CAT Exam  >  CAT Tests  >  Level-wise Tests for CAT  >  Test Level 3: Mixtures and Alligations - CAT MCQ

Test Level 3: Mixtures and Alligations - CAT MCQ


Test Description

10 Questions MCQ Test Level-wise Tests for CAT - Test Level 3: Mixtures and Alligations

Test Level 3: Mixtures and Alligations for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 3: Mixtures and Alligations questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Mixtures and Alligations MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Mixtures and Alligations below.
Solutions of Test Level 3: Mixtures and Alligations questions in English are available as part of our Level-wise Tests for CAT for CAT & Test Level 3: Mixtures and Alligations solutions in Hindi for Level-wise Tests for CAT course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test Level 3: Mixtures and Alligations | 10 questions in 20 minutes | Mock test for CAT preparation | Free important questions MCQ to study Level-wise Tests for CAT for CAT Exam | Download free PDF with solutions
Test Level 3: Mixtures and Alligations - Question 1

Directions: Read the information given below and answer the question.
Sweetness of various items relative to sucrose, whose sweetness is taken as 1, is:
Lactose: 0.16, Maltose: 0.32, Glucose: 0.74, Sucrose: 1.00, Fructose: 1.70, Saccharin: 675.00
Approximately how many times is sucrose sweeter than a mixture consisting of glucose, sucrose and fructose in the ratio 1 : 2 : 3?

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 1

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


 

Test Level 3: Mixtures and Alligations - Question 2

A total of 750 candidates appeared in an examination. 70% of the boys and 65% of the girls passed the examination. If the total pass percentage was 68%, find the number of boys who appeared in the examination.

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 2

Applying the rule of alligation:

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

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test Level 3: Mixtures and Alligations - Question 3

Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what respective ratio should these metals be mixed, so that the mixture is 15 times as heavy as water?

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 3

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

Test Level 3: Mixtures and Alligations - Question 4

Mixtures A and B are contained in two separate vessels. Mixture A contains ingredients P, Q and R in a ratio of 3 : 5 : 2 and mixture B contains ingredients P and Q in a ratio of 4 : 5. We have to make 540 litres of a new mixture by adding the mixtures A and B in a ratio of 1 : 2. What will be the quantity of ingredient P in the final mixture?

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 4

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.

Test Level 3: Mixtures and Alligations - Question 5

The weights of two boxes A and B containing sugar are in the ratio 4 : 5. Some amount of sugar is added to both the boxes and it is found that the weight of box A increases by 10% and the total weight of boxes A and B together becomes 82.8 kg. If this weight is 15% more than the previous weight, then find the percentage with which the weight of box B increases.

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 5

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

 

 

 

 

 

 

 

 

 

 

Test Level 3: Mixtures and Alligations - Question 6

The percentage amounts of an acid in three solutions form a geometric progression. If we mix the first, second and third solutions in the ratio 3 : 2 : 1 by weight, we get a solution containing 22% acid. On the other hand, if we mix the solutions in the ratio 2 : 3 : 4 by weight, we obtain a solution containing 32% acid. Find the percentage of acid in each solution.

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 6

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%.

Test Level 3: Mixtures and Alligations - Question 7

There are two solutions of HCl in water. The first is 40% strong and the second is 60% strong. The two solutions are mixed. Some amount of pure water is added and a 20% strong solution is obtained. If instead of adding pure water, the same amount of an 80% strong solution had been added to the mixture, then a 70% strong solution would have been obtained. What was the initial quantity of the 40% and the 60% solutions?

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 7

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.

Test Level 3: Mixtures and Alligations - Question 8

There are three mixtures. The first mixture contains 30% salt and 70% pepper, the second contains 10% pepper and 90% chilly, and the third contains 15% salt, 60% chilly and the rest pepper. They are mixed to obtain a new mixture which contains 40% chilly. The percentage of pepper in the new mixture cannot be more than

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 8

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 

Test Level 3: Mixtures and Alligations - Question 9

There are two qualities of gold alloys 22 carat and 23 carat having different prices per gram, their volumes being 130 gm and 180 gm, respectively. After equal quantities were taken from both the alloys, the alloy removed from the 22-carat alloy was added to the 23-carat alloy and vice versa. The resulting two types of alloys now have the same price per gram. Find the quantity of alloy drawn out from each type. It is to be assumed that 100% gold is 24 carats.

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 9

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.

Test Level 3: Mixtures and Alligations - Question 10

P and Q are two different mouth fresheners, prepared by mixing supari and saunf in the proportions 9 : 4 and 9 : 17, respectively. If equal quantities of the mouth fresheners are mixed to form a third kind of mouth freshener R, find the ratio of saunf to supari in R.

Detailed Solution for Test Level 3: Mixtures and Alligations - Question 10

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

5 docs|272 tests
Information about Test Level 3: Mixtures and Alligations Page
In this test you can find the Exam questions for Test Level 3: Mixtures and Alligations solved & explained in the simplest way possible. Besides giving Questions and answers for Test Level 3: Mixtures and Alligations , EduRev gives you an ample number of Online tests for practice

Top Courses for CAT

Download as PDF

Top Courses for CAT