A man buys milk at the rate of 5 rupees per litre and mixes it with water. By selling the mixture at Rs 4 a litre he gains 25 percent. How much water did each litre of the mixture contain?
Answer –a) 9/25 Explanation : By rule of allegation Ratio of water: milk = 9:16, so in one litre water will be = 9/25
A mixture containing milk and water in the ratio 3:2 and another mixture contains them in the ratio 4:5. How many litres of the later must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of milk and water?
Answer – d) 5.2/5 litre Explanation : milk = 3*3/5 = 9/5 litre and water = 3*2/5 = 6/5 litre (in first mixture) milk = 4k/9 and water = 5k/9 litres in second mixture, so 9/5 + 4k/9 = 6/5 + 5k/9, we get k = 27/5 litre
Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the ratio in which the contents of both the vessels must be mixed to get a new mixture containing milk and water in the ratio 3:2.
Answer – a) 2:1 Explanation : let the ratio be k:1 then in first mixture, milk = 7k/10 and water = 3k/10 and in second mixture, milk = 2/5 and water = 3/5 [7k/10 + 2/5]/[3k/10 3/5] = 3/2 K = 2, so ratio will be 2:1
In 80 litre mixture of milk and water, water content is 40 percent. The trader gives 20 litre of the mixture to the customer and adds 20 litres of water to the mixture. What is the final ratio of milk and water in the mixture?
Answer – b) 9:11 Explanation : milk = 48 and water = 32 litre initially then milk = 48 – 20*3/5 = 36 and water = 32 – 20*2/5 + 20 = 44 so ratio = 9:11
70 litres of a mixture of milk and water contains 20% water. How much water should be added so that the mixture has 28% water?
Answer –c) 70/9 litre Explanation : milk = 56 litre and water = 14 litre. Let x litre of water is added the, (14 + x)/(70 + x) = 28/100
Rice worth Rs. 110 per kg and Rs. 95 per kg are mixed with a third variety in the ratio 1:1:2. If the mixture is worth Rs. 115 per kg, the price of the third variety per kg will be
Answer –b) 127.5 Explanation : First two types of rice are mixed in 1:1 so total cost for 2 kg of rice is 205, so average price = 102.5 So, x – 115 = 12.5, x = 127.5
A trader has 60 kg of pulses, one part of which is sold at 8% profit and the rest is sold at 14% profit. He gains 12% on whole. What is the quantity sold at 14% profit?
Answer – c) 40kg Explanation : So ratio will be 1:2, so quantity sold at 14% profit = 2/3*60 = 40kg
Two cans of 60 and 80 litres are filled with the mixtures of milk and water. The proportion of milk and water in the cans being 5:7 and 9:7 respectively. If the contents of the two cans are mixed and 30 litres of the water is added to the whole, then find the ratio of milk and water in the final mixture?
Answer – b) 7:10 Explanation : milk = 60*5/12 = 25 and water = 60*7/12 = 35 milk = 80*9/16 = 45 and water = 80*7/16 = 35 milk = 70 and water = 70 + 30 = 100
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.
Answer – b) 401/544 Explanation : Milk = 2/5 + 3/7 + 4/9 and water = 3/5 + 4/7 + 5/9 so ratio will be 401/544
In two alloys copper and zinc are in the ratio of 1:4 and 3:1 respectively. 20 kg of first alloy and 32 kg of second alloy and some pure zinc are melted together.An alloy is obtained in which the ratio of copper and zinc was 3:5. Find the quantity of zinc melted.
Answer – c) 68/3 kg Explanation : Copper = 1/5*20 + 3/4*32 = 28kg zinc = 4/5*20 + 1/4*32 = 24kg now x kg of zinc is added, so [28/24 + x] = 3/5. X = 68/3 kg
An alloy contains iron, copper and zinc in the ratio of 3:4:2. Another alloy contains copper, zinc and tin in the ratio of 10:5:3. If equal quantities of both alloys are melted, then weight of tin per kg in the new alloy
Answer – c) 1/12 kg Explanation : I:C:Z = 3:4:2 (in first alloy) and C:Z:T = 10:5:3 Equal quantities is taken. So, I:C:Z = 6:8:4 in first alloy and C:Z:T = 10:5:3 I = 6
C = 8 + 10 = 18
Z = 4+5 = 9
T = 3
So weight of tin = 3/36 = 1/12
8 litres are drawn from a flask containing milk and then filled with water. The operation is performed 3 more times. The ratio of the quantity of milk left and total solution is 81/625. How much milk the flask initially holds?
Answer – b) 20ltr Explanation : let initial quantity be Q, and final quantity be F F = Q*(1 – 8/Q)^4
81/625 = (18/Q)^4
3/5 = 1 – 8/Q
Q = 20
A 40 litre mixture contains milk and water in the ratio of 3:2. 20 litres of the mixture is drawn of and filled with pure milk. This operation is repeated one more time. At the end what is the ratio of milk and water in the resulting mixture?
Answer – d) 9:1 Explanation : milk = 40*3/5 = 24 and water = 16 litres initially milk = 24 – 20*3/5 + 20 = 32 – 20*4/5 + 20 = 36 water = 16 – 20*2/5 = 8 – 20*1/5 = 4
Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the ratio in which the contents of both the vessels must be mixed to get a new mixture containing milk and water in the ratio 3:2.
Answer – a) 2:1 Explanation : Let the ratio be k:1 then in first mixture, milk = 7k/10 and water = 3k/10 and in second mixture, milk = 2/5 and water = 3/5 [7k/10 + 2/5]/[3k/10 3/5] = 3/2 K = 2, so ratio will be 2:1
How many Kgs of rice A costing rupees 20 per kg must be mixed with 20 kg of rice B costing rupees 32 per kg, so that after selling them at 35 rupees per kg, he gets a profit of 25%.
Answer – a) 10 kg Explanation : by rule of alligation, 20 32
…….28………..
4 8
So, x = 10
How many litres of water must be added to 60 litre mixture that contains milk and water in the 7:3 such that the resulting mixture has 50% water in it?
Answer – c) 24 Explanation : milk = (7/10)*60 = 42 and water = 18 so water must be added = 42 – 18 = 24
A sample of x litre is replaced from a container containing milk and water in the ratio 2:3 by pure milk. If the container hold 30 litres of the mixture, and after the operation proportion of milk and water is same. Find the value of X?
Answer – b) 5 Explanation : milk = 30*2/5 = 12 and water = 30*3/5 = 18 milk = 12 – x*2/5 + x and water = 18 – x*3/5 equate both equation and we get x = 5
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.
Answer – c) 7:4 Explanation : Milk in 1 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 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
Answer – b) 2ltr Explanation : (6*5000 + 7*6000)/(13 + w) = 4800 (w is the amount of water added)
How many kilograms of rice costing Rs. 9 per kg must be mixed with 27kg of rice costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per kg?
Answer – a) 63 Explanation : By rule of allegation: – 9 7
….8.4…..
1.4 0.6
So, x/27 = 1.4/0.6, we get x = 63 kg
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