UrbanPro
Login Ask & Answer Success Stories Paid Courses Free Classes Tuition & Classes Fees Write a Review All Categories Help Center
Post your Learning Need Signup as a Tutor
✕

Popular Cities

Bangalore

Hyderabad

Delhi

Mumbai

Chennai

Pune

Kolkata

Gurgaon

Ahmedabad

Jaipur

Noida

✕

Take Class 10 Tuition from the Best Tutors

  • Affordable fees
  • 1-1 or Group class
  • Flexible Timings
  • Verified Tutors

Book a Free Demo

Class 10 > Learn Mathematics > A motor boat whose speed is 18 km/h in still water...
✕

Search in

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Asked by Last Modified  

1 Answer

Learn Mathematics +1 Class IX-X Tuition

Follow 1
Answer

Please enter your answer

Math Magician

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour. Now 24/(18-X) - 1 =24/ (18+X) . OR solving the equation—X2 +48X -18×18 =0 , OR (X-6)(X+54) OR ...
read more

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

Boat speed in Still water 18km/hour. So boat speed in down stream =( 18+X) km/hour  ,boat speed in up stream =(18 -X)km/hour. Time in up stream – time in down stream =1 hour.  Now  24/(18-X)   - 1  =24/ (18+X)  .

OR    solving the equation—X2 +48X -18×18 =0  , OR   (X-6)(X+54)  OR   X=6 km/hour speed of stream                        

read less
Comments
Dislike Bookmark

Related Questions

The perimeter of one face of a wooden cube is 20 cm. Find its weight if 1 cm3 of wood weighs 8.25 g.
Let a be the side of cube,Perimeter of a wooden cube=20cm=4aWe know, Perimeter of a cube = 4a Therefore, 4a = 20cm a = 5cm=
Kanak
0 0
5
Define pie chart and frequency polygon?
Pie chart: If you consider a circle and it into some several parts. for each part you allot some percentage of items okay.
Ambar
0 0
7
Name the axes where the two coordinate meet?
The point of intersection of the axes is called the origin.
Bimal
What are the measures of central tendency?
Mean, median and Mode are the measures of central tendency.
Sitaram
What is the formula for arithmetic mean for unclassified data?
Arithmetic mean of unclassified data x1, x2, x3, x4, ... xn = X X= x1+ x2 + x3 + x4+ ... + xn —--—--——--—--—--——--—--— . n
Raj

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Quadratic Equation Completing Square
Find the x-intercepts of y = 4x2 – 2x – 5. First off, remember that finding the x-intercepts means setting y equal to zero and solving for the x-values, so this question is really asking...

Kiran Kumar HJ

0 0
0
Summary of 10th maths CBSE (imp topic is covered)
Hello friendsJust read all those stuff and try to remind it and if u have doubts in any of them then go through book and either call me 9710368667It's really helpful: 1)real number :for finding HCF and...

Deepak Kumar

0 0
0
Rotation of a Shape
* Rotation is nothing but rotating an object through an angle of 90 degrees, 180 degrees, 270 degrees either clockwise or anticlockwise. * For rotating 90 degrees clockwise and anticlockwise the steps...

Rasheedahamed F

0 0
0
Common Tangent

Rahul J.

1 0
0
Class 10th CBSE

Vikas Mehra

0 0
0

Recommended Articles

Meet Swati, a Hindi Tutor from Bangalore

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Quest Academy - Institute of the month

Quest Academy is a professional Bangalore based NEET and JEE (Main + Advanced) training institute. The academy was incorporated in 2015 to cater to the needs of students, who aim to crack competitive exams by connecting with the best brains around. The institute helps students enhance their skills and capabilities through...

Read full article >

Meet Sandhya R, a B.Sc tutor from Bangalore

Sandhya is a proactive educationalist. She conducts classes for CBSE, PUC, ICSE, I.B. and IGCSE. Having a 6-year experience in teaching, she connects with her students and provides tutoring as per their understanding. She mentors her students personally and strives them to achieve their goals with ease. Being an enthusiastic...

Read full article >

Meet Raghunandan.G.H, a B. Tech Tutor from...

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Read full article >

Find Class 10 Tuition near you
Class 10 Tuition in Delhi
Class 10 Tuition in Bangalore
Class 10 Tuition in Hyderabad
Class 10 Tuition in Kolkata
Class 10 Tuition in Mumbai
Class 10 Tuition in Chennai
Class 10 Tuition in Pune
Class 10 Tuition in Noida
Class 10 Tuition in Lucknow
Class 10 Tuition in Jaipur

Looking for Class 10 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you
X

Looking for Class 10 Tuition Classes?

The best tutors for Class 10 Tuition Classes are on UrbanPro

  • Select the best Tutor
  • Book & Attend a Free Demo
  • Pay and start Learning

Take Class 10 Tuition with the Best Tutors

The best Tutors for Class 10 Tuition Classes are on UrbanPro

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more

ThinkVidya Learning Pvt Ltd © 2010-2024All Rights Reserved