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16 horses can haul a load of lumber in 24 minutes. 12 horses

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Re: 16 horses can haul a load of lumber in 24 minutes. 12 horses [#permalink]

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New post 28 Mar 2016, 07:48
ZaydenBond wrote:
daviesj wrote:
(Question Source: VeritasPrep)

This is what VeritasPrep has to say about OA.

The correct response is (B).
The answer is always “yes.” Let’s consider how “big” this lumber-haul job is first. To load the lumber, it took 16 horses x 24 minutes = 384 horse-minutes of work. If mules do it, it takes 48 mules x 16 minutes = 768 mule-minutes of work. Notice that 384 is half of 768! That means that one horse can do the work of two mules.

In 14 minutes, 12 horses will do 12 x 14 = 168 horse-minutes of work. That leaves 384 – 168 = 216 horse-minutes of work left to do.

To complete the job, we have 12 horses and 12 mules. The 12 mules do the work of 6 horses, so 12 horses and 12 mules will do the work at the same rate 18 horses would:

18 horses x ___ minutes =216 horse-minutes of work

216/18 = 12 minutes to complete the remaining work.

If you chose (A), this information is not sufficient. Without knowing how much slower mules work than horses, we cannot answer the question.

If you chose (C), the second statement alone is sufficient.

If you chose (D), the first statement doesn’t tell us anything about the mule’s work-rate.

If you chose (E), the second statement is sufficient because we are given the rates of both animals. You may want to get more practice on challenging work-and-rate word problems to review this difficult concept in more detail.


The answer is always “yes.” Let’s consider how “big” this lumber-haul job is first. To load the lumber, it took 16 horses x 24 minutes = 384 horse-minutes of work. If mules do it, it takes 48 mules x 16 minutes = 768 mule-minutes of work. Notice that 384 is half of 768! That means that one horse can do the work of two mules.

This makes no sense what so ever. There are 48 mules doing work in 16 minutes, as opposed to 16 horses doing the work in 24 minutes. How exactly does that mean that one horse does the work of two mules?


Think about it this way:

Say 10 children work on painting a fence and finish it off in 10 mins. This means the fence needs 100 child mins to get done. So if 1 child were to paint the entire fence, she would take 100 mins. If 100 children were to work on the fence, they would finish it in 1 min.
Now, say when 4 adults are working on the fence, they take 5 mins to paint it. Hence the same fence takes only 20 adult mins to paint. So if 1 adult were to paint it, he would take 20 mins.
If 1 child were to paint it, she would take 100 mins but if one adult were to paint it, he would take only 20 mins. So a child takes 5 times as much time as a adult so an adult is equivalent to 5 children.
Does this help?
The solution uses the same concept.
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Re: 16 horses can haul a load of lumber in 24 minutes. 12 horses [#permalink]

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New post 01 Aug 2016, 03:01
daviesj wrote:
16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 14 minutes, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

(1) Mules work more slowly than horses.
(2) 48 mules can haul the same load of lumber in 16 minutes.


(1) Mules work more slowly than horses.
Nothing can be inferred without knowing the actual rate of mules
INSUFFICIENT

(2) 48 mules can haul the same load of lumber in 16 minutes
The rate of horse is known. the rate of mule is known.
Mixed rates of horse and mules can be find easily using suitable calculations
SUFFICIENT

ANSWER IS B
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Re: 16 horses can haul a load of lumber in 24 minutes. 12 horses started h [#permalink]

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New post 13 Nov 2016, 23:43
help help help!!!! :(:(
i am solving it as follows but i am stuck:

16H=1/24 thus, 1H does 1/(16*24) L/min, so 12 horses do 12H=1/(16*2) L/min

in 14 mins, the 12H will do 7/16 L, thus remains 9/16L.

so, 12H+12M= ((1/16*2)+(12M))*15=.....

i am stuck here can someone please help with this approach?
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Re: 16 horses can haul a load of lumber in 24 minutes. 12 horses started h [#permalink]

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New post 08 Jul 2017, 04:16
Cuco wrote:
chetan2u wrote:
hycday wrote:
Hi,

Would the following logic be ok during the test or no :

without even computing anything, statement A doesnt give any real info, only 'slower' but it can be much slower, or a lot slower.
statement B does give the rate of the mule, and thus can allow us to make some computation and answer the question (but we wont since we don't need the actual answer).
thus, only statement B is helpful.
done in 20seconds.

would this be correct or can there be a case where only statement A is helpful ? such as a situation where the only fact of knowing that the mule are slower can tell us the answer?

thank you !



hi, ....
you cannot take this as a routine. It may not have mattered here with the kind of numbers we have...
but it will not be true if i change only one value in the original question ..

16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 1 minute, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

(1) Mules work more slowly than horses.
(2) 48 mules can haul the same load of lumber in 16 minutes.

please find answer for this


Actually, hycday's logic still works with the changes you made to the original question.... With the given information we can still calculate the time it would take 12 horses and 12 mules working together to finish the job, hence B is sufficient. Whether it takes more, or less, than 15 min is indifferent, the fact is that with the information provided in B we can calculate the time it would take the animals to finish the job, so B is sufficient.
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16 horses can haul a load of lumber in 24 minutes. 12 horses started h [#permalink]

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New post 19 Oct 2017, 05:37
Answer is B.

Let h - time taken by single horse to load (in minutes)
m - time taken by single mule to load (in minutes)

Given: 16 horses -> 24 mins => time taken by single horse is (16 * 24) = 384 minutes
so amount of job completed by 1 horse in 1 min => 1/384
amount of job completed by 12 horse in 1 min => 12/384
amount of job completed by 12 horses in 14 min => (14 * 12/384) = 9/16

So before mules join (7/16) of work completed,

now , amount of job completed by 1 mule in 1 min => 1/m
amount of job completed by 12 mules in 1 min => 12/m

now the question is remaining job, (9/16) will be completed less than 15 mins?

(9/16)/ ( (12/384) + (12/m)) < 15 ? ---------------------- (1)

Statement 1:
mules work slower than horses,

now if considering hypothetical situation, mules at same speed as horse., them m = 384(time taken by single horse to complete full job)
Eqn --- (1) becomes , (9/16) / ((12/384) + (12/384)) = 9 minutes (which is less than 15 mins)
so, if mules work at same speed as horses, remaining job will be completed in 9 minutes, but given mules work slower than horses, we don't know how slow, so time taken could be < 15 or >= 15 minutes, we can't say -> Insufficient.

Statement 2:
48 mules can haul the same load of lumber in 16 minutes.
so time taken by single mule to do whole job = (16 * 48) minutes (which is the value of "m")

We can substitute the value of "m" in equation --- -(1) and can find remaining job time taken, and accordingly we will be able to say <15 or >= 15

Sufficient.

So answer B
16 horses can haul a load of lumber in 24 minutes. 12 horses started h   [#permalink] 19 Oct 2017, 05:37

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