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 By: achieve_goal (offline)  Thursday, December 27 2012 @ 01:46 AM CST (Read 1339 times)
achieve_goal

Pleas find the questions in the files attached...

Thank you!

Regular Member

Registered: 09/14/11
Posts: 102

 By: jo sarah (offline)  Thursday, December 27 2012 @ 08:06 AM CST
jo sarah

hi,

In the end:
.................Peter..........................Colin
...........4u..........13p...............13u.........6p.........................[using the ratio given for what they had in the end]
.........................-2p...........................+2p+6.......................[Colin ate 6 and gave 1/4 to Peter....so 'put back' -- the 6p must be 3/4, so 1/4 = 2p]
........+4u+2............................-4u.......................................[Peter ate 2, then gave 1/2 to Colin, so he was left with half which is 4u]

So, in beginning,
.........8u+2.........11p...............9u.........8p+6......................[note that each boy took half of each kind]

so, 8u+2 = 9u; that is, 1u = 2 sweets
and, 11p = 8p+6; that is, 3p = 6 cookies, so that 1p = 2 cookies

Therefore, each boy had 22 cookies and 18 sweets at first. [ANS]

answer to 001: [i use algebra here]
Let there be r red balls and b blue ones originally.
so, (r - 1) / b = 3/5; that is, 3b = 5r - 5........................................[1st scenario]
and, r / (b - 3) = 5/6; that is, 6r = 5b - 15, or, 5b = 6r + 15..........[2nd scenario]
so that, 15b = 25r - 25.................[1]
and, 15b = 18r + 45................[2]
which gives, 7r = 70, that is, r = 10
and so, b = (50 - 5) / 3 = 15

there were 10 red and 15 blue at first.
so when you add another 2 red, total = 12 + 15 = 27
And therefore, fraction of red = 12/27 = 4/9. [ANS]

Hope this helps.

Regular Member

Registered: 03/20/12
Posts: 111

 By: achieve_goal (offline)  Sunday, December 30 2012 @ 02:18 AM CST
achieve_goal