Mixtures and Alligations Questions for CAT with Answers PDF

Ratio given = 1 : 2 : 3
Sum of the parts of Ratios = 1 + 2 + 3 = 6
Required sweetness = 

Applying the rule of alligation:

Hence, number of boys who appeared in the examination = 750 × 3/5 = 450

If water weighs x kg, then:
Weight of gold = 19x
Weight of copper = 9x
Weight of mixture desired = 15x
If we mix 'a' kg of the first mixture and 'b' kg of the second mixture,
Then

⇒ 19a + 9b = 15a + 15b

Start making the mixture with 1 litre of A and 2 litres of B.
In 1 litre of A, the quantities of ingredients are as follows.

In 2 litres of B, the quantities of ingredients are as follows.

Therefore, quantity of ingredient P in 3 litres of mixture will be

The quantity of ingredient P in 540 litres of mixture = 
This is our required solution.

By alligation method:


 

x - 15 = 4
x = 19

The given problem is summarised as follows:

That is, 6 kg of solution contains (a/100) x (3 + 2r + r²) kg of acid.
% of acid in this case = ((a/100) x (3 + 2r + r²) x 100)/6 = a/6(3 + 2r + r²)
It is however given that
(a/6) x (3 + 2r + r²) = 22
a x (3 + 2r + r2) = 22 x 6   ... (1)
Now, let us consider the second case where the ratio is 2 : 3 : 4.

9 kg of mixture contains (a/100) x (2 + 3r + 4r²) kg of acid.
% of acid in this case = ((a/100) x (2 + 3r + 4r²)  100)/9 = (a/9)  (2 + 3r + 4r2)
This is equal to 32.
Or (a/9) x (2 + 3r + 4r2) = 32
a x (2 + 3r + 4r2) = 32  9                        … (2)
Dividing (2) by (1), we get
(a(2 + 3r + 4r2))/(a(3 + 2r + r2))  = (9 x 32)/(6 x 22) = 24/11
4r2 – 3r – 10 = 0
i.e. (r – 2) (4r + 5) = 0
r = 2 or – 5/4
Since –5/4 is inadmissible, r = 2.
Substituting this value in (1) or (2), we get a = 12.
Hence, the percentages are 12%, 24% and 48%.

Mixture = HCl : Water
Quantity taken out
First 40% = (2 : 3)x kg
Second 60% = (3 : 2)y kg
Let us suppose, we add z kg of water.
When we mixed, we get 20% strong solution.

⇒ 5x + 10y - 5z = 0
x + 2y - z = 0 .......(1)
Again, on adding z kg of 80% solution, we get a 70% solution.

⇒ 15x + 5y - 5z = 0
3x + y - z = 0 ……..(2)
Since variables are more and equations are less, required solution cannot be determined.

Assume that 100x, 100y and 100z respectively of the three mixtures are mixed.

So, 40% of 100(x + y + z) = 90y + 60z
40x + 40y + 40z = 90y + 60z
or 40x – 50y – 20z = 0
4x – 5y – 2z = 0 … (i)
Let us assume that the new mixture contains k% pepper.
To find k: If k% of 100(x + y + z) = 70x + 10y + 25z
i.e. x(70 – k) + y(10 – k) + z(25 – k) = 0 … (ii)
Using (i) and (ii), we can find the ratio x : y : z by the rule of cross multiplication.
-5 -2 4 -5
10 – k 25 – k 70 – k 10 – k

Since x, y and z are to be positive;
3k – 105 > 0
6k – 240 > 0
–9k + 390 > 0
∴ k > 35, k > 40, and k < 
∴ 40 < k < 
The percentage of pepper cannot be more than 

Let g grams be removed from each alloy.
As the price of the resulting alloy is the same for both, suppose p% alloy was taken from both alloys.
Quantity of gold metal in the first alloy after interchanging (in gm) = ((22/24) x 130 - (22/24) x g + (23/24) x g) = (p/100) x 130 ... (i)
Or, p/100 = ((22/24) x 130 - (22/24) x g + (23/24) x g)/130 ... (ii)
Similarly, on working for the second alloy, we get
p/100 = ((23/24) x 180 - (23/24) x g + (22/24) x g)/180 ... (iii)
On equating the RHS of (ii) and (iii), we get
((22/24) x 130 - (22/24) x g + (23/24) x g)/130 = ((23/24) x 180 - (23/24) x g + (22/24) x g)/180
18(2860 + g) = 13(4140 - g)
51,480 + 18g = 53,820 - 13g
31g = 2340
g = 75.48 
Thus, 75.48 gm is drawn out from each type of alloy.
Thus, answer option b is correct.

In mouth freshener P, quantity of supari = 9/13 qnd quantity of saunf = 4/13
In mouth freshener Q, quantity of supari = 9/26 and quantity of saunf = 17/26
When we add these two mouth fresheners, total quantity of supari = 
Total quantity of saunf =
So, ratio of saunf to supari in R = 25 : 27