answer any of this.
Answers
Sorry bro I AM Not sure about the first one so I am just giving answers of 2 and 3 one.
Answer 2.Let the tap with smaller diameter fills the tank alone in x hours
Let the tap with larger diameter fills the tank alone in (x – 10) hours.
In 1 hour, the tap with a smaller diameter can fill 1/x part of the tank.
In 1 hour, the tap with larger diameter can fill 1/(x – 10) part of the tank.
The tank is filled up in 75/8 hours.
Thus, in 1 hour the taps fill 8/75 part of the tank.
1/x + 1/(x-10) = 8/75
(x-10) + x / x(x-10) = 8/75
2x – 10/x(x-10) = 8/75
75 (2x-10) = 8(x2-10x) by cross multiplication
150x – 750 = 8x2 – 80x
8x2 − 230x + 750 = 0
4x2−115x + 375 = 0
4x2 − 100x −15x + 375 = 0
4x(x−25)−15(x−25) = 0
(4x−15)(x−25) = 0
4x−15 = 0 or x – 25 = 0
x = 15/4 or x = 25
Case 1: When x = 15/4
Then x – 10 = 15/4 – 10
⇒ 15-40/4
⇒ -25/4
Time can never be negative so x = 15/4 is not possible.
Case 2: When x = 25 then
x – 10 = 25 – 10 = 15
∴ The tap of smaller diameter can separately fill the tank in 25 hours and the time taken by the larger tap to fill the tank = ( 25 – 10 ) = 15 hours.
Answer 3.let the speed of the stream be = u kmph
Let the speed of the boat be = v kmph
speed upstream will be = v – u kmph
speed downstream will be = v + u kmph
30 km upstream in time duration = 30 / (v – u) hrs
44 km down stream in time duration = 44 / (v + u) hrs
44 / (v + u) + 30 / (v -u) = 10 hrs — (1)
Similarly,
40 /(v – u) + 55 / (v +u) = 13 hrs Multiply with 3/4:
30/(v-u) + 165 / 4(v+u) = 39/4 — (2)
Now (1) – (2)
⇒ [44 – 165/4] / (v+u) = 10 – 39/4 = 1/4
⇒ v + u = 11 — (3)
Substitute this in (1) to get: 44/11 + 30/(v-u) = 10
⇒ v – u = 30/6 = 5 — (4)
Solving (3) and (4) ,
we get :
v = 8 kmph and
u = 3 kmph
I hope it will help you!!!