed to make further progress. There was evidence of the wrong formula being used for the volume of a cone despite this being given on the formula sheet at the front of the paper; formulae for the volume of a cylinder or surface area of a cone were commonly seen. It was common to see the volume of the large cone being found correctly, and then halved for the volume of the frustum.Q4.The most common error here was to substitute 2x for the radius but to forget to use brackets so ending up with 2x2 rather than 4x2. This error was condoned for the two method marks as the candidate was automatically penalised at the accuracy stage. The use of 9 rather than 9x was not condoned. Many candidates correctly substituted into the formula for the volume of a cylinder but then failed to equate to the formula for the volume of the sphere. Occasionally the formula for the surface area of a sphere rather than that for the volume of a sphere was used.Q5.The most successful candidates structured their working clearly, often annotating the diagram to show different sections to match their calculations. Some identified that as the trough was a prism, it was not essential to consider volume but worked with the cross-section areas instead. Large numbers with zeros led to many arithmetical errors and many candidates did not recognise that they had to consider the rate of leakage. These errors along with problems converting between minutes and hours meant that many candidates presented final answers which were far too large. Candidates need to be encouraged to make use of estimation and consider the reasonableness of any answer reached. Perhaps most importantly, candidates need to practice solving unstructured problems and compare the efficiency of a variety of approaches so that they can select appropriate methods to use.Q6.Many candidates had difficulty associating the net with a volume calculation. There were many surface area calculations seen. An area of 6 was often seen, sometimes on the diagram and for this 1 mark was awarded. Some candidates got as far as 6 x 7 but could not arrive at the correct answer. Many errors in basic multiplication were seen leading to inaccurate answers. There were a significant number of candidates who could find the area of the triangular cross section but could go no further. There were also many who gave 84 in the working but were unable to take the last step to the correct answer. The correct units were often seen and managed to gain the candidates 1 mark. The question produced a spread of marks but pleasingly the modal mark was 4.Q7.This question was well attempted by most candidates but as many candidates achieve M1A1 as achieved M0A0. The most common error was to do 5 12 10 and forget to divide by 2. Another repeated error was to calculate the surface area of the shape. Weaker candidates simply added all the dimensions.Q8.Many candidates did not seem to realise that the dimensions of the container and the dimensions of the box were given in different units and failed to convert m to cm or cm to m. Candidates who tried to convert the units for a volume usually failed a typical incorrect conversion was 200 m3 = 20000 cm3.Those who converted a length were usually more successful. The most common approach to the question was for candidates to find the volume of the container and the volume of the box and to then divide the volume of the container by the volume of the box. A few candidates attempted the division the wrong way round, usually if their volumes were 200 and 40000, as they divided the larger number by the smaller. Working out 500 1000 400 to find the volume of the container often resulted in an answer with too few zeros and the division by 40000 also caused problems. Many candidates attempted to give an answer of 5000 from incorrect working.Some candidates did not find the two volumes but found the number of boxes that would fit into each dimension of the container and were then left with the relatively straightforward calculation, 25 20 10. Candidates who used this method were generally more successful.Q9.This question was well attempted by most candidate, though poor arithmetic led to many candidates scoring M2A0. Most candidates attempted to solve the problem by calculating the volumes but a very common error was to incorrectly calculate the volume of the carton as 9000, instead of 90 000, leading incorrectly to the answer 9. Where candidates chose to solve the problem by calculating the number of boxes that would fit along each edge, the most common error was to simply state the values 5, 6 and 3 and not calculate 5 6 3. Other repeated errors were to calculate the surface areas or to divide by 10 instead of 1000. Weaker candidates added dimensions showing no understanding of volume.Q10.Most candidates attempted this by finding the sum of the areas of the constituent cuboids. The most common successful approach was 9 2 8 + 5 5 8 although many also 'saw' a 7cm by 8 cm by 5 cm cuboid. Approaches which involved finding the correct cross sectional area were rarer. A common error was to find the volume of the bottom 7cm by 8cm by 5cm cuboid correctly, but then to confuse some of the measurements of the remnant 4cm by 2 cm by 8 cm cuboid. A minority of candidates simply worked out 9 7 as their cross-sectional area. A few confused volume with an attempt at the surface area. Q11.Many candidates established the correct area of a circle with 78.5 seen. Some responses did not go on to use 1000 correctly to calculate the height.Q12.There were two slightly different approaches to this question depending on whether the candidate treated the problem as involving the 'holistic' formula r2h or as an analogy with the prism formula 'cross-sectional area length'. Not all candidates could convert litre to millilitres and hence cm3. Many found the height correct to 1 decimal place by trial and improvement. This was acceptable for full marks. The most common errors were to use 2r in place of r2h and to use the formula for the total surface area instead of the volume formula. Q13.Candidates struggled to bring both elements of this question together. One mark was earned by the one third of candidates who wrote an expression for either the area of the circle or the surface area of the hemisphere and 2 marks were gained by candidates who did both of these. Very few candidates could go one step further and give the answer as 75, although 235.5 was also accepted as many candidates ignored the instruction to leave answers in terms of , and attempted calculations using an interesting variety of decimal approximations. A significant number of candidates used the formula for the volume of a sphere, once again confusing area and volume. %%% WorkingAnswerMarkNotes 1 hour 45 mins 6 M1 for method to find volume of pond, eg (1.3 + 0.5) 2 1 (= 1.8) M1 for method to find the volume of water emptied in 30 minutes, eg 1 2 0.2 (= 0.4), 100 200 20 (= 400000) A1 for correct rate, eg 0.8 m/hr, 0.4 m in 30 minutes M1 for correct method to find total time taken to empty the pond, eg "1.8" "0.8" M1 for method to find extra time, eg 2 hrs 15 minutes 30 minutes A1 for 1.75 hours, 1 hours, 1 hour 45 mins or 105 mins OR M1 for method to find volume of water emptied in 30 minutes,.eg. 1 2 0.2 (= 0.4), 100 200 20 (= 400000) M1 for method to work out rate of water loss eg. "0.4" 2 A1 for correct rate, eg 0.8 m/hr M1 for correct method to work out remaining volume of water eg. (1.1 + 0.3) 2 1 (= 1.4) M1 for method to work out time, eg "1.4" "0.8" A1 for 1.75 hours, 1 hours, 1 hour 45 mins or 105 mins NB working could be in 3D or in 2D and in metres or cm throughout Q1.Q2. %%% WorkingAnswerMarkNotes 1180 3 M1 for a correct method to find the area of the cross section M1 (dep) for a complete correct method for the volume of the prism A1 cao ORM1 for a correct method to find the volume of one cuboid M1 (dep) for a complete correct method for the volume of the prism A1 cao Q3. %%% WorkingAnswerMarkNotes 13 152 40 13 7.52 20 8250 4 B1 for 15cm as diameter or 7.5 cm as radius of smaller cone (may be marked on diagram or used in a formula) M1 for a numerical expression for the volume of one cone eg. 13 152 40 (=9424...) or 13 7.52 20 (=1178...) M1 for 13 152 40 oe 13 7.52 20 oe A1 for answer in the range 8240 8250 ORB1 for 23 M1 for a numerical expression for the volume of the large cone eg. 13 152 40 (=9424...) M1 volume of frustrum = 78 13 152 40 oe A1 for answer in the range 8240 8250 Q4.Question%%% WorkingAnswerMarkNotes Vol cylinder = (2x)2 9x = 36x3 36x3 = 43 r3 r3 = 27x3 3x 3 M1 for sub. into r2h eg. (2x)2 9x oe M1 for (2x)2 9x = 43 r3 oe A1 oe eg. NB : For both method marks condone missing brackets around the 2x Q5.QuestionWorkingAnswerMarkNotes 45 200 minutes 6 M1 for 120 2030(= 72000) M1 for "72000"120 A1 for 600cm3 min oe M1 for (120 + 80) 40 30 (= 120000) M1 for "120000"600" A1 for 200 minutes or 3 hours 20 mins oe SC B1 for 4 hours Q6.Question%%% WorkingAnswerMarkNotes (4 x 3) x 7 = 6 x 7 42 cm3 4 M2 for 4 37 oe (M1 for (4 3) or 4 3 7 or 6 seen as an area or 7 'cross sectional area' or 84 seen) A1 for 42 B1 (indep) for cm3 Q7. %%% WorkingAnswerMarkNotes 5 12 10 = 300 2 M1 for 5 12 ( 10) oe A1 cao Q8.Question%%% WorkingAnswerMarkNotes 500 1000 400 = 200 000 000 20 50 40 = 40 000 200 000 000 40 000 = 5000 OR (500 20) (1000 50) (400 40) = 25 20 10 = 5000 Proof 4 B1 for a correct unit conversion, could be seen on the diagram or in working M1 for 500 1000 400 or 200 000 000 or 20 50 40 or 40 000 or 5 10 4 or 200 or 0.2 0.5 0.4 or 0.04 M1(dep) for '200 000 000' '40 000' C1 for fully correct working leading to final answer of 5000 OR B1 for a correct unit conversion, could be seen on the diagram or in working M1 for (500 20) or (1000 50) or (400 40) or at least two of 25, 20, 10 seen M1(dep) for '25' '20' '10' C1 for fully correct working leading to final answer of 5000 Q9. %%% WorkingAnswerMarkNotes 90 3 M1 for one division (eg 60 10), may be implied by correct number of marks on the diagram or correct number on one edge of diagram or eg 610, or by two of 6, 5 and 3 seen M1 for (60 10) (50 10) (30 10) A1 cao OR M1 for 10 10 10 or 60 50 30 M1 for (60 50 30) (10 10 10) A1 cao Question%%% WorkingAnswerMarkNotes 9 7 4 5 = 43 43 8 OR 9 7 8 = 504 54 8 = 160 504 - 160 OR 9 2 + 5 5 43 8 OR 9 2 8 = 144 5 5 8 = 200 144 + 200 OR 7 5 8 = 280 2 4 8 = 64 280 + 64 344 3 M1 9 7 4 5 or 43 M1 "43" 8 A1 cao OR M1 9 7 8 or 504 or 54 8 or 160 M1 9 7 8 or 504 - 54 8 or 160 A1 cao OR M1 9 2 + 5 5 or 43 M1 "43" 8 A1 cao OR M1 9 2 8 or 144 or 5 5 8 or 200 M1 9 2 8 or 144 + 5 5 8 or 200 A1 cao OR M1 7 5 8 or 280 or 2 4 8 or 64 M1 7 5 8 or 280 + 2 4 8 or 64 A1 cao Q10.Q11. %%% WorkingAnswerMarkNotes 12.7 3 M1 for 3.142 5 5 oe or 3.142 5 5 'h' (=78.5 78.55) M1 for 1000 (3.142 5 5) A1 for 12.7 12.8 NB: multiples of acceptable for M marks Question%%% WorkingAnswerMarkNotes litre = 500ml 500 = 42 hh = 500 ( 42 ) 9.95 5 B1 litre = 500ml or 500 seen M1 42 h (= 50.2..h )or 42 (= 50.2..) M1 ft "500" = 42 h oe M1 (h =) "500" ( 42 ) oe A1 9.9 10.0 Q12.Q13. %%% WorkingAnswerMarkNotes 75 3 M1 for (4 52 ) 2 oe M1 for 52 oe A1 for 75 accept 235.5 Condone the use of = 3.14