Mettl Mixture and Allegation Quiz 1

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

Video solution:

Suppose the barrel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = (2 – \frac{2x}{8} + x) litre

Quantity of glucose in new mixture = (6 – \frac{6x}{8}) litres

-x + 4x + 3x = 16

=> 6x = 16

=> x= \frac{8}{3}

So, part of the mixture replaced = (\frac{8}{3} \times \frac{1}{8})=\frac{1}{3}


Video solution:

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:

(12-7):(7-5)=5:2

Suppose the barrel initially contains 5x and 3x of mixtures water and tang respectively.

Quantity of water in mixture left = (5x – \frac{5}{8} \times 4)litres

= (5x – \frac{5}{2}) litres

Quantity of tang in mixture left = (3x – \frac{3}{8}  \times 4) litres

= (3x – \frac{3}{2}) litres

\frac{(5x – \frac{5}{2})}{((3x – \frac{3}{2}) +9) = \frac{9}{11}

=> 110x -54x = 45+55

=> x = 1.78 

So, the barrel contained 5x = 8.92 litres of water.

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

40-x / x-35 = 8/7

280 – 7x = 8x -280

15 x= 37.33

Profit % = 37.50 – 37.33 / 37.33 x 100

=0.455%

Let C.P. of 1 litre liquor be Re. 1

Then, S.P. of 1 litre of mixture = Re. 1, Gain = 20%.

C.P. of 1 litre mixture = Re.(100/120 x 1) = 5/6

By the rule of alligation, we have:

C.P. of 1 litre of liquor                                 C.P. of 1 litre of water

               Rs 1                                                                Rs 0

                                              Mean Price

                                                     5/6

                 ⅚                                                                      ⅙

Ratio of milk to water = 5/6 : 1/6 = 5 : 1.

Hence, percentage of water in the solution = (1/6 x 100)% = 16.6%

Let the capacity of the pot be 'P' litres.

Quantity of milk in the mixture before adding milk = 4/9 (P - 8)

After adding milk, quantity of milk in the mixture = 6/11 P.

6P/11 - 8 = 4/9(P - 8)

10P = 792 - 352 => P = 44.

Cost price of this set of sugar =80×13.5= Rs. 1080

Amount of sugar bought at 16 per kg =120 kg

Cost price of this set of sugar =120×16= Rs. 1920

Total cost price =1920+1080= Rs. 3000

Total weight of sugar =80+120=200 kg

∴ To gain 16% profit, Selling Price = 100/116 ​ ×3000=3480

Selling price per kg = 3480/200 ​ = Rs. 17.40