Question

plz answer the questions both
plzzzz​

Answer 1

28) Let one tap fill the tank in x hrs.

Therefore, other tap fills the tank in (x + 3) hrs.

Work done by both the taps in one hour is

1/x + 1/(x+3) = 13/40

(2x + 3)40 = 13(x2 + 3x)

13x2 – 41x – 120 = 0

(13x + 24)(x – 5) = 0

x = 5 (rejecting the negative value)

Hence, one tap takes 5 hrs and another 8 hrs separately to fill the tank.

29) Let the time taken by the smaller tap to fill the tank = x hours

time taken by larger tap = x - 9

In 1 hour, the smaller tap will fill 1/x of tank

In 1 hour, the larger tap will fill 1/(x-9) of tank.

In 1 hour both the tank will the tank = ⅙

In 1 hour both the tank will fill the tank= 1/x + 1/x-9

⅙ = (x-9)+ (x)/(x) (x-9)

⅙ = 2x-9/x²-9x

6(2x-9) = x²-9x

12x-54= x²-9x

x²- 9x-22x +54= 0

x²- 21x+54= 0

x²-18x-3x +54=0

x(x-18)-3(x-18)=0

(x-18) (x-3)=0

x= 18, x=3

We take x= 18

Smaller tap(x)= 18 h

Larger tap (x-9)=18-9= 9h

Hence, the time taken by the smaller tap to fill the tank = 18 hrs & the time taken by the larger tap to fill the tank = 9 h.

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