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Group 2

Reflection

Question: What are some of the challenges which you faced while you were trying to solve the task given?

Answer: We had to think out of the box and we had difficulties changing the seating arrangements as we did not know which one could contain more people. Other than that, we took some time to find out the gap between 2 seats as we wanted an ideal gap for the audiences to be seated comfortably during the performance.

Question: How did you overcome it?

Answer:With much teamwork using Google Docs,and overcoming tons of problems, our group tried to think out of the box and use creative methods to break the question into many simpler ones in order to have a solution smoothly.We also thought of worst case scenarios so we stepped into the audience’s shoes to ensure they have a comfortable seat each.We must also use real life examples by mimicking an audience’s seating posture and leg resting position.

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In my opinion, i agree that we require more team work. We could have been more organised together to allow for maximum efficency. However, we still completed it in the end.

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Challenge faced : 1)cannot find accurate number
2)cannot decide on the amount of spacing
solutions: 1) we use the closest estimation
2) we come up with constructive idea

strength of solution: we used the elimination method to find the best Area for improvement: come up with more solution
We have learnt that we have to plan out everything specifically, step by step

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We have difficulties calculating the measurements ,we also had many different opinions on what the price of the ticket should be and we also had different opinions on what the size should be in between the chairs .We decided to just round off to the closest number and decide to just listen and chose the best opinion.

The price of the ticket is reasonable
there are a spaces to allow people to walk
the amount earn will be more than enough
there are may seats for enough people to seat
the amount of days taken to raise enough funds is only 2 days

we learned how to calculate for fund raising
and answet

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The idea of the gaps between the chairs are very accessible. People can feel more comfortable when walking in and out of the seats and when seating.The thing we want you guys to improve on is the price of the tickets.Tickets for the VIP seats,which costs $68, is quite expensive and not very reasonable, causing people not to buy

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Our solutions are pretty precise but we should be quicker in our speed
I have learnt that math can help us solve real world problems.
We should be more co-operative and complete the task more efficiently

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Grp 9Number of rows: 26 Number of columns: 95cm gap between each chair2m width walkwayTotal number of chairs : 468Price of 1 ticket(First 3 rows):$28Price of 1 ticket(Middle 10 rows):$25Price of 1 ticket(Last 13 rows)$18Number of shows:2

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Grp 9Number of rows: 26 Number of columns: 95cm gap between each chair2m width walkwayTotal number of chairs : 468Price of 1 ticket(First 3 rows):$28Price of 1 ticket(Middle 10 rows):$25Price of 1 ticket(Last 13 rows)$18Number of shows:2

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Strengths of solution:
In real life context, the price is quite reasonable and the number rows and chairs is possible to acquire and assemble in the school hall.
Areas of improvement:
We could have created more leg and gap space and counteracted the lesser number of chairs by increasing the price of a chair, thus still amounting to over $10 000 in one show.
What have we learnt?:
Secondary students have the ability to organize charity concerts, and also that organizing charity concerts is time consuming and tedious. One must be resilient and well disciplined in order to follow it through.

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teamwork is highly needed

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PLAN 1

Length of hall-1200cm

Breadth of hall-1000cm

Chair-40cm X 40cm=1600 cm2

Number of chairs that can be placed in a row=23

Chair arrangement

15 Rows

Each Row-23 chairs

Gap between row 1 and row 2= 30cm

Total number of seats=23X15

=345 seats

Each person occupies=40cmX70cm

Pricing of tickets

VIP seats-Row A to Row C(nearest to the stage)=$68/seat

Non-VIP seats-Row D to Row F=$50/seat

Row G-K= $28/seat

Row L-N= $15/seat

Row P-Q= $10/seat

Total amount of money raised=$12213

Only 1 show is performed.

Reason: front rows will have a better view of the performance so they should pay more. We left 30cm for the gap between 2 chairs because

there needs to have a gap where the audience can rest their legs, seat comfortably and they can walk in and out easily. The total amount earned is $12,213 because some audiences may not turn up for the performances so the fund earned may not reach the target of $10,000.

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teamwork is very important for a good solution

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Solution 2
Space between stage and 1st seating audience=3m
Seating for audience in length =6m
Seating for audience in breadth=7m
Leg space=15cm
There are 14 rows. There are 10 chairs in each row.
1st 4 rows=$40/pax
next 4 rows=$30/pax
last 3 rows=$20/pax
total:$6800
we need at least two shows to fund up at $10000

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Solution 2

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Group 3( Jabriel, Yu Ying, Jiang Rui & Ga Wen)

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Group 3( Jabriel, Yu Ying, Jiang Rui & Ga Wen)

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Method 2:
Space between the chairs=20cm
space between the chairs and the stage =1m
space between the chairs and the back of the hall =1m
space between the walls at the sides=1m
so there is 20 rows and 14 columns
total number of chairs= 280 chairs
1 ticket price=  $20
amount of funds raised a day= $20*280
=$5600
amount of days needed to raise $10 000= $10 000 divide by $5600
=2 days(round off to whole number)

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leg space=40cm, each chair occupies 80cm. separate the vertical part of the seats into 3 portions. the portion nearest to the stage costs $50, while the middle portion costs $40, and the portion farthest from the stage costs $30. There will be 2 performances, as each performance would get $5400 in total. thus, 2 performances would cost $10800. There would be a total of 45 seats in each portion, and a vertical 1m gap in the middle.

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Solution 1:
Space between stage and 1st seating audience= 2m
Seating arrangement for audience in length= 10m
leg space: 20cm
one row: 320cm
area of sitting space: 48m sq
middle walkway: 2m
one side 287 seats
total:5

1st 10 rows of seat= 30/pax
=$4200
2nd 15 rows of seat= 20/pax
=$4200
3rd 16 rows of seat =10/pax
=$2240
Total amt of money raised=$4200+$4200+$2240
=$10640
Only 1 show is needed to raise $10000.

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Solution 1
no. of rows = 22
no. of columns =  25
total no. of chairs = 22 x 25 = 550
Price of one ticket = $10000 divided by $550 =$19
No. of shows = 4

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Solution 1
Total area = 1200cm*1000cm
=1200000cm
Leg space=17cm
Gap between chairs= 7cm
Space occupied by 1 chair (horizontal) : 47cm
Space occupied by 1 chair (Vertical) : 57cm
No. of chairs (Vertical) =1200/57
=21/1/19 (estimate to 21)
No. of chairs(Horizontal) = 1000/47
21/13/47(estimate to 21)
Total no. of chairs = 21*21
=441
In order to earn $10000, each person must at least pay
=$10000/441
=  $23( Estimation)
= $30(Round off to nearest 10)
Each person pay $30 so that more money can be donated.
1 show is needed.

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Some people in our group were not co-operating with us halfway. End up they helped out by giving suggestions. We made full use of the space in the hall for the guests to seat.The price are reasonable,the closer the seats to the hall,the price is higher as it allow a better view of the performance.We can have bigger leg spaces so that the audience can enjoy the performance more comfortably.We learnt how to work better as a team and organize seats.

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Number of chairs per row = (1000cm-100cm-60cm)/40cm
=21 chairs (w/ 1m gap at one side and 60cm middle of row)
Number of rows = (1200cm-100cm)/(40cm+20cm)
=19 (w/ 50cm gap from stage and back of hall and gap between rows)
Pricing:
1st row (VIP): $35
2nd row - 7th row (Standard): $29
8th row - 19th row (Back): $22.50
Amount raised in one show = $35*21+$29*6*21+$22.50*12*21
=$10059

Only one show is needed to raise at least $10 000.

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answer 1(group 3 Jabriel,Yu Ying,Ga Wen,JiangRui)

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answer 1(group 3 Jabriel,Yu Ying,Ga Wen,JiangRui)

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length of row=10 m-1 m
=9 m
length of column=12 m-1 m
=11 m
length between chair=10 cm
length of chair and gap=40 cm +10 cm
=50 cm
=0.5 m
no. of chair in a row=9 m/0.5 m
=18
no. of chair in a column=11 m/0.5 m
=22
total number of chair=18*22
=396
assume if each price is $25.50
396*$25.50=$10098
number of day needed to complete the fund=$10000/$10098
=0.990...(days)
=1(days)(round off to nearest whole number)
Reasons:
1 m to the path for the audience to walk
and the 10 cm is for the audience to walk between the chairs