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Question 35204This question is from textbook Mathematics for Elementary Teachers
: Okay I'm confused. Here is the problem:
A girl bought some pencils, erasers, and paper clips at the stationary store. The pencils cost 10 cents each, the erasers cost 5 cents each, and the clips cost 2 for 1 cent. If she bought 100 items altogether at a total cost of $1, how many of each item did she buy?
I set up the variables as such:
X=the number of pencils
Y=the number of erasers
Z=the number of paper clips
THen set up the problem
X+Y+Z= 100
.10X + .05Y +.01 Z/2 = 1.00
I tried solving for X, but it still left two variables. So I tried to substitue for X then solve for Y. Needless to say I came up with bunk. My sister, nephew and I spent 3 hours on this problem and can't seem to solve it. My homework is due next week and as it is the last 3 weeks of school I am trying to get it out of the way. ANy help you can give would be greatly appreciated.
Sincerely Jennifer Smith
This question is from textbook Mathematics for Elementary Teachers
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! GOOD TO SEE YOUR ATTEMPT..SEE MY COMMENTS BELOW
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A girl bought some pencils, erasers, and paper clips at the stationary store. The pencils cost 10 cents each, the erasers cost 5 cents each, and the clips cost 2 for 1 cent. If she bought 100 items altogether at a total cost of $1, how many of each item did she buy?
I set up the variables as such:
X=the number of pencils
GOOD
Y=the number of erasers
GOOD
Z=the number of paper clips
GOOD
THen set up the problem
X+Y+Z= 100
VERY GOOD....CALL IT EQN.I
.10X + .05Y +.01 Z/2 = 1.00
EXCELLENT..YOU SHOWED GOOD MATURITY IN CONVERTING ALL O SAME CURRENCY THAT IS DOLLARS!!!KEEP IT UP!!BUT IN GENERAL WE ARE MORE COMFORTABLE WITH INTEGERS THAN DECIMALS OR FRACTIONS.SO BETTER WRITE IT AS
10X+5Y+Z/2=100....MULTIPLY BY 2 THROUGH OUT TO GET RID OF Z/2..
20X+10Y+Z=200...........CALL THIS EQN.II..............
I tried solving for X, but it still left two variables.
VERY CORRECT..THERE ARE 3 UNKNOWNS AND ONLY 2 EQNS..SO YOU CANNOT FIND UNIQUE SOLUTION UNLESS....????...LET US SEE...THERE IS A HIDDEN CONDITION..OBVIOUS BUT NOT MENTIONED...KNOW WHAT??IT IS CALLED "ONION PEEL" APPROACH.YOU HAVE TO PEEL THE ONION TO GET THE REAL STUFF!HERE YOU HAVE TO UNVEIL THE HIDDEN FACT........ IT IS THAT THE ANSWERS FOR X,Y AND Z SHOULD BE INTEGERS AND THEY CANNOT BE NEGATIVE!!!WE CANNOT BUY HALF PENCIL OR -2 ERASERS???IS IT NOT??LET US SEE WHETHER IT HELPS
LET US DO EQN.II-EQN.I....WE GET
20X+10Y+Z-X-Y-Z=200-100=100......
19X+9Y=100................EQN.III
NOW LET US PATIENTLY TRY OUR ARITHMATIC..PUTTING X=1,2,3,4,5..THAT IS ALL AS IF X IS MORE THAN 5 ,Y WILL BECOME NEGATIVE.
X=1....19*1+9Y=100....OR....9Y=100-19=81...OR...X=9..EUREKA! GOT IT!!BUT LET US SEE IF THERE ARE ANY OTHER ANSWERS.
X=2......9Y=100-38=62....NOT POSSIBLE
X=3......9Y=100-57=43....NOT POSSIBLE
X=4......9Y=100-76=24....NOT POSSIBLE
X=5......9Y=100-95=5.....NOT POSSIBLE
NO SO THE ONLY SOLUTION IS X=1,Y=9 AND Z=100-1-9=90...GOT IT??
So I tried to substitue for X then solve for Y. Needless to say I came up with bunk. My sister, nephew and I spent 3 hours on this problem and can't seem to solve it.
DONT DESPAIR THAT FAST...THE MORAL OF THE STORY IS THAT SOME TIMES WE MISS
THE OBVIOUS INFRONT OF US(IS IT BECAUSE OF PROXIMITY?OR IT IS TAKEN FOR GRANTED?)AND SEARCH FOR ILLUSIONS!!YOU HAVE DONE EXTREMELY WELL TO COME THIS FAR AND YOU ARE SURE TO SUCCEED WITH THIS APPROACH AND DETERMINATION.KEEP IT UP!
My homework is due next week and as it is the last 3 weeks of school I am trying to get it out of the way. ANy help you can give would be greatly appreciated.
Sincerely Jennifer Smith
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