3L water is taken out from a vessel full of water and substituted by pure milk. This process is repeated two more times. Finally the ratio of milk and water in the solution becomes 1728:27. Find the volume of the original solution.
Answer: C Final quantity of water = y(1 - y/x)n => x(1 - 3/x)3 => F/x = {(x-3)/x}3 = 27/1728 => (x-3)/x = 3/12 => x= 4L.
Q. No. 2:
Several litres of acid were drawn off from a 54 L vessel full of acid and an equal amount of water is added. Again the same volume of mixture was drawn off and replaced by water. As a result, the vessel contained 24L of pure acid. How much acid was drawn off initially?
A lump of two metals weighing 18gm is worth Rs 87 but if their weights be interchanged, it would be worth Rs 78.60. If the price of one metal be Rs 6.70 per gm, find the weight of the other metal in the mixture.
Answer: A Cost of (18gm of 1st metal + 18gm of 2nd metal) = Rs 165.60 Cost of (1gm of 1st metal + 1gm of 2nd metal) = Rs 9.20 Cost of 1 gm of 2nd metal = (9.70 - 6.70) = Rs 2.50 Mean price = Rs (87/18) = Rs (29/6)per gm. CP of 1gm of first metal = Rs 6.70 and CP f 1gm of second metal = Rs 2.50 Thus, By alligation = Quantity of 1st metal / quantity of 2nd metal = (14/6) / (56/30) => 5:4 In 9gm mix 2nd metal = 4gm In 18 gm mix 2nd metal =8gm.
Q. No. 4:
One type of liquid contains 25% of milk, the other contains 30% of milk. A can filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of the milk in the new mixture.
Answer: D Thus by Alligation rule , (30-x) / (x-25) = 6/4 => x= 27.
Q. No. 5:
In two alloys, copper and zinc are related in the ratios of 4:1 and 1:3. 10kg of 1st alloy, 16 kg of 2nd alloy and some of pure copper are melted together. An alloy was obtained in which the ratio of copper to zinc was 3:2. Find the weight of the new alloy.
Answer: B Let the quantity of pure copper be xkg. Then, x + (4/5*10) + (1/4*16) = 3/5* (10+16+x) => 12+x = 3/5* (26+x) => x= 9kg. Weight of the new alloy = 9+10+16 = 35 kg
Q. No. 6:
Fresh fruits contains 72% water and dry fruit contains 20% water. How much dry fruit from 100kg of fresh fruit can be obtained?
Answer: A Let x kg of dry fruit be obtained from 100kg of fresh fruit. Now pulp in fresh fruit = pulp in dry fruit => 28/100 *100 = 80/100 * x => x = 35 kg