Numeracy, Data & IT Homework: Detailed Solutions and Analysis

This assignment solution covers numeracy, data analysis, and IT skills. The numeracy section involves fractions, percentages, and basic calculations, including finding equivalent fractions, calculating percentages of books in a library, and determining discounts. The data analysis section interprets a table of Olympic medal data, comparing countries' performances and identifying trends. The IT section focuses on using Excel functions, such as RANK and IF, to analyze data. The solution includes detailed explanations and step-by-step instructions for each question, providing a comprehensive guide for students studying numeracy, data analysis, and IT.

Numeracy, Data & IT

TABLE OF CONTENTS
PART 1 – USING NUMERACY....................................................................................................4
Q1................................................................................................................................................4
Q2................................................................................................................................................4
Q3................................................................................................................................................4
Q4................................................................................................................................................5
Q5................................................................................................................................................5
Q6................................................................................................................................................6
Q7................................................................................................................................................6
Q8. ..............................................................................................................................................6
Q9................................................................................................................................................7
Q10. ............................................................................................................................................7
PART 2 – USING DATA................................................................................................................8
f)..................................................................................................................................................8
g).................................................................................................................................................8
h).................................................................................................................................................9
i)..................................................................................................................................................9
j)..................................................................................................................................................9
Question 11...............................................................................................................................10
a)................................................................................................................................................10
b)...............................................................................................................................................10
c)................................................................................................................................................10
d)...............................................................................................................................................11
PART 3 – USING IT.....................................................................................................................11
Question 12...............................................................................................................................11
Question 13...............................................................................................................................12
a)................................................................................................................................................12
b)...............................................................................................................................................13
c)................................................................................................................................................14

d)...............................................................................................................................................15
e)................................................................................................................................................15
Question 14...............................................................................................................................16
a)................................................................................................................................................16
b)...............................................................................................................................................17
c)................................................................................................................................................17
d)...............................................................................................................................................18
Question 15...............................................................................................................................19
a)................................................................................................................................................19
b)...............................................................................................................................................20
c)................................................................................................................................................21
d)...............................................................................................................................................22
Question 16...............................................................................................................................22
a)................................................................................................................................................22
b)...............................................................................................................................................23
REFERENCES..............................................................................................................................25

PART 1 – USING NUMERACY
Q1.
a.) Numerator : Numerator is a Latin word that means 'number'. It is the part of a fraction that
show at above the line of fraction to show the number of parts of the whole. For example, a
fraction 3/9 in that 3 is the numerator.
b.) Denominator : Denominator is also a Latin word that means 'namer'. It is the part of a fraction
that show at below the line of fraction for work as the divisor of the numerator (Loc, Tong and
Chau, 2017). It represents the equal parts of the number in an item. For ex. A fraction 3/9 in that
9 is the denominator.
Q2.
Numerator 24 18
Denominator 40 42
The Simplest form 24/40 = 3/5 18/42 = 3/7
The Simplest forms of fraction means only 1 is the common factor of its numerator and
denominator. For this, first find the HCF ( the highest common factor) of the fraction, then
divide both numerator & denominator by HCF & then fraction is reduced at its lowest term.
Q3.
a.)
Numerator 2 3 5
Denominator 3 4 6
Equivalent fractions
with a denominator of
12
4 3 2
New fraction 2/3 * 4/4 = 8/12 3/4 * 3/3 = 9/12 5/6 * 2/2 = 10/12
Two or more fractions are equal to the same fraction when simplified is said to be equivalent. In
thus process both numerator & denominator is simplified with same value (Barahmand, 2020).

b.)
Total books in library 60,000
Business books 14,000
Healthcare books 22,000
Psychology and law books 12,000
Remaining books 12,000
Computing books are 2/3 of remaining books 12,000 * 2/3 = 8,000
Computing books % 8,000/60,000 * 100 = 13.33%
In above solution, computing books are calculated. For this first calculate the remaining total
books of library then from remaining books calculate computing books that are 2/3rd of the
remaining books.
Q4.
No. Of pairs of shoes bought by Liz 2
She gave three crisp £50 notes
Total money she gave £150
Money she returned £10.50
Total price of 2 pairs of shoes £139.50
Price of 1 pair of shoes £69.75
In this solution, Liz purchase two pairs of running shoes. She pay three crisp £50 notes and given
change of £10.50. According to that given information, first calculate the total price of two pair
of shoes that is 139.50 by total money she gave 50×3=150 less given change of 10.50. From
139.50 calculate the price of each pair of shoes that is 69.75 by dividing 139.50/2.
Q5.
a.) 240.50×19.54 = 4699.37

In this solution, answer is given in two significant figures.
b.) 52100 = 5.21×1000
= 5.21×10^4
Represent the number 52100 as a power of 10
Q6.
a.)
Discount offer 30%
Patty and her 2 siblings pay total amount of
(70% payment)
£210
1% = (£210 / £70) £3
100% £300
Total savings £300 - £210 = £90
Patty and her 2 siblings means total 3 persons had a discount of 30% means that they paid 70%.
To find the total savings they made, first calculate the total cost (100%) that is total of discount
(30%) and payment (70%) they made (Dodoo, Gustavsson and Tettey, 2017). If 1% is equal to
[210/70 = £3] then 100% the gym offers will be £300 then total savings thay made is 300-210 =
90.
b.)
Average saving per person 30
Average saving of per person is calculated by total saving (90) divided by number of total person
(3).
Q7.
a.)
0.75 – 0.78 + (0.66) 0.63
b.)

The Largest number 0.1
Q8.
Total no. of men 90
Total no. of women 60
Total no. of people 150
Total people said yes 90
From total person women said yes 3/10 * 90 = 27
Total men said yes 90 – 27 = 63
Total men said no 90 – 63 = 27
In this solution, calculate the % of men said no. 90 men and 60 women are there i.e. total 150
persons are there from those 150 altogether 3/5 of people said yes that is 150*3/5 = 90. From
those 90 people calculating the 3/10 of women who said yes that is 27 that means total men said
yes is 90-27=63. Total men said no (27) are calculated by total men (90) less the number of men
said yes (63).
Q9.
Conference time in Birmingham is at 10:30 am.
For that, she has to leave her home at 08:00 am.
Described the steps below :
Time to reach Euston Rail station from her home is 1 hour, so she will be station at 09:00 am
Birmingham trains run at the following times:
5 minutes past the hour, 25 minutes past the hour and 45 minutes past the hour i.e. she will get
her train at 9.05 am.
Journey of train from Euston to Birmingham is 1 hour & 10 min. so she will reach to
Birmingham at 10:15 am.
Meeting venue in Birmingham is a 5-minute walk from the station according to that she will
reach the venue at 10:20 am.

According to that, she will reach exactly 10 minutes before the meeting.
Important note : If Annabelle wants to overcome 10 minutes for that she leaves at 8:10 am from
her home, then she will reach at station 9:10 am. According to the timings of train, she will get
her train at 9:25 am & at the end she will reach at venue 10 minutes late (Rawson, 2019).
Q10.
Weight of a Shredded wheat box is 0.35 kg
Weight of a Weetabix box is 0.36 kg
Weetabix box is heavier by 0.01 kg
In the above solution, calculate the weight of heavier box by using information given in the
question. Weight of Shredded wheat box is 0.35 kg & weight of a Weetabix box is 9/25 i.e. 0.36
kg, so the weight of Weetabix box is more than Shredded wheat box (Baidoo, 2019).
PART 2 – USING DATA
f)
Team Total Games Gold Silver Bronze Total
Australia 26 147 163 187 497
China 10 227 165 151 543
France 28 212 241 260 713
Germany 24 275 313 349 937
Great Britain 28 263 295 289 847
Hungary 26 175 147 169 491
Italy 27 206 178 193 577
Soviet Union 10 440 357 325 1122
Sweden 27 147 170 179 496
United States 27 1,022 794 704 2,520

In the given table, the country Great Britain has gold medal (263), silver medal (295) and bronze
medal (289). The country Germany & Soviet Union has gold medal (275, 440), silver medal
(313, 357) & bronze medal (349, 325). Apart from united states, the country Germany and Soviet
Union has more gold medals, more silver medals and more bronze medals than Great britain.
g)
Team Comparing no. of total games with overall no.
of medals received
Australia 497/26 = 19.11
China 543/10 = 54.3
France 713/28 = 25.46
Germany 937/24 = 39.04
Great Britain 847/28 = 30.25
Hungary 491/26 = 18.88
Italy 577/27 = 21.37
Soviet Union 1122/10 = 112.2
Sweden 496/27 = 18.37
United States 2,520/27 = 93.33
In the above table, it shows that the comparing of total no. of medals with total no. of games in
country participated & it gives the result that country United States gives the best performance
i.e. the highest no. of medals received per game (93.33).
h)
Two reasons of why a country like Jamaica does not feature in the top 10 medals:

 Lack of resources : the country has lack of resources for practicing such as trainer,
practicing ground, players, system of rules and so on. Due to lack of this resources
country is not participating in olympic games.
 Low population : the country has low population, due to this reason they do not
participate in other games other than athletes. They perform well in athletes and mostly
participate in that games (Farahani, Rabiee and Vatanpour, 2019).
i)
From the above table, it shows that country Soviet Union has close competitor of United States.
The answer is justified by table given below. In that table it show that in bronze medal category
country United States is far outperformed from its closest competitor that is country Soviet
Union.
Gold Silver Bronze
1,022 794 704
440 357 325
1,022/440 = 2.32 794/357 = 2.22 704/325 = 2.17
j)
Team Gold Silver Bronze Range (highest-
lowest)
Australia 147 163 187 187-147 = 40
China 227 165 151 227-151 = 76
France 212 241 260 260-212 = 48
Germany 275 313 349 349-275 = 74
Great Britain 263 295 289 295-263 = 32
Hungary 175 147 169 175-147 = 28
Italy 206 178 193 206-178 = 28

Soviet Union 440 357 325 440-325 = 115
Sweden 147 170 179 179-147 = 32
United States 1,022 794 704 1,022-704 = 318
Through the above working table, it is shown that country Hungary, Italy, Sweden & Great
Britain had the most evenly distributed number of gold, silver & bronze medals i.e. least smallest
range (Pak, 2020).
Question 11
a)
The lowest number of overall medals out of the ten countries is Hungary. It is because the
country has awarded only 491 medals which is minimum among all the others. The lowest
number is calculated in excel using the simple minimum function whose formula is
=Minimum(number1, number2, number3.....) (Sinaga and Simarmata, 2020).
b)
The name of countries competed in the least number of games are as follows:
 China
 Soviet Union
This is basically two countries which involve in only 10 total games thus considered as the least
number of games countries. This is also easily calculated in excel with the use of only minimum
formula.
c)
Mode is basically the value which appears most frequently and maximum time in a data set.
Whenever, a particular data arises more times than it is considered as mode. In the given case,
the mode of number of games countries participated is 27 as it arises three time. Such as Sweden,
United States and Italy are three countries which are participated in total 27 games.
d)
In order to identify the countries which got more silver medals than bronze medals, there is a
need proper comparisons. After comparison the list of countries are as follows:
Countries Silver Medal Bronze Medal

China 165 > 151
Great Britain 295 > 289
Soviet Union 357 > 325
United States 794 > 704
So, on this basis there are four countries whose silver medals are higher than bronze. This is
easily calculated in excel application with the use of IF formula. The formula of IF is involve
=IF(logical test, value if true, value if false)
Here, the logical test involve condition, value if true is the result and value if false is zero.
PART 3 – USING IT
Question 12
The spreadsheet shown below which is covered in excel file attached with report file. Excel are
as follows:
Question 13
a)
Excel rank function basically helps in identifying the returns of a numerical value as compared to
a list of other numeric value. The steps need to follow to calculate rank is as follows: