Two riders have distance between them $118$ km and they are moving towards each other to meet . B starts an hour later by A. A travels $7$km in hour while B travels $16$km every three hours. How many kilometers(km) will A have already traveled, once they meet each other?
Could anyone help me with this puzzle?
Let us assume the time taken by them to meet for A=$x$ and for B=$y$. So now we know speed.time=distance so now we can generate 2 equations $7x+\frac{16y}{3}=118$...(1) and B starts 1 hour late $x-y=1$ ..(2) solving them we get $x=10,y=9$ . Hope its clear.so distance travelled by $A=7.t=7.10=70km$
Note that $$\mbox{speed·time = distance}$$ So, the distance $d$ by A after time $t$ is $$d = 7t.$$ Since both A and B have to meet and B started 118km away an hour later, the position of B will be given by $$118 - d = \frac{16}{3}(t-1).$$
Thus, you now have a system of two equations and two unknowns, I guess you can continue from there.
The speed of B is $16/3=5\frac{1}{3}$ km per hour. When he starts, A has already driven 7 km. You can therefore model the driving of the 2 by the equation
$7x+7=5\frac{1}{3}x$
do you understand why this is an equation?
Where the left hand side represents A, the RHS represents B, and x is the amount of time (in hours).
Can you do the rest?
