Name T = full pool
X fills a pool in a days ==> 1 day X fills: T/a
Y fills a pool in b days ==> 1 day Y fills: T/b
Z fills a pool in c days ==> 1 day Z fills: T/c

1 day (X+Y+Z) together fill: T(1/a + 1/b + 1/c)
d days (X+Y+Z) together fill: T

==> d = Tx1 / T(1/a+1/b+1/c) = abc/(ab+bc+ca)
==> d = abc/(ab+bc+ca)

Statement 1: d < c ==> Correct because three hoses together fill faster than one hose does

Statement 2: d > b ==> Wrong because d may be less than or greater than b. Please note that the question is MUST BE TRUE.

Statement 3: c/3 < d < a/3 ==> Correct

* Because (ab+bc+ca) < 3ab. [Please note that a > b > c]
==> d = abc/(ab+bc+ca) > abc/3ab ==> d > c/3

* Because (ab+bc+ca) > 3bc [ab > bc; bc = bc; ac > bc ==> ab+bc+ca > 3bc]
==> d = abc/(ab+bc+ca) < abc/3bc ==> d < a/3

Hence, C is correct.

"Stolen" question from GMAT Prep:


Discussed here: in-a-certain-bathtub-both-the-cold-water-and-the-hot-water-127878.html
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Thanks for the explanation, however, I don't understand how you get this inequality: \((ab+bc+ca) < 3ab\)

I tried to solve it but was unsuccesful

Here is where I got stuck:

\(\frac{c}{3} < d = \frac{abc}{ab+ac+cb}\)
\(c^2b + abc + ac^2 < 3 abc\)
\(c (b+a) < 2ab\) ???

Hi szDave
Because a > b > c
So:
(1) ab = ab
(2) ab > bc (because a > c ==> a*b > c*b)
(3) ab > ca (because b > c ==> b*a > c*a)

(1) + (2) + (3) = 3ab > ab + bc + ca
This is the key for this question.

Hope it helps.

Regards.

I used a conceptual PLUS VIC ( variables in choices MGMAT) approach.

I. d<c - We KNOW this is True because 3 hoses working together MUST BE faster than one hose by itself! B&D out!!
II. d>b - This is conceptual as well because we can think of many instances where combining 3 hoses/machines etc. would be faster than ANY individual machine, that's kinda the benefit of combining your rates to increase efficiency so....D & E are out!!

Now we have a 50/50 chance between A & C! better than 20% eh?

III. c/3<d<a/3 - With this option I knew I could try it algebraically but it's very easy to get tangled up in "alphabet soup" (for me), so I went with VIC! **Plus the thing about MUST BE TRUE options is that all you have to do is find 1 option that is to the contrary and you are good to go!** I plugged in 10, 8, and 6, but you can plug in any values and you will see that the principal holds.

Stolen question from GMAT Prep again. Prep companies need to start getting more creative

Express solution

Statement 1 is correct because the three hoses together fill faster than one hose does ALWAYS.

Now, the second statement is NOT always true because 'd' may be less than or greater than 'b', it depends on the values of 'a' and 'c'.

Third condition is ALWAYS true, 3 times the average must always be between the extremes.

Therefore C is correct

Hope this helps
Cheers
J

This question can be answered without really resorting to any calculations at all.

I. d<c. This has to be true, because the three hoses working together will take less time to fill the pool than hose z working alone. CORRECT.
II. d>b. This has to be false, because the time taken by the three hoses working together cannot be more than the time taken by hose y working alone. INCORRECT.
III. c/3<d<a/3. If all three hoses worked at the rate of the slowest (i.e. hose x which takes a days), then the time taken to fill the pool would be a/3. Since the other two hoses (y and z) are faster than x, the time taken has to be less than a/3. So d<a/3.
If all three hoses worked at the rate of the fastest (i.e. hose z which takes c days), then the time taken to fill the pool would be c/3. As the other two hoses (x and y) are slower than z, the time taken has to be more than c/3. So d>c/3.
CORRECT.

(C) it is.
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Hi there,
I´m sorry but I don`t quite understand the III statement.

- In the question stem we are given that: d < c < b < a
- In III statement we are given that c/3 < d < a/3. Therefore, c < 3d < a

How is it that when you plugged in d = 6, c = 8, and a = 10, the principle held true?

Can you please show me where I am worng?

all together work faster than each individually, thus I is always true, and we can eliminate B and D.
II - same thing as said before, but d can never be greater than individual rate. thus, II is not correct, and we can eliminate E.
III - I used some testing:
a=4, b=2, c=1.
1/4+1/2+1/1 = 1+2+4/4 = 7/4 or d=4/7
c/3 = 1/3
d=4/7
a/3 = 4/3

c/3 < d < a/3
true.
thus, we can eliminate A and select C.

I don't think you quite understand the concept...you can plug in a, b, and C. D is deducted from a,b,and C. you cannot just plug in value for D.

Hello friends, my 2 cents
a>b>c>d => 3>2>1>d
abc/(ab+bc+ca) = 6/(6+2+3) = 6/11 ~ 0.6 ~> 3>2>1>0.6
(iii) c/3 = 1/3 = 0.33 & a/3 = 3/3 = 1 => 0.33 < 0.6 < 1 => c/3 < d < a/3

Roman numeral I is true. When all three hoses work together, the number of days it takes must be less than the number of days it takes for any individual hose to complete the job by itself. Using the same argument, we see that Roman numeral II can’t be true.

If each hose works as fast as hose z, it will take exactly c/3 days. However, since they do not, and since hose z is the fastest, they must take more than c/3 days. That is, d > c/3. Similarly, if each hose works as fast as hose x, it will take exactly a/3 days. However, since they do not, and since hose x is the slowest, they must take less than a/3 days. That is, d < a/3. We see that c/3 < d < a/3, which means Roman numeral III is true also.

Answer: C
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OFFICIAL VERITAS SOLUTION
Solution: C

Explanation: The best way to solve this question is with a conceptual understanding of work rate relationships (algebra and number picking are both cumbersome). If a, b, and c represent the times of the three hoses, then their time together, d, must be less than the smallest of the three times, c. For instance, if Bill takes \(3\) hours to mow a lawn, Steve takes \(4\) hours to mow that lawn, and John takes \(5\) hours to mow that lawn, together they must take less than \(3\) hours to mow a lawn! Therefore it must be true that \(d<c\)
so you can eliminate (B) and (D).

The second statement \(d>b\)
must be false for the same logic just given previously (d must be less than each a, b, and c) so you can also eliminate (D).

Lastly, you must decide if statement III must be true. If all three times were the same (say they were all the longest time a) then it would take them \(\frac{a}{3}\)
days to fill the pool. But b and c are faster hoses than a (their times are less) so the time together (d) for the three MUST be less than \(\frac{a}{3}\)
. Likewise, imagine if they are all three the same as the fastest hose with a time of c, then their time would be \(\frac{c}{3}\)
. The other two hoses are slower so the time together (d) must be greater than \(\frac{c}{3}\)
so it must be true that\(\frac{c}{3}<d<\frac{a}{3}\)

. You could prove this with some tedious number picking or some VERY tedious algebra, but the best way is to understand the inverse relationships between times and rates and apply it abstractly.

Answer is (C).
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I feel number plugging could be an easier approach for this solution.
If we assign numbers then -> a>b>c could be 8>4>2 (these numbers so the calculations are easier)

To calculate d -> 1/d= 1/8+1/4+1/2 = 7/8 so then d = 8/7

To the question: Which of the following must be true?

I. d<c is 8/7<2 or 8<14 which is true

II. d>b 8/7>4 or 8>28 which is not true

III. c/3<d<a/3 2/3<8/7<8/3 or can be written as 2<24/7<8/3 or can be written as 14<24<56 which is true

so I and III must be true

As given a>b>c
so let a = 10, b = 4, c =2

So the total work done d together : (a*b*c)/(a+b+c) = 80/16 = 5
here, d > b

But b supports the answer in both the cases.
But let us take a = 4, b = 3, c = 2
So total work done: 24/9 = 8/3
now d<b

I didnt want to get into solving algebraic expressions involving work done in 1 day and so on

So I plugged in values (15,10,5), then calculated LCM, did the calculations and checked for relations mentioned and ultimately got this one right praying that there wont be weird values that would destroy my assumption of values/plugging in of values

I want to know, how to make sure (for this question) that the values that I have plugged in are correct

Or

Is plugging values for this one a wrong approach?

Help!
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