(sigh, now maths is so difficult that i can't even do pri sch and sec sch maths even though the fact is that i pass PSLE and O Levels already)...
There are 300 pens in boxes X, Y and Z
Box Y has 50% of the number of pens in box Z
1/3 of the number of pens are taken out from box X, number of pens treble in Y and put in 40 more pens in box Z.
Now, each box has the same amount of pens
Find the current total number of pens.
Originally posted by noob321:(sigh, now maths is so difficult that i can't even do pri sch and sec sch maths even though the fact is that i pass PSLE and O Levels already)...
There are 300 pens in boxes X, Y and Z
Box Y has 50% of the number of pens in box Z
1/3 of the number of pens are taken out from box X, number of pens treble in Y and put in 40 more pens in box Z.
Now, each box has the same amount of pens
Find the current total number of pens.
Don't fret. There are even JC questions appearing in P6 Maths, haha.
http://www.exampaper.com.sg/miss-loi-the-tutor/psle-1-higher-education-0
Back to your question.
Let us ignore X 1st and look at Y and Z.
Originally, we have:
Y l___l
Z l___l___l
Currently, we have:
Y l___l___l___l
Z l___l___l___l
This implies that 1 unit is actually 40 pens.
Since all three boxes have the same amount of pens now, we have 9 units of pen ( X is also 3 units now ).
Total pens now = 40 X 9 units = 360 pens.
pro!