Mixtures & Alligations CAT Previous Year Questions with Answer PDF

Every time 20% of the mixture is replace with a pure adulterant (here cocoa), the concentration of the mixture becomes 80% of the initial concentration.

Similarly when a proportion p of a mixture is replaced with pure adulterant, the concentration of the mixture becomes (1 - p) times the previous concentration. (0 ≤ p ≤ 1)

Initially there was pure coffee, so the strength of the mixture is 100% or 1 (in terms of proportion)
 

Let a proportion p of the mixture be replaced each time… Since after repeating the process twice, we have the concentration of coffee as   at the end of the two replacements.

Or, 20% of the mixture is replaced each time.

The ratio of cocoa powder in the mixture P and Q will then be 
Therefore, the ratio is 5 : 9

Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.
In sugar syrup, 100% is sugar syrup.
These two are mixed in the ratio 1:3.

Required ratio = 50 : 350 = 1 : 7

We know that the process of shifting half volumes is happening three times overall.

So it would be wise to assume the initial volume to be some multiple of 8 so that we don’t have to deal with fractions later on.

The ratio of Sugar Syrup to Water in the second container is 5 : 6.

Required weight of C = 
= 84 kg

In Liquid L1 - 1L = 1000 grams
So, 1000 mL = 1000 grams
1 mL = 1 gram
x mL = x grams
In Liquid L2 - 1L = 800 grams
1000 mL = 800 grams
1mL = 0.8 grams
(500-x) mL = (500-x) x 0.8 grams
Total mass = x + 400 - 0.8x = 480 grams
0.2x = 80 grams
x = 400 grams
Therefore, Liquid 1 has 400 mL and Liquid 2 has 500 - 400 = 100 Ml
Therefore, Percentage of Liquid 1 = (400/500) x 100 = 80%

Vessels A, B and C contains salt solution of strengths 10%, 22% and 32% respectively
It is also given that the amount of salt solution = 500 ml
So, Vessels A, B and C contains salt of 50 grams, 110 grams and 160 grams respectively
100 ml of Solution is transferred from A to B:
A would have 400 ml, B would have 600 ml of solution
Amount salt from A which is transferred to B = (Initial salt amount/5) = 10 grams
So, Total salt in B = 110 + 10 = 120 grams (After first transfer)
Total salt in A = 40 grams (After first transfer)
Now, 100 ml from Vessel B is transferred to Vessel C
So similarly, (1/6)th of salt would transfer from B to C
Total Salt in B = 120 - 20 = 100 grams (After second transfer)
Total Salt in C = 160 + 20 = 180 grams (After second transfer)
Now, 100 ml from Vessel C is transferred to Vessel A
So similarly, (1/6)thof salt would transfer from C to A
Total Salt in C = 160 - 30 = 130 grams (After third transfer)
Total Salt in A = 40 + 30 = 70 grams (After third transfer)
So, Vessel A contains 70 grams Salt in 500 ml solution
Strength of Salt Solution in Vessel A = 14%

Given, The amount of B in mixture ≤ Amount of A in mixture
Selling Price = Rs. 264 and Profit = 10 %
264 = 1.1 × CP
CP = (264/1.1) = Rs. 240 for 10 litres
CP per litre = Rs. 24
We also know that, CP per litre of A = CP per litre of B + 8 Let CP per litre of B = x So, CP per litre of A = x + 8

Given Amount of B ≤ Amount of A,
Maximum possible cost of B occurs when B = A
We need to choose values for A and B in such a way that the quantities remain same.But A should be 8 more than B
Therefore, A should have a CP per litre of Rs. 28 and B should have a CP per litre of Rs 20

B cannot be assigned any more than this as the amount of B would become more than A
Example: Let us assign B to be Rs. 21 and A to be Rs. 29 which would result in a ratio of 3:5 where B amounts more than A, which doesn’t satisfy the condition
Max possible cost of Paint B = Rs. 20 per litre

Let the wholesaler bought walnuts and peanuts at Rs 3x and Rs x respectively
He sold 8kgs of Peanuts to the shopkeeper at 10% Profit
Cost price of the Peanuts bought by the shopkeeper = Rs. 1.1x per kg
Similarly, he sold 16 kgs of Walnuts to the shopkeeper at 20% Profit
Cost price of the Walnuts bought by the shopkeeper = 1.2 × 3x = Rs. 3.6x per Kg
He lost 5 kgs of Walnuts and 3 Kgs of Peanuts in transit
Remaining = 16 - 5 = 11 kgs of Walnuts & 8-3 = 5 kgs of Peanuts
He mixes them together and sells them at Rs. 166 per kg, making an overall Profit of 25%
Selling Price = 5/4 × Cost Price
Overall Cost Price = Rs.(3.6 × 16x + 1.1 × 8x)
Overall Selling Price = Rs. 16 × 166
Overall Selling Price = (5/4) x Overall Cost Price
(3.6 × 16x + 1.1 × 8x) × (5/4) = 16 × 166
( ( 57.6 × 8.8) x ) × 5/4 = 16 × 166
x = Rs. 32
Cost price per Kg of Walnuts bought by the Wholesaler = Rs. 3x = 3 × 32 = >Rs. 96

Given that two drums each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. Drums 1 and 2 are mixed in the ratio 3 : 4 and in the final mixture A and B are in the ratio 13 : 7. For these kinds of questions do not consider separately as A and B. Deal with either A or B as a share of overall
Here A is (18/25)  of overall
In D2, let us assume the proportion of A with respect to the overall mixture is x. 3 parts of D1 is mixed with parts of D2 to give proportion of A of which is (13/20)th of the overall mixture. From here on, it is weighted averages.

In drum 2, A is (239/400) so B should be the remaining. Therefore, A : B = 239 : 161.

Given that 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume

This M is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution ,then the unknown concentration of S has to be found
 
Equal quantities of mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%
The mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%. This 31.25% is 11.25% more than this 20%. This mixture must be 11.25% more than 31.25% so mixture is equal to 42.5%
 
This mixture 42.5% is mixed in the ratio 1:3
 
Using allegations we can find this 22.5% which should be in the ratio 1:3 so the other one is 7.5%
S = 42.5 + 7.5
S = 50%
Here to find the solution, do the second thing first and then go to the first thing
The second one is mixed in the ratio 1 : 1 so that the final thing should be bang in the middle

Given that a jar contains a mixture of 175 ml water and 700 ml alcohol.
It is given that 10% of the mixture is removed and it is substituted by water of the same amount and the process is repeated once again
Now we have to find the percentage of water in the mixture.
Since the mixture is removed and substituted with water, we can deal with alcohol and the second step we can find how much amount of alcohol is retained and not about how much amount of alcohol is removed
As 10% of alcohol is removed, 90% of alcohol is retained
So alcohol remaining = 700 × 90% × 90%
⇒  700 × 0.9 × 0.9 = 567
We totally have 875 ml overall mixture and of this 567 ml is alcohol.
Remaining 875 – 567 = 308 is the amount of water.

We have to find the percentage of water in the mixture i.e. 308/875
Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30%
Hence 35.2% is the percentage of water in the given mixture.

Given that the strength of the salt solution is p% if 100 ml of the solution contains p grams of salt
It is also given that three salt solutions A , B , C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.
So  = 0.2 ⇒ A + 2B + 3C = 1.2 ...(1)
If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%
So,  = 0.3 ⇒ 3A +2B + C = 1.8 ...(2)
It is given that 4^th solution D is produced by mixing B and C in the ratio 2 : 7
So D = 
We have to find the ratio of the strength of D : A
Subtracting equations 1 and 2, we get
2A – 2C = 0.6 or A – C = 0.3
Since we could not find anything from the above methods, we can eliminate the number part and get the ratio going
A + 2B + 3C = 1.2
3A + 2B + C = 1.8
So let us multiply eqn 1 and 2 with 3 and 2,
3A + 6B + 9C = 6A + 4B + 2C
2B + 7C = 3A
It is given that D = 

Hence D = A/3
Therefore the ratio D : A = 1 : 3.

Bottle 1 and bottle 2 contains mixture of milk and water in the ratio 7 : 2 and 9 : 4 respectively.
We have to find by what ratio bottle 1 and Bottle 2 should be combined to obtain a mixture of milk and water in the ratio 3 : 1
For this we can consider milk to the whole mixture or else water to the whole mixture.
⇒ 
Taking LCM we get 

LCM is 52 so 

Bottle 1 and 2 are mixed in the ratio 27 : 13.

There are three mixtures ,the first having water and liquid A in the ratio 1 : 2,
so we can consider only the liquid part which is 2/3.
The second having the water and liquid in the ratio 1 : 3 = 3/4.
The third having the liquid and water in the ratio 1 : 4 = 4/5


Hence by this we can infer that more water than liquid B is true