T Nation

Math Problem


#1

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.


#2

[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.

SCENARIO B
The only other scenario I can think of currently that might be intended would be four groups of 96 screws: where one group has 1/4 in use; another group has 5/8 in use; another group has 3/4 in use; and the remaining group has 1/2 in use.

The fastest way to solve this is to recognize that if each group has 96 screws then the 1/4, 5/8, 3/4, 1/2 are proportionate and the number of screws is irrelevant (as long as all the fractions are whole numbers of screws). Add up the fractions; divide by 4; and then subtract the result from 1. (1/4)+(5/8)+(3/4)+(1/2) = 17/8 ; 17/8 divided by 4 = 17/32 ; 1-(17/32) = 15/32.

=====
I think the intended answer is 15/32.

EDIT – actually, an even faster way to solve this would be to recognize that since the fractional parts are all proportionate and since they would average to 1/2 if the 5/8 were 1/2 instead – the extra 1/8 divided by 4 is 1/32, and then since it’s how many are not being used subtract the 1/32 from a 1/2 instead of adding it. That’s 15/32. But that requires a high degree of confidence in one’s logic to do that without using another method to check the result.


#3

Yep, I got the same thing but, 15/32 was not a possible answer. I cannot remember what the choices were but two had 16 as the denominator one was a decimal and another was…can not remember.

I think there was an error.


#4

I would do:

1/4+3/4=1
1/2+5/8=1 1/8

So you have 4 jars and 2 1/8 used so subtract and you have 1 7/8 left.


#5

[quote]maverick88 wrote:
Yep, I got the same thing but, 15/32 was not a possible answer. I cannot remember what the choices were but two had 16 as the denominator one was a decimal and another was…can not remember.

I think there was an error.
[/quote]
I can’t find my calculator, and I did the following quickly so I might have made a mistake. But I think 15/32 = 0.46875 (.46875 if no leading zero prior to the decimal point). If that was the decimal, then the decimal was probably the intended answer.

(Of course, the whole thing could have been messed up for all I know.)


#6

[quote]debraD wrote:
I would do:

1/4+3/4=1
1/2+5/8=1 1/8

So you have 4 jars and 2 1/8 used so subtract and you have 1 7/8 left.

[/quote]
Sure, but the problem seems to say the answer is supposed to be fraction of the total. That’s why I thought divide by 4. So using your numbers, 2 1/8 used and 1 7/8 left – divide both parts by 4 – 17/32 of the total used and 15/32 of the total left.


#7

I agree with 15/32.

As each jar holds the same number of screws, we only need to consider the fraction of unused in each jar, so…
1/4(3/4 + 3/8 + 1/4 + 1/2) = 1/4(6/8 + 3/8 + 2/8 + 4/8) = 1/4(15/8) = 15/32


#8

15/32 is correct answer.


#9

You guys are looking at this all wrong. 1/4, 1/2, 3/4, and 5/8 are all standard screw sizes. Subtract these sizes from the totality of screw sizes available in the world, and you are left with the percentage of screws not being used. I would assume that this includes both standard and metric sizes.


#10

2/5


#11

What a horribly worded question.


#12

[quote]spar4tee wrote:
What a horribly worded question.[/quote]

Yup.


#13

[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]
4 jars= 32/8 screws which can be used.
The screws in use are (2/8)+(5/8)+(6/8)+(4/8)=17/8 (parenthesis because writing it without them is eye fucking me)

Now we can either just be in a total screws not being used=32/8-17/8 (then simplify to get total screws not being used=15/8) or write it as an equation.

The equation form would be

32/8=17/8+S
32/8-17/8=17/8-17/8+S
15/8=S (the total amount of screws not being used)

That’s all there is to it other than the question being worded in a way that makes me want to put a shotgun in my mouth, and pull the trigger. Also, what is this for?


#14

[quote]James Brown wrote:
2/5[/quote]

Yep
I get that too.
All depends how you read the question.


#15

The four jars are different sizes. The fractions given are the fraction of the volume of the jar that is being used up by the 96 screws and is thus fairly extraneous to the solution of the problem. The answer is that 1/1 of the screws are not being used since they are all in jars (either a screw is in a jar or it is being used, not both).


#16

[quote]Renovator wrote:

[quote]James Brown wrote:
2/5[/quote]

Yep
I get that too.
All depends how you read the question.[/quote]

How?


#17

[quote]Silyak wrote:
The four jars are different sizes. The fractions given are the fraction of the volume of the jar that is being used up by the 96 screws and is thus fairly extraneous to the solution of the problem. The answer is that 1/1 of the screws are not being used since they are all in jars (either a screw is in a jar or it is being used, not both). [/quote]
If 1/1 or 1 or 1.0 or 100% or equivalent was one of the choices, then this could very well be the intended answer.


#18

[quote]Destrength 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]
4 jars= 32/8 screws which can be used.
The screws in use are (2/8)+(5/8)+(6/8)+(4/8)=17/8 (parenthesis because writing it without them is eye fucking me)

Now we can either just be in a total screws not being used=32/8-17/8 (then simplify to get total screws not being used=15/8) or write it as an equation.

The equation form would be

32/8=17/8+S
32/8-17/8=17/8-17/8+S
15/8=S (the total amount of screws not being used)

That’s all there is to it other than the question being worded in a way that makes me want to put a shotgun in my mouth, and pull the trigger. Also, what is this for?[/quote]
But the question part of the question is “…what is the fraction of total screws not being used?”. The fraction of the total would be the fraction of all four jars, not the fraction of one jar. 15/8 divided by 4 = 15/32 which is the answer the OP, myself, and a few other people came up with.


#19

[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?


#20

[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 .