Oxygen was being pumped into a tank, and the speed at which the oxygen

Question

Oxygen was being pumped into a tank, and the speed at which the oxygen was being pumped was increasing at a constant, linear rate with respect to time. At the instant the pump was started, time 0 seconds, the oxygen was being pumped at P0 cubic units per second. 20 seconds later, 480 cubic units of oxygen had been pumped into the tank, and the oxygen was being pumped at P20 cubic units per second.

In the table, select values of P0 and P20 that are together consistent with the information provided. Make only two selections, one in each column.

Manav_24102000 · Accepted Answer

There is a shorter method of solving this using an AP.

Let us assume the first term is x and the common difference is the increasing linear rate, i.e., y .

For 0-1 seconds, the rate is x
For 1-2 seconds the rate is x+y

Now, 480 is the sum of the series

Therefore,

Sum of an AP = n/2 (First term + last term)

480 = 20/2 (First term + Last term)

First term + Last term = 48

Hence, find the numbers the sum of which is 48 and y>x.

catinabox · Answer

At t=0-1 secs, let's assume O2 is being pumed at x units/sec
Since the rate increases linearly,
At t=1-2 secs, the rate at which O2 is being pumped is say, x+y units/sec
At t=2-3 secs, it's x+2y
...
At t= 19-20 secs, it's x+19y

At the end of 20 secs, a total of 20x + 20*19*y/2 units of oxygen have been pumped.

20x + 190y = 480

The only options from the answers that fit this are x = 5, y =2

Therefore at t=0, P0= 5.
At t=20, P20 = 5+19*2= 43.

The question is a bit convulated and I had to make simplifying assumptions to map a linearly increasing rate into concepts that can be solved on the GMAT level. The right concept for this question is setting up an integration equation :/

Elite097 · Answer

Explanation
The initial rate at which the oxygen was being pumped = Po
The rate at which the oxygen was being pumped after 20 seconds = P20
Since the speed at which the oxygen was being pumped was increasing at a constant linear rate, the average rate at which the oxygen was being pumped for 20 seconds =
(Po + P20) /2
The amount of air pumped into a tank in 20 seconds = Time x The average rate at which the oxygen was being pumped for 20 seconds
480 = 20 x(Po + P20)/2
480 10 x (Po + P20)
48 = (Po + P20)
Thus, we need to select 2 answer choices such that (Po + P20) = 48. Additionally, since the speed at which the oxygen was being pumped was increasing at a constant linear rate Po must be less than P20, Po= 5 and P20 = 43 are the only answer choices that satisfy the given conditions.
Hence, for "Po" column, "5", and for "P20" column, "43" is the correct combination of the answer choices.

shashi1277 · Answer

Given: The speed at which the oxygen was being pumped was increasing at a constant, linear rate with respect to time.
Average rate = (P0+P20)/2
Therefore, Quantity of oxygen after P20 would be rate x time
480= (P0+P20)/2 x 20
P0+P20=48
P20=43 & P0=5

pappal · Answer

these type of questions can very easily be solved if we remember the simple equations of time, velocity(or rate) and acceleration(increase in velocity or rate)
v=u + at-------i & s=ut+1/2*a*t^2---------ii
in this case: u=P0, v=P20, t is the time and s is the quantity of oxygen pumped(the distance travelled in the original equation) & a is the increase in the rate per second or the acceleration.
so P20=P0+20*a or a=(P20-P0)/20
and 480=20*P0+1/2*a*20^2
480=20P0+1/2*(P20-P0)/20*400
on solving we get P0+P20=48
only values that satisfy are  P0=5 & P20=43

Oxygen was being pumped into a tank, and the speed at which the oxygen

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Oxygen was being pumped into a tank, and the speed at which the oxygen

Prep Toolkit

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