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In a certain zoo, 9L of water can sustain 5 monkeys

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In a certain zoo, 9L of water can sustain 5 monkeys  [#permalink]

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New post Updated on: 21 May 2017, 02:17
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In a certain zoo, 9 Liters of water can sustain 5 monkeys. What is the maximum number of monkeys 16 Liters of water can sustain?

A. 3
B. 8
C. 9
D. 28
E. 29

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Originally posted by Kritesh on 17 May 2017, 13:15.
Last edited by Kritesh on 21 May 2017, 02:17, edited 1 time in total.
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In a certain zoo, 9L of water can sustain 5 monkeys  [#permalink]

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New post 18 May 2017, 07:35
Kritesh wrote:
In a certain zoo, 9 Liters of water can sustain 5 monkeys. What is the maximum number of monkeys 16 Liters of water can sustain?

A. 3
B. 8
C. 9
D. 28
E. 29

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This maximization problem creates a common quandary: should we round our result up or down?

Keeping track of the multiplier can help.

Set up ratios and a proportion.
Let \(X\)= number of monkeys

9 L of water can sustain 5 monkeys: \(\frac{9L}{5m}\)

16 L of water can sustain X monkeys: \(\frac{16L}{Xm}\)

\(\frac{9L}{5m}\) = \(\frac{16L}{Xm}\)

You can cross multiply to solve,* but whether to round the result up or down may be hazy.

Try finding the multiplier for the numerator. Then use the multiplier on the denominator.

Multiplier, derived from numerator to numerator: by what factor do we multiply \(9\) to get to \(16\)?
Multiplier:\(\frac{16}{9}\approx1.77\)

Now go the other way. Multiply \(5m\) by \(1.77\)
\((5m*1.77)=8.85m\)

We cannot have a part (0.85) of a monkey. Using a multiplier on what we know is a quantity of monkeys should make it easier to see that
16L of water can sustain only 8 monkeys

Answer B

* if you want to cross multiply: \(\frac{9}{5}\) = \(\frac{16}{X}\)

Cross multiply to get 9X = 80

Careful here: 9 goes into 80 only 8 times. 9*9=81, too high - there's not enough water for the 9th monkey.

Answer B
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Re: In a certain zoo, 9L of water can sustain 5 monkeys  [#permalink]

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New post 20 May 2017, 01:10
Number of people/animals etc 'N' always vary Directly with the amount of resources 'R'.
(more the people more the resources required and vice versa).

When two quantities vary Directly, Ratio of change of first (or percentage change in first) has to be equal to the ratio of change of second (or percentage change in second)..

So N1/N2 = R1/R2.
given R1=9, R2=16, N1=5, find N2

5/N2 = 9/16. Solve to get N2=80/9 = 8.88..

We cannot round it up to 9, as any number more than 8.88.. would require more than 16 litres of water.
So MAX 8 monkeys can be sustained.

Hence answer is B
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Re: In a certain zoo, 9L of water can sustain 5 monkeys  [#permalink]

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New post 20 May 2017, 03:28
In a certain zoo, 9 Liters of water can sustain 5 monkeys. What is the maximum number of monkeys 16 Liters of water can sustain?

9 Liters of water is used by 5 monkeys
Let Maximum number of monkeys who can use 16 Liters of water = X
9 Liters ----- 5 monkeys
16 Liters ---- X
By Cross multiplying we get, \(\frac{16*5}{9}\) = \(\frac{80}{9}\) = 8.888 (9x9 = 81, 80 is little less than 81, therefore maximum of 8 monkeys can use 16 Liters of water)
Therefore Maximum number of monkeys who can use 16 Liters of water = 8
Answer B...

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Re: In a certain zoo, 9L of water can sustain 5 monkeys  [#permalink]

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New post 26 Aug 2018, 09:07
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In a certain zoo, 9L of water can sustain 5 monkeys &nbs [#permalink] 26 Aug 2018, 09:07
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