Question

please give answer me
please​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, Bigger tap can fill the tank in x min.

So,

→ Efficiency of Bigger tap = (1/x) / min.

and, Than,

→ Smaller tap will fill the tank in (x + 3) min.

→ Efficiency of Smaller tap = 1/(x+3) / min.

it is given that,

→ Both tap together fill the tank in = 3(1/13) = (40/13) min.

→ Efficiency of both = (13/40) / min.

A/q,

→ 1/x + 1/(x + 3) = (13/40)

→ (x + 3 + x)/x(x+3) = 13/40

→ 40(2x + 3) = 13(x² + 3x)

→ 80x + 120 = 13x² + 39x

→ 13x² + 39x - 80x - 120 = 0

→ 13x² - 41x - 120 = 0

→ 13x² - 65x + 24x - 120 = 0

→ 13x(x - 5) + 24(x - 5) = 0

→ (x - 5) (13x + 24) = 0

→ x = 5 or (-24/13) . { - ve value Not Possible. }

Hence,

→ Bigger tap can fill the tank in = 5 min.

→ Smaller tap can fill the tank in = 8 min.

Answer 2

Let the smaller tap can fill the tank in " x " min.

⇒ Efficiency of smaller tap = 1 / x

So Bigger tap can fill the tank in " x + 3 "min.

⇒ Efficiency of bigger tap = 1 / ( x + 3 )

Now it is given " Both taps can fill the tank in 3(1/13) min = 40/13 min"

⇒ Efficiency of both taps = 13/40

Now ,

$\bf \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{13}{40}\\\\\implies \bf \dfrac{x+3+x}{x(x+3)}=\dfrac{13}{40}\\\\\implies \bf 80x+120=13x^2+39x\\\\\implies \bf 13x^2-41x-120=0\\\\\implies \bf 13x^2-65x+24x-120=0\\\\\implies \bf 13x(x-5)+24(x-5)=0\\\\\implies \bf (x-5)(13x+24)=0\\\\\implies x=5\ \;,\;\;x=\dfrac{-24}{13}$

Time can't be negative.

⇒ x = 5 min

So  ,

Bigger tank can fill tank in x min = 5 min

Smaller tank can fill tank in " x + 3 " min = 8 min