Hey people, u r invited to answer the below question ;-)
Each sweet has a wrapper
Each sweet is sold for $1
3 wrappers can be exchanged for 1 sweet
Question:
How many sweets can you get for $15?
A.0
B.15
C.20
D.21
E.22
Put on your thinking cap and think carefully!
Answer to be out soon ;-)
Pik Wah, Olive Hill Serdang
This is our annual family gathering, held every November to celebrate my aunt’s birthday. She never fail to invite all our relatives to join...
-
Tried this Michelin-approved seafood noodles restaurant... This place is usually crowded during weekend and lunch/dinner hours. The menu is ... -
“L.C.L.Y. is actually, in chinese "lan si, lan yong", which means a big lousy show-off when he/she knows when he/she can't do ...
-
Another foodie bucket list checked! Since last year, I have been hearing alot about this neighbourhood curry omurice. As it is not easy to g...
13 comments:
$15 can get maximum of 22 sweets.
correct or not??
15= 5 free sweets
5 free sweets = 1 free sweets
15+5+1= 24
Formula to get A1 in maths
15+5+1=24
21 sweets
rm15 buy 15 sweets
15 wrappers change 5 sweets
5 wrappers change 1 sweet
actually i prefer zero... wukakakka
22
buy 15 get 5
3 from 5 get another 1
1 + remaining 2 got other one
end up with extra wrapper!
22!!! wukakak thanks pink u pandai
I don't have a thinking cap...
can i still qualify to answer this question??
semua salah...
i tunjuk formula:
let x= no of sweets
Solution: Using the Multiplication Principle,
multiply each side of the equation
by (1/4).
(1/4)4x = (1/4)9
The variable is now isolated.
x = 24
x your statement is even wrong!
5 free sweets = 1 free sweets
how can that be true? blek!
5354939411967
@_@
the funny thing is all of these commentors are the same people i talk to everyday! LOL!
nux, we even discussed and debated on it in msn wor! =D
Post a Comment