GMAT Paper Tests
The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is $$3+\frac{4}{\sqrt{5-x}}$$ parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?
A. 2.4 ft
B. 2.5 ft
C. 2.8 ft
D. 3.0 ft
E. 3.2 ft
OA E
The concentration of a certain chemical in a full water tank
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At a depth of x- feet
$$Concentration=3+\frac{4}{\sqrt{5-x}}\ parts\ per\ millions$$
At what depth (feet) is the concentration = 6 parts per million
$$6-3=\frac{4}{\sqrt{5-x}}$$
$$3=\frac{4}{\sqrt{5-x}}$$
$$\frac{\left(3.\sqrt{5-x}\right)}{3}=\frac{4}{3}$$
$$\left(\sqrt{5-x}\right)^2=\left(\frac{4}{3}\right)^2$$
$$5-x=\frac{4^2}{3^2}$$
$$5-x=\frac{16}{9}$$
$$-x=\frac{16}{9}-5$$
$$-x=\frac{16}{9}-\frac{5}{1}$$
$$-x=\frac{16-45}{9}=-\frac{29}{9}$$
$$x=3.2$$
$$Concentration=3+\frac{4}{\sqrt{5-x}}\ parts\ per\ millions$$
At what depth (feet) is the concentration = 6 parts per million
$$6-3=\frac{4}{\sqrt{5-x}}$$
$$3=\frac{4}{\sqrt{5-x}}$$
$$\frac{\left(3.\sqrt{5-x}\right)}{3}=\frac{4}{3}$$
$$\left(\sqrt{5-x}\right)^2=\left(\frac{4}{3}\right)^2$$
$$5-x=\frac{4^2}{3^2}$$
$$5-x=\frac{16}{9}$$
$$-x=\frac{16}{9}-5$$
$$-x=\frac{16}{9}-\frac{5}{1}$$
$$-x=\frac{16-45}{9}=-\frac{29}{9}$$
$$x=3.2$$
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AAPL wrote:GMAT Paper Tests
The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is $$3+\frac{4}{\sqrt{5-x}}$$ parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?
A. 2.4 ft
B. 2.5 ft
C. 2.8 ft
D. 3.0 ft
E. 3.2 ft
OA E
Solution:
6 = 3 + 4/√(5 - x)
3 = 4/√(5 - x)
Squaring both sides, we have:
9 = 16/(5 - x)
45 - 9x = 16
29 = 9x
3.2 ≈ x
Answer: E
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