Logo
    Logo

    Subject

    Quantitative Aptitude

    Topic

    Speed & Distance

    Two boats A and B are stationed at two points Q and R on a flowing river. The direction of flow of river is from Q to R. When the boats A and B moved towards each other they met at P, Which is at a distance of 20 m from the point R. When A moves towards B and B moves away from A they met at S which is at a distance of 40 m from the point R. Boat B’s speed in still water is 5 times the speed at which river is flowing. Find the ratio of the speed of boat A to boat B

    ক)
    6:1
    খ)
    8:1
    গ)
    5:1
    ঘ)
    7:1
    ঙ)
    None of these

    Explanation

    ∵ Speed × Time = Distance Let the distance between point Q and P is x m Let the speed of boat A and boat B and river is a, b and r m/s respectively. Therefore, We can say from the question that when the boats A and B moved towards each other they met at P, Which is at a distance of 20 m from the point R. So, ⇒xa  +  r  =  20b  −  r  \Rightarrow \frac{{x}}{{a\; + \;r}}\; = \;\frac{{20}}{{b\;-\;r}}{\rm{\;}} - - - - - - - - - - - - - - - - - - - - - - - eqn (1) We can also say that, when A moves towards B and B moves away from A they met at S which is at a distance of 40 m from the point R. hence similarly, ⇒x  +  60a  +  r  =  40b  +  r  \Rightarrow \frac{{x\; + \;60}}{{a\; + \;r}}\; = \;\frac{{40}}{{b\; + \;r}}{\rm{\;}} - - - - - - - - - - - - - - - - - eqn (2) Dividing eqn (1) by eqn (2) we get, ⇒2xx+60  =  b+rb−r  \Rightarrow \frac{{2x}}{{x + 60}}\; = \;\frac{{b+r}}{{b-r}}{\rm{\;}}  - - - - - - - - - - - - - - - - - - eqn (3) It is given in the question that Boat B’s speed in still water is 5 times the speed at which river is flowing ⇒ b = 5r Substituting in eqn 3, 2xx  +  60=b  +  rb  −  r\frac{{2x}}{{x\; + \;60}} = \frac{{b\; + \;r}}{{b\; - \;r}} ⇒2xx+60  =  5r+r5r−r  \Rightarrow \frac{{2x}}{{x + 60}}\; = \;\frac{{5r+r}}{{5r-r}}{\rm{\;}} ⇒2xx  +  60=1.5\Rightarrow \frac{{2x}}{{x\; + \;60}} = 1.5 ⇒ 2x = 1.5x + 90 ⇒ 0.5 x = 90 ⇒ x = 180 m Distance between the points Q and R = 180 + 20 = 200 m Putting value of x = 180 in eqn 1, ⇒xa  +  r  =  20b  −  r  \Rightarrow \frac{{x}}{{a\; + \;r}}\; = \;\frac{{20}}{{b\;-\;r}}{\rm{\;}} ⇒18020=a  +  rb  −  r\Rightarrow \frac{{180}}{{20}} = \frac{{a\; + \;r}}{{b\; - \;r}} ⇒a+rb−r=9\Rightarrow \frac{{a+r}}{{b-r}}=9 ⇒a  +  r5r  −  r=9\Rightarrow \frac{{a\; + \;r}}{{5r\; - \;r}} = 9 ⇒a+r4r=9\Rightarrow \frac{{a+r}}{{4r}} = 9 ⇒ a = 35r Ratio of speed of boat A : boat B = 35r : 5 r = 7:1

    Related questions

    A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:Speed of a boat in standing water is 9 kmph and the speed of the stream is 5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is:two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1/10 km and 9 km respectively. The time taken by slower train to cross the faster train in seconds is -A man can row three-quarters of a kilometre against the stream in 11×(1/4) minutes and down the stream in 7×(1/2) minutes. The speed (in km/hr) of the man in still water is:In a 300 m race A beats B by 5 metre or 6 seconds. B's time over the course is -A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?A boat covers a certain distance downstream in 1 hour, while it comes back in 1x(1/2) hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?A man buys tk 20 shares paying 9% dividend.  The man wants to have an interest of 12% on his money. The market value of each share is-

    Find Tutors

    View all Tutors