Directions: Refer the following data to answer the questions given below.
Solution:
The best way to solve this problem is to go from the answer choices.
The mixture of 60 litres has in it 24 litres of milk and 36 litres of water. (2 : 3 :: milk : water)
When you remove x litres from it, you will remove 0.4 x litres of milk and 0.6 x litres of water from it.
Take choice (2). According to this choice, x = 10.
So, when one removes, 10 litres of the mixture, one is removing 4 litres of milk and 6 litres of water.
Therefore, there will be 20 litres of milk and 30 litres of water in the container.
Now, when you add 10 litres of milk, you will have 30 litres of milk and 30 litres of water – i.e. milk and water are in equal proportion.
Directions: Refer the following data to answer the questions given below.
Directions: Refer the following data to answer the questions given below.
Solution:
Let the number of horses = x
Then the number of pigeons = 80 – x.
Each pigeon has 2 legs and each horse has 4 legs.
Therefore, total number of legs =4x+2(80−x)=260
⇒4x+160–2x=260
⇒2x=100
⇒x= 50.
Directions: Refer the following data to answer the questions given below.
Solution:
The mixture contains 40% milk and 60% water in it. That is 4.8 litres of milk and 7.2 litres of water.
Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.That is we will end up with 6 litres of milk and 6 litres of water.
Water gets reduced by 1.2 litres.
To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove
1.2/0.6 litres of the mixture = 2litres.
Directions: Refer the following data to answer the questions given below.
Solution:
If the selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%, then the cost price of the mixture = 30/1.2 = Rs.25/kg.
We need to find out the ratio in which the three varieties are mixed to obtain a mixture costing Rs.25 /kg.
Let variety A cost Rs.20/kg, variety B cost Rs.24 / kg and variety C cost Rs.30/kg. The mean desired price falls between B and C.
Step 1:
Find out the ratio Qa:Qc using alligation rule. Qa/Qc=(30−25)/(25−20)=1/1
Step 2:
Find out the ratio Qb:Qc using alligation rule. Qb/Qc=(30−25/25−24)=5/1
Step 3:
Qc is found by adding the value of Qc in step 1 and step 2 = 1 + 1 = 2
Therefore, the required ratio = 1 : 5 : 2
If there are 2 kgs of the third variety in the mixture, then there will be 5 kgs of the second variety in the mixture.
Note: When appearing in actual exam problems of this kind are to be skipped, at least in the first go. If you were able to solve decent number of other problems, then you should look at this problem.