Quantity , 1: Value of y. A vessel contains 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Initial contents: Water $= 2.5 \, 	ext{L}$,Milk $= 10 \, 	ext{L}$. Total $= 12.5 \, 	ext{L}$.
After removing $20 \%$ $(1/5)$: Remaining Water

Vedclass · Accepted Answer

Quantity $I =$ Quantity $II$

Vedclass · Answer

(D) Initial contents: Water $= 2.5 \, 	ext{L}$,Milk $= 10 \, 	ext{L}$. Total $= 12.5 \, 	ext{L}$.
After removing $20 \%$ $(1/5)$: Remaining Water $= 2.5 - (0.2 	imes 2.5) = 2 \, 	ext{L}$. Remaining Milk $= 10 - (0.2 	imes 10) = 8 \, 	ext{L}$.
Initial ratio (Water:Milk) $= 2:8 = 1:4$.
Adding $x$ liters of water to reverse the ratio to $4:1$: $\frac{2+x}{8} = \frac{4}{1} \Rightarrow 2+x = 32 \Rightarrow x = 30 \, 	ext{L}$.
Now,contents are Water $= 32 \, 	ext{L}$,Milk $= 8 \, 	ext{L}$.
Adding $y$ liters of milk to reverse the ratio back to $1:4$: $\frac{32}{8+y} = \frac{1}{4} \Rightarrow 128 = 8+y \Rightarrow y = 120 \, 	ext{L}$.
Since $y = 120$ and $Quantity \, 2 = 120$,$Quantity \, I = Quantity \, II$.

Similar Questions

If $\alpha$ and $\beta$ are roots of the equation $x^2 + px + \frac{3p}{4} = 0,$ such that $|\alpha - \beta| = \sqrt{10},$ then $p$ belongs to the set

If $1 - i$ is a root of the equation ${x^2} - ax + b = 0$, then $b = $

Solve the given two equations and select the correct answer from the given options:
$I. 6x^{2} + 29x + 35 = 0$
$II. 3y^{2} + 19y + 30 = 0$

Solve the given two equations and select the correct option.
$I. 3x^2 + 5x - 2 = 0$
$II. 2y^2 - 7y + 5 = 0$

If the roots of the equation $bx^2 + cx + a = 0$ are imaginary,then for all real values of $x$,the expression $3b^2x^2 + 6bcx + 2c^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module