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MT1302
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Manav_24102000
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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.
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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
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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
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