This problem involves calculating the original speed given changes in speed and distance traveled over the same time period.
Using the formula Distance = Speed × Time:
$D = S \times T$
$D = S \times 30$ (Equation 1)
The speed is reduced by $\frac{1}{15}$th. The new speed is:
New Speed $= S - \frac{1}{15}S = \frac{15S - S}{15} = \frac{14}{15}S$ km/h.
In this modified journey, the distance covered is 10 km less than the original distance.
New Distance $= D - 10$ km.
The time taken remains the same: $T = 30$ hours.
Using the formula Distance = Speed × Time for the modified journey:
New Distance $= \text{New Speed} \times T$
$D - 10 = (\frac{14}{15}S) \times 30$
$D - 10 = 14S \times 2$
$D - 10 = 28S$ (Equation 2)
Now, substitute the value of $D$ from Equation 1 into Equation 2:
$30S - 10 = 28S$
Rearrange the terms to solve for $S$:
$30S - 28S = 10$
$2S = 10$
$S = \frac{10}{2}$
$S = 5$ km/h
The original speed is 5 km/h.
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