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(a)

If Ken was left with u sweets in the end, he had 4u chocolates in the end given the ratio 1:4.

Since Ken gave u sweets to James, James was left with (u – 17) sweets in the end and he had 9(u – 17) chocolates in the end given the ratio 1:9.

Since Ken had eaten 37 chocolates,

4u + 37 = 9(u – 17)

4u + 37 = 9u – 153

5u = 190

u =38

So Ken gave 38 sweets to James.

(b) Ken received (4u + 37) chocolates from James,

4u + 37 = 4*38 + 37 = 189

189 * 2 = 378

James bought 378 chocolates.

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Thanks for the solution.

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(a) If James bought 4u chocolates, he gave 2u chocolates and left with 2u. Ken ate 37 chocolates and left with (2u – 37 ) chocolates in the end.

As Ken in the end had sweet : chocolate = 1 : 4

Ken had (2u – 37) /4 sweet in the end.

And James had ( (2u – 37 ) / 4 ) – 17 sweets in the end and given the ratio sweet : chocolate = 1 : 9

9( ((2u -37)/4) – 17 ) = 2u

9((2u – 37) /4 ) – 153 = 2u

18u – 333 – 612 = 8u

10u = 945

u = 94.5

(189 – 37)/4 = 152/4 = 38

James gave 38 sweets to Ken.

(b) 4u = 378

James bought 378 chocolate .

Check: 378/2 = 189

189 – 37 = 152

38 : 152 = 1 : 4

38 – 17 = 21

21 : 189 = 1 : 9 Checked

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Lol… I read this question until I go bonkers.  It’s gotta be one of those questions that turn humans into pretzels.  I have to slowly read your solution to grasp it.  Thank you!

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Personally I think that teachers who set this kind of question should be censured. Nobody (even imaginary people) should be forced to eat 37 chocolates just for one maths problem. 😀

And turning humans into pretzels is just carboloading even more. 😀

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Yes, MOH will grill this person alive as the one who ate so many sweets and chocolates will kana diabetes!

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