XAT Mixtures and Solutions Questions

Let's five small drums have a capacity of 1 unit capacity each and one bigger drum has that of 2 units capacity. We need to consider two cases here. One with minimum volume and the other with maximum volume.

Case 1: Minimum value is possible if the bigger drum is half filled. So, total volume of water = 1 + 2 * (1/2) + 2 * (2/3) + 1 = 26/6 ~ 4.3

Case 2: Maximum value is possible if the bigger drum is completely full. So, total volume of water = 2 + 3 * (1/2) + 2 * (2/3) = 29/6 = 4.833

In any case volume of water is more than 4 units and less than 5 units. Hence, exactly 5 smaller drums are adequate to store the water.

Liquids A and B are in the ratio 5:3. The volume of water is one-third the total mixture. 
Let us assume the volume of the total mixture to be 24x.

Volume of liquid A = 10x
Volume of liquid B = 6x
Volume of water = 8x

The mixture is passed through some medium that absorbs water completely and some quantity of liquid A. 
Water absorbed = 8x
Let the amount of liquid A absorbed be y.

8x+y = 200
=> y = 200-8x -----(1)
It has been given that (10x-y)/6x  = 7/9

Substituting (1), we get,
(10x-200+8x)/6x = 7/9
(18x-200)/6x = 7/9
162x-1800=42x
120x = 1800
=> x = 1800/120
Amount of water absorbed = 8*1800/120 = 120 ml.
Therefore, option E is the right answer. 

Let the quantities of the chemicals X and Y, mixed to produce product M be $$5c$$ and $$4c$$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $$\frac{3}{4} \times 5c = \frac{15 c}{4}$$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $$\frac{2}{3} \times 4c = \frac{8 c}{3}$$

Quantity of B in M = $$\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $$9c = 864$$ => $$c = \frac{864}{9} = 96$$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $$\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$$

$$\therefore$$ Quantity of water added to M = $$1232 - 864 = 368$$ units

Let the capacities of bucket of water carried by Tina, Mina, Gina, Lina and Bina respectively be $$W_T, W_M, W_G, W_L, W_B$$

It is given that : $$W_T > W_M > W_G > W_L > W_B$$

Let they spill $$x$$ litres of water from the bucket.

Thus, %age of water spilled by them respectively be

= $$\frac{x}{W_T} \times 100 $$, $$\frac{x}{W_M} \times 100$$, $$\frac{x}{W_G} \times 100$$, $$\frac{x}{W_L} \times 100$$, $$\frac{x}{W_B} \times 100$$

$$\because W_T > W_M > W_G > W_L > W_B$$

$$\therefore \frac{x}{W_T} \times 100$$ < $$ \frac{x}{W_M} \times 100$$ < $$\frac{x}{W_G} \times 100$$ < $$\frac{x}{W_L} \times 100$$ < $$\frac{x}{W_B} \times 100$$

Thus, Bina lost maximum amount of water as a percentage of the bucket capacity.

Let volume of jug = $$v$$ litre

After first replacement, volume of orange juice = $$(10 - v)$$ litre

Volume of pineapple juice = $$v$$ litre

After second replacement, volume of orange juice remaining 

= $$(10 - v) - (\frac{10 - v}{10} v) = \frac{(10 - v)^2}{10}$$

Volume of pineapple juice remaining = $$v - \frac{v^2}{10} = \frac{v (10 - v)}{10}$$

Total volume of pineapple juice = $$\frac{v (10 - v)}{10} + v = \frac{20v - v^2}{10}$$

It is given that container has equal volumes of both juices.

=> $$\frac{(10 - v)^2}{10} = \frac{20v - v^2}{10}$$

$$100 + v^2 - 20v = 20v -v^2$$

$$2v^2 -40v +100=0$$

=> $$v^2 - 20v + 50 = 0$$

=> $$v = 17.07 , 2.93$$

$$\because$$ Container is 10 litres, => $$v \neq 17.07$$

$$\therefore v = 2.93$$ litres

50 ml of potency 1 solution is equivalent to 1 tablet, 50 ml of potency 2 solution i equivalent to 2 tablets and so on.

Hence, 10 ml of potency 1 solution is equivalent to

= $$\frac{10}{50} = \frac{1}{5}$$

Similarly, 15 ml of potency 2 solution corresponds to = $$\frac{15}{50} \times 2 = \frac{3}{5}$$

and 30 ml of potency 4 solution corresponds to = $$\frac{30}{50} \times 4 = \frac{12}{5}$$

$$\therefore$$ Required dosage 

= $$\frac{1}{5} + \frac{3}{5} + \frac{12}{5}$$

= $$\frac{16}{5} = 3.2$$ tablets

Let Ram replaces $$x$$ litres of 12 % sol. with 39 % solution

Now, quality of 12 % solution in 27 litre = $$\frac{12}{100} \times 27$$

=> After replacing we have volume of 12 % solution

= $$(\frac{12}{100} \times 27) - (\frac{12 x}{100}) + (\frac{39 x}{100})$$

= $$\frac{324 + 27 x}{100}$$

This is equal to 27 litre of 21 % solution.

=> $$\frac{324 + 27 x}{100} = \frac{21}{100} \times 27$$

=> $$27x = 567 - 324 = 243$$

=> $$x = \frac{243}{27} = 9$$

Alternate Solution:

The mixture of two solutions 12% alcohol and 39% alcohol should yield 21% alcohol solution. 

Using allegation:  

The ratio = 18/9 =2:1

So the amount of 39% solution required = 27*1/(2+1) =9

Let the volume of the solution with 30 % acid content lie between $$v_1$$ and $$v_2$$, where we get a 20% acid solution for $$v_1$$

For $$v_2$$, we get a 25 % acid solution as the resultant mixture.

=> $$15 \% (200) + 30 \% (v_1) = 20 \% (200 + v_1)$$

=> $$30 + 0.3 v_1 = 40 + 0.2 v_1$$

=> $$0.1 v_1 = 10$$ => $$v_1 = 10 \times 10 = 100$$ litres

Similarly, $$15 \% (200) + 30 \% (v_2) = 25 \% (200 + v_2)$$

=> $$30 + 0.3 v_2 = 50 + 0.25 v_2$$

=> $$0.05 v_2 = 20$$ => $$v_2 = 20 \times 20 = 400$$ litres

$$\therefore$$ For the acid content in the resultant mixture to lie between 20 % and 25 %, the volume of the 30 % concentration acid solution must lie between 100 litres and 400 litres.