29
Q:

# A can contains a mixture of two liquids A and B in the ratio  7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

 A) 10 B) 20 C) 21 D) 25

Explanation:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left =$\inline (7x-\frac{7}{12}\times 9)=(7x-\frac{21}{4})$ litres.

Quantity of B in mixture left =  $\inline (5x-\frac{5}{12}\times 9)=(5x-\frac{15}{4})$litres.

$\inline \frac{(7x-\frac{21}{4})}{(5x-\frac{15}{4}+9)}=\frac{7}{9}$

$\inline \Rightarrow \frac{28x-21}{20x+21}=\frac{7}{9}$

$\inline \Rightarrow 252x-189=140x+147$

$\inline \Rightarrow x=3$

So, the can contained 21 litres of A.

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

466
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

1247
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

920
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

1038
Q:

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