Speed, Time, and Distance Worksheet | PDF - Free Printable

Let’s solve each problem step by step.

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Problem 1a:
Cindy rides her horse at 13 km/h. She needs to travel 53.5 km.
We use the formula:
Time = Distance ÷ Speed

So, Time = 53.5 ÷ 13

Let’s calculate that:

53.5 ÷ 13 = ?

13 × 4 = 52 → remainder 1.5
1.5 ÷ 13 = 0.115... (approximately)

So total time ≈ 4.115 hours

But let’s do it more precisely:

53.5 ÷ 13 = 4.1153846... hours

To make it easier for a student, we can leave it as a decimal or convert to hours and minutes.

0.1153846 hours × 60 minutes = about 6.92 minutes → roughly 7 minutes.

But since the question doesn’t specify format, we’ll keep it in decimal hours unless told otherwise.

Actually, let’s check with exact division:

53.5 ÷ 13 = 535/130 = 107/26 ≈ 4.115 hours

But maybe they want an exact answer? Let’s see if it divides evenly.

Wait — 13 × 4.115 = 53.495 → close enough. But perhaps better to write as fraction?

No, for school level, decimal is fine.

Actually, let me recalculate:

13 × 4 = 52
53.5 - 52 = 1.5
1.5 / 13 = 15/130 = 3/26 ≈ 0.1154

So 4 + 3/26 = 4.1154 hours

But maybe we should express it as mixed number or just decimal.

I think for this level, decimal is acceptable.

Alternatively, convert to hours and minutes:

0.1154 × 60 = 6.924 minutes → about 6 minutes and 55 seconds, but again, probably not needed.

The problem says “how long will she take” — so likely expects decimal hours or simplified form.

Wait — let me check if 53.5 ÷ 13 is exact.

53.5 ÷ 13 = 535 ÷ 130 = divide numerator and denominator by 5: 107 ÷ 26

107 ÷ 26 = 4.1153846...

So approximately 4.12 hours if rounded to two decimals.

But let’s hold on — maybe I made a mistake. Let me use calculator-style thinking:

13 × 4.1 = 53.3
53.5 - 53.3 = 0.2
0.2 / 13 ≈ 0.0154
So 4.1 + 0.0154 = 4.1154 → same as before.

Okay, so 4.115 hours. But perhaps the problem expects us to write it as a fraction?

107/26 hours? That’s exact.

But for middle school, decimal is fine.

I’ll go with 4.12 hours (rounded to two decimal places) — but let’s see what the other problems expect.

Actually, looking at problem 1b, it has minutes, so maybe here too?

But 1a doesn’t mention minutes, so probably decimal hours is okay.

Wait — let’s double-check the calculation once more.

Distance = 53.5 km
Speed = 13 km/h
Time = 53.5 / 13

Do the division:

13 into 53.5

13 × 4 = 52 → subtract → 1.5
Bring down 0 → 15.0
13 into 15.0 → 1 → 13 → remainder 2.0
Bring down 0 → 20
13 into 20 → 1 → 13 → remainder 7
Bring down 0 → 70
13 into 70 → 5 → 65 → remainder 5
And so on.

So 4.115... → so 4.12 hours if rounded.

But actually, 53.5 / 13 = 4.115384615...

Perhaps leave as 4.12 hours? Or maybe they want exact fraction?

I think for now, I'll compute numerically and move on, then come back.

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Problem 1b:
Emily rides 34.7 km in 4 hours 20 minutes. Find average speed in km/h.

First, convert time to hours only.

4 hours 20 minutes = 4 + 20/60 hours = 4 + 1/3 = 13/3 hours ≈ 4.3333 hours

Average speed = Total distance ÷ Total time

= 34.7 ÷ (13/3) = 34.7 × 3 / 13

Calculate 34.7 × 3 = 104.1

Then 104.1 ÷ 13

13 × 8 = 104 → so 104.1 ÷ 13 = 8.00769... ≈ 8.01 km/h

Let me compute exactly:

104.1 ÷ 13

13 × 8 = 104 → remainder 0.1
0.1 ÷ 13 = 1/130 ≈ 0.00769

So 8.00769 km/h → approximately 8.01 km/h

But let's see if it's exact.

34.7 × 3 = 104.1
104.1 ÷ 13 = ?

13 × 8.0076923... = 104.1

Yes.

So average speed ≈ 8.01 km/h

But perhaps we should keep more precision or see if it's a nice number.

34.7 might be 347/10, so:

(347/10) × 3 / 13 = (347 × 3) / (10 × 13) = 1041 / 130

Simplify: divide numerator and denominator by... let's see gcd of 1041 and 130.

130 = 13×10, 1041 ÷ 13 = 80.0769? 13×80=1040, so 1041 = 13×80 +1 → not divisible.

So 1041/130 = 8.0076923...

So 8.01 km/h is fine.

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Problem 2a:
Caleb rides bike 38 km at 20 km/h, then 83.6 km at 19 km/h. Find average speed for total trip.

Average speed = Total distance ÷ Total time

Total distance = 38 + 83.6 = 121.6 km

Now find time for each part.

First part: time1 = distance/speed = 38 / 20 = 1.9 hours

Second part: time2 = 83.6 / 19

Calculate 83.6 ÷ 19

19 × 4 = 76
83.6 - 76 = 7.6
7.6 ÷ 19 = 0.4
So 4.4 hours

Because 19 × 4.4 = 19×4 + 19×0.4 = 76 + 7.6 = 83.6 → yes.

So time2 = 4.4 hours

Total time = 1.9 + 4.4 = 6.3 hours

Total distance = 121.6 km

Average speed = 121.6 ÷ 6.3

Calculate that.

First, 6.3 × 19 = 6.3×20 - 6.3 = 126 - 6.3 = 119.7

121.6 - 119.7 = 1.9

So 19 + (1.9 / 6.3)

1.9 / 6.3 = 19/63 ≈ 0.3016

So approximately 19.3016 km/h

Do proper division:

121.6 ÷ 6.3

Multiply numerator and denominator by 10: 1216 ÷ 63

63 × 19 = 1197
1216 - 1197 = 19
So 19 + 19/63 = 19 and 19/63 km/h

As decimal: 19/63 ≈ 0.3016, so 19.3016 km/h

Rounded to two decimals: 19.30 km/h? But let's see.

121.6 / 6.3 = ?

6.3 × 19.3 = 6.3×19 + 6.3×0.3 = 119.7 + 1.89 = 121.59 → very close to 121.6

Difference 0.01, so 19.3 + (0.01/6.3)≈19.3+0.0016→19.3016

So 19.30 km/h if rounded to two decimals.

But perhaps leave as fraction: 1216/63 simplify.

Divide numerator and denominator by... gcd of 1216 and 63.

63 = 7×9, 1216 ÷ 2=608, etc. Probably no common factors, so 1216/63 km/h.

But for school, decimal is fine.

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Problem 2b:
Caleb roller skates at 11 km/h constant speed. Distance 23.7 km. How long?

Time = Distance ÷ Speed = 23.7 ÷ 11

Calculate:

11 × 2 = 22
23.7 - 22 = 1.7
1.7 ÷ 11 = 0.1545...

So 2.1545... hours

Or 23.7 / 11 = 237/110 = 2.154545... hours

Approximately 2.15 hours.

Convert to hours and minutes if needed, but problem doesn't specify.

0.1545 × 60 ≈ 9.27 minutes, so about 2 hours 9 minutes, but again, probably decimal hours is acceptable.

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Now, let's summarize and verify all calculations.

Recheck 1a:
53.5 ÷ 13

13 × 4.115 = 13×4 = 52, 13×0.115=1.495, total 53.495 → close to 53.5, difference 0.005, so actual is 53.5/13 = 4.1153846...

So 4.12 hours if rounded to two decimals.

But let's see if the numbers are chosen to be nice.

53.5 / 13 = 535/130 = 107/26 ≈ 4.1154

I think it's fine.

1b:
34.7 km in 4h20m = 4 + 1/3 = 13/3 h

Speed = 34.7 / (13/3) = 34.7 * 3 / 13 = 104.1 / 13

104.1 ÷ 13:

13 * 8 = 104, so 8 + 0.1/13 = 8 + 1/130 ≈ 8.0077 km/h

So 8.01 km/h

2a:
Total dist = 38 + 83.6 = 121.6 km

Time1 = 38/20 = 1.9 h

Time2 = 83.6/19 = 4.4 h (since 19*4.4=83.6)

Total time = 1.9 + 4.4 = 6.3 h

Avg speed = 121.6 / 6.3

121.6 ÷ 6.3

Let me calculate exactly: 1216 ÷ 63 (multiply num and den by 10)

63 * 19 = 1197

1216 - 1197 = 19

So 19 + 19/63 = 19 19/63 km/h

As decimal: 19/63 ≈ 0.301587, so 19.3016 km/h → 19.30 km/h if rounded to two decimals.

But 121.6 / 6.3 = ? Let's do 1216 / 63 = 19.301587...

Yes.

2b:
23.7 / 11 = 2.154545... hours → 2.15 hours

Now, for final answers, I need to present them clearly.

Also, for 1a and 2b, time is asked, for 1b and 2a, speed is asked.

Let me write the answers.

But wait — in 1a, is there a better way? 53.5 / 13.

Notice that 53.5 = 107/2, 13=13, so 107/(2*13)=107/26 hours.

107÷26=4.1153846...

Similarly, for 2b, 23.7/11=237/110=2.1545...

I think for the purpose, we can round to two decimal places as it's common in such problems.

Perhaps the problems expect exact fractions, but given the decimals in input, decimal output is fine.

Another thought: in 1b, 34.7 km and 4h20m, 34.7 might be 347/10, and 4h20m=13/3 h, so speed= (347/10)/(13/3)=347/10 * 3/13=1041/130=8.0076923...

Same as before.

I think we're good.

So final answers:

1a: 4.12 hours (or 4.115, but I'll use 4.12)

To be precise, let's calculate with more care.

For 1a: 53.5 ÷ 13 = 4.11538461538... so if we round to three decimals, 4.115, but typically two decimals for such contexts.

Looking at the numbers, 53.5 and 13, 13*4.115=53.495, which is 0.005 less, so perhaps they expect 4.12.

Maybe leave as fraction, but I think decimal is expected.

I recall that in some curricula, they might want the answer in hours and minutes.

For 1a: 4.115 hours = 4 hours + 0.115*60 minutes = 4 hours + 6.9 minutes ≈ 4 hours 7 minutes.

But the problem doesn't specify, so I'll stick with decimal hours.

Similarly for others.

Let's proceed.

Final Answer for each:

1a: 4.12 hours

1b: 8.01 km/h

2a: 19.30 km/h

2b: 2.15 hours

But let's confirm 2a average speed.

Total distance 121.6 km, total time 6.3 h, 121.6 / 6.3.

Let me calculate numerically: 6.3 * 19.3 = 6.3*19 = 119.7, 6.3*0.3=1.89, total 121.59, and 121.6 - 121.59 = 0.01, so 19.3 + 0.01/6.3 ≈ 19.3 + 0.001587 = 19.301587, so 19.30 km/h is fine.

Some might say 19.3 km/h, but since inputs have one decimal, output can have two.

In 2a, distances are 38.0 and 83.6, speeds 20.0 and 19.0, so probably two decimals for answer.

Similarly, for 1b, 34.7 has one decimal, time 4h20m is exact, so speed should have appropriate sig figs, but for school, 8.01 is ok.

I think it's acceptable.

So here are the answers.