CAT 2025 Slot 1 Quant Question Paper

First function $$f\left(x\right)=x^2-4cx+8$$

For this function $$a>0$$, so minimum value will occur at $$x=-\dfrac{b}{2a}=-\left(-\dfrac{4c}{2}\right)=2c$$

So, the minimum value of the function is = $$2c^2-4c\left(2c\right)+8c=-4c^2+8c$$

Second function $$g(x)=-x^{2}+3cx-2c$$

For this function $$a<0$$, so maximum value will occur at $$x=-\dfrac{b}{2a}=-\dfrac{\left(-3c\right)}{2}=\dfrac{3c}{2}$$

So, the maximum value of the function is = $$-\left(\dfrac{3c}{2}\right)^2+3c\left(\dfrac{3c}{2}\right)-2c=\dfrac{9c^2}{4}-2c$$

So, as per the given condition,

$$\dfrac{9c^2}{4}-2c<-4c^2+8c$$

or, $$\dfrac{9c^2}{4}+4c^2<8c+2c$$

or, $$\dfrac{25c^2}{4}<10c$$

or, $$\dfrac{5c^2}{4}<2c$$

or, $$5c^2<8c$$

or, $$5c^2-8c<0$$

or, $$c\left(c-\dfrac{8}{5}\right)<0$$

or, $$0<c<\dfrac{8}{5}$$

So, the value of $$c$$ which lies in this range is $$\dfrac{1}{2}$$

According to question, Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours.

Given, in the first part time required is 30 minutes.

Also, the time taken to cover these four parts follow arithmetic progression.

So, let us say the times be 30 minutes, (30+d) minutes, (30+2d) minutes and (30+3d) minutes

So, $$30+\left(30+d\right)+\left(30+2d\right)+\left(30+3d\right)=180$$

or, $$6d=180-120$$

or, $$6d=60$$

or, $$d=\dfrac{60}{6}=10$$ minutes.

So, the time required in these four parts are 30 minutes, 40 minutes, 50 minutes and 60 minutes respectively.

Now, in the first part, speed is 960 metres per minute.

Speed in the four parts is also in arithmetic progression.

Let's say the speeds be $$(960+x)$$,$$(960+2x)$$,$$(960+3x)$$ metres per minute

So, total distance covered = $$960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60$$ metres 

So, $$960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60=224000$$

or, $$960\left(30+40+50+60\right)+40x+\left(2x\right)\left(50\right)+\left(3x\right)\left(60\right)=224000$$

or, $$172800+320x=224000$$

or, $$320x=224000-172800=51200$$

or, $$x=\dfrac{51200}{320}=160$$

So, the speed in the fourth part is $$960+3x=960+3\times\ 160=1440$$ metres per minute

Time in the fourth part is 60 minutes

So, distance covered in the fourth part = $$1440\times\ 60=86400$$ metres

So, option D is the correct answer.

According to question, in the given 3-digit number N, the digits are non-zero and distinct.

So, the possible digits in 3-digit number = $$2,3,5,6,7,8$$

It is also given that only one of the digits is a prime number.

So, the minimum possible value of N = 268

(2 is the smallest prime digit, and the non-prime digits has to be 6 and 8)

Now, $$268=4\times\ 67=2^2\times\ 67$$

So, the number of factors = $$\left(2+1\right)\left(1+1\right)=3\times\ 2=6$$

For n=3,4,5 and $$x\in[n,n+1)$$ we have $$\lfloor x\rfloor=n$$, so the equation

$$\lfloor x^2\rfloor=\lfloor x\rfloor^2=n^2$$

$$x^2\in[n^2,n^2+1)$$, i.e. $$x\in[n,\sqrt{n^2+1})$$

Thus for $$3\le x\le6$$

$$S=[3,\sqrt{10})\ \cup\ [4,\sqrt{17})\ \cup\ [5,\sqrt{26})\ \cup{6}$$

Option B and C have $$\sqrt{10}$$ included, which is not part of the original set. And Option D has $$\sqrt{18}$$. So, it is not possible. 

Option A is the answer.

Let the number of stocks B hold be $$x$$

So, number of stocks C hold is $$20-x$$

So, $$10\times\ 120+90\times\ x+150\times\ \left(20-x\right)=3300$$

or, $$1200+90\ x+150\left(20-x\right)=3300$$

or, $$1200+90\ x+3000-150x=3300$$

or, $$4200-60x=3300$$

or, $$60x=900$$

or, $$x=\dfrac{900}{60}=15$$

So, the number of shares of stock that B hold is 15.

Given expression is $$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$$,

$$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}$$ = $$\left\{3^1\times3^4\times3^9...\times3^{400}\right\}$$

Sum of square of n natural numbers is $$\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}$$

= $$3^{\dfrac{\left(20\cdot21\cdot41\right)}{6}}$$ = $$3^{2870}$$

On right hand side of inequlaity we have $$\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$$

 = $$3^{21}\times3^{22}\times...\times3^{20+m}$$ = $$3^{21+22+...+20+m}$$

Using the sum of the first (n) natural numbers,

$$1+2+\cdots+n = \frac{n(n+1)}{2}$$

$$21 + 22 + \cdots + (20+m)$$

= $$1+2+\cdots+(20+m) - (1+2+\cdots+20)$$

$$1+2+\cdots+(20+m)=\frac{(20+m)(21+m)}{2}$$

$$1+2+\cdots+20 = \frac{20\cdot21}{2} = 210$$

So, $$21+22+\cdots+(20+m)$$

= $$\frac{(20+m)(21+m)}{2} - 210$$

Expanding, $$(20+m)(21+m)=m^2+41m+420$$

Thus, $$\frac{m^2+41m+420}{2}-210$$

$$= \frac{m^2+41m}{2}$$

Since the bases are equal, we must compare the powers.

$$2870<\frac{m^2+41m}{2} \Rightarrow 5740<m(m+41) $$

Here, we can put in the option to check the minimum value that satisfies the inequality.

56: We get 5740<5264. This is false

57: We get 5740<5586. This is false

58: We get 5740<5742. This is the minimum possible value.

For base of log in range $$1/4\in(0,1)$$ and $$\log_{1/4}(x)>0$$ is true only if 0<x<1. 

For integer n, $$x=n^2-7n+11$$ is an integer, so it cannot lie strictly between 0 and 1.

So, there is no integer value for which this inequality is satisfied.

It is given, $$x>y\geq3 $$ and $$x+y<14$$

Now for $$y=3$$, the values of $$x$$ can be 4,5,6,7,8,9,10  (7 cases)

Then for $$y=4$$, the values of $$x$$ can be 5,6,7,8,9 (5 cases)

Then for $$y=5$$, the values of $$x$$ can be 6,7,8 (3 cases)

Then for $$y=6$$, the values of $$x$$ will be 7 (1 case)

So, total number of cases =1+3+5+7=16 cases

Let the principal be ₹ P and rate of interest be r%.

Now, $$13920=P+\dfrac{P\times\ r\times\ 3}{100}$$

or, $$13920-P=\dfrac{P\times\ r\times\ 3}{100}$$ ---->(1)

Also, $$18960=P+\dfrac{P\times\ r\times\ 13}{100\times\ 2}$$

$$18960-P=\dfrac{P\times\ r\times13}{100\times\ 2}$$ ----->(2)

Dividing eqn(1) by eqn(2),

$$\dfrac{13920-P}{18960-P}=\dfrac{3}{\frac{12}{2}}=\dfrac{6}{13}$$

or, $$\left(13920-P\right)13=\left(18960-P\right)6$$

or, $$13920\times\ 13-13P=18960\times\ 6-6P$$

or, $$13920\times\ 13-18960\times\ 6=13P-6P$$

or, $$180960-113760=7P$$

or, $$67200=7P$$

or, $$P=\dfrac{67200}{7}=9600$$

Putting this in equation (1),

$$13920-9600=\dfrac{9600\times\ r\times\ 3}{100}$$

or, $$4320=96\times\ 3r$$

or, $$r=\dfrac{4320}{96\times\ 3}=15$$

So, rate percent is $$15\%$$

Now if the same sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, amount = $$9600\left(1+\dfrac{\frac{15}{2}}{100}\right)^4=9600\left(1+\frac{7.5}{100}\right)^4$$

So, interest = $$9600\left(1+\frac{7.5}{100}\right)^4-9600$$

= Rs 3220.50

= Rs 3221

So, the total interest earned is Rs 3221.

200 L with (30%) acid $$\Rightarrow$$ acid = $$0.30\times200=60 L$$

Replace 20% with water: remaining acid $$=60\times(1-0.20)=60\times0.8=48 L$$.

Replace 10% with pure acid: After removing 10% of the mixture, the acid becomes $$48\times0.9=43.2 L$$, then adding back 20 L pure acid $$\Rightarrow acid =43.2+20=63.2 L$$.

Replace 15 with water: acid left $$= 63.2\times(1-0.15)=63.2\times0.85=53.72 L$$.

Final concentration $$= \dfrac{53.72}{200}=0.2686\approx26.86%$$.

The answer is 27%

Let say number of girls be $$g$$ and number of boys be $$b$$.

If 40% of the girls left, remaining number of girls = $$0.6g$$

Also if 60% of the boys left, remaining number of boys = $$0.4b$$

or, $$0.6g=0.4b+8$$

or, $$6g=4b+80$$

or, $$3g=2b+40$$

So, the possible values of (b,g) are: (13,22),(16,24),(19,26),(22,28),(25,30),.....

Now, $$0.6g$$ and $$0.4b$$ has to be an integer.

So, for this $$g$$ and $$b$$ has to be a multiple of 5

So, $$b=25$$ and $$g=30$$

So, minimum possible number of students = $$25+30=55$$

Number of ways to choose a sandwich = $$^5C_1$$ ways

Number of ways to choose a bread = $$^4C_1$$ ways

Number of ways to choose bread size = $$^2C_1$$ ways

Number of ways to choose sauces = $$^6C_0+^6C_1+^6C_2=1+6+15=22$$ ways

So, number of different ways = $$^5C_1\times\ ^4C_1\times\ ^2C_1\times\ 22=5\times\ 4\times\ 2\times\ 22=880$$ ways.

The sum of all the elements in the given set = Sum of first 29 odd numbers = $$29^2$$ = 841

Let's assume that k is the $$m_{th}$$ term. Sum of terms less than k = sum of first (m-1) odd numbers = $$(m-1)^2$$ 

$$841-m^2=(m-1)^2$$

$$841 - m^2 = m^2 - 2m + 1$$

$$840 - 2m^2 + 2m = 0$$

$$m^2 - m - 420 = 0$$

$$(m - 21)(m + 20) = 0$$

m = 21 -20

m = 20. And the 20th term is 2*m+1 = 41

Let's assume that Total work = 1

If the work is entirely carried by each one of them it would cost 

Arun: $$2160 \times 24 = 51840$$

Varun: $$2400 \times 21 = 50400$$

Tarun: $$2160 \times 15 = 32400$$

The cheapest worker is Tarun. He can finish the work in 15 days. But we need the work completed in 10 days.

Tarun does $$10 \times \dfrac{1}{15} = \dfrac{2}{3}$$

$$1 - \dfrac{2}{3} = \dfrac{1}{3}$$

Complete the remaining (1/3) work using the next cheapest worker, which is Varun.

Days needed: $$\dfrac{x}{21} = \dfrac{1}{3} \quad\Rightarrow\quad x = 7$$

Total cost = Amount paid for Tarun + Amount paid for Varun = $$10 \times 2160 + 7 \times 2400 = 38400$$

Let the amounts invested in Stocks be S, Bonds B, and, Gold be G

Given that  $$S + B + G = 100000, B = 0.25G$$

$$0.10S + 0.06B + 0.08G = 8200$$

Substitute S = 100000 - B - G = 100000 - 1.25G

 $$0.10(100000 - 1.25G) + 0.06(0.25G) + 0.08G = 8200$$

$$10000 - 0.125G + 0.015G + 0.08G = 8200$$

$$10000 - 0.03G = 8200$$

$$ 0.03G = 1800 \Rightarrow G = 60000$$

$$B = 0.25G = 15000$$

Gain from bonds = $$0.06 \times 15000 = 900$$

Given, $$a-6b+6c=4$$ --->(1)

and, $$6a+3b-3c=50$$ ---->(2)

Multiplying eqn (1) with $$x$$, $$ax-6bx+6cx=4x$$

And multiplying eqn (2) with $$y$$, $$6ay+3by-3cy=50y$$

So, $$x+6y=2$$ and $$3y-6x=3$$

So, $$x+6y=2$$ ---->(3) and $$-2x+y=1$$ ---->(4)

Multiplying eqn (4) with 6 and subtracting from eqn (3),

$$13x=-4$$

So, $$x=-\dfrac{4}{13}$$

Putting the value of $$x$$ in equation (4),

$$y=\dfrac{5}{13}$$

So, the final answer is $$4x+5y$$

=$$4\left(-\frac{4}{13}\right)+50\left(\frac{5}{13}\right)=-\frac{16}{13}+\frac{250}{13}=\frac{234}{13}=18$$

So, correct answer is $$18$$.

Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).

For parallelogram (PQRS), S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5,4)

Diagonal SQ passes through S(5,4) and Q(1,-5).

Slope m = $$\dfrac{-5-4}{1-5} = \dfrac{9}{4}$$

Equation of SQ is  $$y - 4 = \dfrac{9}{4}(x - 5)$$

At the x-axis, y = 0: $$-4 = \dfrac{9}{4}(a - 5)$$

$$-16 = 9(a - 5)$$

$$9a = 29$$

$$a = \dfrac{29}{9}$$

So, $$a = {\dfrac{29}{9}}$$

The radius of the circle is given to be $$6\sqrt{2}$$ cm. We are also given that the ratio of CA to CB is 3:2. So, if we assume the value of CA to be 3x, then the value of CB becomes 2x. We are also given that $$\angle PQC=45^{0}$$ and since PCQ is an isosceles triangle, $$\angle CPQ=45^{0}$$ and $$\angle PCQ=90^{0}$$.

So, the triangle PCQ is right-angled at C. The value of PQ can be calculated using Pythagoras' theorem as,

$$PQ^2\ =\ PC^2\ +\ QC^2$$

$$PQ^2\ =\ \left(6\sqrt{\ 2}\right)^2\ +\ \left(6\sqrt{\ 2}\right)^2\ =\ 72\ +\ 72\ =\ 144$$

$$PQ\ =\ 12$$ cm

$$AQ\ =\ \dfrac{PQ}{2}\ =\ \dfrac{12}{2}\ =\ 6$$ cm

In triangle ACQ,

$$AC^2\ +\ AQ^2\ =\ CQ^2$$

$$\left(3x\right)^2\ +\ \left(6\right)^2\ =\ \left(6\sqrt{\ 2}\right)^2$$

$$9x^2\ +\ 36\ =\ 72$$

$$9x^2\ =\ 36$$

$$x^2\ =\ 4$$

$$x\ =\ 2$$

$$CB\ =\ 2x\ =\ 2\times2\ =\ 4$$ cm

In triangle BCS,

$$BC^2\ +\ BS^2\ =\ CS^2$$

$$\left(4\right)^2\ +\ \left(BS\right)^2\ =\ \left(6\sqrt{\ 2}\right)^2$$

$$16\ +\ \left(BS\right)^2\ =\ 72$$

$$BS^2\ =\ 56$$

$$BS\ =\ 2\sqrt{\ 14}$$ cm

$$RS\ =\ 2\times\ BS\ =\ 2\times\ 2\sqrt{\ 14}\ =\ 4\sqrt{\ 14}$$ cm

The area of the quadrilateral PQRS can be calculated as,

$$Area=\ \dfrac{1}{2}\times\ AB\ \times\ \left(PQ\ +\ RS\right)\ =\ \dfrac{1}{2}\times\ 10\left(12\ +\ 4\sqrt{\ 14}\right)\ =\ 20\left(3\ +\ \sqrt{\ 14}\right)cm^2$$

Hence, the correct answer is option C.

Let M=13k, A=9k. 

After 21 students were moved $$\frac{13k-21}{9k+21}=\frac{19}{14}\Rightarrow 14(13k-21)=19(9k+21)$$

So $$182k-294=171k+399\Rightarrow 11k=693\Rightarrow k=63$$

So, the number of students in the morning and afternoon shifts are $$819-21=798, 567+21=588$$ respectively.

Let's assume that $$3t$$ and $$8t$$ students joined the respective sessions. 

$$\dfrac{798+3t}{588+8t}=\dfrac{5}{4}\Rightarrow 4(798+3t)=5(588+8t)$$

So $$3192+12t=2940+40t\Rightarrow 28t=252\Rightarrow t=9$$

Number of new students $$=11t=99$$

Let the diagonals of the rhombus be $$d_1$$ and $$d_2$$

We know that the $$\text{Area} = \frac{1}{2} d_1 d_2 = 396 \implies d_1 d_2 = 792$$

For a rhombus with side a=36, the diagonals intersect at right angle. Giving a right-angle triangle with the side as hypotenuse.

$$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \implies \frac{d_1^2 + d_2^2}{4} = 36^2 = 1296$$

$$d_1^2 + d_2^2 = 5184$$

We want $$|d_1 - d_2|$$ 

$$(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2 = 5184 - 2 \cdot 792 = 5184 - 1584 = 3600$$

$$|d_1 - d_2| = \sqrt{3600} = 60$$

The given quadratic equation is $$x^2-5x+k=0$$

Now, discriminant $$D=5^2-4k=25-4k$$

Now, it is given the equation must have integer roots.

So, $$25-4k$$ has to be a perfect square.

We need to find non-negative integer values of $$k$$

Now, for $$k=0$$, $$D=25-4\times\ 0=25$$, is a perfect square

For $$k=4$$, $$D=25-4\times4=25-16=9$$, is a perfect square

For $$k=6$$, $$D=25-4\times\ 6=1$$, is a perfect square

So, there are three non negative integer values of $$k$$.

So, correct answer is $$3$$.

Cost price of each chair = 100

For 13 chairs, total cost = $$13 \times 100 = 1300$$

Profit = 26%, so total revenue

$$ 1.26 \times 1300 = 1638$$

We were told that this amount is equal to the discounted price of 12 chairs. So the discounted selling price per chair = $$\text{SP}_{\text{disc}} = \frac{1638}{12} = 136.5$$

Discount offered = 22%, so:

MP = $$\frac{136.5}{0.78}$$ = 175