There are two alloys A and B. Alloys A contains metals P, R and S. Alloy B contains metals P and Q. Both alloys contains 60% of metal P. They are mixed to form a third alloy. the percentage of metal Q in the third alloy is half of that of metal P in it. The ratio of percentage of R and S in Alloy A is 3:1. Find the ratio of quantities of R in Alloy A and Q in Alloy B, which are mixed to form the third alloy.
Answer: D As, both the alloys A and B contains 60% of metal P, Alloys C will also contains 60% of metal P. Metal Q forms 30%. P---------->Alloy-A(60%)----------Alloy-B(60%)-------------Alloy-C(60%) Q---------->Alloy-A(__)------------Alloy-B(40%)-------------Alloy-C(30%) R---------->Alloy-A(30%)----------Alloy-B(___)--------------Alloy-C(___) S---------->Alloy-A(10%)----------Alloy-B(___)--------------Alloy-C(___)
Let the weights of Alloy A and alloy B be x and y. Let the percentage of R and S in alloy A is 30 % and 10% respectively. Weight of Q in alloy B must be equal to that is alloy C as alloy A does not contain Q. 40y/100 = 30(x+y)/100 => y= 3x. Ratio of quantities of R in alloy A and Q in alloy B => 30x/100 : 40y/100 => 1:4
Q. No. 32:
How many kilograms of tea worth Rs 25 per kg must be blended with 30 kg of tea worth Rs 30 per kg so that by selling the blended variety at Rs 30 per kg there should be a gain of 10% ?
Answer: A There are two varieties of tea. One worth Rs 25 per kg and other worth Rs 30 per kg. Now, by selling the blended variety at Rs 30 per kg, profit should be 10%. So, the cost price of the blended tea should be 30/1.1 = Rs 300/11 per kg. By alligation rule, 25...........................30 .........300/11............. x.............................y We get, x/y = 30/25 Amount of tea of rs 30 per kg to be used is 30kg. If the amount of tea worth Rs 25 per kg is a. 30/25 = a/30 a = 36 kg.
Q. No. 33:
On test-tube contains some acid and another test-tube contains an equal quantity of water. To prepare a solution, 20g of the acid is poured into the second test-tube. Then, two-third of the so-formed solution is poured from the second tube into the first. If the fluid in the first test-tube is four times that in the second, what quantity of water was taken initially ?
Answer: D From options :- Let the tube contains 100g of Acid and 100 gm water. 1st operation = Acid (100-20)gm and Water (100+20)gm II nd operation = Acid (80 + 120*2/3) and Water (120/3 = 40 gm) Ist is 4 times the 2nd.
Q. No. 34:
A petrol pump owner mixed leaded and unleaded petrol in such a way that the mixture contains 10% unleaded petrol. What quantity of leaded petrol should added to one litre mixture so that the percentage of unleaded petrol becomes 5% ?
Answer: C By alligation rule, 90%.............................100% ...................95%............... 5.................................5 => 5:5 = 1:1 = 1000 ml.
Q. No. 35:
There are two vessels of equal capacity, one full of milk, and second one-third full of water. The second vessel is, then filled up out of the first, the contents of the second are then poured back into the first till it is full and then again the contents of the first are poured back into the second till it is full. What is the proportion of milk in the second vessel, if capacity of the vessel is 20L ?
Answer: B Amount of liquid left after n operations, when the container originally contains x units of liquid from which y units is taken out each time is x(1- y/x)n units. Here, x= 20L, y = 20*2/3 = 40/3 and n= 3. => 20 (1- (40/3)/20)3 = 20 (1- 2/3)3 = 20/27.
Q. No. 36:
A rabbit on a controlled diet is fed daily 300g of a mixture of two food, food X and food Y. Food X contains 10% protein and Food Y contains 15% protein. If the rabbits diet provides exactly 38 g of protein daily, how many grams of Food X are in the mixture ?
Answer: D Let there are a gm of food X and (300-a)gm of Food Y. Them a*10% + (300-a)*15% = 38 => 10a/100 + (300-a)*15/100 = 38 => 10a+4500-15a = 3800 => a = 140 gm of food X.