Mixtures and Alligation

Initial mixture: 70 L total, 10% water = 7 L water, 63 L wine. Add x L water. New mixture: (7+x) L water, 63 L wine, total (70+x) L. We want water to be 37%: (7+x)/(70+x) = 0.37. Cross-multiplying: 7+x = 0.37(70+x) = 25.9 + 0.37x. So 7+x = 25.9 + 0.37x, giving 0.63x = 18.9, so x = 30. Verification: 37 L water / 100 L total = 37%.

Let there be x ml CO in the mixture. Hence, there will be (1000 – x) ml CO₂ . The reaction of CO₂ with red hot charcoal may be given as,CO₂ (g) + C(s) → 2CO(g)from reaction we have, x = 400 ml Therefore volume of CO = 400 ml and hence the volume of CO₂ = (1000 – 400 ) ml = 600 ml Hence, this is the correct option.

The equiv. of H⁺ in 30 ml of (N / 2) HCl = (10 / 2) 10^–3 The equiv. of H⁺ in 30 ml of (N / 10) HNO3 = (30 / 10) * 10^–3 The equiv. of H⁺ in 75 ml of (N / 10) HNO3 = (75 / 5) * 10^–3. Hence, total equiv. of H⁺ (5 + 3 + 15) * 10^–3 = 23 10^–3 Total volume of solution = 115 ml Hence, normality of H⁺ in the resulting mixture = [(23 * 10^–3 * 10^3) / 115) (N) = (N / 5) = 0.2 (N). Hence, this is a correct option.

Molecular mass of the mixture = 2 * 38.3 = 76.6 Let, mass of NO₂ in 100 g of mixture = a g. Moles of NO₂ + Moles of N₂O₄ = Moles of mixture i.e., (a / 46) + ((100 – a ) / 92) = (100 / 76.6) a = 30.1 g. Moles of NO₂ in the mixture =( 20.1 / 46) = 0.437 (approx.). Hence, this is a correct option.

Quantity of lemon juice in 100 ml = 40mlQuantity of lemon juice in 1 ml = 40/100Quantity of lemon juice in 500 ml = 40/100 x 500 = 200 mlwhen we want to make it 60 % we have to add more lemon juice .Suppose x ml of lemoon juice is added now 

Quantity of lemonade will be =( 500 + x) ml  Quantity of lemon juice will be = ( 200 + x ) ml Now quantity of lemon juice in  ( 500 + x ) ml is ( 200 + x )  Quantity of lemon juice in 1 ml is ( 500 + x ) / ( 200 + x )  Quantity of lemon juice in 100 ml is=                                                  { ( 500 + x ) / ( 200 + x ) } x 100}%The % of lemon juice is 60%(200 + x) / ( 500 + x ) x 100 = 60(200 + x )100 = 60 x ( 500 + x )   20000 + 100 x = 30000 + 60 x   100 x - 60 x = 30000 - 20000  40 x = 10000

     x = 10000 /40

    x = 250 ml

Let the quantity of milk purchased  = 100 ml  so the cost of milk at the rate of Rs. 4 per iltre is = 4 x 100 =                                                                                                    Rs400                     Quantity of water added 20%                         it means 20 ml of water in 100 ml                      Now quantity of milk after adding water = 100 + 20                                                                                         = 120 ml      Selling price of  milk at the rate of Rs 5 is 5 x 120= Rs.600

            Selling price > cost price            so gain      Gain = Selling Price - Cost Price               = Rs. 600 - Rs. 400                =Rs. 200       Gain %   =   Gain /Cost Price x 100                        = 200 / 400 x 100                              =      50 %

Let quantity of pure aftershave lotion= 6kg and quantity of pure water=4kg Therefore, total quantity of the mixture=10kg Again let x kg of mixture is taken out of 10 kg of mixture. Now, the amount of aftershave lotion left=(6-(6x/10)) kg And the amount of water left=(4-(4x/10)) kg Thus, the amount of water after adding x kg of water becomes (4-(4x/10)+x)kg=(4+(6x/10)) kg According to the question, (6-(6x/10))=(4+(6x/10)) 12x/10=2 x=10/6 Therefore, 1/6 of the10 kg mixture is taken out. It is the correct answer.

Let the quantity of the wine in the cask originally be x litres. Then, the quantity of wine left in the cask after 4 operations = [x(1-8/x)4]litres So, [x(1-8/x)4] /x = 16/81 on solving, x = 24 Therefore, 24 litres is the correct answer.

Let initially, there be 4V litres of A and V litres of B in the solution. Now, when 10 litres of solution is taken out, we have (4V - (4/5) x 10) lkitres or 4V - 8 litres of A left in solution. In the 10 litres of solution taken out, 2 litres of B will go out. Thus, B left in solution = (V - 2) litres. After 10 litres of B is added, B in the solution = (V - 2 + 10) litres = (V + 8) litres. Now, (quantity of liquid A in solution) / (quantity of liquid B in solution) = (4 V - 8) / (V + 8) = 2 /3. On solving, we get V = 4. Thus, initial quantity of A in the solution = 4 V = 4 x 4 litres = 16 litres.

Quantity of milk in 189 litres of mixture = (7 x 189) / 9 = 147 litres Quantity of water = (189-147) litres = 42 litres Let x litres of water be mixed to get the required ratio. 3 : 2 = 147 / (42 + x) 3 / 2 = 3x + 126 294 = 3x 3x = 294 – 126 = 168 x = 168 / 3 = 56 litres

The pot contains 50 litres of water. As 5 litres of water is taken out and mixed with wine, the ratio of water : wine becomes 5 : 45, i.e. 1 : 9. Again, the same process is repeated. But, this time in 5 litres of solution, wine and water both are replaced in a ratio of 1 : 9. So, the present wine in the pot = (5 - 0.5 + 5) as 5 litre of wine is mixed in the pot again, i.e. wine becomes 9.5 litres. At the same time, the water remains as (45 - 4.5), i.e. 40.5. So, the ratio of wine : water = 9.5 : 40.5 => 19 : 81. So, option 2 is correct.

Assume X kg from box A and Y kg from box B are to be taken. So, from box A: Rice = 5X/8 and Dal = 3X/8 and from box B : Rice = 7Y/10 and Dal = 3Y/10 So, (5X/8 + 7Y/10) = 6 => 25X + 28Y = 240 (3X/8 + 3Y/10)= 3 => 15X + 12Y = 120 Solving both the equations: X = 4; Y = 5 So, 4 kg from box A and 5 kg from box B are to be taken. So, option 1 is correct.

It is 110 litres. Quantity of sugar in the mixture is 2/10 x 220 = 44 litres. Quantity of salt in the mixture is 3/10 x 220 = 66 litres. Hence, the total quantity of sugar and salt is 44 + 66 = 110 litres. This answer is correct.

(Alcohol is 20% of 5 litres, i. e. one litre. When 1 litre water is added, mixture becomes 6 litres, in which alcohol is 1 litre. Therefore, strength of alcohol = 1/6 X 100% = 50/3%)

(Since the total part of milk & water in each is same, we add the part of milk & water separately. Required ratio is = (2 + 4 + 3) : (3 + 1 + 2) = 9 : 6 = 3:2)