You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**math_2778****Member**- Registered: 2006-02-17
- Posts: 3

I need help with the following problem, I know it should be easier but for some reason I cannot get the right answer. I know the answer (It's supposed to be 'e') but I need the explanation of the process:

John and Mark went to pick up some apples on the morning. John picked up 2/3 more apples than Mark. They went back on the afternoon and at the end of the day John had 59 apples, and Mark had 14 apples less than John. ¿Which one of the following is not true?

a)In the afternoon Mark picked up 23 apples.

b)John and Mark picked up the same amount of apples during the afternoon.

c)Mark picked up 12 apples in the morning.

d)John picked up 24 more apples than Mark in the morning.

e)John picked up 39 apples in the morning.

I would appreciate it if you could help me,

Thanks,

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Well you got me. I can derive 3 equations from that and there are six variables. The problem is it gives no relationship about the apples they picked in the afternoon. Perhaps you can't actually solve it, but you can use the relationships to prove if a certain condition is not possible.

Try inserting the possible scenario's into the equations and see what happens.

But wait, are we assuming that they picked apples at a consant rate?

*Last edited by mikau (2006-02-17 11:47:10)*

A logarithm is just a misspelled algorithm.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Don't feel bad mikau, I couldn't solve this one either. Missing information? I tried the constant rate scenario and got numerical results, but they did not agree with the stipulated parameters. For example; If John consitently picked 5/3 what Mark picked he would end up with 65 of the 104 total picked and Mark would be left with 39. So the constant rate possiblity is definitely out.

I really don't think that this problem is solvable with just the information in the post.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Nice observation.

As far as I can tell, I don't see whats wrong with condition e. If john picked 39 apples in the morning, then mark picked 23.4 in the morning, so he'd have to pick 21.6 apples in the afternoon. As far as I know there is no law against picking 0.4 of an apple in the morning, and the other 0.6 in the afternoon. "sorry man! darn union says I can't pick more then 23.4 apples without taking a break!"

Anyways, if john picked 39 apples in the morning, then he picked 20 in the afternoon. Mark picked 23.4 in the morning, and 21.6 in the afternoon. This complies with every stipulation the problem made and every condition is met. So I don't see why it can't be true.

*Last edited by mikau (2006-02-17 12:36:37)*

A logarithm is just a misspelled algorithm.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

whoah! I think I got it. But its not e. Its d.

d)John picked up 24 more apples than Mark in the morning.

This means J = M + 24

We also know J = 5/3 M

Solving this system of equations gives us J = 60. John could not have picked 60 apples in the morning and ended up with 59 apples at the end of the day. Unless picking apples in the afternoon implies sitting under the tree and eating an apple. (boink) EUREKA!

*Last edited by mikau (2006-02-17 12:46:34)*

A logarithm is just a misspelled algorithm.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Nice mikau! I know what I did wrong now. I tried to solve the problem with just the "problem" stated. I didn't even try to plug in the solutions. Lunkhead!

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I did the same thing. Then I started plugging the values in and lo and behold the solution. But what I'm wondering is math_2778 said the answer is supposed to be e as if that were the book's answer. We'll see when he gets back.

A logarithm is just a misspelled algorithm.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

math_2778 probably said e because;

If John picked 39;

39 = 5Mark/3

Mark = 23.4 (again we run into the fraction fruit thing)

Maybe, contrary to how it reads, there are more than one wrong answer. Again, I smell a typing error somewhere.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

perhaps, but most of those answers other then e produce fractional apples. Hmmm... frapples?

A logarithm is just a misspelled algorithm.

Offline

**math_2778****Member**- Registered: 2006-02-17
- Posts: 3

Thanks a lot for your help you all. Maybe 'e' is not even the right question, I tried what some of you did, assuming a,b,c and d where right and using thos values in the equation but there seems to be something wrong:

Using 'c' and 'd':

c)Mark picked up 12 apples in the morning.

d)John picked up 24 more apples than Mark in the morning.

John picked up 2/3 more apples than Mark:

J=2/3M+M

J=5/3M

J= (2/3)(12)+12

J=20, but since John picked up 24 more apples than Mark the answer should be 36, right?, So 'e' would be false, but also 'a' and 'b'.

I took this problem from a practice test, so I might just don't worry about it and solve the other problems

Thanks a lot for your help, I really appreciate it.:)

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I'm not sure what your doing there. Your making both conditions true at once. You should only be considering each scenario individually. Not both at once.

It seems to me d is the only scenerio that doesn't make sense, cause it would mean john would have to pick -1 apples in the afternoon. You said the answer is supposed to be e. Is that the answer the book gave?

A logarithm is just a misspelled algorithm.

Offline

**math_2778****Member**- Registered: 2006-02-17
- Posts: 3

Yes, that's supposed to be the answer for the practice test. Since 'e' is the only one that is not true (supposedly), I am asumming a,b,c and d are true, that's why I considered those scenarios at once

Offline

Pages: **1**