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# From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued  similarly for the third time, the amount of milk left after the third replacement:

 A) 45L B) 36.45L C) 40.5L D) 42.5L

Explanation:

General Formula:

Final or reduced concentration = initial concentration $\inline \dpi{100} \fn_cm \left ( 1-\frac{amount\: being\: replaced \: in \: each\: operation}{total\: amount} \right )^{n}$

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.

Therefore,    $\inline \dpi{100} \fn_cm 50\times \left ( 1-\frac{5}{50} \right )\times \left ( 1-\frac{5}{50} \right )\times \left ( 1-\frac{5}{50} \right )$

$\inline \dpi{100} \fn_cm 50\times \frac{45}{50}\times \frac{45}{50}\times \frac{45}{50}=50\times \left ( \frac{45}{50} \right )^{3}=50\times \left ( \frac{9}{10} \right )^{3}$

= 36.45 L

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• Related Questions

A Container has a capacity of 48 litres and is full of milk. 4 liters of milk is taken out of it and replaced by same quantity of water. Again 4 liters of the mixture is taken out and replaced by the same quantity of water . The process is repeated five times. How much milk is left in the mixture ?

$\inline \dpi{100} \fn_cm \frac{Milk\: Left\: in\: the\: container\: after\: the\: operations}{Original\: quantity\: of\: milk\: in\: the\: container\: }$ = $\inline \dpi{100} \fn_cm (\frac{Total\: quantity\: of\: milk - quantity\: drawn\: each\: time}{Total\: mixture })^n$

$\inline \dpi{100} \fn_cm \frac{Milk\: left\: in\: the\: container}{48 }=[\frac{48-4}{5}]^5$

Milk left in the container = $\inline \dpi{100} \fn_cm [\frac{44}{48}]^5\times 48$

$\inline \dpi{100} \fn_cm \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times 48$

$\inline \dpi{100} \fn_cm \frac{161051}{5184}$ = $\inline \dpi{100} \fn_cm 31\frac{347}{5184}$ liters

Subject: Alligation or Mixture - Quantitative Aptitude - Arithmetic Ability

8

In what proportion water must be added to spirit to gain 20% by selling it at the cost price ?

Let the C.P of spirit be = Rs.10 per litre

S.P of the mixture = Rs. 10 per litre

Profit = 20

$\inline \dpi{100} \fn_cm \therefore$ C.P of the mixture = Rs. $\inline \dpi{100} \fn_cm \frac{10\times 100}{120}$ = Rs. $\inline \dpi{100} \fn_cm \frac{25}{3}$ per litre

$\inline \dpi{100} \fn_cm \frac{Quantity\: of\: water}{Quantity\: of\: spirit}=\frac{5}{3}\times \frac{3}{25}=\frac{1}{5}$

$\inline \dpi{100} \fn_cm \therefore$Ratio of water and spirit = 1 : 5

Subject: Alligation or Mixture - Quantitative Aptitude - Arithmetic Ability

3

A sum of Rs.118 was divided among 50 boys and girls such that each boy received Rs.2.60 and each girl Rs.1.80. Find the number of boys and girls ?

Average money received by each = $\inline \dpi{100} \fn_cm \frac{118}{50}$ = Rs. 2.36

$\inline \dpi{100} \fn_cm \therefore$Ratio of No.of boys and girls = 56 : 24 = 7 : 3

$\inline \dpi{100} \fn_cm \therefore$ Number of boys = $\inline \dpi{100} \fn_cm 50\times \frac{7}{10}$ = 35

Number of girls = 50 - 35 = 15

Subject: Alligation or Mixture - Quantitative Aptitude - Arithmetic Ability

10

A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km /hr. Find the distance travelled on foot ?

$\inline \dpi{100} \fn_cm \frac{Time\: taken\: on\: foot}{Time\: taken\: on\: bicycle}=\frac{32}{24}=\frac{4}{3}$

Thus out of 7 hrs in all, he took 4 hrs to travel on foot

$\inline \dpi{100} \fn_cm \therefore$ Distance covered on foot in 4 hrs = ( 4 x 8) = 32 km

Subject: Alligation or Mixture - Quantitative Aptitude - Arithmetic Ability

5

In what ratio should two varieties of tea at Rs. 60 per kg and Rs.120 per kg be mixed together so that by selling the mixture at Rs.96 per kg, a profit of Rs.20% is obtained ?

Let us first calculate the cost of price of the mixture. the selling price of the mixture is given as Rs.96 and the profit is given as 20%. So, Cost price = $\inline \dpi{100} \fn_cm 96\times \frac{100}{120}$ = Rs. 80

Now, let us apply allegation method :

= 40 : 20 =  2 : 1