Using Numeracy, Data, and IT: Detailed Solutions and Explanations
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Homework Assignment
AI Summary
This assignment solution covers various aspects of numeracy, data analysis, and information technology. It includes solving numerical problems involving fractions, percentages, and significant figures, as well as data interpretation using Olympic medal data to calculate ranges, averages, and comparisons between countries. The IT section demonstrates the use of Excel to rank teams based on medal counts. The solution provides detailed, step-by-step explanations for each question, making it a comprehensive guide for students studying numeracy, data, and IT applications.

USING NUMERACY,
DATA & IT
1
DATA & IT
1
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Table of Contents
Introduction......................................................................................................................................3
MAIN BODY...................................................................................................................................3
PART 1 USING OF NUMERACY.............................................................................................3
Question 1................................................................................................................................3
Question 2 ...............................................................................................................................4
Question 3 ...............................................................................................................................4
Question 4 ...............................................................................................................................5
Question 5................................................................................................................................5
Question 6:...............................................................................................................................6
Question 7................................................................................................................................7
Question 8................................................................................................................................7
Question 9................................................................................................................................8
Question 10..............................................................................................................................9
PART 2 USING DATA.............................................................................................................10
Question 11............................................................................................................................10
PART 3 USING IT....................................................................................................................14
Question 12............................................................................................................................14
Question 13............................................................................................................................14
Question 14:...........................................................................................................................17
Question 15............................................................................................................................18
Question 16............................................................................................................................19
CONCLUSION..............................................................................................................................21
References:.....................................................................................................................................22
Books and journals:....................................................................................................................22
2
Introduction......................................................................................................................................3
MAIN BODY...................................................................................................................................3
PART 1 USING OF NUMERACY.............................................................................................3
Question 1................................................................................................................................3
Question 2 ...............................................................................................................................4
Question 3 ...............................................................................................................................4
Question 4 ...............................................................................................................................5
Question 5................................................................................................................................5
Question 6:...............................................................................................................................6
Question 7................................................................................................................................7
Question 8................................................................................................................................7
Question 9................................................................................................................................8
Question 10..............................................................................................................................9
PART 2 USING DATA.............................................................................................................10
Question 11............................................................................................................................10
PART 3 USING IT....................................................................................................................14
Question 12............................................................................................................................14
Question 13............................................................................................................................14
Question 14:...........................................................................................................................17
Question 15............................................................................................................................18
Question 16............................................................................................................................19
CONCLUSION..............................................................................................................................21
References:.....................................................................................................................................22
Books and journals:....................................................................................................................22
2

Introduction
Numeracy means working with numbers. It uses mathematics concept while doing task in it.
Numeracy is a part of daily life. It is done by housewife, children and in an organization all of
them operates their working in it. It plays a vital role in ever body lives. It is learned in each and
every stage (O’Donoghue, 2002). From a child to the CEO of a company all somewhat do the
function of this. Numeracy includes number, fractions, decimals, geometrics, probability and
statistics.
Data are the raw input which is converted to get the output in the form of information. It includes
raw facts, figure and values that a person gather and then by skills convert them into a useful
content. In mathematics data is known as a scratch of values. There are two type of data
quantitative data and qualitative data. This data is actually regenerated to gather useful message
for the actual calculation.
IT stands for information technology. With the rapid change in the technology the concept of IT
increases a lot (Coles and Copeland, 2014). It becomes such a useful part that working without it
is difficult. In mathematics it play an important role that the calculation done by it is accurate and
the person or an organization can rely on it. IT takes the help of digital devices to perform their
task.
MAIN BODY
PART 1 USING OF NUMERACY
Question 1
Solution:
a) Numerator: The upper part of the fraction is called as numerator. It is that part which indicate
the distribution of the total number in equal fraction. Latin word” Enumerate” has derives the
word numerator.
b) Denominator: The bottom part of the fraction is term as denominator. It indicates the total part
the fraction holds. “Nomnen” is a Latin word through which denominator is been comes.
Let us understand these two fraction with the help of the example:
2
8
3
Numeracy means working with numbers. It uses mathematics concept while doing task in it.
Numeracy is a part of daily life. It is done by housewife, children and in an organization all of
them operates their working in it. It plays a vital role in ever body lives. It is learned in each and
every stage (O’Donoghue, 2002). From a child to the CEO of a company all somewhat do the
function of this. Numeracy includes number, fractions, decimals, geometrics, probability and
statistics.
Data are the raw input which is converted to get the output in the form of information. It includes
raw facts, figure and values that a person gather and then by skills convert them into a useful
content. In mathematics data is known as a scratch of values. There are two type of data
quantitative data and qualitative data. This data is actually regenerated to gather useful message
for the actual calculation.
IT stands for information technology. With the rapid change in the technology the concept of IT
increases a lot (Coles and Copeland, 2014). It becomes such a useful part that working without it
is difficult. In mathematics it play an important role that the calculation done by it is accurate and
the person or an organization can rely on it. IT takes the help of digital devices to perform their
task.
MAIN BODY
PART 1 USING OF NUMERACY
Question 1
Solution:
a) Numerator: The upper part of the fraction is called as numerator. It is that part which indicate
the distribution of the total number in equal fraction. Latin word” Enumerate” has derives the
word numerator.
b) Denominator: The bottom part of the fraction is term as denominator. It indicates the total part
the fraction holds. “Nomnen” is a Latin word through which denominator is been comes.
Let us understand these two fraction with the help of the example:
2
8
3
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In this 2 is the numerator and 8 is the denominator.
Question 2
Solution:
a) 24 / 30
The simplest form is 4 / 5 because these two number can easily be divided by 6.
b) 18 / 42
The simplest form is 3 / 7 because these two number comes in the table 6.
Question 3
Solution:
a) Equivalent fraction:
2 /3 with denominator 12 as equivalent fraction
Multiply the above by 4 s that denominator 12 can be get
= 2 * 4 / 3 * 4
= 8 / 12
equivalent fraction will 8 / 12
3 / 4 as equivalent fraction with denominator 12 it will be:
Multiply the numerator and denominator by 3
= 3 * 3 / 4 * 3
= 9 / 12
5 / 6 as equivalent fraction
Propagate the fraction by 2
= 5 * 2 / 6 * 2
= 10 / 12
b) Total books in library = 60,000
Business books = 14,000
Healthcare books = 22,000
Psychology ad law = 12,000
Remaining books are 12000 (6000 – (14000 + 22000 + 1200))
4
Question 2
Solution:
a) 24 / 30
The simplest form is 4 / 5 because these two number can easily be divided by 6.
b) 18 / 42
The simplest form is 3 / 7 because these two number comes in the table 6.
Question 3
Solution:
a) Equivalent fraction:
2 /3 with denominator 12 as equivalent fraction
Multiply the above by 4 s that denominator 12 can be get
= 2 * 4 / 3 * 4
= 8 / 12
equivalent fraction will 8 / 12
3 / 4 as equivalent fraction with denominator 12 it will be:
Multiply the numerator and denominator by 3
= 3 * 3 / 4 * 3
= 9 / 12
5 / 6 as equivalent fraction
Propagate the fraction by 2
= 5 * 2 / 6 * 2
= 10 / 12
b) Total books in library = 60,000
Business books = 14,000
Healthcare books = 22,000
Psychology ad law = 12,000
Remaining books are 12000 (6000 – (14000 + 22000 + 1200))
4
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2/3rd books are of computer books
To calculate the actual number of computing books
= (Remaining books * 2 /3 )
= 12000 *2 / 3
= 8000
The percentage of computing books out of total books
Formula is:
(total number of computer books / total books) * 100
= (8000 / 60000) *100
= 13.33 %
Question 4
solution:
Total number of shoes Liza bought = 2
Total money spent given to shopkeeper
= £50 * 3
= £150
Total change she gets is £10.50
Actual money spent in purchase shoes are:
= (Money given to shopkeeper – total change gets)
= £ 150 - £ 10.50
= £ 139.50
Price per pair of shoes is;
= (money spent on purchase / total number of shoes)
= £ 139.50 / 2
= £ 69.75.
Question 5
Solution:
a) 240.50 * 19.54
5
To calculate the actual number of computing books
= (Remaining books * 2 /3 )
= 12000 *2 / 3
= 8000
The percentage of computing books out of total books
Formula is:
(total number of computer books / total books) * 100
= (8000 / 60000) *100
= 13.33 %
Question 4
solution:
Total number of shoes Liza bought = 2
Total money spent given to shopkeeper
= £50 * 3
= £150
Total change she gets is £10.50
Actual money spent in purchase shoes are:
= (Money given to shopkeeper – total change gets)
= £ 150 - £ 10.50
= £ 139.50
Price per pair of shoes is;
= (money spent on purchase / total number of shoes)
= £ 139.50 / 2
= £ 69.75.
Question 5
Solution:
a) 240.50 * 19.54
5

Multiplying the above = 4699.37
In two significant figure will be 4700
b) 52100 in power of 10
a * 10b
a = 52100
Multiply it by 10000
New number in decimal will be = 5.2100
There are 4 decimals after 5
b= 4
Number in 10 powers
a * 10b
a = 5.21
b = 4
new numeric will be:
5.21 * 104
5.22
Question 6:
Solution:
a) Discount rate = 30 % per person
Total number participants = 3
Total sum paid = £ 210
Discounted amount =
= (total money paid * discount rate) / 100
= (£ 210 * 30) / 100
= £ 63
Saving out of £ 210 is of £ 63.
b) Total Euro paid as an individual is 70
Total saving by an individual is
6
In two significant figure will be 4700
b) 52100 in power of 10
a * 10b
a = 52100
Multiply it by 10000
New number in decimal will be = 5.2100
There are 4 decimals after 5
b= 4
Number in 10 powers
a * 10b
a = 5.21
b = 4
new numeric will be:
5.21 * 104
5.22
Question 6:
Solution:
a) Discount rate = 30 % per person
Total number participants = 3
Total sum paid = £ 210
Discounted amount =
= (total money paid * discount rate) / 100
= (£ 210 * 30) / 100
= £ 63
Saving out of £ 210 is of £ 63.
b) Total Euro paid as an individual is 70
Total saving by an individual is
6
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= sum paid as individual * discount rate
= (70 * 30) / 100
= £ 21
Question 7
Solution:
a) Applying BODMAS method in 3 /4 – 7 / 9 + 2 / 3
Add 7 / 9 + 2 / 3
= 7 + 6 / 9
= 13 / 9
Left over fraction
= 3 / 4 – 13 / 9
= 27 – 52 / 36
= 25 / 36
b) Largest among 0.1, 0.02, 0.003, 0.0004, 0.00005 is 0.1
0.1 is largest let's calculating the above decimals by 100
The decimal obtained will be as 10, 2, 0.3, 0.04, 0.005
So from the above calculation it is clear that 0.1 is the largest.
Question 8
Solution:
Number of men = 90
Number of women = 60
Total people = number of men + women
= 90+ 60
= 150
Total number of people who watched the movie = total people * 3 / 5
= 150 * 3 / 5
= 90
Number of women who watched the movie
= total number of person who watched * 3 /10
7
= (70 * 30) / 100
= £ 21
Question 7
Solution:
a) Applying BODMAS method in 3 /4 – 7 / 9 + 2 / 3
Add 7 / 9 + 2 / 3
= 7 + 6 / 9
= 13 / 9
Left over fraction
= 3 / 4 – 13 / 9
= 27 – 52 / 36
= 25 / 36
b) Largest among 0.1, 0.02, 0.003, 0.0004, 0.00005 is 0.1
0.1 is largest let's calculating the above decimals by 100
The decimal obtained will be as 10, 2, 0.3, 0.04, 0.005
So from the above calculation it is clear that 0.1 is the largest.
Question 8
Solution:
Number of men = 90
Number of women = 60
Total people = number of men + women
= 90+ 60
= 150
Total number of people who watched the movie = total people * 3 / 5
= 150 * 3 / 5
= 90
Number of women who watched the movie
= total number of person who watched * 3 /10
7
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= 90 * 3 / 10
= 27
Number of men who watch movie
= total people who watched – total number of women who watched
= 90 – 27
= 63
Number of men who did not watch is = 90 – 63
= 27
Percentage of men who said didn't watch is =
= (men did not watch / total men) * 100
= (27 / 90) * 100
= 22.22 %
Question 9
Solution: The reverse calculation technique will be chose to find the exact time.
Annabelle lives in Bermondsey which is the part of London.
The time at which she want to deliver her speech is 10:30 a.m. at Birmingham.
She left his home to reach Euston station from where she will take train to reach
at the destination city Birmingham takes her an hour (1 hour).
The train from Euston Station to Birmingham takes one hour and ten minutes (7/6
hours).
The meeting destination from station is a 5-minute (1/6-hour) walk.
So the total traveling time from her house to meeting point is 1 hr + 7/6 hr + 1/12 hr = 27/12 hr
= 2 hours 15 minutes.
To find the exact time subtract the total time that is been took to reached at the destination from
the planned time
= (10 hours 30 minutes) – (2 hours 15 minutes) = 8 hours 15 minutes.
The train from Euston to Birmingham arrives at 5 minutes after the hour, 25 minutes after
the hour, and 45 minutes after the hour.
8
= 27
Number of men who watch movie
= total people who watched – total number of women who watched
= 90 – 27
= 63
Number of men who did not watch is = 90 – 63
= 27
Percentage of men who said didn't watch is =
= (men did not watch / total men) * 100
= (27 / 90) * 100
= 22.22 %
Question 9
Solution: The reverse calculation technique will be chose to find the exact time.
Annabelle lives in Bermondsey which is the part of London.
The time at which she want to deliver her speech is 10:30 a.m. at Birmingham.
She left his home to reach Euston station from where she will take train to reach
at the destination city Birmingham takes her an hour (1 hour).
The train from Euston Station to Birmingham takes one hour and ten minutes (7/6
hours).
The meeting destination from station is a 5-minute (1/6-hour) walk.
So the total traveling time from her house to meeting point is 1 hr + 7/6 hr + 1/12 hr = 27/12 hr
= 2 hours 15 minutes.
To find the exact time subtract the total time that is been took to reached at the destination from
the planned time
= (10 hours 30 minutes) – (2 hours 15 minutes) = 8 hours 15 minutes.
The train from Euston to Birmingham arrives at 5 minutes after the hour, 25 minutes after
the hour, and 45 minutes after the hour.
8

Question 10
Solution:
Weight of Shredded wheat box = 0.35 kg
Weight of Weeta box = 9 / 25 kg
= 0.36 kg
Among these two box Weeta box is heavier by 0.1 kg.
9
Solution:
Weight of Shredded wheat box = 0.35 kg
Weight of Weeta box = 9 / 25 kg
= 0.36 kg
Among these two box Weeta box is heavier by 0.1 kg.
9
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PART 2 USING DATA
Question 11
Solution:
a) Hungary has won 491 medals. So Hungary has won the lowest number of medal then rest
countries.
b) China and Soviet Union both has competed in 10 games which is least. So they are the
countries with least number of games played.
c) The mode is 27in the column total games because it is repeating 3 times.
d) Range = highest gold medal – lowest number of gold medal
Highest number of gold medal = 1022
Lowest number of gold medal = 147
Range = 875
e) Countries won silver medal more bronze medal:
Participating countries Silver Bronze
10
Question 11
Solution:
a) Hungary has won 491 medals. So Hungary has won the lowest number of medal then rest
countries.
b) China and Soviet Union both has competed in 10 games which is least. So they are the
countries with least number of games played.
c) The mode is 27in the column total games because it is repeating 3 times.
d) Range = highest gold medal – lowest number of gold medal
Highest number of gold medal = 1022
Lowest number of gold medal = 147
Range = 875
e) Countries won silver medal more bronze medal:
Participating countries Silver Bronze
10
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US 794 704
China 165 151
Soviet Union 357 325
Great Britain 295 289
Total four countries are there.
f) Countries wining more medal then Great Britain are:
Team Gold
medal
Silver
medal
Bronze
medal
Soviet Union 440 357 325
Germany 275 313 349
Great Britain has won 263 gold medal but Germany won12 more medal and Soviet Union
177 more gold medal.
Great Britain receive total 295 medal but Germany got 18 more and Soviet Union has 62
more silver medal.
Great Britain has total 289 number of bronze medal but Germany has 60 more and Soviet
Union has 36 more bronze medal then the Great Britain.
Overall there is two countries who has won the more medals then Great Britain.
g) Number of medal receive by countries in per game are:
Medal per game won= Total Medal won / Total game played
Country Total
game
Medal
won
Per games
medal won
United state 27 2520 93.33
China 10 543 54.3
Sweden 27 496 18.37
11
China 165 151
Soviet Union 357 325
Great Britain 295 289
Total four countries are there.
f) Countries wining more medal then Great Britain are:
Team Gold
medal
Silver
medal
Bronze
medal
Soviet Union 440 357 325
Germany 275 313 349
Great Britain has won 263 gold medal but Germany won12 more medal and Soviet Union
177 more gold medal.
Great Britain receive total 295 medal but Germany got 18 more and Soviet Union has 62
more silver medal.
Great Britain has total 289 number of bronze medal but Germany has 60 more and Soviet
Union has 36 more bronze medal then the Great Britain.
Overall there is two countries who has won the more medals then Great Britain.
g) Number of medal receive by countries in per game are:
Medal per game won= Total Medal won / Total game played
Country Total
game
Medal
won
Per games
medal won
United state 27 2520 93.33
China 10 543 54.3
Sweden 27 496 18.37
11

Soviet union 10 1122 112.2
Italy 27 577 21.37
Hungary 26 491 18.88
Great Britain 28 847 30.25
Germany 24 937 39.04
France 28 713 25.46
Australia 26 497 19.12
US is the team which participate in highest number of game and also won the highest medal per
game.
h)
Reason for Jamaica to not rank in top 10 are:
Jamaica has struggle a lot to win medal in per game, because they are good in athletics.
Another reason is that they don’t have players who can participate in different category of
game.
I)
The closest competitor of United State is Soviet Union
Gold category
Country Gold Medal Silver medal Bronze medal
United State 1022 794 704
Soviet Union 440 357 325
Difference 582 437 379
So from the above data it is clear that United State has performed extremely good in Gold
category.
12
Italy 27 577 21.37
Hungary 26 491 18.88
Great Britain 28 847 30.25
Germany 24 937 39.04
France 28 713 25.46
Australia 26 497 19.12
US is the team which participate in highest number of game and also won the highest medal per
game.
h)
Reason for Jamaica to not rank in top 10 are:
Jamaica has struggle a lot to win medal in per game, because they are good in athletics.
Another reason is that they don’t have players who can participate in different category of
game.
I)
The closest competitor of United State is Soviet Union
Gold category
Country Gold Medal Silver medal Bronze medal
United State 1022 794 704
Soviet Union 440 357 325
Difference 582 437 379
So from the above data it is clear that United State has performed extremely good in Gold
category.
12
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