What is Alligation method in maths?

What is Alligation method in maths?

Alligation is an old and practical method of solving arithmetic problems related to mixtures of ingredients. Alligation medial is merely a matter of finding a weighted mean. Alligation alternate is more complicated and involves organizing the ingredients into high and low pairs which are then traded off.

How do you solve an Alligation problem?

By the alligation Rule, milk and water are in the ratio of 5: 1. quantity of milk in the mixture = 5 ×16 = 80 litres. Ex3: 300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% of the solution?

What is the ratio of water mixed with milk to gain 16 2 3?

Mixture and Alligation #19

19. In what ratio must water be mixed with milk to gain 1623% 16 2 3 % on selling the mixture at cost price?
A. 1 : 4 B. 6 : 1
C. 1 : 6 D. 4 : 1

How do you solve mixture ratio problems?

Solution using conceptual reasoning: Milk in the beginning is seven eighth of 40 liters, that is, 35 liters and rest 5 liters was water. Milk amount is not changed and by the desired ratio it has to be 3 times the water in the new mixture. This volume of water in the new mixture is then, liters.

How many gallons of 15% sugar solution must be mixed with 5 gallons of a 40% sugar solution to make a 30% solution?

Your answer, in that case, is that gallons of a 15% solution plus 5 gallons of a 40% solution are required to make 15 gallons of a 30% solution. 2.5 + 2 = 4.5 gallons of solution from the 15% and 40% solution. .

What quantity of a 60% acid solution must be mixed with a 30% solution to produce 300 ml of a 50% solution?

SOLUTION: What quantity of a 60% acid solution must be mixed with a 30% solution to produce 300mL of a 50% solution? x=200 ml of 60% is used. 300-200=100 ml of 30% is used.

What quantity of a 70% acid solution must be mixed with a 40% solution to produce 600 mL of a 60% solution?

SOLUTION: what quantity of a 70% acid solution must be mixed with a 40% solution to produce 600mL of a 60% solution? Question 350378: what quantity of a 70% acid solution must be mixed with a 40% solution to produce 600mL of a 60% solution? X=400 ml. of 70% solution is used.

How many liters of a 20% alcohol solution must be mixed with 60 liters of a 90% solution to get a 30% solution?

x=liters of 20% alcohol solution to be mixed with 60 liters of 90% alcohol solution. ans: 80 liters of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution.

What amount of pure acid must be added to 300 mL of a 50 acid solution to produce a 60 acid solution?

SOLUTION: what quantity of pure acid must be added to 300mL of a 50% acid solution to produce a 60% acid solution? Question 1123843: what quantity of pure acid must be added to 300mL of a 50% acid solution to produce a 60% acid solution? = 75 mL.

How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?

SOLUTION: How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution that is 36% acid? Question 1068232: How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution that is 36% acid? x=grams of pure acid. x=10 gms.

How much pure acid must be added to 50ml of a 35% acid solution to produce a mixture that is 75% acid?

How much pure acid must be added to 50 mL of a 35% acid solution to produce a mixture that is 75% acid? (See Figure 2.58.) FIGURE 2.58 Mixing solutions. We need to add 80 mL of pure acid to the 35% acid solution to make a solution that is 75% acid. Now try Exercise 31.

How many liters is a 20% solution?

500 = 20x + 400 Subtract 400 from both sides. 100 = 20x Divide both sides by 20. Therefore, 5 liters of a 20% solution must be added to 10 liters of a 50% solution to make a 40% solution.

How many liters l of a 20% alcohol solution must be mixed with 60 l of a 90% solution to get a 80% solution?

Each L of the 20% solution contributes 0.2 L of alcohol and thus a total of 0.2x L of alcohol, while the 60 L of 90% solution contributes 54 L of alcohol. Solve for x: So you will need 30 L of the 20% solution.

How many liters of a 30% alcohol solution should be added to 40 liters of a 60% alcohol solution to prepare a 50% solution?

75. How many litres of a 30% alcohol solution should be added to 40 litres of a 60% alcohol solution to prepare a 50% solution? \ The ratio in which 30% alcohol and 60% alcohol should be mixed to get 50% alcohol is 1 : 2.

How many liters of a solution that is 30% alcohol must be mixed with 80 liters of a solution that is 90% alcohol to get a solution that is 60% alcohol?

That means you need 45 liters of the 30% solution.

How many liters of distilled water must be added to 80 liters of a 60 acid solution to obtain a 50% acid solution?

Answer: how many liters of distilled water must be added to 80 liters of 60% acid solution to obtain a 50 % acid solution? (80*0.60)=48 L of 100% acid. To make 50% acid, it must be diluted to have a total volume of 48/(0.5)=96 L.

How much of a 10% alcohol solution must be mixed with 20 gallons of a 15% alcohol solution to obtain a solution that is exactly 13% alcohol?

Question: How Much Of A 10% Alcohol Solution Must Be Mixed With 20 Gallons Of A 15% Alcohol Solution To Obtain A Solution That Is Exactly 13% Alcohol? 13 Gallons.

How many liters of a 70 alcohol solution must be added to 50 liters of a 40 alcohol solution?

TAGS. How many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution? Hence Sol 1 will be 25.

How much alcohol must be added to 24 gallons of a 14% solution of alcohol in order to produce a 20% solution?

2 * x = 4.8 – 3.36. simplify to get . 8 * x = 1.44. this says that you need to add 1.8 gallons of pure alcohol to get a solution that is 20% alcohol.

How much water should we add to 5 gallons of 18% acid solution to dilute it to a concentration of 10 %?

4 gallons of water is needed to dilute to a concentration of 10%. Step-by-step explanation: It is given that, The 5 gallons of water contains 18% of acid solution.

How many gallons of a 5% acid solution must be mixed with 5 gal of a 10% solution to obtain a 7% solution?

You will need to add 7.5 gallons of 5% acid solution to 5 gallons of 10% acid solution to obtain 12.5 gallons of 7% acid solution.

32 liters

How much pure acid should be mixed with 2 gallons of a 40 acid solution to get a 70 acid solution?

SOLUTION: how much pure acid should be mixed with 2 gallons of a 40% acid solution in order to get a 70% acid solution. x + 2(0.4) = (2+x)(0.7) Simplify and solve for x. x+0.8 = 1.4+0.7x Subtract 0.7x from both sides. 0.3x+0.8 = 1.4 Now subtract 0.8 from both sides.

How much pure acid should be mixed with 5 gallons of a 50 acid solution?

1 Expert Answer You need 7.5 cgallons of pure acid.

How much pure acid should be mixed with 2 gallons of a 70 acid solution to get a 90 acid solution?

of pure acid. You need add ‘x’ gallons of 100% pure acid to make a 90% solution. Answer: Add 2 gallons of pure acid to 2 gallons of 80% acid to make a 90% acid solution.

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