Question

solve it bros please send me fast
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Answer 1

Let the time taken by smaller tap be x hours and larger be x-2 hours.

ATQ..

1/x +1/x-2 = 1÷15/8hours

(x-2)+x÷ x²-2x = 8/15hours

2x-2÷x²-2x=8/15

By cross multipliacation:

8x²-16x=30x-30

4x²-23x+15=0

4x²-20x-3x+15=0

4x(x-5)-3(x-5)=0

(4x-3)(x-5)=0

reject x=3/4

Hence x=5

# time taken by smaller tap= 5 hours

# time taken by larger tap= (x-2)hours

                                            =3 hours

please mark as brainliest...        

Answer 2

$\mathfrak{\huge{Answer:}}$

Let the time taken by the tap of larger diameter (Let this be = $\sf{T_{1}}$) be = x

Then, the time taken by the tap of smaller diameter (Let this be = $\sf{T_{2}}$ )will be = x + 2

Now, the one hour work of $\sf{T_{1}}$ = $\tt{\frac{1}{x}}$

One hour work of $\sf{T_{2}}$ = $\tt{\frac{1}{x +2}}$

Given is that they finish their work together in $\tt{1\frac{7}{8}}$ hours.

The one hour work of $\sf{T_{1}}$ and $\sf{T_{2}}$ together = $\tt{\frac{8}{15}}$

According to the question, put all this information in proper sequence and then solve :

$\tt{\frac{1}{x} + \frac{1}{x+2} = \frac{8}{15}}$

=》 $\tt{\frac{x + 2 + x}{x^{2} + 2x} = \frac{8}{15}}$

=》 $\tt{30x + 30 = 8x^{2} + 16x}$

=》 $\tt{8x^{2} - 14x - 30 = 0}$

Equation is formed. Now, put the values in the quadratic formula and then solve.

$\mathfrak{Quadratic\:Formula}$ = $\tt{\frac{-b ± \sqrt{b^{2} - 4ac}}{2a}}$

Now, here, keep the values of the variable as :

a = 8

b = ( -14 )

c = ( - 30 )

Now, solve the equation:

$\tt{\frac{- (-14) \pm \sqrt{(-14)^{2} - 4 (8)(-30)}}{2(8)}}$

=》 $\tt{\frac{14 \pm \sqrt{196 +960}}{16}}$

=》 $\tt{\frac{14 \pm \sqrt{1156}}{16}}$

=》 $\tt{\frac{14 \pm 34}{16}}$

=》 $\tt{\frac{14 + 34}{16} \: or\: \frac{14 - 34}{16}}$

=》 $\tt{3\:or\:\frac{(-5)}{8}}$

Since Time > 0, the second answer isn't correct.

Thus, x = Time = 3 hours

Time of another tap = x + 2 = 5 hours

Answer : $\bold{3\:hours\:and\:5\:hours}$