Each piglet in a litter is fed exactly one-half pound of a : GMAT Data Sufficiency (DS)
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# Each piglet in a litter is fed exactly one-half pound of a

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13 Aug 2009, 02:42
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Each piglet in a litter is fed exactly one-half pound of a mixture of oats and barley. The ratio of the amount of barley to that of oats varies from piglet to piglet, but each piglet is fed some of both grains. how many piglets are there in the litter?

(1) Piglet A was fed exactly 1/4 of the oats today
(2) Piglet A was fed exactly 1/6 of the barley today
[Reveal] Spoiler: OA

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Each piglet in a litter is fed exactly one-half pound of a mixture of oats and barley. The ratio of the amount of barley to that of oats varies from piglet to piglet, but each piglet is fed some of both grains. how many piglets are there in the litter?

1) Piglet A was fed exactly 1/4 of the oats today
2) Piglet A was fed exactly 1/6 of the barley today

SOL:

St1:
Just knowing how much of the oats Piglet A was fed today is not sufficient to answer how many piglets are there in the litter. If Piglet A was fed 1/4 of the barley as well then we could conclude that there are a total of 4 piglets. But if the amount of barley is more or less then we can't be sure.
=> NOT SUFFICIENT

St2:
By similar reasoning,
=> NOT SUFFICIENT

St1 & St2 Together:
Since Piglet A was fed 1/4 of the Oats and 1/6 of the barley, we can say for sure that there can be a minimum of 4 piglets and a maximum of 6 piglets.

Using statements 1 & 2 we get the equation => 1/4*O + 1/6*B = 1/2
=> 3O + 2B = 6 ................... I

We know that the number of piglets is between 4 and 6 and that for every piglet the total of (O + B) required is 0.5 pound where O & B cannot be zero. With this knowledge and the assumption that nothing of oats and barley should be left after the piglets have been fed, lets form another equation:
If number of piglets is 4, we would require (O + B)*4 = 0.5 * 4 = 2
=> O + B = 2 ...................... IIA
Solving I & IIA we get,
O = 2, B = 0 => Not valid since B cannot be 0.

If number of piglets is 5, we would require (O + B)*5 = 0.5 * 5 = 2.5
=> O + B = 2.5 .................... IIB
Solving I & IIB we get,
O = 1, B = 1.5 => Valid

If number of piglets is 6, we would require (O + B)*6 = 0.5 * 6 = 3
=> O + B = 3 ...................... IIC
Solving I & IIC we get,
O = 0, B = 3 => Not valid since O cannot be 0.
=> SUFFICIENT

ANS: C

PS: If all the oats and barley doesnt need to be fed to the piglets then the answer to this question would be E, as the number of piglets could be 4 or 5.
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Neither statement is sufficient alone. Together, if the pig had 1/6 of the barley and 1/4 of the oats, the fraction of the total food it had was somewhere between 1/6 and 1/4. Since each pig gets an equal amount of food, each pig gets 1/n of the total amount of food, where n must be an integer. So 1/6 < 1/n < 1/4, and n must be 5.
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05 Sep 2009, 08:41
Hi everyone,

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?

Thanks,

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05 Sep 2009, 09:47
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restricting the no between 4 and 6 because some piglet has 1/4 of oats and 1/6 of something else does not make senseand cannot be correct
can someone post OA..
from 2 statements one can get 3o+2b=6..(i)
and if t is total no of piglets total food=t*1/2=t/2 pounds..
also total food=b+o=t/2....(ii)
two eq but three variables o,b and t so ans should be E
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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.

chetan2u wrote:
restricting the no between 4 and 6 because some piglet has 1/4 of oats and 1/6 of something else does not make senseand cannot be correct
can someone post OA..
from 2 statements one can get 3o+2b=6..(i)
and if t is total no of piglets total food=t*1/2=t/2 pounds..
also total food=b+o=t/2....(ii)
two eq but three variables o,b and t so ans should be E

You aren't using one crucial piece of information: t must be an integer (and o and b must be positive). When you have restrictions on your unknowns, then counting equations and unknowns can be very misleading (see Q123 in the DS section of OG12 for another example). If the solution I posted above doesn't make sense to you, you can proceed algebraically, using the equations you wrote above. First, multiply the equation b + o = t/2 by 2 on both sides, and subtract from the equation 3o + 2b = 6:

3o+2b = 6
- 2o + 2b = t
o = 6 - t

Now, o is clearly positive, so the right side of the equation above must be positive, and t < 6.

Next, multiply the equation b + o = t/2 by 3 on both sides, and subtract the equation 3o + 2b = 6 from it:

3o + 3b = 3t/2
- 3o + 2b = 6
b = 1.5t - 6
2b/3 = t - 4

Now b is clearly positive, so the right side of the above equation must be positive, and t > 4.

Since 4 < t < 6, and t is an integer, t = 5. The answer is C.
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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.

chetan2u wrote:
restricting the no between 4 and 6 because some piglet has 1/4 of oats and 1/6 of something else does not make senseand cannot be correct
can someone post OA..
from 2 statements one can get 3o+2b=6..(i)
and if t is total no of piglets total food=t*1/2=t/2 pounds..
also total food=b+o=t/2....(ii)
two eq but three variables o,b and t so ans should be E

You aren't using one crucial piece of information: t must be an integer (and o and b must be positive). When you have restrictions on your unknowns, then counting equations and unknowns can be very misleading (see Q123 in the DS section of OG12 for another example). If the solution I posted above doesn't make sense to you, you can proceed algebraically, using the equations you wrote above. First, multiply the equation b + o = t/2 by 2 on both sides, and subtract from the equation 3o + 2b = 6:

3o+2b = 6
- 2o + 2b = t
o = 6 - t

Now, o is clearly positive, so the right side of the equation above must be positive, and t < 6.

Next, multiply the equation b + o = t/2 by 3 on both sides, and subtract the equation 3o + 2b = 6 from it:

3o + 3b = 3t/2
- 3o + 2b = 6
b = 1.5t - 6
2b/3 = t - 4

Now b is clearly positive, so the right side of the above equation must be positive, and t > 4.

Since 4 < t < 6, and t is an integer, t = 5. The answer is C.

Thanks, I totally overlooked the fact that the amount of B and O for each pig had to add up to 0.5lbs.
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14 Oct 2009, 19:40
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Another way to find the No of pig is 5:
We need to know how much (O + B) there is (in pounds).

As 1/4 O + 1/6 B = 1/2 => O + 2/3 B = 2 (both sides multiplied with 4) but O + 2/3 B < O + B, then O + B > 2, which means there must be more than 4 pigs to eat more than 2 pounds.

Similarly, 1/4 O + 1/6 B = 1/2 => 3/2 O + B = 3 (both sides multiplied with 6) but 3/2 O + B > O + B, then O + B < 3, which means that the food is not enough for 6 pigs.

The No of pigs must be integer, then it is definitely 5.

However this explanation also uses the same method as in above posts.
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24 Jan 2010, 22:23
Finding the limit of the number of pigs to be between 4 and 6 was key.
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05 Oct 2010, 13:11
My issue with this problem is that it says the pigs get a half pound mixture of Oats and Barley in differing ratios. It seems all of these solutions assume that the oats and barley mix for each pig has the same mixture ratios. Please explain. Thanks!
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15 Oct 2010, 22:24
This is how I read the problem.
O + B = t ( Let t be the total weight of Oats and Barley, this has to be multiple of 1/2lb, since each piglet require 1/2 lbs ... ratio may be different )
3O + 2B = 6 ( comming from o/4 + b/6 = 1/2 )
Now, here is the meat,
O can be multiple of 1/2 and B can be multiple of 1/2 ONLY since O+B = Multiple of 1/2
Now the only way to make o and b multiples of 1/2 is by,
O = 1 and B = 3/2
Total max number of Pigs = 2(from O) + 3( from B ) = 5
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05 Apr 2011, 06:53
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St 1 : InSuf
piglet A fed 1/4 of oats.
St 2 : InSuf
piglet B fed 1/6 of barley.

St 1+2 : Suff

There are total 10 parts of Oats & Barley
Piglet A fed 2 parts out of 10. These 2 parts weigh 0.5 pound.
So 10 parts weigh 2.5 pounds.
Each piglet fed 0.5 pound so we can say that 2.5/0.5 = 5 piglets
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05 Apr 2011, 18:11
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@subhashghosh: As requested by you, I will try and explain the logic used in this question.

For now, forget the data given in the question. Consider this:

10% of Oats and 20% of Barley was fed to piglet A.

My question is - Of the total food (Oats + Barley) what percentage was fed to piglet A?
It will depend on the ratio of Oats and Barley. If the mixture had only oats, the piglet was fed 10% of total food. If the mixture had only Barley, the piglet was fed 20% of the mixture. If the mixture had half oats and half barley, the piglet was fed 15% of the mixture. If the mixture had 1 part Oats for every 4 parts Barley, the piglet was fed 18% of the mixture (it is just weighted average). Whatever the case, the piglet was fed more than 10% of total food and less than 20% total food if the mixture had both Oats and Barley.

If this sounds good, consider data given in the question – the piglet was fed 25% Oats and 16.66% Barley (1/4 Oats and 1/6 Barley). So definitely, the piglet was fed more than 1/6 of the total mixture and less than ¼ of the total mixture (as reasoned above). 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.
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1/4 + 1/6 = half pound ....(1)
Combination of half pounds are : (1/4, 1/6) (1/4, 1/6) (1/4, 1/6) (1/4, 1/6) Remaining is (1/6,1/6)
1/6 in itself is not half pound (refer 1). Therefore, 1/6 barley + 1/6 barley must be half pound. Therefore, 5 combination of half pounds. C is the answer.
However, dont assume that the last piglet is going to eat barley only. Just look for how many half pounds you can make in total - that must be no. of piglets.
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Re: Each piglet in a litter is fed exactly [#permalink]

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04 Jan 2013, 03:05
VeritasPrepKarishma wrote:
dkj1984 wrote:
Hi Karishma,

I understood your explanation for the general part,however,with reference to this problem,I am still a little confused.

Can u please elaborate on the usage of the weighted average method for this problem.

Thanks!

Regards,

Ok, let me explain using a different example.
Say a meal consists of a sandwich and a milkshake. You eat 1/2 of the sandwich and drink 1/2 of the milkshake. Can I say you have had 1/2 of the meal? Sure, right?
If you eat only 1/4 of the sandwich and drink 1/4 of the milkshake, then you would have had only 1/4 of the meal.
What happens in case you eat 1/2 of the sandwich but drink only 1/4 of the milkshake? In that case, you have had less than 1/2 of the meal but certainly more than 1/4 of the meal, right?

So when piglet A is fed 1/4 of the Oats and 1/6 of the Barley, it is fed less than 1/4 of the total food but more than 1/6 of the total food.

Another thing to consider here is that number of piglets has to be a positive integer, say 'n'. Now, since it is given that each piglet gets the same amount of food and there are n piglets, each piglet will get 1/n of the total food. So piglet A must have got 1/n of the total food too.

This 1/n must lie between 1/4 and 1/6. Only 1/5 lies between 1/4 and 1/6 (such that n is a positive integer). Hence n must be 5.

What might be the level of this problem?
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05 Jan 2013, 01:57
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Sachin9 wrote:
What might be the level of this problem?

Certainly above 700.
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