Math expert come in and help

samich

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Uncle math cmi, this P6 question how to solve thanks.

At a party, 87 blue balloons and green balloons line one side of the wall. There are at least 3 blue balloons between any 2 green balloons. What is the greatest possible number of green balloons along the wall?
 

wutawa

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G + B = 87
GBBBG
4xA + 1 = 87
A = 21.5
G = 22, B = 65
 
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mackenic

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22.

The pattern will repeat after every 4 balloons ( 1G+3B), hence you use 87/4, which gives you 21 r3.
With the remainder 3balloons, you can still have 1 more. So the total is 21+1 = 22 green balloons
 

Song Bro

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Since Blue needs to be min,
Need Green (3 Blues) follow by green = 4 balloon per set (Green, Blue, Blue, Blue)
87/4= 21 r 3, (r3 is green "1", blue, blue)
2 blues of r3 is not considered as it does not have a "green ending",
So 21+1 = 22

Dun know if correct or not.
 

hardwater

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87 = 1G + X (3B+1G)
87 - 1G = X(3B+1G)
86 = X(4)
86/4 = X
21.5 = X --> 21 (round down as balloon can only be whole unit) + 1 (the first green balloon) --> 22
 

zamowk

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Final Answer

The greatest possible number of green balloons along the wall is
22
 

coolmyth

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22?

Dont count first green. 86/4 = 21 r 2.

Since must have at least 3 blue between two green, meaning the remaining 2 must be blue.

Kym? :LOL:
 

zueinder

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Uncle math cmi, this P6 question how to solve thanks.

At a party, 87 blue balloons and green balloons line one side of the wall. There are at least 3 blue balloons between any 2 green balloons. What is the greatest possible number of green balloons along the wall?
If cannot use 'algebra' then can only do by group.

min. 3 blue balloon between any green balloon, so it means, the first ballon has no blue balloon on one side.

So must draw diagram with 3 repeating pattern to show that it will repeat with 3B 1G, which requires 4 balloons.
(ie. the diagram will be 1G [3B 1G] [3B 1G] [3B 1G]

So to figure how many times this pattern repeats there are, take away the first green ballon (87-1),
you have 86 balloons for this.

86 / 4 = 21.5 ~~ 21 (round up to 21)

Greatest number of green ballons = 1 + 21 = 22

This is how I'll solve it if is Pri 6 standard in my time. But apparently now got algebra so can use symbols, no need write so many statements.

Probably will be broken english if is actual P6 but in my time is still transitioning to computer age so teachers all machiamp very strict about our statements must make sense.
 

ninjaghost

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If cannot use 'algebra' then can only do by group.

min. 3 blue balloon between any green balloon, so it means, the first ballon has no blue balloon on one side.

So must draw diagram with 3 repeating pattern to show that it will repeat with 3B 1G, which requires 4 balloons.
(ie. the diagram will be 1G [3B 1G] [3B 1G] [3B 1G]

So to figure how many times this pattern repeats there are, take away the first green ballon (87-1),
you have 86 balloons for this.

86 / 4 = 21.5 ~~ 21 (round up to 21)

Greatest number of green ballons = 1 + 21 = 22

This is how I'll solve it if is Pri 6 standard in my time. But apparently now got algebra so can use symbols, no need write so many statements.

Probably will be broken english if is actual P6 but in my time is still transitioning to computer age so teachers all machiamp very strict about our statements must make sense.
the problem is algebra only being taught in secondary.

Drawing bar or written statement, people able to understand the thinking process of pupils. If me, i will make a mistake to include 1st Green balloons, and become 87/5. Many maths teachers don just give marks because of given a correct answer, they also see the process how one get the answer. I think every students in sg going thru this stage right?
 
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