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Each piglet in a litter is fed exactly one-half pound of a mixture of

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New post 19 Feb 2018, 01:47
Great Question.
Let n = number of piglets.
Let 4x = total oats in pounds, 6y = total barley in pounds
Piglet A: \(x(1/4th) + y(1/6th) = 1.5\) ----(1)
Remaining piglets: \(3x + 5y = 1.5(n-1)\)
= \(3x + 3y + 2y = 1.5n - 1.5\)
= 3(x+y) + 2y = 1.5n - 1.5
= \(3(1.5) + 2y = 1.5n - 1.5\) (from --(1))
= \(1.5n = 6 + 2y\)
= \(n = 2(y+3)/1.5\)
=> \(n = 4(y+3)/3\)
Now note, n is an integer, also we have 4, (y+3)/3 which is a fraction,
multiplication of 4 and (y+3)/3 a fraction can yield integer, only if the fraction is multiple of 1.25
if \((y+3)/3\) = 1.25 => y = 0.75
if \((y+3)/3\) = 2.5 => y = 4.5 (but y cannot be greater than 1.5)
also y cannot be zero and make (y+3)/3 = 1 ((as given prompt, each piglet is fed some grain))
so \((y+3)/3\) must be 1.25 and n = 5 => sufficient (C)
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New post 12 Sep 2018, 12:26
hellosanthosh2k2 wrote:
Great Question.
Let n = number of piglets.
Let 4x = total oats in pounds, 6y = total barley in pounds
Piglet A: \(x(1/4th) + y(1/6th) = 1.5\) ----(1)
Remaining piglets: \(3x + 5y = 1.5(n-1)\)
= \(3x + 3y + 2y = 1.5n - 1.5\)
= 3(x+y) + 2y = 1.5n - 1.5
= \(3(1.5) + 2y = 1.5n - 1.5\) (from --(1))
= \(1.5n = 6 + 2y\)
= \(n = 2(y+3)/1.5\)
=> \(n = 4(y+3)/3\)
Now note, n is an integer, also we have 4, (y+3)/3 which is a fraction,
multiplication of 4 and (y+3)/3 a fraction can yield integer, only if the fraction is multiple of 1.25
if \((y+3)/3\) = 1.25 => y = 0.75
if \((y+3)/3\) = 2.5 => y = 4.5 (but y cannot be greater than 1.5)
also y cannot be zero and make (y+3)/3 = 1 ((as given prompt, each piglet is fed some grain))
so \((y+3)/3\) must be 1.25 and n = 5 => sufficient (C)



Hello hellosanthosh2k2,

How did u get this \(x(1/4th) + y(1/6th) = 1.5\) . shouldn't it be 0.5.

Well i used a approach similar to yours, here is what i did

Say total piglets is x

say piglet a was fed\(\frac{1}{4}\) of and\(\frac{1}{6}\) of , then each piglet was fed\(\frac{1}{4}\)+\(\frac{1}{6}\) = \(\frac{5}{12}.\)

Remaining food is\(\frac{3}{4}+\frac{5}{6}\) =\(\frac{19}{12}\)

So the Remaining food was equally divided among (x-1) piglets with each getting \(\frac{5}{12}\)

so \(\frac{5}{12}* (x-1)= \frac{19}{12)\)

So x is some value.
Hence C is sufficient.

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New post 15 Sep 2018, 05:42
Let's say ratio of oats to total mixture in each piglet A,B,C,D,... be a,b,c,d,.. respectively.
considering condition 1 : (1/2)*a=(1/4)*[(1/2)*(a)+(1/2)*(b)+...so on]
this equation gets down to 3a=b+c+d+..so on -------------(I)
this equation does not restrict us to any number of piglets so condition 1 by itself is not sufficient

considering only condition 2: on similar lines
(1/2)*(1-a)=(1/6)*[(1/2)*(1-a)+(1/2)*(1-b)+...so on]
5(1-a)=(1-b)+(1-c)+...so on -------------------(II)
this equation does not restrict us to any number of piglets so condition 2 by itself is not sufficient
Adding I and II
5-2a=[b+c+d+..so on]+[(1-b)+(1-c)+...so on]=1+1+1+1+..so on
As R.H.S of above equation is an integer, L.H.S of above equation gives a=1/2
So there are 5 piglets.
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New post 21 Nov 2018, 02:47
Hi VeritasKarishma,

solving the problem like this, you are assuming that all the grains available must be fed to piglets. This is not highlighted in the question stem. If this assumption were relaxed, you cannot be sure about the total amount of piglets (which can be from 1 to 5).

What am I getting wrong?

Thank you very much!
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New post 21 Nov 2018, 21:30
Leuro wrote:
Hi VeritasKarishma,

solving the problem like this, you are assuming that all the grains available must be fed to piglets. This is not highlighted in the question stem. If this assumption were relaxed, you cannot be sure about the total amount of piglets (which can be from 1 to 5).

What am I getting wrong?

Thank you very much!


Hi Leuro,

No, we are not assuming that all the available grains must be fed. Look, it has to do with the wording of the question.

"Each piglet in a litter is fed exactly one-half pound of a mixture of oats and barley." - So this means that 1/2 pound is fed to each piglet. So if there are 10 piglets, 5 pounds in total is fed today. This 5 pounds will be a mix of oats and barley perhaps 2 pounds oats and 3 pounds barley or whatever.

(1) Piglet A was fed exactly 1/4 of the oats today
So we have a fixed amount of 5 pounds that was fed today out of which 2 pounds was oats. Piglet A was fed 1/4 of the oats.
There is no question of anything remaining. We don't have that total there is 100 pounds of mix etc. Since everyday 1/2 pound is fed to each piglet, that is the quantity we are talking about.

I hope this helps.
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New post 07 Feb 2019, 00:56
Quote:
The total food mixture was split equally among all the piglets. Number of piglets has to be an integer, say n. Each piglet gets the same amount of food i.e. 1/n of the total food. But each piglet also gets less than ¼ of total food and more than 1/6 of total food. The only integral value for n such that 1/6 < 1/n < ¼ is 5.


Hi VeritasKarishma, I did not understand this bit, can you kindly elaborate on this part please?
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New post 14 Sep 2019, 21:52
IanStewart wrote:
dhushan wrote:
Looking over the answer to the solution, I am confused as to why their needs to be a minimum and maximum of 4 to 6 piglets respectively. Given that their is 3/4 of the Oats and 5/6 of the Barley, why could their not be 7 piglet where each received 3/28 of the Oats and 5/42 of the Barley? Or following the same logic, even more than 7 piglets?


You're not using one crucial piece of information - each pig ate the same total amount of food. We know that one pig had 1/4 of the oats and 1/6 of the barley. If each other pig had 3/28 of the oats, then each other pig had less oats than the first pig, and if each other pig had 5/42 of the barley, then each other pig had less barley than the first pig. If the other pigs had less oats and less barley than the first pig, there's no way they could have had the same total amount of food.



Dear IanStewart,

Would you please further elaborate the highlighted part? What if one of the other 7 pigs (total 8 pigs) had 3/28 of the oats (which is less oats than the first pig), but for example 9/42 of the barley, which is more than the first pig, then it would be fine, because the restriction is that just as long as it doesn't eat more than half a pound, it'd fit the requirements. Therefore, could it still be the case that there could be 7 piglets? Thank you!
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