CAT 2025 Slot 3 Quant Question Paper

The total number of products is $$120+135+150+165= 570$$. Therefore, the total cost must have been $$570\times 240 = 136800$$

Let the marked price for the products be $$X$$. The January, February, March, and April, selling prices would respectively be $$0.8X$$, $$0.9X$$, $$0.95X$$, and $$X$$; based on the discounts given on each of the months.

The revenue earned from selling the products in each of the months would be: $$120*0.8X + 135*0.9X + 150*0.95X + 165*X = 525X$$.

The profit earned would be: Revenue -  Cost, and therefore, 

$$525X - 136800 = 138825$$

$$525X = 275625$$

$$X = \dfrac{275625}{525} = 525$$

The correct answer is option B, Rs. 525.

Let the per hour productivity of each member from teams A, B, and C, be $$a$$, $$b$$, and $$c$$ units respectively.

Let $$W$$ be the total units of work that each of the teams does. We have:

$$W = 40*5*a = 50*8*b = 4*10*c$$

This gives, $$a= \dfrac{W}{200}$$,  $$b= \dfrac{W}{400}$$,  and $$c = \dfrac{W}{40}$$

When two members from team A, three members from team B, and one member from team C together start the job, the total per hour productivity for the first 23 hours will be;

$$\dfrac{2W}{200} + \dfrac{3W}{400} + \dfrac{W}{40} = \dfrac{17W}{400}$$.

In 23 hours, the work done will be $$\dfrac{23*17W}{400} = \dfrac{391W}{400}$$

The remaining work, after 23 hours, is $$W - \dfrac{391W}{400} = \dfrac{9}{400}$$

The current per hour efficiency of the group that comprises members from only teams A and B, is $$\dfrac{2W}{200} + \dfrac{3W}{400} = \dfrac{7W}{400}$$

In the next hour, the members from A and B will finish $$\dfrac{7W}{400}$$ units of work and $$\dfrac{9W}{400}-\dfrac{7W}{400} = \dfrac{2W}{400}$$ units of work would remain.

This work has to be finished by newly added members from team B in one hour, therefore, the number of new members required from team B would be $$\dfrac{2W}{400} \div \dfrac{W}{400} = 2$$

Option B is the correct answer.

Let the total number of students who chose science, arts, and commerce streams be $$S$$, $$A$$, and $$C$$, respectively.

We have,

$$S+A+C = 1500$$, such that, $$C = 1500-S-A$$ .....(1)

Also, $$1100S + 1000A + 800C = 15,50,000$$,  which can be simplified by dividing by $$100$$;

$$11S + 10A + 8C = 15500$$ .....(2)

Substituting the value of $$C$$ from equation (1) in equation (2), we have

$$11S + 10A + 8(1500-S-A) = 15500$$

$$\Rightarrow 3S + 2A = 3500$$

Since $$S\leq A$$, the maximum value of $$S$$ (if possible), would occur when $$S=A$$, therefore,

$$5S = 3500$$,  or $$S= 700$$.

Let the first term of the arithmetic progression be $$a$$, and the common difference be $$d$$.

We have:

$$(a+3d) + (a+6d) + (a+9d) = 99$$

$$\Rightarrow 3a + 18d = 99$$  or  $$a + 6d = 33$$ .....(1)

We are also told that the sum of the first fourteen terms is $$497$$,

$$\dfrac{14}{2}(2a + (14-1)d) = 497$$

$$\Rightarrow (2a+13d) = \dfrac{497}{7}$$

$$\Rightarrow 2a+13d = 71$$ .....(2)

Solving equations (1) and (2), we get;

$$a= 3$$ and $$d=5$$

The first five terms would therefore be: $$3, 8, 13, 18, 23$$, and their sum would be $$3+8+13+18+23 = 65$$.

Let the walking speed of Ankita be $$5x$$, implying that her running speed (which is $$40\%$$ more than her walking speed), is $$1.4\times 5x = 7x$$.

Thus, the ratio of her walking and running speeds is $$5:7$$. Therefore, the ratio of the time taken by Ankita to cover a fixed distance walking and running would by $$7:5$$.

She takes 3 hours 30 minutes, or $$3.5$$ hours to walk from B to C. Since Ankita is running from B to C in the second scenario, her time will reduce inversely to the ratio of her speed. The time taken by her to run from B to C would be $$\dfrac{3.5}{7}\times 5 = 2.5$$ hours.

She takes 3 hours 30 minutes, or $$3.5$$ hours to run from A to B. Since Ankita is walking from A to B in the second scenario, her time will increase inversely to the ratio of her speed. The time taken by her to walk from A to B would be $$\dfrac{3.5}{5}\times 7 = 4.9$$ hours.

Therefore, the total time it takes for Ankita in the second scenario is $$4.9+2.5 = 7.4$$ hours. This gives $$7.4*60 = 444$$ minutes.

Let the thousands, hundreds, tens, and units digits of the number be $$a$$, $$b$$, $$c$$, and $$d$$, respectively.

The sum of the digits in the thousands, hundreds, and tens places is 15;  $$a+b+c = 15$$ .....(1)

The sum of the digits in the hundreds, tens, and units places is 16;  $$b+c+d = 16$$ .....(2)

Solving equations (1) and (2), we get $$d = a+1$$.

The digit in the tens place is 6 more than the digit in the units place;  $$c=d+6 = a+1+6 = a+7$$ .....(3)

Substituting the value of $$c$$ in terms of $$a$$ in equation (1), we get;  $$a+b+a+7 = 15$$  or  $$b=8-2a$$

Thus, the number is:  $$\underline{a}\text{ }\underline{8-2a}\text{ }\underline{a+7}\text{ }\underline{a+1}$$

We know that since the number is a four digit number, $$a\geq 1$$. But all the four digits have to be less than or equal to $$9$$ but greater than or equal to $$0$$, thus,

$$8-2a\geq 0$$  gives  $$a\leq 4$$  and  $$a+7\leq 9$$  gives  $$a\leq 2$$.

Thus, $$a$$ can be $$1$$ or $$2$$.

The two possible values for the number, therefore, are, $$1682$$ and $$2493$$, when we put $$a=1$$ and $$a=2$$, respectively.

The difference between the only two possible values is, $$2493 - 1682 = 811$$

Option A is the correct answer.

Let the initial speed of Rahul be $$x$$ kilometres per hour. Since he travels the distance usually in $$6$$ hours, (5 pm to 11 pm), the total distance should be $$6\times x = 6x$$ kilometres.

Let the distance after which Rahul stops for some duration in both scenarios be $$y$$ kilometres.

In the first scenario, he stops for $$20$$ minutes or $$\dfrac{1}{3}$$ hours, therefore, the travel time would be $$6-\dfrac{1}{3} = \dfrac{17}{3}$$ hours. We have, 

$$\dfrac{k}{x} + \dfrac{6x-k}{x+3} = \dfrac{17}{3}$$ 

$$\Rightarrow \dfrac{6x^2+3k}{x^2+3x} = \dfrac{17}{3}$$

$$\Rightarrow 18x^2 + 9k = 17x^2 + 51x$$

$$\Rightarrow x^2 = 51x - 9k$$ .....(1)

Similarly, in the second scenario, he stops for $$20+10=30$$ minutes or $$\dfrac{1}{2}$$ hours, therefore, the travel time would be $$6-\dfrac{1}{2} = \dfrac{11}{2}$$ hours. We have, 

$$\dfrac{k}{x} + \dfrac{6x-k}{x+5} = \dfrac{11}{2}$$

$$\Rightarrow \dfrac{6x^2+5k}{x^2+5x} = \dfrac{11}{2}$$

$$\Rightarrow 12x^2 + 10k = 11x^2 + 55x$$

$$\Rightarrow x^2 = 55x - 10k$$ .....(2)

From equations (1) and (2), we have;

$$55x - 10k = 51x - 9k$$  or  

$$4x = k$$.

Substituting the value of $$k$$ in equation (1), we have

$$x^2 = 51x - 36x$$  or, since $$x$$ is positive, $$x= 15$$. Therefore, option B is the correct answer.

Let the rate of flow of water from pipe A be $$4x$$ units per hour. From the ratio of the rates of flow of water (4:9:36), we have that the rate of flow of water from pipe B should be $$9x$$ units per hour, and from pipe C should be $$36x$$ units per hour.

Since pipe A fills the empty tank in 15 hours, the capacity of the tank must be $$15\times 4x = 60x$$ units.

When all three pipes work together, their combined capacity would be $$4x+9x+36x = 49x$$ units per hour.

Therefore, the time it would take for the three pipes to fill the empty tank working together is:  $$\dfrac{60x}{49x}$$ hours

In minutes, this time is equivalent to $$\dfrac{60x}{49x}\times 60 \approx 73.47$$ minutes.

Option A is the closest, and is the correct answer.

Let $$(x^2+3x)$$ be equal to $$k$$. We have, 

$$f(x)= k(k+2) = k^2+2k$$ 

Therefore, $$\sqrt{f(x)+1} = \sqrt{k^2+2k+1} = \sqrt{(k+1)^2} = k+1 = 9701$$

We get $$k=9700$$.

Thus, $$x^2+3x=9700$$ or $$x^2+3x-9700 = 0$$

Since $$x$$ is real, the discriminant of the above quadratic has to be greater than or equal to zero.

We find that $$3^2 + 4*9700 \geq 0$$ and therefore the quadratic has real roots.

The sum of the roots will be $$-\dfrac{b}{a} = -\dfrac{3}{1} = -3$$

Option D is the correct answer.

Let $$y= \dfrac{2x-3}{2x^{2}+4x-6}$$

$$\Rightarrow 2yx^2 + 4yx - 6y = 2x - 3$$

$$\Rightarrow 2yx^2 + (4y-2)x - 6y + 3 = 0$$ 

Equation (1) is a quadratic in $$x$$, where $$x$$ is real. Therefore, the discriminant of the quadratic has to be greater than or equal to zero.

$$(4y-2)^2 + 4\times 2y\times (6y-3) \geq 0$$

$$16y^2 + 4 - 16y - 24y + 48y^2 \geq 0$$

$$64y^2 - 40y + 4 \geq 0$$

$$16y^2 - 10y + 1 \geq 0$$ 

The roots of the quadratic above will be $$\dfrac{10\pm 6}{32} = \dfrac{1}{2}$$ or $$\dfrac{1}{8}$$

Since the coefficient of $$y^2$$ is positive, the quadratic will be less than zero in the range $$\left(\dfrac{1}{8},\dfrac{1}{2}\right)$$

The quadratic will be greater than or equal to zero otherwise.

Therefore, the domain the quadratic, or possible values of $$y$$, which is the range of $$f(x)$$, will be,

$$\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty \right)$$

Option C is the correct answer.

$$(10^{50} + 10^{25} - 123)$$ can be written as $$10^{50} + (10^{25}-123)$$

$$1000-123$$ gives $$877$$. 
$$10000-123$$ gives $$9877$$.
$$100000-123$$ gives $$99877$$.
$$1000000-123$$ gives $$999877$$.
And so on.

Considering $$10^n-123$$, which has a total of $$n$$ digits; in general, of the $$n$$ total digits; the digit $$7$$ is repeating $$2$$ times, digit $$8$$ is repeating $$1$$ time, and digit $$9$$ is repeating $$(n-3)$$ times.

Also, when added to $$10^{50}$$, the cumulative increase in the sum of the digits of the resultant number would be $$1$$, which would come from the leftmost digit of the result. All other newly added digits will be zero.

Thus, the sum of the digits of the result will be given by;

$$[7*2] + [1*8] + [(25-3)*9] + [25*0] + [1*1] = 14+8+198+1 = 221$$

The correct answer is option B.

The altitudes $$CM$$ and $$BN$$ will be equal as the triangle $$\triangle ABC$$ is isosceles. 

Further, in $$\triangle AOB$$, using Pythagoras Theorem, we have

$$AO^2 + OB^2 = AB^2$$, therefore, $$AO^2 = 50^2 - 40^2$$

Thus, $$AO^2 = 900$$ or $$AO = 30$$ cm. 

We know that the ratio of altitudes in a triangle is the inverse of the ratio of sides; therefore, the ratio of altitudes of the triangle will be;

$$\dfrac{1}{50} : \dfrac{1}{50} : \dfrac{1}{80} = 8 : 8 : 5$$

Therefore, $$CM = BN = 8x$$ and $$AO = 5x$$.

Since $$5x = 30$$, we have $$x= 6$$, and the sum of altitudes, in centimetres, will be $$8x+8x+5x = 21x = 21*6 = 126$$.

$$\left(x+\dfrac{1}{x}\right)^2 = x^2+\dfrac{1}{x^2} + 2 = 25+2 = 27$$

Therefore, $$\left(x+\dfrac{1}{x}\right) = \sqrt{27} = 3\sqrt{3}$$

Also, $$\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} + 3\left(x+\dfrac{1}{x}\right)$$

Therefore, $$x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 9\sqrt{3} = 72\sqrt{3}$$

Also, $$\left(x^2+\dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2$$

Therefore, $$x^4 + \dfrac{1}{x^4} = (25)^2 - 2 = 623$$

Lastly, $$\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + x + \dfrac{1}{x}$$

Therefore, $$x^7+\dfrac{1}{x^7} = 623*( 72\sqrt{3}) - 3\sqrt{3} = 44853\sqrt{3}$$

Option A is the correct answer.

Based on the information provided, we can make the following diagram;

We know that the sum of all entries in the Venn diagram should be $$150$$. Therefore,

$$111+ 75-2k + 40 - (k-x) - k + 0 = 150$$

$$76 = 4k-x$$, where $$x\geq 1$$

We must maximise the number of students who chose Physics but not Mathematics. This number is equal to $$75-2k$$. To maximise this we must minimise $$k$$. 

We have $$4k = 76+x$$ or $$k = 19 + \dfrac{x}{4}$$

Since $$x$$ and $$k$$ are both integers, $$x$$ should be divisible by $$4$$. To minimise $$k$$, we will take the minimum value of $$x$$, which is $$4$$.

Therefore, $$k = 19 + \dfrac{4}{4} = 20$$.

Thus, the maximum possible number of students who chose physics but not mathematics, is $$75-2*20 = 35$$.

Since the base of the logarithm has to be greater than zero and cannot be $$1$$, $$x$$ has to be greater than $$3$$ and cannot be $$4$$. Also, since $$x^2-9>0$$, $$x$$ will be greater than $$3$$.

The equation can be rewritten as;

$$\log_{x-3}{(x^{2}-9)}-\log_{x-3}{(x+1)} = 2$$

Or,

$$\log_{x-3}{\dfrac{x^2-9}{x+1}} = 2$$

$$\Rightarrow \dfrac{(x+3)(x-3)}{x+1} = (x-3)^2$$

$$\Rightarrow \dfrac{(x+3)}{x+1} = (x-3)$$

$$\Rightarrow (x+3) = (x-3)(x+1)$$

$$\Rightarrow x+3 = x^2 - 3x + x - 3$$

$$\Rightarrow x^2 - 3x - 6 = 0$$

The roots of the quadratic above are $$\dfrac{3\pm \sqrt{3^2 + 24}}{2}$$

The negative value of $$x$$ will not be possible, the positive value that satisfies is $$\dfrac{3+ \sqrt{33}}{2}$$

The correct answer is option D.

Let the average salary of managers be $$x$$ and let the average salary of engineers be $$y$$. The total salary of all the employees will be $$(5+25)*60000 = 1800000$$

We have, $$5x+25y = 1800000$$ .... (1)

If the average salary of all the employees increases by $$5\%$$, the total salary of all the employees will also increase by $$5\%$$, because the total number of employees remains the same. The new total salary will be, $$1800000\times 1.05 = 1890000$$

Also, the average salary of all the managers has increased by $$20\%$$ and has become $$1.2x$$, we have

$$5*1.2x + 25y = 1890000$$,  or  $$6x + 25y  = 1890000$$ .... (2)

Subtracting equation (1) from equation (2), we get,

$$x = 90000$$

Which gives $$25y = 1800000 - 90000*5$$  or  $$y = \dfrac{1350000}{25} = 54000$$

Therefore, the correct answer is option C.

We extend $$BC$$ and $$AD$$ to meet at $$E$$, and based on the information provided, can construct the diagram as follows:

$$DC = AT =x$$ and $$BT = 2x$$ such that $$AB = 3DC$$. $$\angle A= 90^{\circ}$$ and triangles $$ABE$$ and $$DCE$$ are similar. 

Since $$AD$$ is equal to the diameter of the circle, and $$AE = 3DE$$, we get $$DE = 3$$ cm.

In right-angled triangle $$DCE$$, we have $$CE = \sqrt{x^2+9}$$

In right-angled triangle $$BCT$$, we have $$BC = \sqrt{4x^2+36} = 2\sqrt{x^2+9}$$

Since $$BE = BC+CE$$, we have $$BE = 3\sqrt{x^2+9}$$

The radius of the incircle of a right-angled triangle is given by $$\dfrac{s_1 + s_2 - \text{hypotenuse}}{2}$$

Therefore, $$3 = \dfrac{3x+9-3\sqrt{x^2+9}}{2}$$

Which gives,

$$-x-1 = -\sqrt{x^2+9}$$  or  $$x^2+2x+1 = x^2 + 9$$  or  $$x=4$$.

Lastly, the area of the trapezium would be $$\dfrac{1}{2}\times (x+3x) \times 6 = 12x = 12\times 4 = 48$$ square centimetres.

$$12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20 ^{12x} \times 243^{3x-6}$$

On rewriting after prime factorisation, we get,

$$\cancel{2^{24x}}\times 3^{12x} \times 2^{48x+24}\times 5^{2y} = 2^{12z}\times \cancel{2^{24x}} \times 5^{12x} \times 3^{15x-30}$$

Since LHS = RHS, the corresponding powers must be equal. We have,

Powers of $$3$$:  $$12x = 15x-30$$  or  $$x=10$$, and

Powers of $$2$$:  $$48x+24 = 12z$$  or  $$4x+2 = z$$  or  $$z = 42$$

Powers of $$5$$:  $$2y = 12x$$  or  $$y= 6x$$  or  $$y=60$$

Therefore, $$x+y+z = 10+42+60 = 112$$

Based on the information provided, we can construct the following diagram;

We have $$AB = AC = 12$$ cm, and $$AD = 8$$ cm. And $$\angle ACB = \angle AEB$$

Since the triangle is isosceles, $$\angle ACB = \angle AEB = \angle ABC$$ (or $$\angle ABD$$)

In $$\triangle ABE$$ and $$\triangle ADB$$,

$$\angle BAE =\angle DAB$$ and $$\angle AEB = \angle ABD$$. Therefore, by AA similarity criterion, $$\triangle ABE \sim \triangle ADB$$

Hence, we get,

$$\dfrac{AB}{AD}=\dfrac{AE}{AB}$$  or  $$\dfrac{12}{8} = \dfrac{AE}{12}$$  or  $$AE=18$$ cm.

Vessel A contains $$60$$ litres of pure alcohol, and Vessel B contains $$60$$ litres of pure water. Let the amount taken out from vessel A be $$x$$ litres. The contents in Vessel A now are $$60-x$$ litres of alcohol, and in Vessel B now are $$60$$ litres of water and $$x$$ litres of alcohol.

After mixing, when the same quantity $$x$$ is taken out from Vessel B, the alcohol that would be taken out would be $${\left(\dfrac{x}{60+x}\right)}^{\text{th}}$$ of $$x$$, which is $$\dfrac{x^2}{60+x}$$ litres.

The initial total quantity of $$60$$ litres has been restored in A after replacement. The total amount of alcohol in Vessel A after the process is complete is:

$$60 - x + \dfrac{x^2}{60+x} = \dfrac{(60+x)(60-x) + x^2}{60+x} = \dfrac{3600}{60+x}$$

The total quantity in vessel A is $$60$$ litres where the ratio of alcohol to total quantity is $$\dfrac{15}{15+4} = \dfrac{15}{19}$$

Therefore, 

$$\dfrac{3600\div (60+x)}{60} = \dfrac{15}{19}$$

$$\Rightarrow \dfrac{60}{60+x} = \dfrac{15}{19}$$

$$\Rightarrow x = 16$$

Thus, the quantity taken out and replaced in the two iterations is $$16$$ litres.

Let the initial total number of coins in box A be $$17x$$, since the ratio of the number of coins in boxes A and B is $$17:7$$, the initial total number of coins in box B would be $$7x$$.

After $$108$$ coins are shifted from box A to box B, the number of coins in box A would be $$17x-108$$ and in box B would be $$7x+108$$. We have,

$$\dfrac{17x-108}{7x+108} = \dfrac{37}{20}$$

$$\Rightarrow 340x - 2160 = 259x + 3996$$

$$\Rightarrow 81x = 6156$$  or  $$x= 76$$

Therefore, the total number of coins in the two boxes combined are $$(17+7)*76 = 1824$$, half of which is $$912$$. The number of coins in box B currently is $$7*76+108 = 640$$. The number of coins needed to make the coins equal in both the boxes is, $$912-640 = 272$$.

Thus, $$272$$ coins will be transferred from box A to box B to make the number of coins equal (1:1) in both the boxes.

We have $$p+q+r = 900$$, and $$r$$ is a perfect square lying between $$150$$ and $$500$$.

Let $$q$$ be a non-variable number that does not have a range.

We have $$0.3q \leq p \leq 0.7q$$,  or  $$q+0.3q \leq p+q \leq q+0.7q$$  or  $$1.3q \leq p+q \leq 1.7q$$

Also, we can rewrite $$p+q = 900-r$$

Therefore, $$1.3q \leq 900-r \leq 1.7q$$

Since the extreme values (maximum and minimum) of $$p$$ depend on the value of $$q$$ ($$0.3q$$ and $$0.7q$$), we can maximise or minimise $$r$$ to find the extreme values of $$p$$.

Since $$1.3q \leq 900-r$$, the minimum possible value of $$q$$ would be when $$1.3q = 900-r$$ and when $$r$$ is maximum. The maximum value of $$r$$ is $$484$$. We get, $$1.3q = 900-484$$, or $$q = 320$$.

This gives the minimum possible value of $$p$$ as $$0.3\times 320 = 96$$.

Since $$900-r \leq 1.7q$$. the maximum possible value of $$q$$ would be when $$900-r=1.7q$$ and when $$r$$ is minimum. The minimum value of $$r$$ is $$169$$. We get $$1.7q = 900-169$$ or $$q= 430$$

This gives the maximum possible value of $$p$$ as $$0.7\times 430 = 301$$

The sum of the maximum and minimum values of $$p$$ is $$96+301 = 397$$.