Test Level 3: Mixtures and Alligations - CAT MCQ

The sweetness ratio of the mixture consisting of glucose, sucrose, and fructose in the ratio 1:2:3 can be calculated as follows:

Glucose: 0.74 * 1 = 0.74
Sucrose: 1.00 * 2 = 2.00
Fructose: 1.70 * 3 = 5.10

The sweetness ratio of the mixture is 0.74 : 2.00 : 5.10.

To determine how many times sucrose is sweeter than the mixture, we need to compare the sweetness ratios. Dividing the sweetness ratio of sucrose (1.00) by the sweetness ratio of the mixture (0.74), we get:
1.00 / 0.74 ≈ 1.35
Therefore, sucrose is approximately 1.35 times sweeter than the given mixture.
The closest option to this value is: 1.3
So, the correct answer is: 1.3

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

Let the original weights of P & Q be 4k, 5k

==> Total weight = 9k

Original total weight = 82.8/1.15 = 72 = 9k

==> k=8

==> Original weights of P&Q are 32 kg, 40 kg

1.1*32+x*40 = 82.8

==> 40x = 47.6

==> x=1.19

==> Q’s weight was increased by 19%

1

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