A motorboat going downstream overcame a raft at a point A; T = 60 min later it turned back and after some time passed the raft at a distance l = 6.0 km from the point A. Find the flow velocity assuming the duty of the engine to be constant. [IE Irodov]
Given the duty of the engine remains constant, we will assume that the speed of the boat in still water is v and that of the flow is u, hence the raft will 'flow' with speed u. Now, the downstream velocity will be v + u, while the upstream velocity will be v-u. Here, an underlying deduction is that the speed of the boat must be greater than that of the stream for it to move upstream. For the given information we can write: Distance traveled by raft in the time boat turns back plus the distance traveled by raft by the time boat reaches the raft while travelling upstream should be 6 km. Hence, u(60/60) + u*((v+u)*(60/60) - u(60/60))/(v) = 6 where u(60/60) is the distance it covers before turnaround, and u*((v+u)*(60/60) - u(60/60))/(v+u) is the distance it covers after turnaround. Post turnaround travel time has been reached by subtracting pre-turnaround distance of raft from pre-turnaround distance of boat. Therefore, we have: u + u + u^2/v - u^2/v = 6 That is, u = 3 Km/hr The problem appears tricky for multiple variables with just one equation to determine the values. However, the trick here lies in realizing that the variables eventually cancel each other out leaving the answer required.
Consider a unit square that is kept at all possible arrangements in such a manner that one of the vertices lie on the Y axis while the other always lies on the X axis. Does any of the vertices reach a distance greater than sqrt(2) from the origin?
One can quickly break this problem down by considering the boundary cases. Let's consider a situation wherein the square's side makes an a angle of 45 degrees with the X axis, hence the two vertices on the axes will be at (1/sqrt(2) , 0) and (0, 1/sqrt(2)). In this arrangement, the diagonal of the square will be perpendicular to the axis. Now we know that diagonal of the unit square is of length sqrt(2). Further, the actual distance of the farthermost vertex will be sqrt(x^2 + y^2) where x and y are coordinates of the point. Now as we determined above, the y coordinate for the farthermost point, which is the length of the diagonal, is sqrt(2), therefore, overall distance must be greater than sqrt(2).
What are futures contracts and how are they priced?
Futures contract is fundamentally an arrangement in which two parties agree to transfer an asset at a particular time and at a particular price. For instance, if I were an oil importer who is to import Oil 2 months later and I expect the oil prices to increase steeply, I would 'buy' a futures contract that would fix the transaction price today itself. This arrangement 'hedges' my risk against a surge in prices. The futures price of an asset with the spot price being S is given as: F = Se^(r* t) where r is the risk-free rate and t is the time to maturity. The pricing formula can be deduced using the no-arbitrage principle. To elaborate: if F > Se^(r* t), arbitrageurs can buy the asset and short the futures contract. If F < Se^(r* t), they can short the asset and buy the futures contract. At maturity, both situations would lead to a risk free assured profit. Therefore, for the no-arbitrage principle to hold, F must be equal to Se^(r* t). In the former case, that is, when F > Se^(r* t), one essentially carries the asset until maturity and is also known as cash-and-carry arbitrage.