Pipes and cistern 🌹

Let a,b,c be the respective time for which z1,z2,z3 pipes are opened (b-a)=2(c-b) 3b=2c+a c=6 (5,10,25)===50 10,5,2 10a+5b=38 -a+3b=12 (30+5)b=38+120=158 b=158/35

Solution anyone

Volume of tank = (2/3)(22/7)*10.5^3 = 2425.5 m^3 = 2425500 L Time = 2425500*(2/3)/7.7 = 210000 second = 175/3 hour

5/z + 5/(z + 3) + 5/(z + 8) = 1 5{z^2 + 11z + 24) + 5(z^2 + 8z) + 5(z^2 + 3z) = z(z + 3)(z + 8) 15z^2 + 110z + 120 = z(z^2 + 11z + 24) 15z^2 + 110z + 120 = z^3 + 11z^2 + 24z z^3 - 4z^2 - 86z - 120 = 0 z^3 - 12z^2 + 8z^2 - 96z + 10z - 120 = 0 z^2(z - 12) + 8z(z - 12) + 10(z - 12) = 0 (z - 12)(z^2 + 8z + 10) = 0 z = 12

32 48 - 64 192 6 4 - 3 3x+1*(112-x)=192 2x=80 x=40 B tym= 112-40=72 hrs

Two pipes a and b can fill an empty cistern in 32 and 48 hr and 3rd pipe empty in 64 hrs Initially when the cistern was empty pipe a and c are opened After a few hours Pipe a was turned off and b turned on instantly In all it took 112 hours to fill the cistern For how long was pipe b turned on??

1/2 of the work = 1.2 1/3 rd = 0.8 5/6 th = 2 1.2*x/y+ 8*y/x = 2

12 and 8 ni samjh aa rha please kaise aya please explain

Tw = 2.4 12x/y + 8y/x = 20 Let x/y = z 3z + 2/z = 5 z = 2/3 2:3 Faster = 2.4*5/3=4

Time taken to cross 125m (1/8 km) is 1/(18*8) Time taken to cross 375m (3/8 km) is 3/(36*8) Dist travelled by car: 6*90/(36*8) ≈ 1.8?

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75 A:b = 5:3 5*8 = 40 75-40=35 X(5*5/6 + 5*3/8)=35 X=168/29