[quote]maverick88 wrote:
[quote]Renovator wrote:
[quote]James Brown wrote:
2/5[/quote]
Yep
I get that too.
All depends how you read the question.[/quote]
How?
[/quote]
“SCENARIO A” in my preceding post, but done correctly as per my third attempt (not incorrectly, as per my first and second attempts).
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]maverick88 wrote:
Any help? I can not seem to figure this out.
There are 4 jars each with 96 screws in them. In one 1/4 is being used in another 5/8, in another 3/4, and the last one 1/2 what is the fraction of total screws not being used?
I was thinking that maybe the wording was wrong and it was supposed to be each can hold 96 screws.
[/quote]
I think in order to do much of anything with this, we will have to make some assumptions about what is meant.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use –
No, that cannot be right.
[/quote]
Let’s try SCENARIO A again.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use – sure, what the hell.
Jar 3 has 96 inside and 288 screws outside in use.
Jar 4 has 96 inside and 96 outside in use.
384 screws in the jars not being used; 857.6 total screws (or maybe 858 total screws).
384/858 = 64/143 = just under 0.448
384/857.6 = 30/67 = just under 0.448
Were any of –
384/858
384/857.6
3840/8576
64/143
30/67
0.448
.448
0.447 something
.447 something
– among the choices?
[/quote]
Ah, wait, I screwed up SCENARIO A a little bit.
Jar 2 has 96 inside and 160 screws outside in use; not 57.6 screws outside in use.
384/960 = fraction not in use; not 384/857.6 .
384/960 = 2/5 .
There’s the 2/5 .[/quote]
Why is the denominator 960?
And how do you not get a decimal for 5/8?
[quote]maverick88 wrote:
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]maverick88 wrote:
Any help? I can not seem to figure this out.
There are 4 jars each with 96 screws in them. In one 1/4 is being used in another 5/8, in another 3/4, and the last one 1/2 what is the fraction of total screws not being used?
I was thinking that maybe the wording was wrong and it was supposed to be each can hold 96 screws.
[/quote]
I think in order to do much of anything with this, we will have to make some assumptions about what is meant.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use –
No, that cannot be right.
[/quote]
Let’s try SCENARIO A again.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use – sure, what the hell.
Jar 3 has 96 inside and 288 screws outside in use.
Jar 4 has 96 inside and 96 outside in use.
384 screws in the jars not being used; 857.6 total screws (or maybe 858 total screws).
384/858 = 64/143 = just under 0.448
384/857.6 = 30/67 = just under 0.448
Were any of –
384/858
384/857.6
3840/8576
64/143
30/67
0.448
.448
0.447 something
.447 something
– among the choices?
[/quote]
Ah, wait, I screwed up SCENARIO A a little bit.
Jar 2 has 96 inside and 160 screws outside in use; not 57.6 screws outside in use.
384/960 = fraction not in use; not 384/857.6 .
384/960 = 2/5 .
There’s the 2/5 .[/quote]
Why is the denominator 960?
And how do you not get a decimal for 5/8?
[/quote]
The answer to that question is 15/32 or .46875. Not saying that is what they were asking for but based on how they worded the question that is the answer.
[quote]maverick88 wrote:
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]undoredo wrote:
[quote]maverick88 wrote:
Any help? I can not seem to figure this out.
There are 4 jars each with 96 screws in them. In one 1/4 is being used in another 5/8, in another 3/4, and the last one 1/2 what is the fraction of total screws not being used?
I was thinking that maybe the wording was wrong and it was supposed to be each can hold 96 screws.
[/quote]
I think in order to do much of anything with this, we will have to make some assumptions about what is meant.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use –
No, that cannot be right.
[/quote]
Let’s try SCENARIO A again.
SCENARIO A
Let’s assume each jar literally has 96 screws inside it; the screws “being used” are outside each jar; the fractions being used are fractions of each total inside and outside the jar; and the fraction of total screws not being used is the fraction of all screws inside and outside the jars. In that case:
Jar 1 has 96 inside and 32 outside in use.
Jar 2 has 96 inside and 57.6 screws outside in use – sure, what the hell.
Jar 3 has 96 inside and 288 screws outside in use.
Jar 4 has 96 inside and 96 outside in use.
384 screws in the jars not being used; 857.6 total screws (or maybe 858 total screws).
384/858 = 64/143 = just under 0.448
384/857.6 = 30/67 = just under 0.448
Were any of –
384/858
384/857.6
3840/8576
64/143
30/67
0.448
.448
0.447 something
.447 something
– among the choices?
[/quote]
Ah, wait, I screwed up SCENARIO A a little bit.
Jar 2 has 96 inside and 160 screws outside in use; not 57.6 screws outside in use.
384/960 = fraction not in use; not 384/857.6 .
384/960 = 2/5 .
There’s the 2/5 .[/quote]
Why is the denominator 960?
And how do you not get a decimal for 5/8?
[/quote]
He’s taking it literally, that 96 screws is the fraction unused in each jar. So 96 is 1/4th of the total of #1, 5/8ths of the total of #2, and so on.
Doing it this way, 2/5ths is the answer.
I still think 15/32 should be the answer, and if they want a different answer they need to change the wording of the question.
[quote]undoredo wrote:
Ah, wait, I screwed up SCENARIO A a little bit.
Jar 2 has 96 inside and 160 screws outside in use; not 57.6 screws outside in use.
384/960 = fraction not in use; not 384/857.6 .
384/960 = 2/5 .
There’s the 2/5 .[/quote]
I agree with 2/5.
[quote]maverick88 wrote:
Why is the denominator 960?
And how do you not get a decimal for 5/8?
[/quote]
5/8 is the used portion, so 3/8 is the unused portion that is 96. 96/(3/8)=256 is the total that belongs to Jar 2, so 96 are in the jar and 160 are being used. Figure the rest out in a similar fashion.
Your formula should be x/(96+x)=5/8. Solve for x to get the total screws used for each jar.
Edit: My math is much better than my english.
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.
[quote]maverick88 wrote:
Any help? I can not seem to figure this out.
There are 4 jars each with 96 screws in them. In one 1/4 is being used in another 5/8, in another 3/4, and the last one 1/2 what is the fraction of total screws not being used?
I was thinking that maybe the wording was wrong and it was supposed to be each can hold 96 screws.
[/quote]
Total screws = (96 / 3/4) + (96 / 3/8) + (96 / 1/4) + (96 / 1/2) = 960
Screws not in use = 4 * 96 = 384
Proportion not in use = 384 / 960 = 2/5
Total screws = 4 * 96 = 384
Screws not in use = 96 * (3/4 + 3/8 + 1/4 + 1/2) = 180
Proportion not in use = 180 / 384 = 15/32
AH…YES, I was reading the question wrong. I was reading it as a fraction of 96 was being used.
I finally got 2/5.
[quote]spar4tee wrote:
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.[/quote]
Depending on the context, I disagree. It’s a poor algebra question but a good logic question. If you take this question in its most literal sense, which is the only way you should, there is only one way to answer the question. If this is a sixth grade math question, yes it may be unnecessarily confusing but for an LSAT or even an SAT/ACT the question looks like a good test of logic.
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.[/quote]
Depending on the context, I disagree. It’s a poor algebra question but a good logic question. If you take this question in its most literal sense, which is the only way you should, there is only one way to answer the question. If this is a sixth grade math question, yes it may be unnecessarily confusing but for an LSAT or even an SAT/ACT the question looks like a good test of logic.[/quote]
I disagree. Too much ambiguity for my liking. In context of testing, I’d prefer not to gamble with intended meaning in an environment in which I am unable to argue for my interpretation of said intent.
[quote]spar4tee wrote:
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.[/quote]
Depending on the context, I disagree. It’s a poor algebra question but a good logic question. If you take this question in its most literal sense, which is the only way you should, there is only one way to answer the question. If this is a sixth grade math question, yes it may be unnecessarily confusing but for an LSAT or even an SAT/ACT the question looks like a good test of logic.[/quote]
I disagree. Too much ambiguity for my liking. In context of testing, I’d prefer not to gamble with intended meaning in an environment in which I am unable to argue for my interpretation of said intent.[/quote]
That’s where you err. You’re interpreting something that doesn’t need interpretation. The test maker laid the question right out, but you and many others assumed something else. Furthermore, as to not steer you down the right path, the test maker gave you a free out when 15/32 was not included as a choice.
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.[/quote]
Depending on the context, I disagree. It’s a poor algebra question but a good logic question. If you take this question in its most literal sense, which is the only way you should, there is only one way to answer the question. If this is a sixth grade math question, yes it may be unnecessarily confusing but for an LSAT or even an SAT/ACT the question looks like a good test of logic.[/quote]
I disagree. Too much ambiguity for my liking. In context of testing, I’d prefer not to gamble with intended meaning in an environment in which I am unable to argue for my interpretation of said intent.[/quote]
That’s where you err. You’re interpreting something that doesn’t need interpretation. The test maker laid the question right out, but you and many others assumed something else. Furthermore, as to not steer you down the right path, the test maker gave you a free out when 15/32 was not included as a choice.
[/quote]
I agree with you. I got 15/32 and it was not a possible answer. I also got a few other answers when trying out different solutions. It never even occurred to me to interpret it in a way that would get me 2/5.
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
[quote]spar4tee wrote:
[quote]tedro wrote:
Edit: My math is much better than my english.[/quote]
The same can probably be said of whoever created this problem.[/quote]
Depending on the context, I disagree. It’s a poor algebra question but a good logic question. If you take this question in its most literal sense, which is the only way you should, there is only one way to answer the question. If this is a sixth grade math question, yes it may be unnecessarily confusing but for an LSAT or even an SAT/ACT the question looks like a good test of logic.[/quote]
I disagree. Too much ambiguity for my liking. In context of testing, I’d prefer not to gamble with intended meaning in an environment in which I am unable to argue for my interpretation of said intent.[/quote]
That’s where you err. You’re interpreting something that doesn’t need interpretation. The test maker laid the question right out, but you and many others assumed something else. Furthermore, as to not steer you down the right path, the test maker gave you a free out when 15/32 was not included as a choice.
[/quote]
There’s assuming and there’s taking in content at face value. The question implies 384 screws in total. It goes as far to say “in one… being used”. This implies that, out of the 96 screws that are already in each jar, certain fractions are being partitioned for use. Not ever did it state that there were screws outside of the jars already.
[quote]spar4tee wrote:
There’s assuming and there’s taking in content at face value. The question implies 384 screws in total.
[/quote]
No, it implies 384 screws in jars, 96 in each. It says this explicitly.
It should be intuitive to the reader that a screw can not be “in use” and “in the jar” at the same time and thus 96 remain in the jar, as stated. So the fraction being partitioned must be a part of a whole greater than 96.
Yes it did. It said a certain fraction was in use. How can a screw be in use if it is in the jar? The only assumption you really had to make was that the jars were for storing screws, which should not be too much of a stretch for anyone ready to tackle the most simple algebra problems.
And again, keep in mind that the original question was multiple choice. The fact that 15/32 was not a choice would be an immediate clue that you are not approaching the problem correctly.
[quote]tedro wrote:
[quote]spar4tee wrote:
There’s assuming and there’s taking in content at face value. The question implies 384 screws in total.
[/quote]
No, it implies 384 screws in jars, 96 in each. It says this explicitly.
It should be intuitive to the reader that a screw can not be “in use” and “in the jar” at the same time and thus 96 remain in the jar, as stated. So the fraction being partitioned must be a part of a whole greater than 96.
Yes it did. It said a certain fraction was in use. How can a screw be in use if it is in the jar? The only assumption you really had to make was that the jars were for storing screws, which should not be too much of a stretch for anyone ready to tackle the most simple algebra problems.
And again, keep in mind that the original question was multiple choice. The fact that 15/32 was not a choice would be an immediate clue that you are not approaching the problem correctly.[/quote]
“There are” is present tense. So as the first sentence is read, there are 96 screws in each jar. Next sentence goes “In one” meaning inside of the jar. “1/4 is being used” meaning that, as this is being read, a portion of the screws are being used. The partitioning of the other jars reads the same way. “Being used” does not mean “outside of the jar”. Your logic is contradicted by the repeated use of the word “in”. If the screws are not originating from the jars, then what are “in one”, “in another”, “in another”, and “last one” referencing? If it is referencing a portion of the jar and not the screws, how would that be relevant to question or allow you to find the ratio? If it is referencing a portion outside of the jar, why is the word “in” used?
If the question wasn’t multiple choice, there would be a lot of people getting fucked over. Which is my point.
[quote]spar4tee wrote:
“There are” is present tense. So as the first sentence is read, there are 96 screws in each jar. Next sentence goes “In one” meaning inside of the jar. “1/4 is being used” meaning that, as this is being read, a portion of the screws are being used. The partitioning of the other jars reads the same way.
[/quote]
Agreed through here.
[quote]
“Being used” does not mean “outside of the jar”. Your logic is contradicted by the repeated use of the word “in”. If the screws are not originating from the jars, then what are “in one”, “in another”, “in another”, and “last one” referencing? If it is referencing a portion of the jar and not the screws, how would that be relevant to question or allow you to find the ratio? If it is referencing a portion outside of the jar, why is the word “in” used?[/quote]
Your problem here is that you want to disregard what you already know. Each jar has 96 screws in it. You are forgetting this fact as you proceed through the question. If you remember the basic truth that there are 96 screws in each jar the rest of the question falls in place and can only be interpreted as intended.
Assuming the OP posed the question exactly the same as the test maker, “in” simply has two different meanings here. In the first sentence in means physically inside, and in the second in means a part of a whole. I.e. in jar 1 there are 96 screws inside the jar, and 128 that belong to the jar.
If the writer were to rephrase this question it would become a basic algebra problem, and not the question of logic that I would intend it to be. I’m assuming this question is on a higher level test.
[quote]spar4tee wrote:
If the question wasn’t multiple choice, there would be a lot of people getting fucked over. Which is my point.[/quote]
And my point is that this is much more than a basic arithmetic question. The question is not meant to test your ability to multiply and add fractions, but to deduce your basic logic skills.
[quote]tedro wrote:
[quote]spar4tee wrote:
“There are” is present tense. So as the first sentence is read, there are 96 screws in each jar. Next sentence goes “In one” meaning inside of the jar. “1/4 is being used” meaning that, as this is being read, a portion of the screws are being used. The partitioning of the other jars reads the same way.
[/quote]
Agreed through here.
[quote]
“Being used” does not mean “outside of the jar”. Your logic is contradicted by the repeated use of the word “in”. If the screws are not originating from the jars, then what are “in one”, “in another”, “in another”, and “last one” referencing? If it is referencing a portion of the jar and not the screws, how would that be relevant to question or allow you to find the ratio? If it is referencing a portion outside of the jar, why is the word “in” used?[/quote]
Your problem here is that you want to disregard what you already know. Each jar has 96 screws in it. You are forgetting this fact as you proceed through the question. If you remember the basic truth that there are 96 screws in each jar the rest of the question falls in place and can only be interpreted as intended.
Assuming the OP posed the question exactly the same as the test maker, “in” simply has two different meanings here. In the first sentence in means physically inside, and in the second in means a part of a whole. I.e. in jar 1 there are 96 screws inside the jar, and 128 that belong to the jar.
If the writer were to rephrase this question it would become a basic algebra problem, and not the question of logic that I would intend it to be. I’m assuming this question is on a higher level test.[/quote]
So it’s a matter of whether the scenario is sequential or static. If it static, then yes, each jar always contains 96 screws. This means that the writer is referencing a space that hasn’t been defined in the question that includes the screws in the jars and screws outside of the jar. If it is sequential, then that would simply mean that each jar starts with 96 screws inside of it and screws are being removed to be used. From the use of tense, we can determine that the scenario is static. If what you say is true, I find “in” to be poor usage. While in as a preposition can show inclusion in something abstract, it does not necessarily mean part of an unnamed whole. For instance, the reader could simply think that the writer means that the screws are being planned to be used and not actually in use but and that the screws are still in the jar.