One tank fills in 3 1/13 minutes with the help of two taps. If the tank is filled using one tap, the smaller tube takes 3 minutes longer than the larger tube. So how long will it take to fill each tank individually?
Answers
Answer:
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Answer:
5 mins
Step-by-step explanation:
Let us assume the taps be A and B (B is smaller tap)
Given that
duration taken by taps A&B to fill the tank = 3 1/13 mins = 40/13 mins
=> efficiency of taps A & B together = 13/40 parts (w.r.t minutes ) -------- (1)
Assume that tap A fill the Tank in 'x' mins
=> efficiency of tap A = 1/x parts & time taken by tap B to fill the tank = x+3 mins.(given)
=> effeciency of tap B = 1/(x+3) parts
=> effeciency of the 2 taps together = 1/x + 1/ (x+3)
= x+3+x / (x)(x+3)
= 2x+3 / x²+3x
by (1) ,
2x+3 / x²+3x = 13/40
=> 80x+120 = 13x²+39x
=> 13x²+39x-80x-120 = 0
=> 13x²-41x-120 = 0
=> 13x²-65x+24x-120 = 0
=> 13x(x-5)+24(x-5) = 0
=> (x-5)(13x+24)=0
=> either x-5 =0 or 13x+24 =0
=> x=5 or x = -24/13
Here we have taken x as time, and time cannot be negative
so, omitting the negative value of x, gives x=5 mins
=> Tap A will take 5 mins to fill the tank