# Question

Hi, can someone help on this question. Thank you!

Henry has a total of 34 blue and green stickers. 1/3 of the blue stickers and half of the green stickers are rectangular. How many blue stickers does he have if he has a total of 15 rectangular stickers ?

# Answer

Alternate solution with only 1 unknown.

Blue -> 6u  (6 is chosen because it is common multiple of 2 and 3)

Green -> (34 – 6u)

1/3 of blue -> 2u

1/2 of green -> (17 – 3u)

(17 – 3u) + 2u = 15

1u = 2

Blue -> 6u = 12

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Assume the blue sticker is 3B and green sticker is 2G.

Total stickers :

3B + 2G =  34   —– (1)

Rectangle stickers :

1/3 of blue sticker + 1/2 of green sticker  =  B + G  =  15  —- (2)

(2) x 2                                                                 2B + 2G = 30 —- (3)

(1) – (3)                                                                             B  = 34 – 30  =  4

Total blue stickers  =  3B  =  3 x 4  = 12

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Can u explain why need to x 2 at Step 3? Thanks

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I think it’s fine to ×3 as well. Though u will end up with one extra steps.

Coz it’s not easy for most of primary school kids to remember which one.

I always tell my students to make any of the variable the same, they can choose the one which is easier for them.

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This is called the “Elimination Method” commonly used for solving Simultaneous Equations.

The objective here is to removed the variable “G” when we “combine” equations (1) and (2) together so that only  variable “B” is remained.

Currently, Equation (1) has “2G” while ….. Equation (2) has “1G”, hence we make both of the “same value” by Equation (2) x 2 ….. so that when we do Equation (1) – Equation (3), we are only left with one variable “B”.

Hope my explanation is not confusing.

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I got it. I will explain to my DD now.

Thank you so much !!

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For those who prefer model drawing, the solution steps are much shorter.

Just another alternative to share with.

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