Question 1
|
14,16x9=144,2x8=16
14+144+16=174
a)174 people can be seated in the whare kai.
|
Question 2
|
5x174=870
3x174=522
4x174=696
Roimata and her whanau would need to provide,
Pieces of vegetables- 870
Slices of meat- 522
Pieces of fruit- 696
|
Question 3
|
4people-1pk Biscuit, 4people x 40pk biscuits=160 people+1pk=172people+1pk-6bisuits=174 people/41.5 packs of biscuits.
Roimata will need 42 packs of biscuits and will be left with 6biscuits.
|
Question 4 (a)
|
174x300=52200
Roimata will need upto 52200ml of water.
|
Question 4 (b)
|
52200 divided by 1000=52.2L
52.2 divided by 2=26.1
They would need 27, 2L jugs of water.
|
Question 5
|
Roimata will have to set up 19 tables with the following amounts of fruits and biscuits.
1xtable/14people
4x14=56 fruits
3x14=42 biscuits
16xtables/9people
9x4=36 fruits
9x3=27 biscuits
2xtables/8people
8x4=32 fruits
8x3=24 biscuits
|
No comments:
Post a Comment
Note: only a member of this blog may post a comment.