Math Is Fun Forum

harish

Guest

apple problem.....

Please tell me the method to get the answer....

The problem is as follows:

There has been a mistake at the apple farm
Someone has mixed up the apples. Read carefully what has happened. Can you find a way to help solve the problem.

    * There are 10 baskets containing apples.
    * There are various amounts of apples in each basket ranging from 10 to 20.
    * 9 of the baskets contain apples weighing 4 ounces each.
    * 1 of the baskets contains apples weighing 5 ounces each.
    * All the apples look the same.
    * The equipment you have is a set of scales and an empty basket.
    * It is late and the lorry is waiting to take the apples to market. You only have time to make one measurement using the scales.

Can we find out which basket contains the 5 ounce apples?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: apple problem.....

You'd need to put 1 apple from the first basket on the scales, 2 apples from the second, 3 from the third and so on up to 10 from the last basket on the scales.

That's 55 apples on the scales, so you'd expect the weight to be 220 ounces. But some of the apples are heavier, so the reading will be heavier.

For every ounce heavier than 220 the reading is, there is one 5 ounce apple on the scales. And if you know how many 5 ounce apples are on the scales, you know which basket has the heavy apples.

---
Of course, for practical purposes, it would take you much more time to set up all those apples and do the working out than to just quickly weigh one of each apple 10 times. Plus you wouldn't be able to tell which of the 55 apples on the scales were heavy, so you'd have to throw them all out. So that's around a third of all your apples wasted. You'd be better just selling them all as 4 ounce apples and making a slight loss.

Last edited by mathsyperson (2007-03-26 20:32:57)

---

Why did the vector cross the road?
It wanted to be normal.

harish

Guest

Re: apple problem.....

wow..
thanx a lot for the solution....

ramaniciyer

Member

Registered: 2007-03-25

Posts: 1

Re: apple problem.....

Just a quick correction in the above solution.  The total apples would be 55 and should ideally weigh 220 ounces.  Rest of the solution is perfect.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: apple problem.....

Ah, I see what I must have done there. I bet I thinking that it would be slightly more efficient to weigh none of the first kind of apple, one of the second, 2 of the third and so on, rather than what I have there. My numbers were right for the method that I didn't say.

But yes, you're right. I'll edit that to make it correct (although the questioner probably won't notice, given how old this thread is).

---

Why did the vector cross the road?
It wanted to be normal.

NinewingedTurkey

Member

Registered: 2007-05-27

Posts: 1

Re: apple problem.....

um...this is going to sound stupid and I am probably wrong, but couldn't you just weigh each apple and make a mental note of which basket it came from and it's weight. Once you have done that, if it weighs 4 ounces, put it in the empty basket and if it weighs 5 ounces you know which basket has the 5 ounce apples in it??? I have rpobably missed something...what is it?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: apple problem.....

The question says that you can only use the scales once. Otherwise, yes, that method would be fine. But that would make the problem trivial.

---

Why did the vector cross the road?
It wanted to be normal.

tcowen613

Member

Registered: 2007-07-10

Posts: 1

Re: apple problem.....

The problem really shouldn't read a "set of scales" this really seems to imply a balance not a scale!

Steve-o

Guest

Re: apple problem.....

Not to be picky but the above method implies that you also have a set of weights (not actually listed in the problem).

Is this the accepted solution? Because if you had weights you colud just put the 4oz weight on and try each apple faster than the 'proper' method!

I read it as a question of using the different apples on each end of the scale some how and solve it via balance.

Shari

Guest

Re: apple problem.....

I  think I am missing something - would it not be easier and more logical to put the empty basket on the scales and then add one apple from each basket, one at a time. You would start with the weight of the basket and the weight would increase by 4 ounces for each apple until you got to the basket with the 5 ounce apples when the weight would increase by 5 ounces. You would be using the scales once and you would be only moving at most 10 apples (quicker then 55 apples). As long as you pay attention to which basket each apple comes from you would find out which basket contains the 5 ounce apples (since all 5 ounce apples are in the same basket).

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: apple problem.....

You would be using the scales once...

But you're still making more than one measurement with them, and it's measurements that you're only allowed one of.

---

Why did the vector cross the road?
It wanted to be normal.

2root2

Member

Registered: 2008-10-30

Posts: 1

Re: apple problem.....

I realise that this is a 'maths' problem, but I'll too jump on the 'practicality' of the sollution. I presume that when you are done, these apples all have to be returned to their baskets for this waiting truck. However, we now have a basket with a mixture of 4 and 5 ounch apples. How do we make sure that the five ounch ones end up in the correct basket?

WayneTrudge

Guest

Re: apple problem.....

One of my 11yo students suggested adding one apple from each basket until the weight is not a multiple of 4. Don't need to add one from each basket only until its different.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: apple problem.....

That would work, but again, it counts as using the scales more than once.
The question wants you to make a pile of apples, and work out which set is heavy from just the full weight of that pile.

---

Why did the vector cross the road?
It wanted to be normal.

Newcastle Maths

Guest

Re: apple problem.....

I don't know if this is the same as your solution, but the way I worked it out was that:
If you put 1 of one basket, one from the next, one from the next etc. all into the empty basket, and then weighed that you'd get a number.
Then if you had that as the multiple of five and a remainder, you kept reiterating, reducing the multiple of five by one, then the remainder goes up by five. Do this until you have a remainder which is a multiple of four, and then divide it by four, and you'll get the number of apples weighing four ounces you put in, therefore you can work out which basket was which. I know this is pretty unclear, sorry.
Ex.
For the sake of the example say that we put in six apples weighing four ounces.
So we put in
10 apples weighing five,
9 apples weighing five,
...
6 apples weighing four,
...
1 apple weighing five.

The weight of all these apples then sums to 269.
269/5= 52 r 9
269/5= 51 r 14
269/5= 50 r 19
269/5 = 49 r 24.

So 24/4 is the first integer solution, six. So you know that the basket with the four ounce apples in is the one from which you took six apples.

deepaknmr

Guest

Re: apple problem.....

its better to have a one apple from each basket and wiegh them.
so count would be 10 apples.. with weigh should not be greater than 40.

if weighs> 40, keep removing each apple from basket untill total wiegh remains divisible by 40.
find the odd one which helped in wieghing to 4 divisible count.

Hope this also would solve the ques..

dafasfas

Guest

Re: apple problem.....

its better to have a one apple from each basket and wiegh them.
so count would be 10 apples.. with weigh should not be greater than 40.

if weighs> 40, keep removing each apple from basket untill total wiegh remains divisible by 40.
find the odd one which helped in wieghing to 4 divisible count.

Hope this also would solve the ques..

thats still 2 measurements =.="