Saturday, November 6, 2010

Mathrix

Hey Friends, Remember about old days we usually do those Trains, Man, Bridge, Cycle ..etc questions. Those were called Distance = Rate * Time mathematics queries.

So here again, just for fun......

Since an equation remains true as long as you divide through by the same non-zero element on each side, this formula can be written in different ways:

* To find rate, divide through on both sides by time:

Distance
Rate = -----------
Time

Rate is distance (given in units such as miles, feet, kilometers, meters, etc.) divided by time (hours, minutes, seconds, etc.). Rate can always be written as a fraction that has distance units in the numerator and time units in the denominator, e.g., 25 miles/1 hour.

* To find time, divide through on both sides by rate:

Distance
Time = -----------
Rate


==================================Best Questions==================================
Train A leaves the station traveling at 30 miles per hour. Two hours
later train B leaves the same station travelling in the same
direction at 40 miles per hour. How long does it take for train B to
catch up to train A?


If train A is going 30 miles/hour, then the amount of distance it can
go in a certain time (call it t) is:

30 miles/hour * t hours = 30t miles

The second train starts out 2 hours later, so for it the time it is
travelling is t-2. That menas that train B (going 40 miles per hour)
can go this distance in t-2 hours:

40 miles/hour * (t-2) hours = 40(t-2) miles

The question says that train B catches up with train A (which mean
that they go the same distance), so that means:

30t = 40(t-2)

Solve for t to find the number of hours the trains were traveling.

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Sadharan Byaaj = (Mooldhan*Dar*Time)/100

Chakravraddhi Byaj = Mooldhan(1+dar)^time

Time and Work formula : M1*T1/W1 = M2*T2/W2



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