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Maths Puzzle
X runs 3/2 times as fast as Y. If X gives Y a starts of 300 meter, how must X run before he cateches up with Y?
Read Solution (Total 4)
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- let speed of y = z
then speed of x = 3z/2
if y runs i meter before x catches up
then x have to run i + 300 meter
equating the time we get :
(300+i)/(3z/2)=i/z
so i==600 meters - 11 years agoHelpfull: Yes(2) No(0)
- let speed of X be sx, & speed of Y be sy. then, sx/sy = 3/2. let X catches Y after running x m. then,the time at which X run (300+x) m, at the same time, Y will run x m. speed=dis/time => time= dis/speed.
=>sx/sy = (300+x)/x. =3/2.=> x=500m answer. - 11 years agoHelpfull: Yes(1) No(3)
- Ans- 900
Y=300 meter so X=3/2*Y=3*300/2=450
So Y is already 300 meter so next 300 meter X=450
So y=300+300+300 and X=450+450
So X cateches Y 900 meters - 11 years agoHelpfull: Yes(1) No(1)
- let y speed is k m/s then X have 3/2 k m/s
let Y run x meter and they met
so time is equal but X run 300 +xand Y only x meter
so x/k = (300+x)* 3/2k
ans is 600 m - 11 years agoHelpfull: Yes(0) No(0)
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