(D) The formula for the amount of substance remaining after $n$ operations is $A = x(1 - y/x)^n$,where $x$ is the initial quantity,$y$ is the quantity

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(D) The formula for the amount of substance remaining after $n$ operations is $A = x(1 - y/x)^n$,where $x$ is the initial quantity,$y$ is the quantity replaced,and $n$ is the total number of operations.Here,$x = 80$,$y = 8$,and the process was performed once initially and repeated $3$ more times,so $n = 1 + 3 = 4$.Remaining milk $= 80(1 - 8/80)^4$$= 80(1 - 0.1)^4$$= 80(0.9)^4$$= 80 \times 0.6561$$= 52.488 \text{ litre}$.Rounding to the nearest provided option,the answer is $52.48$.

Competitive Quantitative Aptitude Mixture and Alligation Mixture and Alligation

Medium

$A$ container contained $80$ $litre$ of milk. From this container $8$ $litre$ of milk was taken out and replaced by water. This process was further repeated $3$ times. How much milk is now contained in the container? (in $litre$)

A
$54.23$
B
$54.26$
C
$56.26$
D
$52.48$

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Mixture and Alligation

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Mixture and Alligation

$A$ barrel contains a mixture of wine and water in the ratio $3:1$. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes $1:1$?

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$A$ person has a chemical costing $Rs. 25$ per $litre$. In what ratio should water be mixed in the chemical so that by selling the mixture at $Rs. 20$ per $litre$,he may get a profit of $25\%$?

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Gold is $19$ times as heavy as water and copper is $9$ times as heavy as water. In what ratio should these metals be mixed so that the mixture may be $15$ times as heavy as water?

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Gold is $19$ times as heavy as water and copper is $9$ times. In what ratio should these metals be mixed so that the mixture may be $15$ times as heavy as water?

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Two alloys contain tin and iron in the ratio of $1:2$ and $2:3$. If the two alloys are mixed in the proportion of $3:4$ respectively (by weight),the ratio of tin and iron in the newly formed alloy is:

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$A$ container contained $80$ $litre$ of milk. From this container $8$ $litre$ of milk was taken out and replaced by water. This process was further repeated $3$ times. How much milk is now contained in the container? (in $litre$)

Similar Questions

$A$ barrel contains a mixture of wine and water in the ratio $3:1$. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes $1:1$?

$A$ person has a chemical costing $Rs. 25$ per $litre$. In what ratio should water be mixed in the chemical so that by selling the mixture at $Rs. 20$ per $litre$,he may get a profit of $25\%$?

Gold is $19$ times as heavy as water and copper is $9$ times as heavy as water. In what ratio should these metals be mixed so that the mixture may be $15$ times as heavy as water?

Gold is $19$ times as heavy as water and copper is $9$ times. In what ratio should these metals be mixed so that the mixture may be $15$ times as heavy as water?

Two alloys contain tin and iron in the ratio of $1:2$ and $2:3$. If the two alloys are mixed in the proportion of $3:4$ respectively (by weight),the ratio of tin and iron in the newly formed alloy is:

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