6
Q:

# Tea worth Rs. 126 per kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be

 A) Rs. 169.50 B) Rs.1700 C) Rs. 175.50 D) Rs. 180

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135/2) = Rs.130.50
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.
Hence, price of the third variety = Rs.175.50 per kg.

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

462
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

1217
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

871
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

1008
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