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Either paper 5 marks
38 students took a mock exam. The frequency tree shows some information about the results.
Altogether 25 of the students passed.
(a)Complete the frequency tree.
(b)One of the 38 students is chosen at random.
Work out the probability that the student is a girl who passed.
Show answer
(a) Girls 25, boys who failed 5, girls who passed 17, girls who failed 8
(b) \(\frac{17}{38}\)There are 38 students and 13 of them are boys, so \(38 - 13 = 25\) are girls.
8 of the 13 boys passed, so \(13 - 8 = 5\) boys failed.
25 students passed altogether, so \(25 - 8 = 17\) of them are girls.
That leaves \(25 - 17 = 8\) girls who failed.
Check: \(5 + 8 = 13\) students failed, and \(25 + 13 = 38\).
The completed tree shows 17 girls who passed, out of 38 students.
\(P = \frac{ 17 }{ 38 }\)
As a fraction in its simplest form this is \(\frac{17}{38}\).
- (a) M1girls \(38 - 13 = 25\) or boys who failed \(13 - 8 = 5\)
- (a) M1girls who passed \(25 - 8 = 17\)
- (a) A1all four values correct, girls who failed \(8\)
- (b) M1girls who passed \(17\) over total \(38\)
- (b) A1simplest form \(\frac{17}{38}\)
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Either paper 4 marks
33 students were asked how they travelled to school one morning and whether they were late. The frequency tree shows the results.
(a)One of these students is chosen at random.
Calculate the probability that the student walked and was late.
(b)One of the students who came by bus is chosen at random.
Calculate the probability that this student was late.
Show answer
(a) \(\frac{4}{33}\)
(b) \(\frac{4}{15}\)The bottom of the tree shows 4 students who walked and were late, out of 33 students altogether.
\(P = \frac{ 4 }{ 33 }\)
As a fraction in its simplest form this is \(\frac{4}{33}\).
Only the bus branch counts here, so the total to divide by is 15, not 33.
4 of those 15 students were late.
\(P = \frac{ 4 }{ 15 }\)
As a fraction in its simplest form this is \(\frac{4}{15}\).
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Either paper 4 marks
79 members pay to use a gym.
They all have the standard membership for \(£22\) per month.
Some also have the swimming pool pass for \(£17\) extra per month.
Some also have the fitness classes pass for \(£4\) extra per month.
The frequency tree shows the number of members with each package.
In total, how much per month do the 79 members pay?
Show answer
\(\text{£}2519\)Everyone has the standard membership, so the base charge is paid by all \(79\) members:
\(79 \times £22 = £1738\)
The tree splits first on the swimming pool pass. The branch for those who have it shows \(37\) members:
\(37 \times £17 = £629\)
The fitness classes pass is split across both branches, so add the two groups who have it, \(16\) and \(22\):
\(16 + 22 = 38\) members
\(38 \times £4 = £152\)
Total per month:
\(£1738 + £629 + £152 = £2519\)
- M1base charge for all \(79\) members: \(£1738\)
- M1first extra: \(37\) members at \(£17\) = \(£629\)
- M1second extra: \(38\) members (two groups added) at \(£4\) = \(£152\)
- A1total \(£2519\)
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Either paper 3 marks
A walking group has 52 members. One week it ran a Wednesday walk and a Saturday walk.
The frequency tree shows some information about the members who went on the walks.
27 members went on the Saturday walk.
The Venn diagram is to show the same information. The number who went on both walks has been filled in from the tree. The other regions are labelled P, Q and R.
\(\xi\) = the 52 members, \(W\) = members who went on the Wednesday walk, \(S\) = members who went on the Saturday walk.
(a)Work out the number of members in region P.
(b)Work out the number of members in region Q.
(c)Work out the number of members in region R.
Show answer
(a) \(18\)
(b) \(19\)
(c) \(7\)Each region of the Venn diagram is one walk without the other, so a branch total of the tree cannot be copied across.
Both walks: the leaf for Wednesday and Saturday on the tree gives \(8\) members.
The Wednesday and not Wednesday branches add up to \(52\), so \(15 + 37 = 52\).
Wednesday only: the Wednesday total, \(15\), includes the \(8\) who went on both, so \(15 - 8 = 7\).
Saturday only: everyone who went on the Saturday walk was in Saturday only or in both, so take the Saturday total minus those in both: \(27 - 8 = 19\).
Neither: the four regions add up to \(52\), so \(52 - 7 - 8 - 19 = 18\).
In the Venn diagram \(W\) is Wednesday and \(S\) is Saturday, so Wednesday only is \(7\), both is \(8\), Saturday only is \(19\) and neither is \(18\).
Region P is the region for neither: \(18\) members.
Each region of the Venn diagram is one walk without the other, so a branch total of the tree cannot be copied across.
Both walks: the leaf for Wednesday and Saturday on the tree gives \(8\) members.
The Wednesday and not Wednesday branches add up to \(52\), so \(15 + 37 = 52\).
Wednesday only: the Wednesday total, \(15\), includes the \(8\) who went on both, so \(15 - 8 = 7\).
Saturday only: everyone who went on the Saturday walk was in Saturday only or in both, so take the Saturday total minus those in both: \(27 - 8 = 19\).
Neither: the four regions add up to \(52\), so \(52 - 7 - 8 - 19 = 18\).
In the Venn diagram \(W\) is Wednesday and \(S\) is Saturday, so Wednesday only is \(7\), both is \(8\), Saturday only is \(19\) and neither is \(18\).
Region Q is the region for Saturday only: \(19\) members.
Each region of the Venn diagram is one walk without the other, so a branch total of the tree cannot be copied across.
Both walks: the leaf for Wednesday and Saturday on the tree gives \(8\) members.
The Wednesday and not Wednesday branches add up to \(52\), so \(15 + 37 = 52\).
Wednesday only: the Wednesday total, \(15\), includes the \(8\) who went on both, so \(15 - 8 = 7\).
Saturday only: everyone who went on the Saturday walk was in Saturday only or in both, so take the Saturday total minus those in both: \(27 - 8 = 19\).
Neither: the four regions add up to \(52\), so \(52 - 7 - 8 - 19 = 18\).
In the Venn diagram \(W\) is Wednesday and \(S\) is Saturday, so Wednesday only is \(7\), both is \(8\), Saturday only is \(19\) and neither is \(18\).
Region R is the region for Wednesday only: \(7\) members.
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Either paper 5 marks
54 people were asked how they got to work one morning and whether they arrived late. The frequency tree shows some of the results.
Altogether 16 of them arrived late.
(a)Complete the frequency tree.
(b)One of the 54 people is chosen at random.
Work out the probability that this person went by train and was on time.
Show answer
(a) Walk 17, car and on time 12, train and on time 15, walk and late 6
(b) \(\frac{5}{18}\)The three ways of travelling account for everybody, so \(54 - 19 - 18 = 17\) people walked.
Car: \(19 - 7 = 12\) were on time.
Train: \(18 - 3 = 15\) were on time.
16 people were late altogether, so \(16 - 7 - 3 = 6\) of the walkers were late.
Check the walking branch: \(6 + 11 = 17\).
The completed tree shows 15 people who went by train and were on time, out of 54.
\(P = \frac{ 15 }{ 54 }\)
As a fraction in its simplest form this is \(\frac{5}{18}\).
- (a) M1walkers \(54 - 19 - 18 = 17\)
- (a) M1on time by car \(12\) or by train \(15\), or late walkers \(16 - 7 - 3 = 6\)
- (a) A1all four values correct
- (b) M1train and on time \(15\) over total \(54\)
- (b) A1simplest form \(\frac{5}{18}\)
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Either paper 5 marks
A group of members were asked whether they had renewed.
\(200\) members were asked altogether.
\(40\%\) of the members were adults.
\(80\%\) of the adults had renewed.
(a)Work out how many of the members were juniors.
(b)Work out how many of the adults had renewed.
(c)One of the \(200\) members is chosen at random.
Calculate the probability that this person had renewed.
Show answer
(a) \(120\ \text{members}\)
(b) \(64\)
(c) \(\frac{143}{200}\)\(40\%\) of \(200\) were adults:
\(\frac{ 40 }{100} \times 200 = 80\)
Everyone else was juniors.
\(200 - 80 = 120\)
This percentage is of the adults only, not of everyone.
\(\frac{ 80 }{100} \times 80 = 64\)
Both branches of the tree end in had renewed: \(64\) adults and \(79\) juniors.
\(64 + 79 = 143\)
\(P = \frac{ 143 }{ 200 } = \frac{143}{200}\)
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Either paper 5 marks\(245\) students from Year 10 and Year 11 take part in a competition.
- There are \(147\) more Year 10 students than Year 11 students.
- \(53\) of the students win a medal.
- \(12\) of the students who win a medal are in Year 11.
(a)Calculate the number of Year 10 students.
(b)Use your answer to part (a) to complete the frequency tree.
Show answer
(a) \(196\ \text{students}\)
(b) Year 10 196: Medal 41, No medal 155. Year 11 49: Medal 12, No medal 37Take the extra \(147\) away and the two groups are equal:
\(245 - 147 = 98\)
\(98 \div 2 = 49\), so there are 49 Year 11 students.
There are \(49 + 147 = 196\) Year 10 students.
The number of Year 10 students is the one asked for.
The other group is \(245 - 196 = 49\).
The winners add up to \(53\), and \(12\) of them are in Year 11, so the winners among the Year 10 students number \(53 - 12 = 41\).
Everyone else in each group did not win:
Year 10: \(196 - 41 = 155\)
Year 11: \(49 - 12 = 37\)
- (a) M1one stage of the sharing, \(245 \div 5 = 49\), or taking off the difference in the other wording
- (a) A1196 Year 10 students
- (b) M1the other group, \(245 - 196 = 49\)
- (b) M1winners among the Year 10 students, \(53 - 12 = 41\)
- (b) A1both groups without the prize: 155 and 37
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Either paper 5 marks
Some students travel to school by bus or on foot.
Branch A is the students who came by bus, and branch B is the students who walked.
The number of students who walked is \(2\) times the number who came by bus.
\(\frac{1}{ 4 }\) of the students who walked were late.
(a)Calculate how many students came by bus.
(b)Calculate how many students there were altogether.
(c)Calculate the probability that a randomly chosen student was late.
Give your answer as a fraction in its simplest form.
Show answer
(a) \(26\ \text{students}\)
(b) \(78\ \text{students}\)
(c) \(\frac{11}{39}\)Branch A splits into \(9\) who were late and \(17\) who were on time.
\(9 + 17 = 26\)
\(2\) times as many walked as came by bus:
\(2 \times 26 = 52\)
\(26 + 52 = 78\)
On branch B, \(\frac{1}{ 4 }\) of \(52\) were late:
\(52 \div 4 = 13\)
Altogether \(9 + 13 = 22\) were late.
\(P = \frac{ 22 }{ 78 } = \frac{11}{39}\)
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Either paper 5 marks
62 students took a mock exam. The frequency tree shows some information about the results.
Altogether 37 of the students passed.
(a)Complete the frequency tree.
(b)One of the 62 students is chosen at random.
Calculate the probability that the student is a girl who passed.
Show answer
(a) Girls 30, boys who failed 13, girls who passed 18, girls who failed 12
(b) \(\frac{9}{31}\)There are 62 students and 32 of them are boys, so \(62 - 32 = 30\) are girls.
19 of the 32 boys passed, so \(32 - 19 = 13\) boys failed.
37 students passed altogether, so \(37 - 19 = 18\) of them are girls.
That leaves \(30 - 18 = 12\) girls who failed.
Check: \(13 + 12 = 25\) students failed, and \(37 + 25 = 62\).
The completed tree shows 18 girls who passed, out of 62 students.
\(P = \frac{ 18 }{ 62 }\)
As a fraction in its simplest form this is \(\frac{9}{31}\).
- (a) M1girls \(62 - 32 = 30\) or boys who failed \(32 - 19 = 13\)
- (a) M1girls who passed \(37 - 19 = 18\)
- (a) A1all four values correct, girls who failed \(12\)
- (b) M1girls who passed \(18\) over total \(62\)
- (b) A1simplest form \(\frac{9}{31}\)
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Either paper 4 marks
38 students were asked how they travelled to school one morning and whether they were late. The frequency tree shows the results.
(a)One of these students is chosen at random.
Calculate the probability that the student walked and was late.
(b)One of the students who came by bus is chosen at random.
Calculate the probability that this student was late.
Show answer
(a) \(\frac{7}{38}\)
(b) \(\frac{4}{19}\)The bottom of the tree shows 7 students who walked and were late, out of 38 students altogether.
\(P = \frac{ 7 }{ 38 }\)
As a fraction in its simplest form this is \(\frac{7}{38}\).
Only the bus branch counts here, so the total to divide by is 19, not 38.
4 of those 19 students were late.
\(P = \frac{ 4 }{ 19 }\)
As a fraction in its simplest form this is \(\frac{4}{19}\).