# Question

pls help.

I recommend to use the graph or table if you kid does not understand algebra fully. The first case the Blue is 7 times of (R-30). The second case is the Blue is 50-R/2.

Then, a simple algebra can resolve this.

Hope this helps.

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Hi, everyone.  Just came across this interesting question and attempted to solve it.  But I have gotten a different answer.  Need advice from the experts here to let me know where my mistake is.

My solution:

At first, there are r red pencils and b blue pencils in the box.

Scenario 1, 30 red pencils are removed

Red pencils left -> r-30

Total pencils left -> r-30+b

Total pencils left is 8 times red pencils left

r-30+b -> 8(r-30)

7r-b -> 210

Scenario 2, 50 blue pencils are removed

Blue pencils left -> b-50

Total pencils left -> b-50+r

Total pencils left is 3 times blue pencils left

b-50+r -> 3(b-50)

2b-r -> 100

14r-2b -> 420 (from Scenario 1, multiply by 2)

13r ->520

r -> 40

b->70

Total number of pencils in the box -> 40 + 70 = 110

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 Red Blue Total Scenario 1: 30 red removed 1u 8u – 1u = 7u 8u At first 1u + 30 7u 8u + 30 Scenario 2: 50 blue removed 3p – 1p = 2p 1p 3p At first 2p 1p + 50 3p + 50

Red:

2p = 1u + 30

1u = 2p – 30

Blue:

7u = 1p + 50

7(2p – 30) = 1p + 50

14p – 210 = 1p + 50

13p = 260

1p = 20

Total number of pencils -> 3p + 50 = 60 + 50 = 110

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