Confused by a seemingly too simple solution for inequality word problem
$begingroup$
Between a city and a village lie two bike tracks: 35km of paved road, and 20km of dirt road. Because of the thick forest between them, a cyclist can choose to ride only one.
The speed on the paved road is at least 4km/h faster than that on the dirt road, and the time it takes to complete the dirt track is shorter by a half hour than that of the paved road.
- In what range of time is it possible to complete the dirt track?
- In what range of time would it take for the cyclist to complete both tracks?
I solved the first question, but the second one stumped me:
$$text{Let }v text{ be the speed on the dirt road}\text{Let }x text{ be the added speed on the paved road}$$
$$frac{35}{v+x}=frac{20}{v} + 0.5$$
From here, you can define $x$ in terms of $v$, and use the inequality $x > 4$, to get the correct range of: $$boxed{10km/h < v < 16km/h}$$
I don't know how to definitively solve the second question. The total time it takes to complete both tracks would be:
$$t = frac{35}{v+x} + frac{20}{v}$$
If you substitute $x = 4$ and the range of $v$ from before, you get the correct answer of:
$$boxed{3h < t < 4.5h}$$
...But why is this? This was just a pure guess by my part, as my first intuition was to describe $t$ in terms of $x$ for every speed limit & repeat the same process I did in the first question, but that turns into hell. How do I solve this without making this assumption?
inequality word-problem
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
$endgroup$
add a comment |
$begingroup$
Between a city and a village lie two bike tracks: 35km of paved road, and 20km of dirt road. Because of the thick forest between them, a cyclist can choose to ride only one.
The speed on the paved road is at least 4km/h faster than that on the dirt road, and the time it takes to complete the dirt track is shorter by a half hour than that of the paved road.
- In what range of time is it possible to complete the dirt track?
- In what range of time would it take for the cyclist to complete both tracks?
I solved the first question, but the second one stumped me:
$$text{Let }v text{ be the speed on the dirt road}\text{Let }x text{ be the added speed on the paved road}$$
$$frac{35}{v+x}=frac{20}{v} + 0.5$$
From here, you can define $x$ in terms of $v$, and use the inequality $x > 4$, to get the correct range of: $$boxed{10km/h < v < 16km/h}$$
I don't know how to definitively solve the second question. The total time it takes to complete both tracks would be:
$$t = frac{35}{v+x} + frac{20}{v}$$
If you substitute $x = 4$ and the range of $v$ from before, you get the correct answer of:
$$boxed{3h < t < 4.5h}$$
...But why is this? This was just a pure guess by my part, as my first intuition was to describe $t$ in terms of $x$ for every speed limit & repeat the same process I did in the first question, but that turns into hell. How do I solve this without making this assumption?
inequality word-problem
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
$endgroup$
add a comment |
$begingroup$
Between a city and a village lie two bike tracks: 35km of paved road, and 20km of dirt road. Because of the thick forest between them, a cyclist can choose to ride only one.
The speed on the paved road is at least 4km/h faster than that on the dirt road, and the time it takes to complete the dirt track is shorter by a half hour than that of the paved road.
- In what range of time is it possible to complete the dirt track?
- In what range of time would it take for the cyclist to complete both tracks?
I solved the first question, but the second one stumped me:
$$text{Let }v text{ be the speed on the dirt road}\text{Let }x text{ be the added speed on the paved road}$$
$$frac{35}{v+x}=frac{20}{v} + 0.5$$
From here, you can define $x$ in terms of $v$, and use the inequality $x > 4$, to get the correct range of: $$boxed{10km/h < v < 16km/h}$$
I don't know how to definitively solve the second question. The total time it takes to complete both tracks would be:
$$t = frac{35}{v+x} + frac{20}{v}$$
If you substitute $x = 4$ and the range of $v$ from before, you get the correct answer of:
$$boxed{3h < t < 4.5h}$$
...But why is this? This was just a pure guess by my part, as my first intuition was to describe $t$ in terms of $x$ for every speed limit & repeat the same process I did in the first question, but that turns into hell. How do I solve this without making this assumption?
inequality word-problem
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
$endgroup$
Between a city and a village lie two bike tracks: 35km of paved road, and 20km of dirt road. Because of the thick forest between them, a cyclist can choose to ride only one.
The speed on the paved road is at least 4km/h faster than that on the dirt road, and the time it takes to complete the dirt track is shorter by a half hour than that of the paved road.
- In what range of time is it possible to complete the dirt track?
- In what range of time would it take for the cyclist to complete both tracks?
I solved the first question, but the second one stumped me:
$$text{Let }v text{ be the speed on the dirt road}\text{Let }x text{ be the added speed on the paved road}$$
$$frac{35}{v+x}=frac{20}{v} + 0.5$$
From here, you can define $x$ in terms of $v$, and use the inequality $x > 4$, to get the correct range of: $$boxed{10km/h < v < 16km/h}$$
I don't know how to definitively solve the second question. The total time it takes to complete both tracks would be:
$$t = frac{35}{v+x} + frac{20}{v}$$
If you substitute $x = 4$ and the range of $v$ from before, you get the correct answer of:
$$boxed{3h < t < 4.5h}$$
...But why is this? This was just a pure guess by my part, as my first intuition was to describe $t$ in terms of $x$ for every speed limit & repeat the same process I did in the first question, but that turns into hell. How do I solve this without making this assumption?
inequality word-problem
inequality word-problem
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
asked Jan 4 at 23:18
daedsidogdaedsidog
29517
29517
add a comment |
add a comment |
1 Answer
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$begingroup$
From the first equation, we can solve for $x$:
$$x=frac{30v-v^2}{v+40}$$
Plug this into the expression for $t$:
$$t=frac{35}{v+frac{30v-v^2}{v+40}}+frac{20}{v}$$
Simplify:
$$t=frac{40}{v}+frac 1 2$$
From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
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votes
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oldest
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active
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votes
$begingroup$
From the first equation, we can solve for $x$:
$$x=frac{30v-v^2}{v+40}$$
Plug this into the expression for $t$:
$$t=frac{35}{v+frac{30v-v^2}{v+40}}+frac{20}{v}$$
Simplify:
$$t=frac{40}{v}+frac 1 2$$
From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
From the first equation, we can solve for $x$:
$$x=frac{30v-v^2}{v+40}$$
Plug this into the expression for $t$:
$$t=frac{35}{v+frac{30v-v^2}{v+40}}+frac{20}{v}$$
Simplify:
$$t=frac{40}{v}+frac 1 2$$
From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
From the first equation, we can solve for $x$:
$$x=frac{30v-v^2}{v+40}$$
Plug this into the expression for $t$:
$$t=frac{35}{v+frac{30v-v^2}{v+40}}+frac{20}{v}$$
Simplify:
$$t=frac{40}{v}+frac 1 2$$
From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
From the first equation, we can solve for $x$:
$$x=frac{30v-v^2}{v+40}$$
Plug this into the expression for $t$:
$$t=frac{35}{v+frac{30v-v^2}{v+40}}+frac{20}{v}$$
Simplify:
$$t=frac{40}{v}+frac 1 2$$
From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 4 at 23:26
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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