Question
Neil has three dogs. Two are brown and one is grey. When he feeds the dogs, Neil uses three bowls and gives them out randomly. There are two red bowls and one yellow bowl. This information is shown on the tree diagram below.
There are 49 mice in a pet shop.
30 mice are white.
27 mice are male.
18 mice have short tails.
8 mice are white and have short tails.
11 mice are male and have short tails.
7 mice are male but neither white nor short-tailed.
5 mice have all three characteristics and
2 have none.
Copy the diagram below to your examination script.
One of the dogs is chosen at random.
(i) Find P (the dog is grey and has the yellow bowl).
(ii) Find P (the dog does not get the yellow bowl).[3]
Neil often takes the dogs to the park after they have eaten. He has noticed that the grey dog plays with a stick for a quarter of the time and both brown dogs play with sticks for half of the time. This information is shown on the tree diagram below.
(i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick.
During a trip to the park, one of the dogs is chosen at random.
(ii) Find P (the dog is grey or is playing with a stick, but not both).
(iii) Find P (the dog is grey given that the dog is playing with a stick).
(iv) Find P (the dog is grey and was fed from the yellow bowl and is not playing with a stick).[9]
Complete the diagram, using the information given in the question.[4]
Find (i) \(n(M \cap W)\)
(ii) \(n(M′ \cup S)\)[3]
Two mice are chosen without replacement.
Find P (both mice are short-tailed).[2]
Answer/Explanation
Markscheme
(i) P (a dog is grey and has the yellow bowl)
\( = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}( = 0.111)\) (M1)(A1)(G2)
The (M1) is for multiplying two values along any branch of the tree.(ii) P (dog does not get yellow bowl) \( = \frac{2}{3}\) ( = 0.667 (3sf) or 0.6) (A1)[3 marks]
(i) The tree diagram should show the values \(\frac{1}{2},\frac{1}{2}\) for the brown branch and \(\frac{1}{4},\frac{3}{4}\) in the correct positions for the grey branch. (A1)(A1)(ft)
Follow through if the branches are interchanged.
(ii) P (the dog is grey or is playing with a stick, but not both)
\( = \frac{1}{3} \times \frac{3}{4} + \frac{2}{3} \times \frac{1}{2}\) (M1)
\( = \frac{7}{{12}}\) ( = 0.583) (A1)(ft)(G1)
The (M1) is for showing two correct products (whether added or not). Follow through from b(i). Award (M1) for \( \frac{1}{3} + \frac{1}{4}\) (must be a sum).
(iii) P (dog is grey given that it is playing with stick)
\(\frac{{P(G \cap S)}}{{P(S)}} = \frac{{\frac{1}{3} \times \frac{1}{4}}}{{\left( {\frac{2}{3} \times \frac{1}{2}} \right) + \left( {\frac{1}{3} \times \frac{1}{4}} \right)}}\) or \(\frac{1}{{12}}/\frac{5}{{12}}\) (M1)(A1)(ft)
(M1) for substituted conditional probability formula, (A1) for correct substitutions.
\( = \frac{1}{5}\) ( = 0.2) (A1)(ft)(G2)
(iv) P (grey and fed from yellow bowl and not playing with stick) \( = \frac{1}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{{12}}\) ( = \(\frac{3}{{36}}\) = 0.0833 3sf). (M1)(A1)(ft)(G1)
(M1) is for product of 3 reasonable probability values.[9 marks]
(A1)(A1)(A1)(ft)(A1)(ft)
Award (A1) for 2 (must be in a box), (A1) for 7, (A1)(ft) for 6 and 4, (A1)(ft) for 9 and 13. Observe the assignment of (ft) marks strictly here. Example A common error is likely to be 11 instead of 6 (A0). In this case follow through to 4 and 18 (A1)(ft) for the final pair. Here the 4 follows from the total of 27 for n(M).[4 marks]
(i) \(n(M \cap W) = 14\) (A1)(ft)
(ii) \(n(M’ \cup S) = 22 + 11\) OR \(15 + 18\) (A1)(ft)
= 33 (A1)(ft)
Award (A2) if answer 33 is seen. Award (A1) for any of 22, 11, 15 or 18 seen but 33 absent.[3 marks]
P (both mice short-tailed) \( = \frac{{18}}{{49}} \times \frac{{17}}{{48}} = \frac{{306}}{{352}}\) (= 0.130). (M1)(A1)(ft)(G1)
(Allow alternatives such as 153/1176 or 51/392.) Award (M1) for any of \(\frac{{18}}{{49}}\) and \(\frac{{17}}{{48}}\) or \(\frac{{18}}{{49}} \times \frac{{17}}{{49}}\) or \(\frac{{18}}{{49}} + \frac{{17}}{{48}}\) seen.[2 marks]
Question
Sharon and Lisa share a flat. Sharon cooks dinner three nights out of ten. If Sharon does not cook dinner, then Lisa does. If Sharon cooks dinner the probability that they have pasta is 0.75. If Lisa cooks dinner the probability that they have pasta is 0.12.
A survey was carried out in a year 12 class. The pupils were asked which pop groups they like out of the Rockers (R), the Salseros (S), and the Bluers (B). The results are shown in the following diagram.
Copy and complete the tree diagram to represent this information.
[3]
Find the probability that Lisa cooks dinner and they do not have pasta.[2]
Find the probability that they do not have pasta.[3]
Given that they do not have pasta, find the probability that Lisa cooked dinner.[3]
Write down \(n(R \cap S \cap B)\).[1]
Find \(n(R’)\).[2]
Describe which groups the pupils in the set \(S \cap B\) like.[2]
Use set notation to describe the group of pupils who like the Rockers and the Bluers but do not like the Salseros.[2]
There are 33 pupils in the class.
Find x.[2]
There are 33 pupils in the class.
Find the number of pupils who like the Rockers.[1]
Answer/Explanation
Markscheme
Note: Award (A1) for each correct pair. (A3)[3 marks]
\(0.7 \times 0.88 = 0.616{\text{ }}\left( {\frac{{77}}{{125}},{\text{ }}61.6{\text{ }}\% } \right)\) (M1)(A1)(ft)(G2)
Note: Award (M1) for multiplying the correct probabilities.[2 marks]
\(0.3 \times 0.25 + 0.7 \times 0.88\) (M1)(M1)
Notes: Award (M1) for a relevant two-factor product, could be \(S \times NP\) OR \(L \times NP\).
Award (M1) for summing 2 two-factor products.
\({\text{P}} = 0.691{\text{ }}\left( {\frac{{691}}{{1000}},{\text{ }}69.1{\text{ }}\% } \right)\) (A1)(ft)(G2)
Notes: (ft) from their answer to (b).[3 marks]
\(\frac{{0.616}}{{0.691}}\) (M1)(A1)
Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.
\({\text{P}} = 0.891{\text{ }}\left( {\frac{{616}}{{691}},{\text{ }}89.1{\text{ }}\% } \right)\) (A1)(ft)(G2)[3 marks]
3 (A1)[1 mark]
For 5, 4, 7 (0) seen with no extra values (A1)
16 (A1)(G2)[2 marks]
They like (both) the Salseros (S) and they like the Bluers (B) (A1)(A1)
Note: Award (A1) for “and”, (A1) for the correct groups.[2 marks]
\(R \cap B \cap S’\) (A1)(A1)
Note: Award (A1) for \(R \cap B\), (A1) for \( \cap S’\)[2 marks]
\(21+ 3x = 33\) (M1)
\(x = 4\) (A1)(G2)[2 marks]
17 (A1)(ft)[1 mark]
Question
The Venn diagram below represents the students studying Mathematics (A), Further Mathematics (B) and Physics (C) in a school.
50 students study Mathematics
38 study Physics
20 study Mathematics and Physics but not Further Mathematics
10 study Further Mathematics but not Physics
12 study Further Mathematics and Physics
6 study Physics but not Mathematics
3 study none of these three subjects.
Three propositions are given as
p : It is snowing q : The roads are open r : We will go skiing
Copy and complete the Venn diagram on your answer paper.[3]
Write down the number of students who study Mathematics but not Further Mathematics.[1]
Write down the total number of students in the school.[1]
Write down \(n({\text{B}} \cup {\text{C}})\).[2]
Write the following compound statement in symbolic form.
“It is snowing and the roads are not open.”[2]
Write the following compound statement in words.
\((\neg p \wedge q) \Rightarrow r\)[3]
An incomplete truth table for the compound proposition \((\neg p \wedge q) \Rightarrow r\) is given below.
Copy and complete the truth table on your answer paper.
[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)
Note: Award (A1) for each correct number in the correct position.[3 marks]
28 (A1)(ft)
Note: 20 + their 8.[1 mark]
59 (A1)(ft)[1 mark]
10 + 12 + 20 + 6 (M1)
Note: Award (M1) for use of the correct regions.
= 48 (A1)(ft)(G2)
OR
59 − 8 − 3 (M1)
= 48 (A1)(ft)[2 marks]
\(p \wedge \neg q\) (A1)(A1)
Note: Award (A1) for \(\wedge\), (A1) for both statements in the correct order.[2 marks]
If it is not snowing and the roads are open (then) we will go skiing. (A1)(A1)(A1)
Note: Award (A1) for “if…(then)”, (A1) for “not snowing and the roads are open”, (A1) for “we will go skiing”.[3 marks]
(A1)(A1)(ft)(A1)(ft)
Note: Award (A1) for each correct column.[3 marks]
Question
One day the numbers of customers at three cafés, “Alan’s Diner” ( \(A\) ), “Sarah’s Snackbar” ( \(S\) ) and “Pete’s Eats” ( \(P\) ), were recorded and are given below.
17 were customers of Pete’s Eats only
27 were customers of Sarah’s Snackbar only
15 were customers of Alan’s Diner only
10 were customers of Pete’s Eats and Sarah’s Snackbar but not Alan’s Diner
8 were customers of Pete’s Eats and Alan’s Diner but not Sarah’s Snackbar
Some of the customers in each café were given survey forms to complete to find out if they were satisfied with the standard of service they received.
Draw a Venn Diagram, using sets labelled \(A\) , \(S\) and \(P\) , that shows this information.[3]
There were 48 customers of Pete’s Eats that day. Calculate the number of people who were customers of all three cafés.[2]
There were 50 customers of Sarah’s Snackbar that day. Calculate the total number of people who were customers of Alan’s Diner.[3]
Write down the number of customers of Alan’s Diner that were also customers of Pete’s Eats.[1]
Find \(n[(S \cup P) \cap A’]\).[2]
One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”;[2]
One of the survey forms was chosen at random, find the probability that the form showed “Satisfied” and was completed at Sarah’s Snackbar;[2]
One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”, given that it was completed at Alan’s Diner.[2]
A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.
Write down the null hypothesis, \({{\text{H}}_0}\) , for the \({\chi ^2}\) test.[1]
A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.
Write down the number of degrees of freedom for the test.[1]
A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.
Using your graphic display calculator, find \({\chi ^2}_{calc}\) .[2]
A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.
State, giving a reason, the conclusion to the test.[2]
Answer/Explanation
Markscheme
(A1) for rectangle and three labelled intersecting circles
(A1) for \(15\), \(27\) and \(17\)
(A1) for \(10\) and \(8\) (A3)[3 marks]
\(48 – (8 +10 +17)\) or equivalent (M1)
\( = 13\) (A1)(ft)(G2)[2 marks]
\(50 – (27 +10 +13)\) (M1)
Note: Award (M1) for working seen.
\( = 0\) (A1)
number of elements in A \(= 36\) (A1)(ft)(G3)
Note: Follow through from (b).[3 marks]
\(21\) (A1)(ft)
Note: Follow through from (b) even if no working seen.[1 mark]
\(54\) (M1)(A1)(ft)(G2)
Note: Award (M1) for \(17\), \(10\), \(27\) seen. Follow through from (a).[2 marks]
\(\frac{{40}}{{120}}{\text{ }}\left( {\frac{1}{3}{\text{, }}0.333{\text{, }}33.3\% } \right)\) (A1)(A1)(G2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{{34}}{{120}}{\text{ }}\left( {\frac{{17}}{{60}}{\text{, }}0.283{\text{, }}28.3\% } \right)\) (A1)(A1)(G2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{8}{{28}}{\text{ }}\left( {\frac{2}{7}{\text{, }}0.286{\text{, }}28.6\% } \right)\) (A1)(A1)(G2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
customer satisfaction is independent of café (A1)
Note: Accept “customer satisfaction is not associated with the café”.[1 mark]
\(2\) (A1)[1 mark]
\(0.754\) (G2)
Note: Award (G1)(G1)(AP) for \(0.75\) or for correct answer incorrectly rounded to 3 s.f. or more, (G0) for \(0.7\).[2 marks]
since \({\chi ^2}_{calc} < {\chi ^2}_{crit}5.991 accept (or Do not reject) H0 (R1)(A1)(ft)
Note: Follow through from their value in (e).
OR
Accept (or Do not reject) H0 as \(p\)-value \((0.686) > 0.05\) (R1)(A1)(ft)
Notes: Do not award (A1)(R0). Award the (R1) for comparison of appropriate values.[2 marks]
Question
In a college 450 students were surveyed with the following results
150 have a television
205 have a computer
220 have an iPhone
75 have an iPhone and a computer
60 have a television and a computer
70 have a television and an iPhone
40 have all three.
Draw a Venn diagram to show this information. Use T to represent the set of students who have a television, C the set of students who have a computer and I the set of students who have an iPhone.[4]
Write down the number of students that
(i) have a computer only;
(ii) have an iPhone and a computer but no television.[2]
Write down \(n[T \cap (C \cup I)’]\).[1]
Calculate the number of students who have none of the three.[2]
Two students are chosen at random from the 450 students. Calculate the probability that
(i) neither student has an iPhone;
(ii) only one of the students has an iPhone.[6]
The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.
In the first 10 months they collect 13 050 dollars.
(i) Write down a second equation in x and y to represent this information.
(ii) Write down the value of x and of y .[3]
The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.
In the first 10 months they collect 13 050 dollars.
Calculate the number of months that it will take the students to collect 49 500 dollars.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)(A1)
Notes: Award (A1) for labelled sets T, C, and I included inside an enclosed universal set. (Label U is not essential.) Award (A1) for central entry 40. (A1) for 20, 30 and 35 in the other intersecting regions. (A1) for 60, 110 and 115 or T(150), C(205), I(220).[4 marks]
In parts (b), (c) and (d) follow through from their diagram.
(i) 110 (A1)(ft)
(ii) 35 (A1)(ft) [2 marks]
In parts (b), (c) and (d) follow through from their diagram.
60 (A1)(ft)[2 marks]
In parts (b), (c) and (d) follow through from their diagram.
450 − (60 + 20 + 40 + 30 + 115 + 35 + 110) (M1)
Note: Award (M1) for subtracting all their values from 450.
= 40 (A1)(ft)(G2)[2 marks]
(i) \(\frac{{230}}{{450}} \times \frac{{229}}{{449}}\) (A1)(M1)
Note: Award (A1) for correct fractions, (M1) for multiplying their fractions.
\(\frac{{52670}}{{202050}}\left( {\frac{{5267}}{{20205}},{\text{ 0}}{\text{.261, 26}}{\text{.1% }}} \right)(0.26067…)\) (A1)(G2)
Note: Follow through from their Venn diagram in part (a).
(ii) \(\frac{{220}}{{450}} \times \frac{{230}}{{449}} + \frac{{230}}{{450}} \times \frac{{220}}{{449}}\) (A1)(A1)
Note: Award (A1) for addition of their products, (A1) for two correct products.
OR
\(\frac{{230}}{{450}} \times \frac{{220}}{{449}} \times 2\) (A1)(A1)
Notes: Award (A1) for their product of two fractions multiplied by 2, (A1) for correct product of two fractions multiplied by 2. Award (A0)(A0) if correct product is seen not multiplied by 2.
\(\frac{{2024}}{{4041}}(0.501,{\text{ 50}}{\text{.1% )(0}}{\text{.50086}}…{\text{)}}\) (A1)(G2)
Note: Follow through from their Venn diagram in part (a) and/or their 230 used in part (e)(i).
Note: For consistent use of replacement in parts (i) and (ii) award at most (A0)(M1)(A0) in part (i) and (A1)(ft)(A1)(A1)(ft) in part (ii).[6 marks]
(i) x + 9y = 13050 (A1)
(ii) x = 900 (A1)(ft)
y = 1350 (A1)(ft)
Notes: Follow through from their equation in (f)(i). Do not award (A1)(ft) if answer is negative. Award (M1)(A0) for an attempt at solving simultaneous equations algebraically but incorrect answer obtained. [3 marks]
49500 = 900 + 1350n (A1)(ft)
Notes: Award (A1)(ft) for setting up correct equation. Follow through from candidate’s part (f).
n = 36 (A1)(ft)
The total number of months is 37. (A1)(ft)(G2)
Note: Award (G1) for 36 seen as final answer with no working. The value of n must be a positive integer for the last two (A1)(ft) to be awarded.
OR
49500 = 900 + 1350(n − 1) (A2)(ft)
Notes: Award (A2)(ft) for setting up correct equation. Follow through from candidate’s part (f).
n = 37 (A1)(ft)(G2)
Note: The value of n must be a positive integer for the last (A1)(ft) to be awarded.[3 marks]
Question
A group of \(120\) women in the USA were asked whether they had visited the continents of Europe (\(E\)) or South America (\(S\)) or Asia (\(A\)).
\(7\) had visited all three continents
\(28\) had visited Europe only
\(22\) had visited South America only
\(16\) had visited Asia only
\(15\) had visited Europe and South America but had not visited Asia
\(x\) had visited South America and Asia but had not visited Europe
\(2x\) had visited Europe and Asia but had not visited South America
\(20\) had not visited any of these continents
Draw a Venn diagram, using sets labelled \(E\), \(S\) and \(A\), to show this information.[5]
Calculate the value of \(x\).[2]
Explain, in words, the meaning of \((E \cup S) \cap A’\).[2]
Write down \(n\left( {(E \cup S \cup A)’} \right)\).[1]
Find the probability that a woman selected at random from the group had visited Europe.[2]
Find the probability that a woman selected at random from the group had visited Europe, given that she had visited Asia.[2]
Two women from the group are selected at random.
Find the probability that both women selected had visited South America.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)(A1)(A1)
Notes: Award (A1) for rectangle and three labelled intersecting circles.
Award (A1) for \(7\) in correct place.
Award (A1) for \(28\), \(22\) and \(16\) in the correct places.
Award (A1) for \(15\), \(x\) and \(2x\) in the correct places.
Award (A1) for \(20\) in the correct place.
Accept \(4\) and \(8\) instead of \(x\) and \(2x\).
Do not penalize if \(U\) is omitted from the diagram.[5 marks]
\(3x = 120 – (20 + 28 + 15 + 22 + 7 + 16)\) (M1)
Note: Award (M1) for setting up a correct equation involving \(x\), the \(120\) and values from their diagram.
\(x = 4\) (A1)(ft)(G2)
Note: Follow through from part (a). For the follow through to be awarded \(x\) must be a positive integer.[2 marks]
(Women who had visited) Europe or South America and (but had) not (visited) Asia (A1)(A1)
Notes: Award (A1) for “(visited) Europe or South America” (or both).
Award (A1) for “and (but) had not visited Asia”.
\(E\)(urope) union \(S\)(outh America) intersected with not \(A\)(sia) earns no marks, (A0).[2 marks]
\(20\) (A1)
Note: Award (A0) for the embedded answer of \(n(20)\).[1 mark]
\(\frac{{58}}{{120}}{\text{ }}\left( {\frac{{29}}{{60}},{\text{ 0.483, 48.3% }}} \right){\text{ (0.48333}} \ldots {\text{)}}\) (A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, follow through from their value of \(x\), or their diagram, (A1) for denominator.[2 marks]
\(\frac{{15}}{{35}}{\text{ }}\left( {\frac{3}{7},{\text{ 0.429, 42.9% }}} \right){\text{ (0.428571}} \ldots {\text{)}}\) (A1)(ft)(A1)(ft)(G2)
Note: Award (A1)(ft) for numerator, (A1)(ft) for denominator, follow through from their value of \(x\) or their diagram. [2 marks]
\(\frac{{48}}{{120}} \times \frac{{47}}{{119}}\) (A1)(ft)(M1)
Notes: Award (A1)(ft) for two correct fractions, follow through from their denominator in part (e), follow through the numerator from their answer to part (b) or from their diagram, (M1) for multiplication of their two fractions.
\( = \frac{{2256}}{{14\,280}}\left( {\frac{{94}}{{595}},{\text{ 0.158, 15,8% }}} \right){\text{ (0.157983}} \ldots {\text{)}}\) (A1)(ft)(G2)
Notes: Award (A1)(M1)(A1) for correct fractions, correctly multiplied together with an answer of \(0.16\).
Award (A0)(M1)(A0) for \(\frac{{48}}{{120}} \times \frac{{48}}{{120}} = 0.16\).
Award (G1) for an answer of \(0.16\) with no working seen.[3 marks]
Question
A group of tourists went on safari to a game reserve. The game warden wanted to know how many of the tourists saw Leopard (\(L\)), Cheetah (\(C\)) or Rhino (\(R\)). The results are given as follows.
5 of the tourists saw all three
7 saw Leopard and Rhino
1 saw Cheetah and Leopard but not Rhino
4 saw Leopard only
3 saw Cheetah only
9 saw Rhino only
Draw a Venn diagram to show this information.[4]
There were 25 tourists in the group and every tourist saw at least one of the three types of animal.
Find the number of tourists that saw Cheetah and Rhino but not Leopard.[2]
There were 25 tourists in the group and every tourist saw at least one of the three types of animal.
Calculate the probability that a tourist chosen at random from the group
(i) saw Leopard;
(ii) saw only one of the three types of animal;
(iii) saw only Leopard, given that he saw only one of the three types of animal.[6]
There were 25 tourists in the group and every tourist saw at least one of the three types of animal.
If a tourist chosen at random from the group saw Leopard, find the probability that he also saw Cheetah.[2]
Answer/Explanation
Markscheme
(A1)(A1)(A1)(A1)
Note: Award (A1) for rectangle and three labelled intersecting circles (the rectangle need not be labelled), (A1) for 5, (A1) for 2 and 1, (A1) for 4, 3 and 9.[4 marks]
\(25 – (5 + 2 + 1 + 4 + 3 + 9)\) (M1)
Notes: Award (M1) for their \(5 + 2 + 1 + 4 + 3 + 9\) seen even if total is greater than \(25\).
Do not award (A1)(ft) if their total is greater than \(25\).
\( = 1\) (A1)(ft)(G2)[2 marks]
(i) \(\frac{{12}}{{25}}{\text{ }}(0.48,{\text{ }}48\% )\) (A1)(ft)(A1)(G2)
Notes: Award (A1)(ft) for numerator, (A1) for denominator.
Follow through from Venn diagram.
(ii) \(\frac{{16}}{{25}}{\text{ }}(0.64,{\text{ }}64\% )\) (A1)(A1)(G2)
Notes: Award (A1) for numerator, (A1) for denominator.
There is no follow through; all information is given.
(iii) \(\frac{4}{{16}}{\text{ }}(0.25,{\text{ }}25\% )\)) (A1)(A1)(ft)(G2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator.
Follow through from part (c)(ii) only.[6 marks]
\(\frac{6}{{12}}{\text{ }}(0.5,{\text{ }}50\% )\) (A1)(A1)(ft)(G2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator.
Follow through from Venn diagram.[2 marks]
Question
A group of 100 customers in a restaurant are asked which fruits they like from a choice of mangoes, bananas and kiwi fruits. The results are as follows.
15 like all three fruits
22 like mangoes and bananas
33 like mangoes and kiwi fruits
27 like bananas and kiwi fruits
8 like none of these three fruits
\(x\) like only mangoes
Copy the following Venn diagram and correctly insert all values from the above information.
[3]
The number of customers that like only mangoes is equal to the number of customers that like only kiwi fruits. This number is half of the number of customers that like only bananas.
Complete your Venn diagram from part (a) with this additional information in terms of \(x\).[2]
The number of customers that like only mangoes is equal to the number of customers that like only kiwi fruits. This number is half of the number of customers that like only bananas.
Find the value of \(x\).[2]
The number of customers that like only mangoes is equal to the number of customers that like only kiwi fruits. This number is half of the number of customers that like only bananas.
Write down the number of customers who like
(i) mangoes;
(ii) mangoes or bananas.[2]
The number of customers that like only mangoes is equal to the number of customers that like only kiwi fruits. This number is half of the number of customers that like only bananas.
A customer is chosen at random from the 100 customers. Find the probability that this customer
(i) likes none of the three fruits;
(ii) likes only two of the fruits;
(iii) likes all three fruits given that the customer likes mangoes and bananas.[4]
The number of customers that like only mangoes is equal to the number of customers that like only kiwi fruits. This number is half of the number of customers that like only bananas.
Two customers are chosen at random from the 100 customers. Find the probability that the two customers like none of the three fruits.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)
Notes: Award (A1) for 15 in the correct place.
Award (A1) for 7, 18 and 12 seen in the correct places.
Award (A1) for 8 in the correct place.
Award at most (A0)(A1)(A1) if diagram is missing the rectangle.
(A1)(A1)
Notes: Award (A1) for \(x\) seen in the correct places.
Award (A1) for \(2x\) seen in the correct place.
Award (A0)(A1)(ft) if \(x\) and \(2x\) are replaced by 10 and 20 respectively.
\(2x + x + x + 15 + 8 + 7 + 18 + 12 = 100\;\;\;(4x + 60 = 100{\text{ or equivalent)}}\) (M1)
Note: Award (M1) for equating the sum of the elements of their Venn diagram to \(100\). Equating to \(100\) may be implied.
\((x = ){\text{ }}10\) (A1)(ft)(G2)
Note: Follow through from their Venn diagram. The answer must be a positive integer.
(i) \(50\) (A1)(ft)
(ii) \(82\) (A1)(ft)
Note: Follow through from their answer to part (c) and their Venn diagram.
Award (A0)(ft)(A1)(ft) if answer is \(\frac{{50}}{{100}}\) and \(\frac{{82}}{{100}}\).
(i) \(\frac{8}{{100}}\;\;\;\left( {\frac{2}{{25}};{\text{ }}0.08;{\text{ }}8\% } \right)\) (A1)
Note: Correct answer only. There is no follow through.
(ii) \(\frac{{37}}{{100}}\;\;\;(0.37,{\text{ }}37\% )\) (A1)(ft)
Note: Follow through from their Venn diagram.
(iii) \(\frac{{15}}{{22}}\;\;\;(0.681;{\text{ }}0.682;{\text{ }}68.2\% )\;\;\;(0.681818 \ldots )\) (A1)(A1)(ft)(G2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator, follow through from their Venn diagram. Award (A0)(A0) if answer is given as incorrect reduced fraction without working.
\(\frac{8}{{100}} \times \frac{7}{{99}}\) (A1)(ft)(M1)
Note: Award (A1)(ft) for correct fractions, follow through from their answer to part (e)(i), (M1) for multiplying their fractions.
\( = \frac{{56}}{{9900}}\;\;\;\left( {\frac{{14}}{{2477}},{\text{ }}0.00565656 \ldots ,{\text{ }}0.00566,{\text{ }}0.0056,{\text{ }}0.566\% } \right)\) (A1)(ft)(G2)
Question
A water container is made in the shape of a cylinder with internal height \(h\) cm and internal base radius \(r\) cm.
The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.
The volume of the water container is \(0.5{\text{ }}{{\text{m}}^3}\).
The water container is designed so that the area to be coated is minimized.
One can of water-resistant material coats a surface area of \(2000{\text{ c}}{{\text{m}}^2}\).
Write down a formula for \(A\), the surface area to be coated.[2]
Express this volume in \({\text{c}}{{\text{m}}^3}\).[1]
Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.[1]
Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).[2]
Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).[2]
Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).[3]
Using your answer to part (e), find the value of \(r\) which minimizes \(A\).[3]
Find the value of this minimum area.[2]
Find the least number of cans of water-resistant material that will coat the area in part (g).[3]
Answer/Explanation
Markscheme
\((A = ){\text{ }}\pi {r^2} + 2\pi rh\) (A1)(A1)
Note: Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.[2 marks]
\(500\,000\) (A1)
Notes: Units not required.[1 mark]
\(500\,000 = \pi {r^2}h\) (A1)(ft)
Notes: Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).
Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).[1 mark]
\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\) (A1)(ft)(M1)
Note: Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.
Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).
Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).
\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\) (AG)
Notes: The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.
Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]
\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\) (A1)(ft)(M1)
Note: Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.
Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).
Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).
\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\) (AG)
Notes: The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.
Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]
\(2\pi r – \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\) (A1)(A1)(A1)
Note: Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ – 2}}\), (A1) for \( – {\text{1}}\,{\text{000}}\,{\text{000}}\).[3 marks]
\(2\pi r – \frac{{1\,000\,000}}{{{r^2}}} = 0\) (M1)
Note: Award (M1) for equating their part (e) to zero.
\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\) (M1)
Note: Award (M1) for isolating \(r\).
OR
sketch of derivative function (M1)
with its zero indicated (M1)
\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\) (A1)(ft)(G2)[3 marks]
\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\) (M1)
Note: Award (M1) for correct substitution of their part (f) into the given equation.
\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\) (A1)(ft)(G2)[2 marks]
\(\frac{{27\,679.0 \ldots }}{{2000}}\) (M1)
Note: Award (M1) for dividing their part (g) by 2000.
\( = 13.8395 \ldots \) (A1)(ft)
Notes: Follow through from part (g).
14 (cans) (A1)(ft)(G3)
Notes: Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.[3 marks]
Question
In a company it is found that 25 % of the employees encountered traffic on their way to work. From those who encountered traffic the probability of being late for work is 80 %.
From those who did not encounter traffic, the probability of being late for work is 15 %.
The tree diagram illustrates the information.
The company investigates the different means of transport used by their employees in the past year to travel to work. It was found that the three most common means of transport used to travel to work were public transportation (P ), car (C ) and bicycle (B ).
The company finds that 20 employees travelled by car, 28 travelled by bicycle and 19 travelled by public transportation in the last year.
Some of the information is shown in the Venn diagram.
There are 54 employees in the company.
Write down the value of a.[1]
Write down the value of b.[1]
Use the tree diagram to find the probability that an employee encountered traffic and was late for work.[2]
Use the tree diagram to find the probability that an employee was late for work.[3]
Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.[3]
Find the value of x.[1]
Find the value of y.[1]
Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.[2]
Find \(n\left( {\left( {C \cup B} \right) \cap P’} \right)\).[2]
Answer/Explanation
Markscheme
a = 0.2 (A1)[1 mark]
b = 0.85 (A1)[1 mark]
0.25 × 0.8 (M1)
Note: Award (M1) for a correct product.
\( = 0.2\,\,\,\left( {\,\frac{1}{5},\,\,\,20\% } \right)\) (A1)(G2)[2 marks]
0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)
Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.
\( = 0.313\,\,\,\left( {0.3125,\,\,\,\frac{5}{{16}},\,\,\,31.3\% } \right)\) (A1)(ft)(G3)
Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).[3 marks]
\(\frac{{0.25 \times 0.8}}{{0.25 \times 0.8 + 0.75 \times 0.15}}\) (A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).
\( = 0.64\,\,\,\left( {\frac{{16}}{{25}},\,\,64{\text{% }}} \right)\) (A1)(ft)(G3)
Note: Award final (A1)(ft) only if answer does not exceed 1.[3 marks]
(x =) 3 (A1)[1 Mark]
(y =) 10 (A1)(ft)
Note: Following through from part (c)(i) but only if their x is less than or equal to 13.[1 Mark]
54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)
Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).
= 8 (A1)(ft)(G2)
Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).[2 marks]
6 + 8 + 13 (M1)
Note: Award (M1) for summing 6, 8 and 13.
27 (A1)(G2)[2 marks]