**Mental Section:** Question 1, Question 2, Question 3, Question 5, Question 6, Question 7, Question 8, Question 9, Question 10, Question 11, Question 12

**Written Section:** Question 13, Question 14, Question 15, Question 16, Question 17, Question 18, Question 19, Question 20, Question 21, Question 22, Question 23, Question 24, Question 25, Question 26, Question 27, Question 28

**Question 1**

In a box of 40 Pens, 12 are the colour blue. What proportion of the pens are not blue? Give your answer as a decimal.

**Solution**

If 12 pens are blue, then 40-12=28 pens are not not blue. Thus, the proportion of pens that are not blue is: \frac{28}{40}=\frac{7}{10}=0.7

**Question 2**

400 teachers across the country all claimed travel expenses to go to training seminar. They each travelled 15 miles. The travel expenses are paid at 30 pence per mile. How much collectively did this cost?

**Solution**

400 teachers each travelled 15 miles, therefore they have travelled a combined mileage of 400 \times 15=6000 miles. Each mile travelled costs 30 pence. Hence, the cost is 6000 \times 30=180000 pence in total. However, the question asks for us to give our answer in pounds. Dividing our answer by 100 yields the answer £1800.

**Question 3**

In a class of 25, 5 children play in the school football team. What percentage of the class play in the school football team?

**Solution**

We need to find the proportion of children who play in the team. We know that 5 out of 25 students play in the team. Thus, we write:

\frac{5}{25}Now we could divide 5 by 25 and then multiply the result by 100 to get the answer, but there is a quicker way of doing this. Notice that the denominator (bottom) of the fraction is 25. We know that 25 \times 4 = 100. With fractions, as long as you multiply/divide the numerator and denominator of the fraction by the same number, then the fraction stays the same. Thus, we can multiply the top and bottom of our fraction by 4 to get.

\frac{5}{25}=\frac{5 \times 4}{25 \times 4}=\frac{20}{100}And so we can read off the answer as 20%.

**Question 4**

There were 30 children in the class. 250ml of apple juice was given to each child. How many 1 litre cartons of apple juice were needed?

**Solution**

We first need to find the total volume of liquid required. To do this, we multiply the number 250 by the number of children in the class. This gives us:

250 \times 30 = 7500So we need 7500ml, or 7.5litres of apple juice in total. Since the cartons contain exactly 1 litre of liquid, we need at least 8 apple juice cartons. Thus, the answer is 8.

**Question 5**

At a school fair, a family of 4 people wins £321 in a raffle. The family decides to share the money equally between each of them.

How much will each family member receive?

**Solution**

The calculation we need to do is 321 \div 4. We can compute this by either long/short division or by dividing by 2 twice to get the answer:

321 \div 4=80.25**Question 6**

Two students both attend a sponsored walk to raise money for charity. The each walk 24 miles and are sponsored at 24 pence per mile. How much do they raise in total? Give your answer in pounds and pence.

**Solution**

Each student raises 24 pence per mile that they walk. But each student walks exactly 24 miles, so they each raise 24 \times 24 pence for charity. Since there are two students, we need to multiply this number by 2. Thus, the calculation we need to do is:

24 \times 24 \times 2 = 576 \times 2=1152From the above, we can see that the pair raised 1152 pence between them. But the question asks for the answer in pounds and pence, so we need to divide our answer by 100 to get 11.52.

**Question 7**

What is 325 divided by 0.2?

**Solution**

We need to find 325 \div 0.2. Notice that 0.2=\frac{2}{10}. So dividing by 0.2 is the same as dividing by \frac{2}{10} (since 0.2=\frac{2}{10}). But dividing by a fraction is exactly the same as multiplying by its reciprocal (the flipped fraction). Thus,

325 \div 0.2 =325 \div \frac{2}{10} =325 \times \frac{10}{2} =\frac{3250}{2} =1625

**Question 8**

A student wants to buy £400 worth of dollars for their school trip to New York. At the current rate, you can buy $1.25 for £1. How many dollars will the student receive for £400?

**Solution**

We consider the ratio \pounds 1: \$ 1.25. Since we have £400, we need to multiply 1.25 by 400 to get our answer.

400 \times 1.25 =500Hence, the answer is 500 dollars.

**Question 9**

A school teacher buys 21 pencils, each costing 12 pence. How much does it cost in total? Give your answer in pounds and pence.

**Solution**

We simply compute 21 \times 12. This yields

21 \times 12 =252 pence.

The questions asks for our answer to be given in pounds, so we divide our answer by 100 to get £2.52.

**Question 10**

A PE department has 42 footballs. If 28 of the footballs are blue, what proportion of them are not blue? Give your answer as a decimal to two decimal places.

**Solution**

If 28 footballs are blue then 42-28=14 of the footballs are not blue. So we know that 14 out of a total of 42 footballs are not blue. Thus, the proportion is

\frac{14}{42}=\frac{7}{21}=\frac{1}{3}To get \frac{1}{3} we have used the fact that multiplying/dividing the top and bottom of a fraction by the same number does not change the fraction. We first divided both sides by 2 to simplify the fraction, and then we spotted that we could also divide both sides by 7 to get the fraction in its lowest terms.

Now you might know off the top of your head that \frac{1}{3}=0.\dot{3}=0.333333.... , but if you don’t then you would need to compute 1 \div 3 using short division to get this result.

However, the question asks for the answer to be rounded to two decimal places. Thus, the answer is 0.33.

**Question 11**

In a school the staff are made up of either teachers or Heads of Department. There are 48 teachers in the school. If the Head of Department’s manage a team of 6 teachers, and any 2 Head of Department’s do not manage the same teacher, how many staff are in the school in total?

**Solution**

The total number of staff in the office is going to be the total number of teachers(48) plus the total number of Heads of Department. We first need to calculate the total number of Heads of Department in the office. We know that Heads of Department manage a team of exactly six teachers. Thus, there are 48 \div 6 = 8 Heads of Department in total.

Thus, the total number of staff is 48+8=56.

**Question 12**

A student sends 15 emails to her subject teachers in 5 minutes. On average, how long does it take her to send one email? Give your answer in seconds.

**Solution**

If the student sends 15 text messages in 5 minutes then it takes her 5 \div 15 = \frac{5}{15}=\frac{1}{3} minutes to send one text. But \frac{1}{3}^{\text{rd}} of a minute is \frac{1}{3} \times 60= \frac{60}{3}=20 seconds.

Hence, the answer is 20 seconds.

**Alternative Solution**
Alternatively, we could first work out that five minutes is equal to 300 seconds and then calculate the following:

**Question 13**

A class of ten A-Level maths students individually sat two test papers. The first paper was an algebra paper and the second was calculus. The percentage scores of both papers are shown on the scatter graph below.
What proportion of the class scored higher in algebra than calculus? **Give your answer as a decimal.**

**Solution**

There are ten students in total. We can confirm this by counting that there are a total of ten dots on the graph.

We need to first count the number of students who did better in algebra than in calculus. Notice that the y=x line (grey line) has been added onto the graph. Points that lie exactly on the grey line indicate students who got exactly the same mark in both calculus and algebra. Points that are above the line represent students who performed better in calculus, and points below the line represent students who performed betting in algebra.

All we need to do, therefore, is to count the number of dots below the line; there are 5 in total.

Now all we need to do is to find the proportion in decimal form. There are 5 out of a total of 10, so we write:

\frac{5}{10}=0.5Thus, our answer is **0.5**.

**Question 14**

Calculate the area of the rectangle below:

*(Diagram not drawn accurately) *

**Solution **

To calculate the area we need to multiply the length of the rectangle by the width.

11.5 \times 21.5= 247.25So the answer is **247.25 cm ^{2}**

**Question 15**

The funding over the past three years of two schools, School A and School B, are displayed below.

**Which of the following statements are true?**

**A:** The funding for School A doubled from 2014 to 2015

**B:** The funding for School B saw an increase of 50% from 2014 to 2015

**С:** There was a £20,000 difference in funding between the two schools in 2015

**Solution**

**Statement A**

We read from the bar chart that the funding for School A was £30000 in 2014 and £60000 in 2015. Since 60000=2 \times 30000, we know that statement A is **TRUE**.

**Statement B**

The funding for School B was £40000 in 2014 and £70000 in 2015.

40000\longrightarrow 70000

Let us increase the funding of School B in 2014 by 50% and see if this equals £70000. To increase a number by 50%, we multiply it by 1.5.

40000 \times 1.5=60000

So, if the funding increased by 50% from 2014 to 2015 then the funding in 2015 would be £60000. But, we know from the graph that the 2015 funding was actually £70000, so the statement is **FALSE**.

**Statement C**

In 2015, School A’s funding was £60000 and School B’s funding was £70000.

70000-60000=10000

The difference between the 2015 funding of the schools is £10000, not £20000, so the statement is **FALSE**.

**Thus, the answer is A.**

**Question 16**

The head of a large school wishes to look into reducing work related stress in four different departments. She produces the following table.

What percentage of the School workers suffer from work related stress?

**Give your answer to one decimal place****.**

**Solution**

\text{Proportion of workers suffering from stress} = \frac{\text{Total number of workers suffering from stress}}{\text{total number of workers}}.

We first need to find the total number of workers suffering from work related stress. We can do this by adding up the numbers in the third column:

\text{Total number of workers suffering from stress} = 3+3+2+10=18.

We now need to find the total number of workers. We do this by summing the numbers in the second column:

\text{Total number of workers} = 12+25+14+51=102.

So, we have 18 out of a total of 102 workers that suffer from work related stress. Thus, the proportion is:

\frac{18}{102}=0.1764705...=17.64705... \%

The question asks us to round our answer to one decimal place. Thus, the answer is **17.6%**.

**Question 17**

The table below illustrates the time taken (in minutes) for 50 students to complete a maths problem.

What is the mean amount of time spent on the question? Give your answer in seconds.

**Solution**

\text{Mean amount of time taken}=\dfrac{\text{total number of minutes}}{\text{total number of students}}.

We first need to find the total number of students. We are told in the question that there are 50 students, but we could also sum up the values in the second column instead.

18+27+5=50\text{ students}.

We now need to find the total number of minutes. We need to remember that this is a frequency table; we cannot just simply sum all the values in the first column. Instead, we need to think of it as follows: if 18 people took 1 minute, then they spent a combined total of 1 \times 18=18 minutes; if 27 people took 2 minutes, then they spent a combined total of 2 \times 27=54 minutes; and if 5 people took 3 minutes, then they spent a combined total of 3 \times 5=15 minutes to complete the problem.

Hence, the total number of minutes is 18+54+15=87. Now, we may evaluate the mean to be:

\frac{87}{50}=1.74\text{ minutes} = 1.74 \times 60\text{ seconds}=104.4\text{ seconds}Since the question asks for our answer in seconds only, we round our answer to the nearest second to get **104**.

**Question 18**

Students in a school are given lessons to help reduce the number of a specific type of error which the school calls minors.

The school displayed the information in a chart shown below.

**Which of the following statements are true?**

**A:** Three tenths of the students had less than 10 lessons.

**B:** The highest number of minors a student got was 10.

**C:** The median number of lessons had 1 minor

**Solution**

**Statement A**
We can see from the graph that a total of three students had less than 9 lessons. We can see that one student had a total of 6 lessons, one had 7 lessons and one had 8 lessons. We now need to find out how many students there were in total. To do this, we count all of the points on the scatter graph: there are 11 in total. So, we have 3 out of a possible 11 students that had 9 lessons or less. Thus, the proportion is:

Clearly \frac{3}{11}\; (0.27...) does not equal \frac{3}{10}\; (0.3), so the statement is **FALSE**.

**Statement B**
The number of minors a student got is dictated by the height of the point (y axis). The further to the right a point is, the more lessons that student had. The further to the top a point is, the more minors the student received. Thus, we need to look for the highest point on the graph. We can see that the highest point on the graph represents the student who had only six lessons. That student got 10 minors. Hence, the statement is **TRUE**.

**Statement C**
The median of a data set is given by:

Where n represents the number of data points. In this example, n=11. We get the number 11 by counting the number of points on the graph.

Using the formula, our median is the

\frac{11+1}{2}=6^{\text{th}}\text{ number}Now, we are asked to find the median number of lessons. Since the number of lessons is on the horizontal axis, we need to count the points horizontally. Counting the points horizontally from smallest to largest, we get that the median number of lessons is 10 lessons.

We now read from the graph that the point representing 10 lessons had one minor. Thus, the statement is **TRUE**.

Hence, the answer is** B,C.**

**Question 19**

A teacher wishes to test the effectiveness of two revision methods: Method A and Method B. To do this, a sample of students were asked to state whether they felt their revision had been effective at increasing their understanding of a topic. The results are compiled in the table below.

**Which of the following statements are true?**

**A:** 1/6 of the students using Method 1 reported ‘high’ effectiveness.

**B:** The majority of students in the sample for Method 2 reported ‘low’ effectiveness.

**C:** ¼ of the total students across both samples reported ‘low’ effectiveness.

**Solution**

**Statement A**
We read from the table that the total number of Method 1 students that reported high effectiveness was 5. So we have that 5 out of a total of 30 students reported high effectiveness. Thus, the proportion is:

Hence, statement A is **TRUE.**

**Statement B**
We read from the table that out of the 30 students who used Method 2 only 7 reported low effectiveness. Thus, statement B is **FALSE**.

**Statement C**
In total there are 30+30=60 students across both samples. In total, 5+7=12 students reported low effectiveness. Hence, the proportion is:

But the statement claims that \frac{1}{4}=0.25 of the students reported low effectiveness, so the statement is **FALSE**.

Hence, the answer is **A.**

**Question 20**

The percentage scores of a spelling test sat by year 7 pupils are displayed in the cumulative frequency diagram below:

Which of the following statements are true? A: The median mark is 60 B: The median mark is 30 C: Not one pupil got a score above 90

**Solution**

**Statement A**

To find the median of a cumulative frequency diagram, we first need to find the total number of people. To do this, we simply read off the highest level that the line reaches. We can see from the diagram that the highest point of the line represents a cumulative frequency of 100. Thus, there are 100 people in the sample.

All we need to do now is to divide this number by two \frac{100}{2}=50 and then read off the number of marks that represents a cumulative frequency of 50.

We can see that a mark of 60 represents a cumulative frequency of 50. Thus, the median mark is 60.

Hence, statement A is **TRUE**.

**Statement B**

We know from the above that the median is 60 so this statement is **FALSE**.

**Statement C**

We can see from the graph that the line is flat between 90 and 100 marks. This means that not one person got a mark above 90, so statement C is **TRUE**.

Thus, the answer is **A,C**.

**Question 21**

The revenues for three companies over a three-year period are shown in the bar chart below.

In which year(s) did Company B earn one third of the total revenues of all three companies combined?

**Solution**

To answer this question, we need to calculate the proportion of market share that Company B had compared to the total of all three companies.

**2014**
\text{Total market share}=90.
\text{Company B market share}=90-60=30.

In 2014, Company B had a market share of 30 out of a total of 90. Thus, the proportion is:

\frac{30}{90}=\frac{1}{3}**2015**
\text{Total market share}=60.
\text{Company B market share}=60-50=10.

In 2015, Company B had a market share of 10 out of a total of 60. Thus, the proportion is:

\frac{10}{60}=\frac{1}{6}**2016**
\text{Total market share}=80.
\text{Company B market share}=80-50=30.

In 2015, Company B had a market share of 30 out of a total of 80. Thus, the proportion is:

\frac{30}{80}=\frac{3}{8}The only year in which Company B had one third of the market share across all three companies was in 2014, so our answer is **2014**.

**Question 22**

Some students in two Maths classes had to resit a test. Out of those who were resitting, some were required to resit one test whilst others had to resit both. The results are displayed in the pie charts below. If there are 25 people in class 1 and there are the same number of people resitting the Trigonometry test in both classes, how many students are there in class 2?

**Solution**

We are told that there are 25 students in class 1 and the pie chart tells us that 20% of the students in class 1 are resitting trigonometry. Thus, a total of 25 \times 0.2=5 of the pupils in class 1 are resitting trigonometry.

But the questions tells us that the there are the same number of people resitting trigonometry in both classes, so we know that there are also 5 people resitting trigonometry in class 2.

In fact, we can tell from the pie chart on the right that 12.5% of people in class 2 are resitting trigonometry. This tells us that:

0.125 \times \text{ the number of students in class 2}=5Hence, there are \frac{5}{0.125}=40 students in class 2. Thus, the answer to the question is **40.**

**Question 23**

The marks awarded for a piece of English coursework for two different classes in a school are compared by using box and whisker plots, as shown below.

Which of the following statements are true? A: The median in class A is 30. B: The highest mark was in class B C: The range of marks is bigger in class A

**Solution**

**Statement A**

We read off from the boxplot on the left that the median mark is 30, so the statement is **TRUE**.

**Statement B**

The highest mark is class A was 40 and the highest mark in class B was 45. Hence, the statement is **TRUE**.

**Statement C**

The range of marks in class A is 40-20=20 and the range of marks in class B is 45-10=35, so the statement is **FALSE**.

Thus, the answer to the question is **A,B.**

**Question 24**

The number of boys and girls at different schools is shown in the table below. Each school is broken up into Key Stages 3, 4 and 5.

**Examine the table and then identify the incorrect statement(s) from the options.**

**A:** School C3 has over three times as many pupils as B2.

**B:** School C3 has over three times as many pupils in KS3 compared to school B2.

**C:** School A1 has the highest percentage of boys studying at KS5, when compared to all pupils within the school.

**Solution**

**Statement A:** School C3 has over three times as many pupils as B2.

The number of pupils in B2, is all boys and girls across all three key stages:

123 + 58 + 3 + 167 + 62 + 7 = 420

Multiplying this by 3 gives, 1260.

The number of pupils in C3 is:

312 + 240 +78 + 302 + 256 + 99 = 1287.

As there are 1287 pupils in C3, which is greater the three times that in B2**– statement TRUE.**

—

**Statement B:** School C3 has over three times as many pupils in KS3 compared to school B2.

C3 has 312 boys in KS3 and 302 girls. In total is has 614 students in KS3.

B2 has 123 boys and 167 girls in KS3, a total of 290 students.

290 × 3 = 870, which is greater than the 614 students in C3, meaning that this** statement is FALSE.**

—

**Statement C:** School A1 has the highest percentage of boys studying at KS5, when compared to all pupils within the school.

A1: Total number of pupils = 1112

B2: Total number of pupils = 420

C3: Total number of pupils = 1287

The Percentage for each school can be calculated as:

A1 has the highest Percentage of boys in Key Stage 5,** this statement is TRUE.**

**The means the only incorrect statement is B**

**Question 25**

The bar chart below shows the percentages of students in years 7, 8 and 9 who have had at least one detention this year. 36 out of the 144 students in year 9 have had at least one detention.

How high should the bar chart for year 9 be? Give your answer as a percentage.

**Solution**
All we need to do here is evaluate the proportion. We know that 36 out of the 144 students in year 9 have had at least one detention, so we evaluate the proportion as:

Hence, the answer is **25%**.

**Question 26**

The table below shows the raw scores and Standardised scores, over three tests, for 6 pupils in their summer examinations.

**What is the mean standardised score for pupils aged 11 and older?**

**A:** 37

**B:** 38

**C:** 39

** D:** 40

**Solution:**

__Step 1__

Find the total score for each pupil that is aged 11 and older:

\begin{aligned}\text{B: }&4+3+4=11 \\ \text{C: }&3+5+5=13 \\ \text{F: }&4+4+4=12\end{aligned}

__Step 2__

Lookup the standardised score for each pupil based on their raw score.

Pupil B has a raw score of 11 and a standardised score of 34

Pupil C has a raw score of 13 and a standardised score of 41

Pupil F has a raw score of 12 and a standardised score of 39

__Step 3__

Find the mean of the three standardised scores

(34+41+39) \div 3 = \bold{38}

**Answer = B**

**Question 27**

Two pupils sat the same two tests and their marks were recorded. The tests were weighted as follows: test 1 had a 40% weighting and test 2 had a 60% weighting. The formula to calculate their final mark is given by:

*Final mark = (test1mark x 0.4)+(test2mark x 0.6)*
Which student achieved the highest final mark?

**Solution**
To answer this question, we simply substitute the values in the table into the formula given in the question:

\text{Final mark}=(\text{test 1 mark}\times 0.4 )+(\text{test 2 mark} \times 0.6).

**Pupil A**
\text{Final mark}=(80\times 0.4 )+(75 \times 0.6)=77.

**Pupil B**
\text{Final mark}=(70\times 0.4 )+(82 \times 0.6)=77.2.

It is clear from the above that Pupil B achieved the highest final mark, so our answer is **B**.

**Question 28**

The number of hours 7 students spent on revision in the winter and summer terms were recorded. The results are displayed in the table below.

What fraction of the students spent more hours revising in the summer term than in the winter term?

**Solution**
To begin, we need to go through each student individually and count the number of students who spent more hours on revision in the summer than they did in the winter. From the table, we can see that students A, C, F and G spent more hours on revision in the summer than they did in the winter.

There are 7 students in total. We have identified that 4 out of seven students spent more hours revising in the summer than in the winter. As a proportion, this is:

\frac{4}{7}