(Untitled)

[aikidrake]

so weird out of the blue, my lecturer send us this email...

Can anybody help?

A Question:

Libby is a teddy bear freak.  She has $100 to spend and she wants to buy 101 teddy bears

How many of each bear would libby have to buy to spend exactly $100 and get 101 bears?

Bear A costs $5

Bear B costs $2

Bear C costs 10cents

Can anyone figure this out for me?

Comments

[gaara83]

4 Bear A, 37 Bear B and 60 Bear C.. my colleague helped me to solve it by using Excel.. =P

[shroudemonix]

teach me how to use excel to solve it =x
i realised i'm damn lousy at using excel.

[xiaogou]

Need to install the Solver Add-in that comes with Excel.

Put 0 in cells A1, B1, C1. (this will be the number of A, B and C bears)
D1 is =5*A1+2*B1+0.1*C1 (this will be the total cost of the bears)
E1 is =A+B+C (this will be the total number of bears)

Start the Solver dialog box
Set the cells to be changed to A1:C1
Add the following constraints:
A1:C1 = integer
A1:C1 >= 0
D1 = 100
E1 = 101

Click solve.

[aikidrake]

oooh thks! all the best for the exams tmr! jia yoU!

[imthechemist]

i got it by trial and error. I can only have 2 equations for 3 unknowns, therefore can't be solved by algebra.

[seaweeds]

ya loh .. dat was my problem ... lolll . I needed one more equation .. hahahaha

[milopengz]

just guess a number on one of the unknowns lor...
den solve it by simultaneous...
easier ones to guess would be bear C that costs 10cents... cos it would have to be whole numbers that are multiples of 10's.

[seaweeds]

I tried ... I used

X + Y + Z = 101

5X + 2Y + Z/10 = 100

den assume z = 10

den all wrong liaoz .. lolll

[schaeffer]

it's like so primary school kind of question? =p
find it more fun to figure the fastest way to compute it. lolz

[entropic82]

Here's my 2 cents worth...

The algebraic method seems easier to solve if (the 2 equations are) combined with an additional condition that the number of type-C bears must be in multiple of 10, followed then by trial and error.

Mmm.. I've also tried solving without using algebraic method with the solution as shown below. But it doesn't seem elegant enough (or to be a method to be used at the intended level)..

Re: wtf

[aikidrake]

haha means ppl nice to help me work it out but think trial and error thus far the best considering tht this qn is for like grade 5 lor

Re: wtf

[entropic82]

lol

that's not the only way I like to be stimulated hokay? :P

ur blog entries dame funny. :)