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“One Topic, to rule them all”
Alligation is a method of solving arithmetic problems related to mixtures of ingredients. Please note that alligation method is applied for percentage value, ratio, rate, prices, speed, etc. and not for absolute value. That is whenever per cent, per km, per hour, per kg, are being compared, we can use Alligation.
Common trick for Ratio-Proportion and Mixture Alligation : Almost 50% of the questions are solvable just by going through the options. Just go through the questions I have solved in this article and you will know the approach.
Given SP = Rs. 320/kg, Profit = 20%
Hence CP = 320/1.2 = Rs. 800/3
So the Mean price is Rs. 800/3 per kg
Now you can apply the formula-
Type 1 : Type 2 = 280 – 800/3 : 800/3 – 180 = 2 : 13
Answer : (B)
Both the containers have equal capacity. Let us assume that both containers are of 28 litres. Why 28? Because 28 is the LCM of (3 + 1) and (5 + 2) or 4 and 7. So taking the capacity as 28 litres will make your calculations easier.
In Container 1, we have (3/4)*28 = 21 litres of milk and (1/4)*28 = 7 litres of water.
In Container 1, we have (5/7)*28 = 20 litres of milk and (2/7)*28 = 8 litres of water.
Total milk in both the containers = 21 + 20 = 41
Total water in both the containers = 7 + 8 = 15
Milk : Water = 41 : 15
Answer : (D)
Container 1 has 3 times more milk than water
Container 2 has 2.5 times more milk than water
When the contents of the two containers are mixed, the milk will still be more than water. How much more ? Somewhere between 2.5 and 3 times
(D) is the only option where the quantity of milk is around 2.7 times (i.e. between 2.5 and 3) that of the water.
Milk in vessel A = 4/7
Milk in vessel B = 2/5
Milk in vessel C = 1/2 (because in vessel C, milk and water are present in 1:1 ratio)
You have to mix 4/7 and 2/5, to produce 1/2. Hence 1/2 is the Mean Price.
A : B = (1/2 – 2/5)/(4/7 – 1/2) = 14 : 10 = 7 : 5
Final ratio of the three varieties is 5 : 7 : 9
The question asks us the quantity of third variety of tea in the final mixture. From the above ratio, it is clear that the quantity of the third variety is a multiple of 9. So 45 is the only option possible.
Answer : (D)
Let the three quantities be 4x, 5x and 8x
New quantities are 4x + 5, 5x + 10 and 8x + p
Now 4x + 5 : 5x + 10 : 8x + p = 5 : 7 : 9
(4x + 5)/(5x + 10) = 5/7 and (4x + 5)/(8x + p) = 5/9
Solving 1st equation, we get x = 5
Solving 2nd equation, we get p = 5
In the final mixture the quantity of the third variety is 8x + p = 8*5 + 5 = 45
In this question we will use the below formula
So from the above formula
(Quantity of acid left)/(Quantity of acid in the original mixture) = (1 – 4/20)^2 = 16:25
Answer : (A)
Let the original quantities of A and B be 4x and x
In 10 litres, quantity of A = 4/5 * 10 = 8 litres
In 10 litres, quantity of B = (10 – 8) = 2 litres
New quantities of A and B are 2x and 3x
(Original Quantity of A) – (New quantity of A) = 8 litres[Because after taking out 10 litres of the mixture, the quantity of liquid A reduced by 8 litres] So, 4x – 2x = 8
or x = 4
Hence quantity of liquid A in original mixture = 4*4 = 16 litres
Answer : (C)
Note : In the above question, there were two different ratios 4:1 and 2:3, then too I took the same constant of proportionality for them, i.e. ‘x’ because the following two conditions were met:
1. The volume of mixture did not change (Like in this question 10 litres were replaced, not removed)
2. The two ratios had same no. of parts (4:1 and 2:3 both have 5 parts)
You can take different constant to solve the question, but that will make the calculations little lengthy.
Since the ratio of alcohol and water is 1:4, hence quantities of alcohol and water in the mixture are 3 litres and 12 litres respectively.
Total volume will become 18 litres after adding 3 litres water
% of alcohol = 3/18 * 100 = 50/3%
Answer : (B)
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