Please help. Thanks.

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# Question

# Answer

(a)

If there were 27n pies initially, Adam ate 27n * 2/3 + 2/3 = 18n + 2/3

So the remaining pies were 27n – ( 18 n + 2/3) = 9n -2/3

Bryan ate (9n – 2/3 ) 2/3 + 2/3 = 6n – 4/9 + 6/9 = 6n + 2/9

So the remaining pies were 9n – 2/3 – ( 6n + 2/9) = 3n – 8/9

Finally, Clara ate ( 3n – 8/9) 2/3 + 2/3 = 2n – 16/27 + 18/27 = 2n + 2/27

2n + 2/27 = 3n – 8/9

n = 2/27 + 24/27 = 26/27

So 27n = 26

Initially there were 26 pies.

(b)

Bryan eaten 6n + 2/3 = 6(26/27) + 2/9 = 54/9 = 6 pies

Sorry for making mistake on the answer, corrected.

(a) Another solution by working backwards,

Clara eaten 2/3 of remaining and left with 1/3 which was 2/3 pie that she had eaten at last.

So the number of pies left before Clara started eating was 3* 2/3 = 2 pies.

So number of pies after Bryan eaten 2/3 of remaining pies = 2 2/3 pies or 8/3 pies.

Number of pies left before Bryan started eating = 3 * 8/3 = 8 pies.

So number of pies after Adam eaten 2/3 of remaining pies = 8 2/3 pies.or 26/3 pies.

Number of pies in the beginning = 3 * 26/3 = 26 pies.

(b) Bryan eaten 8 * 2/3 + 2/3 = 18/3 = 6 pies

Thanks. Gosh. This type of question is difficult. I cannot imagine seeing such questions in PSLE….

Just some perspective; when I was in P4 back in 1977, we were already doing algebra and the problems were like this. But the dropout rate was high, so they made it easier for everybody to pass in the next 3 years and beyond. 🙂