Ratio & Proportion - Free Printable

Let’s solve this step by step.

We are told that y ∝ 1/x, which means y is inversely proportional to x.
That means:
> y = k / x, where k is a constant (the same for all pairs in the same table).

So, for each table, we can find k using one known pair of x and y, then use it to find missing values.

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Part 1: Fill in missing values

#### (a) Table:
x | 2 | 3 | 4 | 8
y | ? | 24 | ? | ?

We know when x=3, y=24 → so k = x × y = 3 × 24 = 72

Now fill others:

- x=2 → y = 72/2 = 36
- x=4 → y = 72/4 = 18
- x=8 → y = 72/8 = 9

✔ So row y: 36, 24, 18, 9

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#### (b) Table:
x | ? | 2 | 4 | 5 | 10
y | ? | ? | 15 | ? | ?

We know when x=4, y=15 → k = 4 × 15 = 60

Now fill others:

- x=2 → y = 60/2 = 30
- x=5 → y = 60/5 = 12
- x=10 → y = 60/10 = 6
- Now what about first x? We don’t have y yet — but wait, maybe we missed something?

Actually, looking again — the table has 5 columns. The third column is x=4, y=15 → k=60.

But the first two columns are blank for both x and y? That can’t be right unless there's more info.

Wait — let me re-read the problem.

Looking back at original image description:

In part 1(b), it says:

x | ? | 2 | 4 | 5 | 10
y | ? | ? | 15 | ? | ?

Hmm — only one value given: x=4, y=15 → k=60.

Then:

- x=2 → y=30
- x=5 → y=12
- x=10 → y=6

But what about the first column? Both x and y missing? That seems underdetermined.

Wait — perhaps I misread. Let me check again.

Actually, in many such problems, sometimes the first entry is meant to be filled based on pattern or maybe it’s a typo? But no — let’s think differently.

Perhaps the table is structured as:

Column 1: x=?, y=?
Column 2: x=2, y=?
Column 3: x=4, y=15
Column 4: x=5, y=?
Column 5: x=10, y=?

Since k=60 from column 3, then:

Column 2: x=2 → y=30
Column 4: x=5 → y=12
Column 5: x=10 → y=6

But column 1: both missing? That doesn’t make sense unless... maybe the first x is supposed to be found if y was given? Or vice versa?

Wait — perhaps I made a mistake. Let me look at the original problem again.

Actually, in the user’s message, for part 1(b):

It says:

x | ? | 2 | 4 | 5 | 10
y | ? | ? | 15 | ? | ?

Only one number given: y=15 when x=4.

So k=60.

Then:

For x=2 → y=30
x=5 → y=12
x=10 → y=6

But for the first column, since both x and y are unknown, we cannot determine them uniquely — unless... wait, maybe the first x is meant to be 1? Or perhaps it’s a trick?

No — actually, looking at other tables, they usually give enough info. Maybe I should assume that the first column is also solvable? But with current data, it’s not.

Wait — perhaps I miscounted. Let me write the table clearly:

Position: Col1 Col2 Col3 Col4 Col5
x: ? 2 4 5 10
y: ? ? 15 ? ?

Given only (4,15) → k=60.

So:

Col2: x=2 → y=30
Col4: x=5 → y=12
Col5: x=10 → y=6

Col1: still unknown. But maybe the problem expects us to leave it? Or perhaps there’s a typo?

Wait — another thought: maybe the first x is 1? Then y=60. But that’s assumption.

Alternatively, perhaps the table is meant to have symmetric values? Not necessarily.

I think there might be an error in my reading. Let me check part (c) and (d) to see pattern.

#### (c) Table:
x | 0.5 | 1 | 1.5 | 2
y | 12 | ? | ? | ?

Given x=0.5, y=12 → k = 0.5 × 12 = 6

Then:

x=1 → y=6/1 = 6
x=1.5 → y=6/1.5 = 4
x=2 → y=6/2 = 3

✔ So y row: 12, 6, 4, 3

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#### (d) Table:
x | ? | ? | 4 | 8
y | 16 | 5 | ? | 2.5

Given last column: x=8, y=2.5 → k = 8 × 2.5 = 20

Check with x=4: y should be 20/4 = 5 — but in table, y is blank for x=4? Wait no:

Table says:

x | ? | ? | 4 | 8
y | 16 | 5 | ? | 2.5

So for x=8, y=2.5 → k=20

Then for x=4, y=20/4=5 — but in table, y=5 is already given for second column? Wait no:

Let’s index:

Col1: x=?, y=16
Col2: x=?, y=5
Col3: x=4, y=?
Col4: x=8, y=2.5

From col4: k=8×2.5=20

Then col3: x=4 → y=20/4=5 — but in table, y=5 is listed under col2? Confusion.

Wait — perhaps the y-values are aligned:

Row y: 16, 5, ?, 2.5
Row x: ?, ?, 4, 8

So:

When x=8, y=2.5 → k=20

When x=4, y=20/4=5 — so the third y should be 5.

But in the table, the second y is 5 — which would correspond to some x.

So for y=5, x=k/y=20/5=4 — but x=4 is already in third column.

This suggests that the second column has y=5, so x=20/5=4 — but that conflicts with third column having x=4.

Unless... perhaps the table is:

Columns:

1: x=a, y=16
2: x=b, y=5
3: x=4, y=c
4: x=8, y=2.5

With k=20 from col4.

Then:

Col3: x=4 → y=5 → so c=5

Col2: y=5 → x=20/5=4 — but that would mean x=4 for both col2 and col3? Impossible.

Wait — unless I miscalculated k.

From col4: x=8, y=2.5 → k=20

From col2: y=5 → x=20/5=4

But col3 has x=4 — so if x=4 appears twice, that’s fine only if y is same, but here for col2 y=5, col3 y=?

If x=4 in col3, y=20/4=5 — so y=5 for both col2 and col3? But that would require x to be different? No.

I think there’s a misalignment.

Perhaps the table is:

x: [blank], [blank], 4, 8
y: 16, 5, [blank], 2.5

And we need to find blanks.

From x=8, y=2.5 → k=20

Then for y=16, x=20/16=1.25
For y=5, x=20/5=4 — but x=4 is already used in third column.

Conflict.

Unless the third column x=4 corresponds to y=5, and the second column y=5 corresponds to x=4 — same thing.

But then why list it twice?

Perhaps it's a typo in my understanding.

Another idea: maybe for col2, y=5, so x=4, and for col3, x=4, so y=5 — so it's consistent, but redundant.

Then col1: y=16 → x=20/16=5/4=1.25

Col3: x=4 → y=5

So table becomes:

x: 1.25, 4, 4, 8
y: 16, 5, 5, 2.5

But having x=4 twice with same y is odd, but mathematically ok if it's the same point repeated.

Perhaps the intention is that the second x is not 4.

Let's calculate k from another pair.

From col1: if y=16, and say x=a, k=16a

From col4: k=20

So 16a=20 → a=20/16=5/4=1.25

From col2: y=5, so x=20/5=4

From col3: x=4, y=20/4=5

So yes, it works, even though x=4 appears twice with y=5.

So we'll go with that.

So for (d):

x: 1.25, 4, 4, 8
y: 16, 5, 5, 2.5

But typically tables don't repeat, so perhaps the third x is not 4? But the problem says x=4 for third column.

I think we have to accept it.

So missing values:

First x: 1.25 or 5/4
Second x: 4
Third y: 5

But second x is already determined as 4 from y=5.

In the table, the x-row has first two blanks, then 4, then 8.

Y-row has 16, 5, blank, 2.5.

So:

- For y=16, x=20/16=1.25
- For y=5, x=20/5=4
- For x=4, y=20/4=5
- For x=8, y=2.5 (given)

So the blanks are:

x1 = 1.25, x2 = 4, y3 = 5

But x2=4 and x3=4 — same x value.

Perhaps it's acceptable.

Maybe the problem has a different k.

Another possibility: use y=5 and x=4 from col2 and col3? But they are different columns.

I think it's correct as per calculation.

So for (d): x values: 1.25, 4, 4, 8; y values: 16, 5, 5, 2.5

But to write neatly, perhaps as fractions.

1.25 = 5/4

So x: 5/4, 4, 4, 8
y: 16, 5, 5, 2.5

2.5 = 5/2

Ok.

Now back to (b). In (b), we have only one given pair: x=4, y=15 → k=60

Then:

x=2 → y=30
x=5 → y=12
x=10 → y=6

First column: both x and y missing. How to find?

Perhaps it's a mistake, or perhaps we can assume x=1, then y=60, but not specified.

Maybe the first x is 3? But not given.

Another thought: in some tables, the first entry might be derived from symmetry or something, but unlikely.

Perhaps I missed that in (b), there is another given value.

Let me double-check the user's input.

User wrote for 1(b):

x | ? | 2 | 4 | 5 | 10
y | ? | ? | 15 | ? | ?

Only y=15 is given for x=4.

So k=60.

Then the only way is to leave the first column as unknown, but that can't be.

Unless the first y is given? No.

Perhaps the "?" for y in first column is to be found, but x is also unknown.

I think there might be a typo in the problem, or perhaps in some versions, the first x is 1 or 3.

Maybe from the context, but let's look at part 2 later.

For now, for (b), we can fill the known ones, and for the first column, since no info, perhaps it's not required, but that seems odd.

Another idea: perhaps the table is for x and y, and the first column is x=1, then y=60, but not stated.

I recall that in inverse proportion, sometimes they start from x=1.

But to be safe, let's assume that for (b), the first x is 1, then y=60.

Or perhaps it's x=3, y=20, but arbitrary.

Let's calculate what it should be if we had another point.

Perhaps from the answer choices or standard, but no.

I think for the sake of progress, I'll assume that the first x is 1, so y=60, but I'm not sure.

Let's move to part 2, and come back.

Part 2: Find equation y = k/x for each table

#### (a) Table:
x | 1 | 2 | 4 | 5 | 8
y | 12 | 6 | 3 | 2.4 | 1.5

Find k = x*y for each:

1*12=12
2*6=12
4*3=12
5*2.4=12
8*1.5=12

All give k=12

So equation: y = 12/x

#### (b) Table:
x | 1 | 2 | 3 | 4
y | 2 | 1 | 2/3 | 1/2

Calculate k:

1*2=2
2*1=2
3*(2/3)=2
4*(1/2)=2

All k=2

So equation: y = 2/x

#### (c) Table:
x | 2 | 1/2 | 1 | 4
y | 9/4 | 9 | 9/2 | 9/8

Calculate k:

2 * (9/4) = 18/4 = 9/2 = 4.5
(1/2)*9 = 9/2 = 4.5
1 * (9/2) = 9/2 = 4.5
4 * (9/8) = 36/8 = 9/2 = 4.5

All k=9/2

So equation: y = (9/2)/x = 9/(2x)

#### (d) Table:
x | 1 | 2 | 4 | 8
y | 8 | 4 | 2 | 1

k = x*y:

1*8=8
2*4=8
4*2=8
8*1=8

All k=8

So equation: y = 8/x

Now back to part 1(b). Since in part 2, they have clear tables, for part 1(b), perhaps the first column is intended to be x=1, y=60 or something.

Maybe in (b), the first x is 3, then y=20, but let's see the values.

Another thought: in (b), after x=2,4,5,10, perhaps the first x is 1, as common.

Or perhaps it's x=6, y=10, but not specified.

I think for consistency, since k=60, and no other info, we can't determine the first column, but perhaps in the context, it's expected to be filled with x=1, y=60.

Maybe the "?" for y in first column is to be left, but that doesn't make sense.

Let's look at the original image description again.

User said for 1(b):

x | ? | 2 | 4 | 5 | 10
y | ? | ? | 15 | ? | ?

Perhaps the first y is given? No.

Another idea: maybe the first x is 3, and y=20, because 3*20=60, and 2*30=60, etc.

But why 3?

Perhaps from the sequence, but not clear.

I recall that in some textbooks, they might have x=1 as first.

To resolve, let's assume that for (b), the first x is 1, so y=60.

So x: 1,2,4,5,10
y: 60,30,15,12,6

That makes sense, and k=60 for all.

Probably that's what is intended.

Similarly for (d), we have x=1.25,4,4,8 but perhaps the second x is not 4.

In (d), if y=5 for second column, x=4, and for third column x=4, y=5, so it's the same, but perhaps it's a duplicate or error.

Maybe the third x is different.

Let's calculate k from first column.

In (d), if we take y=16, and say x=a, k=16a

From last column, k=8*2.5=20, so 16a=20, a=1.25

From second column, y=5, so x=20/5=4

From third column, x=4, y=20/4=5

So it's consistent, even if x=4 appears twice.

So we'll keep it.

So summary for part 1:

(a) y: 36, 24, 18, 9

(b) x: 1,2,4,5,10; y: 60,30,15,12,6 [assuming first x=1]

(c) y: 12,6,4,3

(d) x: 1.25,4,4,8; y: 16,5,5,2.5 or in fractions: x: 5/4,4,4,8; y: 16,5,5,5/2

But for (d), the y for third column is 5, which is already in second column, so perhaps it's fine.

Now for the answer, we need to box the final answers.

But the user asked to solve the problem, so probably fill in the blanks.

Also for part 2, we have the equations.

So let's write the answers.

First, for part 1:

(a) Missing y values: for x=2, y=36; x=4, y=18; x=8, y=9. So the y row is 36, 24, 18, 9

(b) Assuming first x=1, then y=60; for x=2, y=30; x=5, y=12; x=10, y=6. So x row: 1,2,4,5,10; y row: 60,30,15,12,6

(c) y row: 12,6,4,3

(d) x row: 5/4, 4, 4, 8; y row: 16,5,5,5/2

But for (d), the third y is 5, which is the same as second y, so perhaps it's correct.

For part 2:

(a) y = 12/x

(b) y = 2/x

(c) y = 9/(2x)

(d) y = 8/x

Now to present the final answer.

Since the user may expect the filled tables, but in text, we can list the missing values.

For part 1, the missing values are:

(a) y values: 36, 18, 9 (since 24 is given)

In (a), given y for x=3 is 24, so missing are for x=2,4,8: 36,18,9

Similarly for others.

For (b), missing x for first column: 1 (assumed), y for first: 60, y for x=2: 30, y for x=5: 12, y for x=10: 6

But x for first is missing, y for first is missing, etc.

Perhaps list all missing entries.

To simplify, for each subpart, provide the completed rows or the missing numbers.

Since the instruction is to solve, and for final answer, perhaps list the answers as per the tables.

But for the box, maybe summarize.

Given the complexity, I'll provide the answers as follows.

Final Answer:

For part 1:

(a) The missing y values are 36, 18, and 9. So the y row is: 36, 24, 18, 9

(b) The missing x value for the first column is 1, and the missing y values are 60, 30, 12, and 6. So x row: 1, 2, 4, 5, 10; y row: 60, 30, 15, 12, 6

(c) The missing y values are 6, 4, and 3. So y row: 12, 6, 4, 3

(d) The missing x values are 5/4 and 4, and the missing y value is 5. So x row: 5/4, 4, 4, 8; y row: 16, 5, 5, 5/2

For part 2:

(a) y = 12/x

(b) y = 2/x

(c) y = 9/(2x)

(d) y = 8/x

But for (d) in part 1, having x=4 twice might be confusing, but mathematically correct.

Perhaps in (d), the second x is not 4; let's double-check.

In (d) table:

x: ? , ? , 4 , 8

y: 16, 5 , ? , 2.5

From y=2.5, x=8, k=20

From y=16, x=20/16=5/4

From y=5, x=20/5=4

From x=4, y=20/4=5

So the second x is 4, and the third x is 4, so it's the same x value for two columns, which is unusual but possible if it's the same data point repeated.

Perhaps the problem has a typo, but based on given, it's correct.

So I'll go with that.

Now for the final answer box, since it's multiple parts, perhaps list them.

Since the user might expect the filled values, but for simplicity, I'll state the answers.

Final Answer:

Part 1:
(a) y-values: 36, 18, 9
(b) x-values: 1, 4 (for the blanks), y-values: 60, 30, 12, 6 — but better to specify.

Perhaps for each subpart, the completed table or the missing numbers.

To save space, for part 1, the missing values are:

(a) 36, 18, 9 for y
(b) 1 for first x, 60 for first y, 30 for second y, 12 for fourth y, 6 for fifth y — but second y is for x=2, which is given as ?, so yes.

In (b), the y for x=2 is missing, which is 30, etc.

For (c) 6,4,3 for y
For (d) 5/4 for first x, 4 for second x, 5 for third y

And for part 2, the equations.

So for the box, I'll write:

\boxed{\begin{array}{l} \text{Part 1:} \\ (a) \ y = 36, 18, 9 \\ (b) \ x = 1, 4; \ y = 60, 30, 12, 6 \\ (c) \ y = 6, 4, 3 \\ (d) \ x = \frac{5}{4}, 4; \ y = 5 \\ \text{Part 2:} \\ (a) \ y = \frac{12}{x} \\ (b) \ y = \frac{2}{x} \\ (c) \ y = \frac{9}{2x} \\ (d) \ y = \frac{8}{x} \end{array}}

But for (b) in part 1, I have x=1 and x=4 for the blanks, but there are two x blanks and four y blanks, but in the table, there are five columns, with x given for three columns (2,4,5,10 — wait, x is given for 2,4,5,10, so only first x missing, and y missing for all except x=4.

In (b): x has ? for first, then 2,4,5,10 — so only one x missing (first), and y has ? for first, second, fourth, fifth — so four y missing.

With k=60, and assuming first x=1, then y=60 for first, y=30 for x=2, y=12 for x=5, y=6 for x=10.

So missing: x1=1, y1=60, y2=30, y4=12, y5=6

Similarly for (d): x1=5/4, x2=4, y3=5

So in the box, for (b) : x=1, y=60,30,12,6

But y=15 is given, so the missing y are 60,30,12,6

For (d) : x=5/4,4, y=5

So let's adjust.

Final Answer:

\boxed{\begin{array}{l} \text{1.} \\ (a) \ y\text{-values: } 36, 18, 9 \\ (b) \ x_1 = 1, \ y_1 = 60, \ y_2 = 30, \ y_4 = 12, \ y_5 = 6 \\ (c) \ y\text{-values: } 6, 4, 3 \\ (d) \ x_1 = \frac{5}{4}, \ x_2 = 4, \ y_3 = 5 \\ \text{2.} \\ (a) \ y = \frac{12}{x} \\ (b) \ y = \frac{2}{x} \\ (c) \ y = \frac{9}{2x} \\ (d) \ y = \frac{8}{x} \end{array}}

This should be accurate.

Note: For 1(b), we assumed the first x is 1, as it's the most reasonable choice to complete the table with integer values and k=60. If the problem intends otherwise, it might be different, but based on standard problems, this is likely.

For 1(d), the values are as calculated.

So I think this is the solution.