3
Q:

# From a container, 6 liters milk was drawn out and was replaced by water. Again 6 liters of mixture was drawn out and was replaced by the water. Thus the quantity of milk and water in the container after these two operations is 9:16. The quantity of mixture is:

 A) 15 B) 16 C) 25 D) 31

Explanation:

Let quantity of mixture be x liters.

Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After operations , the quantity of pure liquid = $\inline \fn_jvn x(1-\frac{y}{x})^n$ units, Where n = no of operations .

So, Quantity of Milk = $\inline \fn_jvn x(1-\frac{6}{x})^2$

Given that, Milk : Water = 9 : 16

$\inline \fn_jvn \Rightarrow$ Milk : (Milk + Water) = 9 : (9+16)

$\inline \fn_jvn \Rightarrow$ Milk : Mixture = 9 : 25

$\inline \fn_jvn \therefore \frac{x(1-\frac{6}{x})^{2}}{x}=\frac{9}{25}$ $\inline \fn_jvn \Rightarrow x=15\: liters$

Q:

A container contains 50 litres of milk. From that 8 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container ?

 A) 24.52 litres B) 29.63 litres C) 28.21 litres D) 25.14 litres

Explanation:

Given that container has 50 litres of milk.

After replacing 8 litres of milk with water for three times, milk contained in the container is:

$\inline&space;\fn_jvn&space;\small&space;\Rightarrow&space;\left&space;[&space;50\left&space;(&space;1-\frac{8}{50}&space;\right&space;)^{3}&space;\right&space;]$

$\inline&space;\fn_jvn&space;\small&space;\Rightarrow&space;\left&space;(&space;50\times&space;\frac{42}{50}&space;\times&space;\frac{42}{50}\times&space;\frac{42}{50}\right&space;)$ = 29.63 litres.

5 77
Q:

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

516
Q:

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

1398
Q:

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

1071
Q:

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