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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.

Please ask your kid to draw and understand the problem….

Hope this helps.

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

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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