HackerRank Mixture and Alligation Quiz -1

In the mixture of 48 litre,

Roohafza= 48*(3/4)=36ltr

 Water=48 \times \frac{1}{4} = 12 ltr

New ratio,

= 12 + 3 : 36 + 4

= 15 : 40 = 3 : 8

Let milk = 4x, water = 5x

According to the question,

4x - 21 \times \frac{4}{9} : 5x - 21 \times \frac{5}{9} + 21 = \frac{8}{13}

\frac{12x - 28}{15x + 28 } = \frac{8}{13}

36x = 448

x = 12.44

Quantity of water = 5 x 12.44 = 62.2 ltr

Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid

x\left ( 1-\frac{y}{x} \right )^{n}

Hence juice in the jar = 30\left ( 1-\frac{3}{30} \right )^{3}

=> 30\left ( 1-\frac{1}{10} \right )^{3} = 21.87 litres.

Wheat worth Rs. 124 per kg and Rs. 130 per kg are mixed in the ratio 1 : 1

So their average price = (124+130)/2=127

Hence let’s consider that the mixture is formed by mixing two varieties of wheat, one at Rs. 127 per kg and the other at Rs. x per kg in the ratio of 1:1

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

Rs. 127                                                                       Rs. x

                                          Mean Price Rs. 150

(x-150)                                                                                23

 

\frac{(x-150)}{23} = 1

x= Rs 173

Video solution:

Let required amount of second solution to be added = m L

\frac{10 \times 25 + 3m}{25 + m} = 8

250 + 3m = 200 + 8m

50 = 5m

m = 10

Given, quantity of the alloy = 210 gm.

If platinum to gold = 1:2

Quantity of platinum = \frac{1}{3}\times 210 =70 gms

Quantity of gold = \frac{2}{3}\times 210 = 140 gms

As per the  question,

\frac{70 + x}{140} = \frac{2}{1}

x = 210

Let initial quantity of liquor = x litre

After a total of 4 operations, quantity of liquor = x\left ( 1-\frac{y}{x} \right )^{n} = x\left ( 1-\frac{10}{x} \right )^{3}

Given that after a total of 3 operations, the ratio of the quantity of liquor left in jar to that of water = 27:37

Hence we can write as \frac{x\left ( 1-\frac{10}{x} \right )^{3}}{x} =\frac{27}{64}

4x + 3x = 40

7x = 40

x = 5.71

 In 4 litres mixture, the quantity of water = 20% of 4 = 0.8 litres and the quantity of tang = 80% of 4 = 3.2 litres

In 20% of the mixture i.e. 20% of 4 = 0.8 litres, the quantity of tang = 80% of 0.8 = 0.64 litres

The quantity of water = 20% of 0.8 = 0.16 litres

Vishakha adds the same quantity of water,

In the new mixture, the quantity of tang = (3.2 – 0.64) = 2.56 litres

The quantity of water = (0.8 – 0.16 + 0.8) = 1.44 litres

Total quantity of mixture = 4 litres

Let she require x litres of water then 40% of (4 + x) = 2.56

1.6 + 0.4x = 2.56

0.4x = 2.56 – 1.6 = 0.96

x = 2.4 litres

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 = \frac{2}{3} \times 60 = 40 kg

Initial quantity of fanta in the bottle= 208 × \frac{11}{16} = 143 litres

Initial quantity of water in the bottle = 208 × \frac{5}{16} = 65 litres 

So, ‘8x’ litres mixtures contains 5.5x litres fanta and 2.5x litres water

According to the question,

\frac{143 – 5.5x}{65 – 2.5x + 3x – 4} = \frac{4}{3}

429 – 16.5x = 244 + 2x

18.5x = 185 ;

 x = 10

So, the final quantity of fanta = 143 – 55 = 88 litres

Final quantity of water = 65 – 25 + 30 – 4 = 66 litres

So, the difference between the final quantities of fanta and water in the bottle = 88 – 66 = 22 litres11