## Mixture and Alligation

###
**Question 1: **

**In a mixture of 75 litres, the ratio of milk to water is 2 : 1. The amount of water to be further added to the mixture so as to make the ratio of the milk to water 1 : 2 will be**

(1) 45 litres

(2) 60 litres

(3) 75 litres

(4) 80 litres

#### Solution:

In 75 litres of the mixture.Let x litres of water be added,

#### Short Trick:

3 units = 75

*l*because 75

*l*was in ratio 2 : 1 initially.

###
**Question 2: **

A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make 20% water in the new mixture?(1) 3 litres

(2) 4 litres

(3) 5 litres

(4) 6 litres

####
**Solution **

(8 x 5 = 40

*l*which is the initial quantity. Hence 1 is also multiplied by 5)

####
**Short Trick:**

(40 + x) x 20% = 4 + xx = 5

*l*

###
**Question 3: **

In three vessels each of 10 *l*capacity, mixture of milk and water is filled, the ratios of milk and water are 2 : 1, 3 : 1 and 3 : 2 in the three respective vessels, if all the three vessels are emptied into a single large vessel, find the proportion of milk and water in the mixture.

(1) 121 : 59

(2) 123 : 59

(3) 125 : 59

(4) 127 : 59

####
**Solution:**

####
**2**^{nd} Method :

^{nd}Method :

L.C.M. of 3, 4 & 5. Hence 3 is multiplied by 4 x 5; 4 is multiplied by 3 x 5; 5 is multiplied by 3 x 4.

###
**Question 4 :**

**Vessels A and B contain mixtures of milk and water in the ratios 4 : 5 and 5 : 1 respectively. In what ratio should quantities of mixture be taken from A and B to form a mixture in which milk and water is in the ratio 5 : 4?**

(1) 2 : 5

(2) 4 : 3

(3) 5 : 2

(4) 2 : 3

####
**Solution: **

**Quantity of milk in vessel A**

**Quantity of milk in vessel B**

**Quantity of milk in final mixture**

So, mixture of vessels A and B are mixed in ratio of 5 : 2.

####
**2**^{nd} Method :

^{nd}Method :

###
**Question 5:**

50 g
of an alloy of gold and silver contains 80% gold (by weight). The
quantity of gold, that is to be mixed up with this alloy, so that it may
contain 95% gold, is(1) 200 g

(2) 150 g

(3) 50 g

(4) 10 g

####
**Solution: **

Quantity of gold in the alloyLet x gm of gold is added, then

####
**Other Method: **

1 unit when multiplied by 50 gives 50g. Hence 3 unit is also multiplied by 50.

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