Discussion
A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
Answer: B
Write \( b \) for the boat's speed in still water and \( r \) for the stream's. Then
\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]
\[ 4 \text{ h } 30 \text{ min} = 4.5 \text{ h} \]
\[ \frac{30}{15 + r} + \frac{30}{15 - r} = 4.5 \]
Combining the fractions over the common denominator \( (15+r)(15-r) = 225 - r^{2} \):
\[ \frac{30(15 - r) + 30(15 + r)}{225 - r^{2}} = \frac{900}{225 - r^{2}} = 4.5 \]
\[ 225 - r^{2} = 200 \quad\Longrightarrow\quad r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/h} \]
The \( r \) terms in the numerator cancel, which is what turns a messy equation into a simple one.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development