First Naukri Mixtures and Allegations Quiz -1

By rule of Allegation,

 

   36              24

 

            X      

  

 (x-24)     (36-x)

 

  \frac{x-24}{36-x}=\frac{1}{2}

 

2x-48=36-x

 

3x=84

 

x=28

 

Total price of Rice= \frac{2100}{1.25}

 

= Rs. 1680

 

Therefore, total quantity of mixture of rice = \frac{1680}{28} =60 kg

 

So, required quantity of rice of Rs. 24 per kg =

 

= 40 kg

He sells the mixture at the cost price of pure milk.

Profit = 33.33%

\frac{profit}{cost}=\frac{1}{3}

Profit is due to water and cost is due to pure milk

So Water : Milk = 1 : 3

Concentration of milk in mixture = 75% = \frac{3}{4}

Using Allegations,

 80%        60%

 

        75%

  15           5

Ratio of 80% and 60% pure milk is 3 : 1

So, quantity of 60% pure milk = \frac{1}{3} \times 45= 15 ltrs

Solution:

Quantity of type A milk= 24 \times \frac{7}{12}=14 ltrs

Quantity of type B milk= 24 \times \frac{5}{12}=10 ltrs

SP of mixture = 24 \times 56 = Rs.1344

CP of mixture when sold at 12% profit = 1344 \times \frac{100}{112}= Rs. 1200

CP of type A milk = 14 \times 45 = Rs.630

CP of type B milk = Rs. (1200 –630) = Rs.570

Per liter CP of type B milk = \frac{570}{10}=Rs. 57

Let the ratio in which 2 mixtures are mixed = x : y.
Percent of water in the final mixture = 36.8%
Percent of water in first mixture = \frac{48-32}{48} \times 100=\frac{100}{3}%

Percent of water in second mixture = \frac{32-20}{32} \times 100=37.5%

By the rule of allegation-

Ratio=\frac{\text{water in second-water in final}}{\text{water in final-water in first}}

x:y=\frac{37.5-36.8}{36.8-\frac{100}{8}}

x:y=\frac{0.7}{\frac{10.4}{3}}

2.1 : 10.4 = 21 : 104

It is given that selling price of mixture is S mix = Rs 2
And Profit percentage on this S mix = 80%

So, cost price of mixture C mix = \frac{100\times 2}{180}=\frac{10}{9}

Cost of milk given is \frac{100}{9}

and we find that cost of mixture is 10% of cost of pure milk
= \frac{\frac{10}{9}}{\frac{100}{9}}\times 100

= 10%

Also, the cost of the mixture is proportional to the quantity of milk in the mixture. Therefore, we can say that a lot of water is mixed in milk to form a mixture due to which the price has gone down to 10% of pure milk, which means 90% is water and 10% milk.

So, the ratio of milk to water is 1 : 9.

In 56 kg of first mixture, Tea = 56 \times \frac{5}{7} [/latex]

= 40 kg and impurity = 56 -40 = 16 kg
In 44 kg of second mixture, Tea = 44 \times \frac{3}{4}


= 33 kg and impurity = 44 –33 = 11 kg
In large container quantity of pure tea = 40 + 33 + 17 = 90 kg
In large container quantity of impurity = 16 + 11 = 27 kg
Required ratio = 90 : 27 = 10 : 3

Solution:

Using alligation concept,

 

Container A                          Container B

 

   (x-5)%                                   (x+10)%

 

                           (x+4)%

 

      6%                                          9%

 

Ratio = A : B = 6 : 9 = 2 : 3

 

Therefore, amount of container B used = \frac{3}{2+3} \times 85 =51 ltrs

On the first night, he drank 60ml and added 60ml water to replace it.
Second night onwards, he drank 20ml more than the previous night. i.e., 60 + 20 = 80 ml
On third night, 80 + 20 = 100 ml
Capacity of bottle = 3L = 3000 ml
Capacity replaced is –
60 + 80 + 100 + 120 + ...... + 2980 + 3000
In the above expression 3000 is removed because on the last night he emptied the bottle without replacement.
So, the expression is –
60 + 80 + 100 + 120 + 140 + .... 2980

This is a arithmetic progression,
T_{n}= a + (n –1) d
2980 = 60 + (n –1) 20
n = 147
Total quantity is calculated by applying the sum formula
S_{n}={a+L}\times \frac{n}{2}

S_{n}={60+2980}\times \frac{147}{2}
=223440 ml

In 20 litres of milk and water solution, quantity of milk = 60% of 20 = 12 litres and the quantity of water = 20 -12 = 8 litres

In the new mixture, the quantity of milk = 28 + 12 = 40 litres and the quantity of water = 8 litres

The ratio of milk to water in the new mixture = 40 : 8 = 5 : 1

When, he takes out 12 litres of mixture then in 12 litres mixture, the quantity of milk

\frac{5}{6} \times 12

and the quantity of water = 12 -10 = 2 litres

The remaining quantity of milk in the mixture = 40 –10 = 30 litres and the remaining quantity of water in the mixture = 8 -2 = 6 litres

Now, again he mixes 4 litres of water then the quantity of water will become = 6 + 4 = 10 litres and the total quantity of mixture will become 30 + 10 = 40 litres in which 30 litres are milk

The reqd. concentration= \frac{30\times 100}{40} =75%

In 200 ml of mixture,

The quantity of rum = 56% of 200 = 112 ml

The quantity of water = 200 –112 = 88 ml

The quantity of another mixture = 300 –200 = 100 ml

In 300 ml, the quantity of rum = 40% of 300 = 120 ml

The quantity of water = 60% of 300 = 180 ml

The quantity of water was added = 180 –88 = 92 ml

The concentration of water in the mixture P = x% of 100 ml = 92 ml

x = 92%