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# X takes 3 hrs more than Y to walk 30 km. But, if X doubles his pace, he is ahead of Y by 1.5 hrs. What is their speed of walking?

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Jun 19, 2018

$\text{v"_"X" = 10/3\ "kmph}$ and $\text{v"_"Y" = "5 kmph}$

#### Explanation:

Speed ($\text{v}$), distance ($\text{d}$) and time ($\text{t}$) are related as

$\text{v" = "d"/"t}$

For first case

$\text{v"_"X" = 30/"t"_"X} \textcolor{w h i t e}{. .}$ and $\textcolor{w h i t e}{. .} \text{v"_"Y" = 30/"t"_"Y}$

It’s given that $\text{t"_"X" = "t"_"Y} + 3$

So, "v"_"X" = 30/("t"_"Y" + 3) color(white)(...) ——(1)

For second case

“X doubles his pace” means X doubles his speed. So speed of X now is $\text{2v"_"X}$

“X is ahead of Y by 1.5 hrs” means now X takes 1.5 hrs less than Y to travel same distance.
So, time tiken by X to travel 30 km is now $\text{t"_"Y} - 1.5$

"2v"_"X" = 30/("t"_"Y" - 1.5)

Substitute $\text{v"_"X}$ from first equation

2 × cancel(30)/("t"_"Y" + 3) = cancel(30)/("t"_"Y" - 1.5)

$\frac{2}{\text{t"_"Y" + 3) = 1/("t"_"Y} - 1.5}$

$\text{2t"_"Y" - 3.0 = "t"_"Y} + 3$

$\textcolor{b l u e}{\text{t"_"Y} = 6}$

• "v"_"X" = 30/"t"_"X" = 30/("t"_"Y" + 3) = 30/(6 + 3) = 30/9 = 10/3

• $\text{v"_"Y" = 30/"t"_"Y} = \frac{30}{6} = 5$

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2
Jun 19, 2018

${v}_{x} = \frac{10}{3}$, ${v}_{y} = 5$

#### Explanation:

Assuming that both $X$ and $Y$ walk at constant speeds ${v}_{x}$ and ${v}_{y}$, the time they need to walk $30$km is given by $t = \frac{30}{v}$. The first sentence translates into

$\frac{30}{v} _ x = \frac{30}{v} _ y + 3 \setminus q \quad \setminus q \quad \left(1\right)$

In fact, the time it takes to $X$ is $\frac{30}{v} _ x$, and it is $3$ more than the time it takes to $Y$, which is $\frac{30}{v} _ y$.

As for the second sentence, if $X$ doubles his speed he will pass from ${v}_{x}$ to $2 {v}_{x}$, and this time his time is $1.5$ less than $Y$'s, thus translating into

$\frac{30}{2 {v}_{x}} = \frac{30}{v} _ y - 1.5 \setminus q \quad \setminus q \quad \left(2\right)$

We can solve the system by subtracting $\left(1\right) - \left(2\right)$, so that $Y$'s time simplifies: we have

$\frac{30}{{v}_{x}} - \frac{30}{2 {v}_{x}} = \left(\frac{30}{v} _ y + 3\right) - \left(\frac{30}{v} _ y - 1.5\right)$

which can be written as

${\cancel{30}}^{15} / \left(\cancel{2} {v}_{x}\right) = \frac{15}{v} _ x = 4.5$

solving for ${v}_{x}$, we have ${v}_{x} = \frac{15}{4.5} = \frac{10}{3}$

Substitute this value for ${v}_{x}$ in $\left(1\right)$ to obtain ${v}_{y}$:

$\frac{30}{\frac{10}{3}} = \frac{30}{v} _ y + 3 \setminus \iff 30 \cdot \frac{3}{10} = \frac{30}{v} _ y + 3 \setminus \iff 9 = \frac{30}{v} _ y + 3$

which leads to $\frac{30}{v} _ y = 6$ and thus ${v}_{y} = 5$

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