Grocer's dilemma
A grocer discovered his beam balance(zodiac sign for Libra) was faulty, So he started a new method for weighing customer's orders. He divides the order into two halves, putting the first half in the left hand of the balance and weights in the right, then do the opposite.
He is not sure if the method is fair to both his customers and himself. And you?
He is not sure if the method is fair to both his customers and himself. And you?
posted by Rajesh Lal at
9:18 AM
11 Comments:
No, it's not fair to either.
If the balance isn't accurate, he needs to check the weights he's using. The balance will show the same weight, right or wrong, on both sides of the scale.
If it's wrong because of friction in the balance mechanism, it will show the wrong (but same) weight on both sides of the scale.
For a beam balance to be faulty,l it would mean that it leans to one side by a certain amount of weight. Let's say it leans left, by 5 pounds (2.268 kg), and a customer has 20 pounds of groceries.
On the first weighing the grocer takes 11 pounds of groceries (because of the nature of groceries it is impossible to split them exactly in half) and puts them on the left side. The beam will balance when 16 pounds worth of weights are put on the right hand side.
Next, he takes the remaining 9 pounds of groceries and puts them on the right hand side of the scale; since the scale leans left only 4 pounds of weight are required on the left hand side to make the scale balance.
16 + 4 = 20; therefore the grocer is using the correct methodology to weigh his customers' groceries. I don't understand why he doesnt just take weights and put them on one side of the scale permanently to make it balance, or what kind of grocery store sells groceries by weight.
Any market with a produce section or a deli sells by weight.
Granted, these days grocers typically use digital scales to determine weight, but the grocer in this word problem might not be able to afford one since he's losing money on all the "use and throw" knives he's going through.
I agree with Daniel that if the error is a constant imbalance to one side *and* that imbalance is not more than either "half" of the groceries.
w = weight of groceries
e = constant balance error
w1 = weight of 1st "half"
w2 = weight of 2nd "half"
w = w1 + w2
The question is "is this true?"
w = w1+e + w2-e
...and it is.
Yes, if it's out but a set amount.
... but with a percentage off, as you would get with an off-centre fulcrum, he cannot trust this method. The customers will always be charged too much.
The real problem is - What's actually wrong with the scale? Does one of the 'baskets' weigh more, or did one of the 'arms' get bent.
... or is there a tiny invisible goblin randomly jumping from side to side?
Just a suggestion Rajesh: please make the questions clear so there aren't so many different answers depending on how the question is intrepreted. This one, like others has several answers depending on undisclosed information.
steve thanks for your suggestion.
it never said how it was faulty. it could be that it is off by one pound every time he tries to weigh it.
SOLUTION - NO
----------------------------------
The answer is NO
A "natural" fault in an otherwise fair balance is always proportional to the weight.
CASE 1 :
The case when there can be an addition error in the measurement is IFF
A weight is added to one side intentionally
or
One of the pan has more weight.
THIS can be fairly dealed by the method acquired by the grocer.
but since we don't know that this is the case as also pointed out by Dan neely only one who came close to the reasoning )
CASE 2
Error due to the beam which is related to the weight.
Assume we have 1800 gms of weight in two lots of 900 gms each
1. 900 gms of order is put on the left pan and was found to be 1000 gms on first weighing
2. When we put 900gms on right pan , due to the error the left pan will need
900 x 900
---------- = 810 gms
1000
So the total weight for which he charges becomes 1000 + 810 = 1810 = 1800 gms (original) + 10 gms (error)
This method does reduce the bias of error but still its not accurate
QED
GOT RIGHT
------------------------
Anonymous guy partially
Ummm...
if he devides it in half, dosent both halfs weigh the same?
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