Question

anyone fast answer this question ??

0 the question please ^_^^_^​

Answer 1

These is the answer
Thanks for asking these question

Answer 2

let assume smaller tap can fill tank in x hours

then larger tab will fill tank in x-10 hours

in 1 hour smaller tap alone can fill = 1/x part of tank

in 1 hour larger tap alone can fill = 1/(x-10) part of tank

in 1 hour both can fill = 1/x + 1/(x-10) part of tank

given

in 75/8 hour both fill = full tank

in 1 hour both will fill = 1/(75/8) = 8/75 part of tank

equating both

1/x + 1/(x-10) = 8/75

(x-10+x)/(x^2-10x) = 8/75

(x-10 + x) × 75 = 8×^2 - 80x

150x - 750 = 8x^2 -80x

8x^2 -80x -150x + 750 = 0

8x^2 - 230x + 750 =0

4x^2 - 115x + 375 =0

4x^2 - 100x - 15x + 375 =0

4x( x - 25) -15(x - 25) = 0

(4x -15)(x -25)=0

x = 15/4, x =25

x = 15/4 is not valid as x - 10 will go negative

so x =25

larger tap time = x - 10 = 25-10 = 15