Solutions to Problem Set 21

Math 101-06

11-1-2017

[Variation]

1. y is directly proportional to x.

Suppose that $x = 14$ when $y
   = 6$ .

Find y when $x = 8$ . Give an exact answer.

y varies directly with x:

$$y = k x.$$

$x = 14$ when $y = 6$ :

$$\eqalign{ 6 & = k \cdot 14 \cr \noalign{\vskip2pt} \dfrac{6}{14} & = k \cr \noalign{\vskip2pt} \dfrac{3}{7} & = k \cr}$$

Hence,

$$y = \dfrac{3}{7} x.$$

Set $x = 8$ :

$$y = \dfrac{3}{7} \cdot 8 = \dfrac{24}{7}.\quad\halmos$$


2. y varies directly with x.

Suppose that $x = 36$ when $y = 42$ .

Find x when $y = 56$ . Give an exact answer.

y varies directly with x:

$$y = k x.$$

$x = 36$ when $y = 42$ :

$$\eqalign{ 42 & = k \cdot 36 \cr \noalign{\vskip2pt} \dfrac{42}{36} & = k \cr \noalign{\vskip2pt} \dfrac{7}{6} & = k \cr}$$

Hence,

$$y = \dfrac{7}{6} x.$$

Set $y = 56$ and solve for x:

$$\eqalign{ 56 & = \dfrac{7}{6} x \cr \noalign{\vskip2pt} \dfrac{6}{7} \cdot 56 & = \dfrac{6}{7} \cdot \dfrac{7}{6} x \cr \noalign{\vskip2pt} 48 & = x \quad\halmos \cr}$$


3. y is inversely proportional to x.

Suppose that $x = 32$ when $y = 6$ .

Find y when $x = 24$ . Give an exact answer.

y is inversely proportional to x:

$$y = \dfrac{k}{x}.$$

$x = 32$ when $y = 6$ :

$$\eqalign{ 6 & = \dfrac{k}{32} \cr \noalign{\vskip2pt} 32 \cdot 6 & = 32 \cdot \dfrac{k}{32} \cr \noalign{\vskip2pt} 192 & = k \cr}$$

Hence,

$$y = \dfrac{192}{x}.$$

Set $x = 24$ :

$$y = \dfrac{192}{24} = 8.\quad\halmos$$


4. y is inversely proportional to x.

Suppose that $x = 40$ when $y = 20$ .

Find x when $y = 16$ . Give an exact answer.

y is inversely proportional to x:

$$y = \dfrac{k}{x}.$$

$x = 40$ when $y = 20$ :

$$\eqalign{ 20 & = \dfrac{k}{40} \cr \noalign{\vskip2pt} 40 \cdot 20 & = 40 \cdot \dfrac{k}{40} \cr \noalign{\vskip2pt} 800 & = k \cr}$$

Hence,

$$y = \dfrac{800}{x}.$$

Set $y = 16$ and solve for x:

$$\eqalign{ 16 & = \dfrac{800}{x} \cr \noalign{\vskip2pt} x \cdot 16 & = x \cdot \dfrac{800}{x} \cr \noalign{\vskip2pt} 16 x & = 800 \cr x & = 50 \quad\halmos \cr}$$


5. y varies directly with $x^2$ .

Suppose that $x = 8$ when $y = 84$ .

Find y when $x = 6$ . Give an exact answer.

y varies directly with $x^2$ :

$$y = k x^2.$$

$x = 8$ when $y = 84$ :

$$\eqalign{ 84 & = k \cdot 8^2 = 64 k \cr \noalign{\vskip2pt} \dfrac{84}{64} & = k \cr \noalign{\vskip2pt} \dfrac{21}{16} & = k \cr}$$

Hence,

$$y = \dfrac{21}{16} x^2.$$

Set $x = 6$ :

$$y = \dfrac{21}{16} \cdot 6^2 = \dfrac{189}{4}.\quad\halmos$$


6. w is directly proportional to x and inversely proportional to y.

Suppose $w = 36$ when $x = 18$ and $y = 24$ .

Find w when $x = 8$ and $y
   = 30$ . Give an exact answer.

w is directly proportional to x and inversely proportional to y:

$$w = \dfrac{k x}{y}.$$

$w = 36$ when $x = 18$ and $y = 24$ :

$$\eqalign{ 36 & = \dfrac{k \cdot 18}{24} \cr \noalign{\vskip2pt} 36 & = \dfrac{3 k}{4} \cr \noalign{\vskip2pt} \dfrac{4}{3} \cdot 36 & = \dfrac{4}{3} \cdot \dfrac{3 k}{4} \cr \noalign{\vskip2pt} 48 & = k \cr}$$

Hence,

$$w = \dfrac{48 x}{y}.$$

Set $x = 8$ and $y = 30$ :

$$w = \dfrac{48 \cdot 8}{30} = \dfrac{64}{5}.\quad\halmos$$


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