CMAT 2024 Slot 2 Question Paper

Number of kurtis (linen and cotton) sold by store D = 10080

number of linen kurtis sold by store B = 8400

$$\frac{\left(10080-8400\right)}{8400}=\frac{1680}{8400}=\frac{1}{5}=20\%$$

The number of linen kurtis sold by stores D and E together = 6300+13440 = 19740

The average is 9870.

The number of cotton kurtis sold by stores D and E together = 3780+10080 = 13860

The average is 6930.

The difference = 9870 - 6930 = 2940

Ratio between number of kurtis (linen and cotton) sold by store C and number of linen kurtis sold by store F = 6720:8232 = 40:49

(A) Parallel if the system has no solution: True. If the two lines are parallel, they do not intersect, implying the system of equations has no solution. This occurs when the slopes of the lines are equal but their intercepts differ.

(D) Intersecting if the system has only one solution: True. If the lines intersect at exactly one point, the system has a unique solution. This happens when the lines have different slopes.

(B) Coincident if the system has a finite number of solutions: Incorrect. If the lines are coincident, the system has infinitely many solutions, not a finite number.

(C) Intersecting if the system has two solutions: Incorrect. Two linear equations can intersect at only one point, meaning the system cannot have two solutions.

 The nature of the solutions depends on how the lines relate geometrically:

Parallel lines → No solution (A is correct).

Intersecting lines → Unique solution (D is correct).

Lines cannot intersect at more than one point, ruling out C, and coincident lines lead to infinitely many solutions, ruling out B.

Therefore, the correct answer is option C.

Given that currently container has 6 parts of water and 10 parts of milk.

Assume the volume of water to be $$6x$$ and that of milk to be $$10x$$

So, the volume of the container is $$16x$$.

Assume we take out $$y$$ litre of mixture and replace it with water

So, the resultant mixture will have 

$$\left(6x-\dfrac{6x}{16x}\times y+y\right)$$ water and 

$$\left(10x-\dfrac{10x}{16x}\times y\right)$$ milk

Given the volume of water is half of the container i.e. $$8x$$

So, $$\left(10x-\dfrac{10x}{16x}\times y\right)=8x$$

$$10x-\dfrac{5}{8}y=8x$$

or, $$y=\dfrac{16x}{5}$$

In terms of container volume

$$\dfrac{y}{16x}=\dfrac{\dfrac{16x}{5}}{16x}=\dfrac{1}{5}$$

Hence, the answer is Option C

We are given that the HCF of two numbers is 21. Let the numbers be X, and Y. Then X and Y can be written as,

X = 21a and Y = 21b where a and b are coprimes.

We know that, 

product of the numbers =  LCM * HCF

X * Y = 3003 * 21

21a * 21b = 3003 * 21

a * b = 143

a * b = 11 * 13

There are two possibilities for (a,b) with a and b as coprimes, one being (1, 143) and the other being (11, 13). We cannot take (1, 143) because, in this case, one of the numbers would be 21, but we are given that both numbers are greater than 21. So, the numbers X and Y are 21 * 11 and 21 * 13, which are 231 and 273. So, the sum of the numbers = 231 + 273 = 504.

Therefore, the correct answer is option A.

Discount is always calculated on the marked price or the regular price. 

A: Regular price = 65, Sale price = 55

Discount = 10

Discount % = $$\dfrac{10}{65}=15.38\%$$

B: Regular price = 60, Sale price = 50

Discount = 10

Discount % = $$\dfrac{10}{60}=16.66\%$$

C: Regular price = 70, Sale price = 50

Discount = 20

Discount % = $$\dfrac{20}{70}=28.57\%$$

(Note: The correct discount percentage should be 28.57%, however, in actual exam, the correct option given was 14.29%)

D: Regular price = 75, Sale price = 65

Discount = 10

Discount % = $$\dfrac{10}{75}=13.33\%$$

Percentage profit = $$\dfrac{\Pr ofit}{Cost\ \Pr ice}\times100$$

(A) Cost price : ₹36, Profit: ₹17

Percentage profit = $$\dfrac{17}{36}\times100$$ = 47.22%

(B) Cost price : ₹50, Profit : ₹24

Percentage profit = $$\dfrac{24}{50}\times100$$ = 48%

(C) Cost price : ₹40, Profit : ₹19

Percentage profit = $$\dfrac{19}{40}\times100$$ = 47.5%

(D) Cost price : ₹60, Profit : ₹29

Percentage profit = $$\dfrac{29}{60}\times100$$ = 48.33%

Thus, percentage profits are in this sequence D > B > C > A

A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles.

If a single marble is chosen at random from the jar, the probability of getting a yellow marble =$$\frac{\text{no of yellow marbles}}{\text{Total no of mables}}$$= $$\frac{4}{10}$$

Probability of getting a green marble = $$\frac{\text{no of green marbles}}{\text{Total no of mables}}$$ = $$\frac{3}{10}$$

Probability of getting either a yellow or a green marble = $$\frac{\text{no of yellow or green marbles}}{\text{Total no of mables}}$$ = $$\frac{4+3}{10}$$ = $$\frac{7}{10}$$

Probability of getting either a red or a yellow marble = $$\frac{\text{no of yellow or red marbles}}{\text{Total no of mables}}$$ = $$\frac{4+1}{10}$$ = $$\frac{5}{10}$$

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

We know the volume of a cylinder of radius $$r$$ and height $$h$$ is $$\pi r^2h$$

Given D = 35 cm, so $$r=17.5$$ cm

Also, $$h=32$$ cm

So, Volume = $$\pi r^2h=\pi\ \left(17.5\right)^2\times32\ cm^3$$

We know that the volume of a cone of radius $$r$$ and height $$h$$ is $$\dfrac{1}{3}\pi r^2h$$

Given the diameter of cone = 4 cm, so the radius = 2 cm

Height = 7 cm

So, the volume of the cone is $$\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi\left(2\right)^2\times7$$

To find the maximum number of cones that can be used, divide the volume of the cylinder by the volume of a cone

Max cones = $$\dfrac{\pi\left(17.5\right)^2\times32}{\dfrac{1}{3}\pi\left(2\right)^2\times7}=1050$$

Hence, the answer is 1050.