Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are given some Mixture and Alligations problems that is important for competitive exams. We have included Some questions that are repeatedly asked in exams !!
Questions Penned by Yogit
- In two alloys copper and zinc are in the ratio of 1:3 and 4:1 respectively. 20 kg of first alloy and 35 kg of second alloy and some quantity of pure zinc is melted together. The final alloy has copper and zinc in the ratio of 5:4. Find the amount of pure zinc melted.
e) None of theseAnswer – b) 4.4
In 1st alloy copper = (1/4)*20 = 5kg and zinc = (3/4)*20 = 15kg
in 2nd alloy copper = (4/5)*35 = 28kg and zinc = (1/5)*35 = 7kg
So, 33/(22+x) = 5/4 (X is the amount of pure zinc added)
- In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs must be mixed so that the mixture can be sold at 1.65rs per kg.
e) None of theseAnswer – c) 11:20:44
By the rule of allegation,
Final ratio = 11:20:44
- A container filled with liquid containing 4 parts of water and 6 parts of milk. How much of mixture must be drawn off and filled with water so that the mixture contains half milk and half water.
e) None of theseAnswer – c) 1/6
Let water = 40ltr and milk is 60ltr.
Water = 40 – x*(2/5) + x and milk = 60 – x*(3/5) [x is the amount of mixture taken out]
Equate both the equation, we get x = 50/3.
Now, mixture drawn off = (50/3)/100 = 1/6
- A trader has 1500 kg of wheat. One part of it is sold at 10 percent profit and other part at 18 percent profit. He gains a total of 16 percent on the whole lot. The quantity sold at 10% is-
e) None of theseAnswer – b) 375
Ratio => 1:3. So quantity sold at 10% = (1/4)*1500 = 375
- Two cans P and Q contains milk and water in the ratio of 3:2 and 7:3 respectively. The ratio in which these two cans be mixed so as to get a new mixture containing milk and water in the ratio 7:4.
e) None of theseAnswer – c) 7:4
Milk in 1st can = 3/5 and water = 2/5. Similarly in second can milk = 7/10 and water = 3/10.
Take the ratio = K:1
(3k/5 + 7/10)/(2k/5 + 3/10) = 7/4
Solve for k, we get k = 7/4. So the ratio is 7:4
- A dishonest seller professes to sell his milk at cost price but he mixes water with milk and gains 25 percent, then find the percentage of milk in the mixture.
e) None of theseAnswer – c) 80%
Suppose initially there is 100ltr of milk costing 100 rupees.
Now he gains 25% means in 100ltr of milk he add 25ltr water, so percentage of milk in the mixture = (100/125)*100 = 80%
- Fresh fruit contains 75 percent water and dry fruit contains 20 percent water. How much dry fruit can be obtained from 150 kg of fresh fruit.
e) None of theseAnswer – c) 47
Dry fruit obtained from 150kg of fresh fruit = (25/100)*150 = (80/100)*x.
Solve for x
- How much water must be added to 50 litre of milk costing 10 rupees per litre so as to bring the cost of milk to 8 rupees per litre.
e) None of theseAnswer – b) 12.5
By using the allegation rule
Water: milk = 1:4 = x:50
- A trader mixes 6ltr of milk costing 5000 rupees with 7ltr of milk costing 6000 rupees per litre. The trader also mixes some quantity of water to the mixture so as to bring the price to 4800 per litre. How many litres of water is added
e) None of theseAnswer – b) 2ltr
(6*5000 + 7*6000)/(13 + w) = 4800 (w is the amount of water added)
- There are three vessels each of 20 litre capacity is filled with the mixture of milk and water. The ratio of milk and water are 2:3, 3:4 and 4:5 respectively. All the vessels are emptied into fourth vessel, then find the ratio of milk and water in the final mixture.
e) None of theseAnswer – b) 401/544
Milk = 2/5 + 3/7 + 4/9 and water = 3/5 + 4/7 + 5/9
so ratio will be 401/544