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Direct and Inverse Proportion: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The recorded equation

Positive readings satisfy $3y=7x$ . Write the relationship as $y=kx$ and identify $k$ .
1. Hint 1
  Direct variation expresses one reading as a fixed multiple of the other.
2. Hint 2
  Divide both sides by the coefficient of $y$ .
Answer

$y=\frac73x$ ; $k=\frac73$ .

Full solution

Divide both sides by the nonzero number $3$ .
$y=\frac73x$
Thus the constant of proportionality is $7/3$ .

For every positive $x$ , the ratio $y/x$ is $7/3$ , which checks that the relationship is direct variation.

Answer

$y=\frac73x$ ; $k=\frac73$ .

Key idea

A constant ratio can be present even when an equation is not initially solved for its output.
Problem 1: Mark complete
Problem 2 The product record

Positive readings satisfy $\frac{xy}{5}=9$ . Write $y$ in terms of $x$ in the form $y=\frac{k}{x}$ and identify $k$ .
1. Hint 1
  Inverse variation keeps the product of the readings fixed.
2. Hint 2
  Clear the constant denominator before dividing by positive $x$ .
Answer

$y=\frac{45}{x}$ ; $k=45$ .

Full solution

Multiply both sides by $5$ .
$xy=45$
Since $x>0$ , division by $x$ is allowed.
$y=\frac{45}{x}$

The product of $x$ and its matching $y$ is $45$ , so $k=45$ is the inverse-variation constant.

Answer

$y=\frac{45}{x}$ ; $k=45$ .

Key idea

Clearing a numerical denominator can expose the fixed product in an inverse relationship.
Problem 2: Mark complete
Problem 3 The juice bottles

A fixed volume of juice exactly fills $44$ bottles that each hold $0.75$ liters. The number of bottles it fills varies inversely with the size of each bottle. Find the constant of proportionality, with its unit, and how many $0.6$ liter bottles the same juice exactly fills.
1. Hint 1
  In inverse variation the product of the two quantities stays fixed.
2. Hint 2
  Multiply the known number of bottles by the known size to get the constant, then use it for the new size.
Answer

$k=33$ liters; $55$ bottles.

Full solution

Let $n$ be the number of bottles and $s$ the size of each bottle in liters.
Inverse variation means $ns=k$ .
The known pair gives
$k=44(0.75)$ $k=33$
The constant is the total volume of juice, $33$ liters.

For bottles of $0.6$ liters,
$n=\frac{33}{0.6}$ $n=55$
The check is $55(0.6)=33$ , the same volume.
Smaller bottles need more of them, as inverse variation predicts.

Answer

$k=33$ liters; $55$ bottles.

Key idea

In inverse variation the constant is the fixed product, here the total volume, and it predicts every other matching value.
Problem 3: Mark complete
Problem 4 The incomplete table

The figure shows a table for a relationship known to be either direct or inverse variation for positive inputs. Complete the two blank entries, identify the kind of variation, and state its equation.

A table of $x$ and $y$ values with two blank entries.

Text description of this figure

A table with two columns, headed x and y, and four rows of equal height, every cell outlined. Row one: x is 3 and y is 30. Row two: x is 5 and y is 18. Row three: x is 9 and the y cell is empty. Row four: the x cell is empty and y is 6. No other columns, labels or equations are shown.
1. Hint 1
  Use the complete rows to test a repeated ratio and a repeated product.
2. Hint 2
  Once the constant is known, each incomplete row must give that same ratio or product.
Answer

Blank $y=10$ ; blank $x=15$ ; inverse variation, $y=\frac{90}{x}$ .

Full solution

The complete rows are $(3,30)$ and $(5,18)$ .
Their ratios are $10$ and $3.6$ , which differ.
Their products are both $90$ , so the relationship is inverse with
$y=\frac{90}{x}$

At $x=9$ ,
$y=\frac{90}{9}$ $y=10$
When $y=6$ ,
$x=\frac{90}{6}$ $x=15$
Both completed rows have product $90$ .

Answer

Blank $y=10$ ; blank $x=15$ ; inverse variation, $y=\frac{90}{x}$ .

Key idea

Complete rows identify a variation constant that then fills incomplete rows.
Problem 4: Mark complete
Problem 5 The added material

The mass of a piece of uniform steel rod varies directly with its length. Adding $5$ meters of the same rod to one piece increases its mass by $4$ kg. Find the proportionality constant in kg per meter and the mass of a $12$ meter piece.
1. Hint 1
  The added length has the same mass per meter as the original piece.
2. Hint 2
  Call the original piece's length $x$ meters, and write the difference between the masses of lengths $x+5$ and $x$ using one constant.
Answer

$0.8$ kg per meter, or $\frac45$ kg per meter; $9.6$ kg.

Full solution

Let $m=kx$ describe mass in kg for length $x$ in meters.
With the original piece $x$ meters long, the increase gives
$k(x+5)-kx=4$ $5k=4$ $k=0.8$

For twelve meters,
$m=0.8(12)$ $m=9.6$
The added five meters have mass $0.8(5)=4$ kg, checking the reported change.

Answer

$0.8$ kg per meter, or $\frac45$ kg per meter; $9.6$ kg.

Key idea

A difference in direct-variation measurements can reveal the same constant as a complete pair.
Problem 5: Mark complete
Problem 6 The production setting

Output $q$ varies jointly with positive settings $a$ and $b$ . When $a=2$ and $b=5$ , the output is $30$ . Find the value of $b$ that gives output $36$ when $a=4$ .
1. Hint 1
  Joint variation uses the product of the two inputs.
2. Hint 2
  Find the constant from the complete record before solving for the new setting.
Answer

$b=3$ .

Full solution

Write $q=kab$ .
The complete record gives
$30=k(2)(5)$ $k=3$
For the required output,
$36=3(4)b$ $36=12b$ $b=3$

The check is
$3(4)(3)=36.$
All settings are positive as required.

Answer

$b=3$ .

Key idea

A joint-variation rule can be solved backward for a missing input after its constant is found.
Problem 6: Mark complete
Problem 7 The unchanged output

Positive output $y$ varies directly with $x$ and inversely with $z$ . A record with $x=4$ and $z=3$ has $y=20$ . A new record must keep $y=20$ while increasing $x$ to $10$ . Find the required $z$ and write the variation equation with its constant filled in.
1. Hint 1
  In combined variation, the ratio of the direct input to the inverse input controls the output.
2. Hint 2
  Use the complete record in $y=kx/z$ , then solve for the new denominator.
Answer

$z=7.5$ , or $\frac{15}{2}$ ; $y=\frac{15x}{z}$ .

Full solution

The record gives
$20=\frac{4k}{3}$ $k=15$
Thus $y=15x/z$ , with $z>0$ .
The new condition gives
$20=\frac{150}{z}$ $20z=150$ $z=7.5$

Both input ratios are $4/3$ and $10/7.5=4/3$ , so the output is unchanged.
Substituting gives $15(10)/7.5=20$ , as required.

Answer

$z=7.5$ , or $\frac{15}{2}$ ; $y=\frac{15x}{z}$ .

Key idea

To hold a combined-variation output fixed, the direct and inverse inputs must be scaled by the same factor, so their ratio stays the same.
Problem 7: Mark complete
Problem 8 The rising entries

The table in the figure has positive $x$ and $y$ values that both increase down the rows. Ana claims this alone shows direct variation. Decide whether the displayed data fit direct variation, inverse variation, or neither, and explain.

A table of three pairs of $x$ and $y$ values.

Text description of this figure

A table with two columns, headed x and y, and three rows of equal height, every cell outlined. Row one: x is 2 and y is 7. Row two: x is 4 and y is 14. Row three: x is 6 and y is 24. No other columns, labels or highlighting are shown.
1. Hint 1
  Increasing together is not the definition of direct variation.
2. Hint 2
  Test the ratios and products across all three rows.
Answer

Neither; the claim is false.

Full solution

The ratios $y/x$ in the table are $3.5$ , $3.5$ , and $4$ .
They are not all equal, so the data do not fit direct variation.

The products are $14$ , $56$ , and $144$ , also not equal.
Thus the data do not fit inverse variation either.
Both quantities increasing is not enough to establish a fixed ratio.

Answer

Neither; the claim is false.

Key idea

The ratio or product must repeat across the data, not merely follow a general increasing or decreasing trend.
Problem 8: Mark complete
Problem 9 The shared product

Two positive pairs $(a,b)$ and $(c,d)$ have both the same ratio of second coordinate to first coordinate and the same product of their coordinates. Lee claims the two pairs must be identical. Is this true? Justify the decision.
1. Hint 1
  Use the common ratio to write each second coordinate as the same multiple of its first.
2. Hint 2
  The common products then compare the squares of the two positive first coordinates.
Answer

Yes; $a=c$ and $b=d$ .

Full solution

Let the shared positive ratio be $k$ .
Then $b=ka$ and $d=kc$ .
Equal products give
$ka^2=kc^2$
Since $k>0$ , divide by $k$ to get
$a^2=c^2$

Both $a$ and $c$ are positive, so they are the same positive square root and $a=c$ .
Then $b=ka=kc=d$ .
Thus Lee is correct.

Answer

Yes; $a=c$ and $b=d$ .

Key idea

For positive coordinates, a shared ratio and a shared product determine the pair uniquely.
Problem 9: Mark complete
Problem 10 The unit labels

A direct relationship is written $y=2x$ when $x$ measures length in meters. Let $u$ be the numerical measure of the same length in centimeters, with $1$ meter equal to $100$ centimeters. Bo claims the same equation becomes $y=2u$ . Decide whether this is correct and give the equation in $u$ .
1. Hint 1
  The numerical measure changes when its unit changes.
2. Hint 2
  Express the meter measure $x$ in terms of the centimeter measure $u$ .
Answer

No; $y=0.02u$ , or $y=\frac{u}{50}$ .

Full solution

Since there are one hundred centimeters per meter,
$x=\frac{u}{100}$
Substituting into the original relationship gives
$y=2\left(\frac{u}{100}\right)$ $y=0.02u$

For example, one meter corresponds to $u=100$ and gives $y=2$ under both correct forms.
The claimed form would give $200$ , so it is incorrect.

Answer

No; $y=0.02u$ , or $y=\frac{u}{50}$ .

Key idea

A proportionality constant depends on the units used for its numerical input.
Problem 10: Mark complete

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