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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