Question

agar tum ma dam ho to ya question solve karo wrong answer is not allowed​

Answer 1

Given :

• A tank is filled completely in 2 hours when both taps are open.
• If only one tap is open, the smaller tap takes 3 hours more than larger tap.

To Find :

• Time taken by smalller tap and larger tap to fill the tank separately.

Solution :

Let the time taken by larger tap to fill the tank be x hours.

$\bold{\sf{\underline{\underline{As\:per\:the\:question}}}}$

Time taken by smaller tap to fill the tank = ( x + 3) hours.

Part to tank filled in 1 hour :

• Larger tap = $\mathtt{\dfrac{1}{x}}$
• Smaller tap = $\mathtt{\dfrac{1}{x+3}}$

Larger + smaller tap in 1 hour :

• $\mathtt{\dfrac{1}{x}\:+\:{\dfrac{1}{x+3}}}$

Given, Both taps are able to fill the tank completely in 2 hrs when opened together.

Which implies, both tanks together will fill $\mathtt{\dfrac{1}{2}}$ part of the tank in a single hour.

Equation :

➟ $\mathtt{\dfrac{1}{x}}$ + $\mathtt{\dfrac{1}{x+3}}$ = $\mathtt{\dfrac{1}{2}}$

➟ $\mathtt{\dfrac{x+3+x}{x(x+3)}}$ = $\mathtt{\dfrac{1}{2}}$

➟ $\mathtt{\dfrac{2x+3}{x^2+3x}}$ = $\mathtt{\dfrac{1}{2}}$

➟ $\mathtt{2(2x+3)\:=\:x^2\:+\:3x}$

➟ $\mathtt{4x+\:6\:=\:x^2\:+\:3x}$

➟ $\mathtt{x^2\:+\:3x\:=\:4x\:+\:6}$

➟ $\mathtt{x^2\:+\:3x\:\:-\:4x\:-6\:=\:0}$

➟ $\mathtt{x^2\:-\:x\:-\:6\:=\:0}$

The equation formed is a Quadratic Equation which we can solve and find the value of x using factorization method.

➟ $\mathtt{x^2\:-\:3x\:+\:2x\:-\:6\:=\:0}$

➟ $\mathtt{x\:(x-3)\:\:+2\:(x-3)\:=\:0}$

➟ $\mathtt{x-3\:=\:0\:\:\:OR\:\:\:x+2\:=\:0}$

➟ $\mathtt{x=3\:\:\:OR\:\:\:x\:=\:-2}$

Since, time cannot be expressed in negative, x = - 2 is not acceptable.

Therefore, we are left with, x = 3.

$\bold{\large{\sf{\therefore{\underline{Time\: taken\: by\:larger\: tap\:<strong> </strong>to\:fill\:the\:tank\: =\:x\: =\: 3<strong> </strong>\:hrs}}}}}$

$\bold{\large{\sf{\therefore{\underline{Time\: taken\: by\: smaller \:tap \:to \:fill \:the\: tank\: = \:x\:+\:3 =3\:+\:3\:=\:6\:hrs}}}}}$

Answer 2

$\huge\underline\textbf{Question}$

A tank fills completely in 2 hours if both the taps are open. If only one of the tap is open at the given time the smaller tap takes 1 hour more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?

$\huge\underline\textbf{Answer}$

$\bf{Given,}$

→ A tank can be filled completely in 2 hours if both the taps are open.

→ If only one of the tap is open,the smaller tap takes 1 hour more than the larger one to fill the tank = Both the taps are filled in 2 hours,smaller tap takes 1 hour more = (2+1) hrs = 3 hrs more

$\bf{Find?}$

→ The time taken by each tap (smaller tap and larger tap) to fill the tank completely.

$\huge\textbf{Solution}$

• Let the time taken by larger tap to fill the tank = x hours

$\bf{A.T.Q}$ (According To Question)

• Time taken by smaller tap to fill the tank = $\bf{(x+3)\:hours}$
• Part of tank filled in 1 hour :

→ Larger tap :

$\frac{1}{x}$

→ Smaller tap :

$\frac{1}{x + 3}$

→ So,larger tap + smaller tap

$= > \frac{1}{x} + \frac{1}{ x + 3}$

• We came to know that,both the taps are filling the tank completely in 2 hours when they are opened together. So,both of the tank will fill 1/2 part of the tank in a single hour.

Solution :-

$\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2} \\ \frac{x + x + 3}{x(x + 3)} = \frac{1}{2} \\ \frac{2x + 3}{ {x}^{2} + 3x} = \frac{1}{2} \\ 2(2x + 3) = 1( {x}^{2} + 3x) \\ 4x + 6 = {x}^{2} + 3x \\ {x}^{2} + 3x = 4x + 6\\ {x}^{2} + 3x - 4x - 6 = 0 \\ {x}^{2} - x - 6 = 0$

• The equation formed above is a quadratic equation.
• So,now we can find the answer with the help of factorization.

${x}^{2} - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ (x + 2)(x - 3) = 0 \\ x + 2 = 0 \: or \: x - 3 = 0 \\ x = - 2 \: or \: x = 3$

• We know,time can't be considered in negative values,i.e. x = -2.
• So,now we are left with x = 3

$\huge\textbf{Final\:Answer}$

→ Time taken to fill the larger tank = 3 hours

→ Time taken to fill the smaller tank = x + 3 = 3 + 3 = 6 hours

Thank You!!