Problem 1:
The total mass of five rugby players is 425kg.
The average (mean) mass of ten ballet dancers is 40kg.
What is the average mass of all fifteen people?
Problem 2:
An ice cream stall sells vanilla, strawberry and chocolate ice creams.
The pie chart illustrates the sales of ice cream today.
The number of vanilla and the number of chocolate ice
creams sold were the same.
The stall sold 60 strawberry ice creams.
How many chocolate ice creams were sold?
Problem 3:
The mean mass of five giant dates was 50g.
Kate ate one, and the mean mass of the four remaining dates was 40g.
What was the mass of the date that Kate ate?
Problem 4:
A tourist attraction, which opens every day, needs 30 000 visitors per day on average to
break even.
Last week there were 120 000 visitors.
What is the number of visitors needed this week to break even over the two-week
period?
Problem 5:
Here is a table of information about the wingspan of bats. Unfortunately, two entries are
missing from the table.
Student
A
B
Data set (wingspans in cm)
13
16
12
10
13
16
13
15
Mean (cm)
13
Student A collected 6 measurements and worked out that their mean was 13cm.
Student B collected 4 measurements.
The overall mean of the combined data was 13.4cm.
Can you work out what the two missing values are?
ANSWERS
10 ballet dancers is 10×40kg=400kg
The average mass of all 15 people is
(425kg+400kg)/15=825kg/15=55kg
1. The mass of the
2.
90 chocolate ice creams were sold.
The angle occupied by the 'chocolate' sector is
1/2(360∘−90∘)=135∘.
1.5 times bigger than the 'strawberry' sector, so the number of
chocolate ice creams sold is 1.5×60=90.
This is
3. The total mass of the original five dates was 5 x 50g = 250g.
The total mass of the four remaining dates was 4 x 40g = 160g.
Therefore the mass of the date that Kate ate was 250g - 160g = 90g.
4. Over a two-week period, the required number of visitors is 14 x 30 000 =
420 000.
Therefore the number required this week is 420 000 - 120 000 = 300 000
visitors
5. Student A's bats have an average wingspan of 13 cm. Student A has 6 bats, and
so the sum of their wingspans divided by 6 is equal to 13.
So the sum of student A's bats' wingspans is 6×13=78
So the missing measurement can be found by subtracting the five known
wingspans from the total: 78−(13+16+12+10+15)=12 cm.
Next, we can find student B's mean using the total mean:
There are 10 bats altogether, and the average wingspan is 13.4 cm, so the total
wingspan is 10×13.4=134 cm.
We already know that student A's bats' had a total wingspan of 78 cm. So student
B's bats' total wingspan must be 134−78=56 cm.
So B's missing measurement can be found by subtracting the three known
wingspans from the total: 56−(13+16+13)=14 cm.
By 'balancing' the numbers
We can think of the average bat wingspan as being the wingspan that each bat
would get if the total amount of wing were shared out equally between all of the
bats.
So total wingspan of student A's bats is the same as the total wingspan of 6 bats
which each had a wingspan of 13 cm. We can write them below each other and
then make the sums 'balance':
Notice that the first 13 has 13 below it, so these are already balanced. 16 is 3
more than 13 and 10 is 3 less than 13, so together 10 and 16 balance out 13 and
13. This is shown below:
12 and 15 do not balance with 13 and 13, because 12 is 1 less than 13 but 15 is
2 more than 13. That is 1 below but 2 above, so we need another 1 below in
order to balance the sums.
So the missing measurement must be 12.
Then, we can find the average for student B.
Altogether there are 10 bats, and if the wingspan was shared out equally, each
would get 13.4 cm. Alternatively, we could share the wingspan out by student.
Student A's bats would each get 13 cm and student B's bats would each get ? cm
(their respective averages). This is shown below:
But student A's bats each get 0.4 less than 13.4, so in total, they contribute
6×0.4 less than their fair share to the total. So student B's bats must contribute
6×0.4 more than their fair share to the sum.
6×0.4 is the same as 4×0.6, so each of student B's 4 bats could have 0.6 cm more
than the average wingspan - which is 13.6 + 0.6 = 14 cm.
Finally, we can find student B's missing measurement:
The 13s are each 1 below 14, and the 16 is 2 above 14. So in the bottom sum,
we have 2 below and 2 above 14 - which balances out to 14! So the sums are
already balanced, so ? must be 14 to maintain this balance.
So student B's missing measurement is 14 cm.