CAT 2025 Slot 2 Quant Question Paper

Let's assume that $$x^{2}+2x-3 = t$$ 

$$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$$ can be written as $$9^t-4(3^{t+1})+27=0$$

$$3^{2t}-12(3^t)+27=0$$

Let's assume that $$3^t = y$$

$$y^2-12y+27=0$$

$$(y-9)(y-3)=0$$

y = 3 or 9 $$\Rightarrow$$ t = 1 or 2

Let's solve when t = 1

$$x^{2}+2x-3 = 1 \Rightarrow x^{2}+2x-4 = 0$$

$$b^2-4ac\ =\ 4+16\ =\ 20$$

Positive, so the equation has real roots.

Product of possible value of x  = -4

Let's solve for t = 2

$$x^{2}+2x-3 = 2 \Rightarrow x^{2}+2x-5 = 0$$

$$b^2-4ac\ =\ 4+20\ =\ 24$$

Positive, so the equation has real roots.

Product of possible value of x = -5

The product of all values = 20

Let the copies sold on days 1 through 9 be $$d_1,\dots,d_9$$

From the averages:

$$d_1+\cdots+d_7=7\times60=420$$

$$d_1+\cdots+d_8=8\times63=504$$

So $$d_8=504-420=84$$. Then $$d_9=84-11=73$$

The sum from day 2 to day 9 is $$8\times66=528$$

$$d_2+\cdots+d_9=(d_1+\cdots+d_8)-d_1+d_9=504-d_1+73=577-d_1$$

Thus $$577-d_1=528 \Rightarrow d_1=49$$

We are asked to solve

$$x^2 - |x+9| + x > 0$$

Split into two cases based on the absolute value.

Case 1: $$x+9 \ge 0 \Rightarrow x \ge -9$$

$$|x+9| = x+9$$

Inequality becomes: $$x^2 - (x+9) + x > 0 \implies x^2 - 9 > 0 \implies (x-3)(x+3) > 0$$

So $$x<-3  or  x>3$$. Combined with $$x\ge -9$$, we get $$x\in [-9,-3) \cup (3,\infty)$$

Case 2: $$x+9 < 0 \Rightarrow x < -9$$

$$|x+9| = -(x+9) = -x - 9$$

Inequality becomes: $$x^2 - (-x-9) + x > 0 \implies x^2 + 2x + 9 > 0$$

The quadratic $$x^2 + 2x + 9$$ has discriminant (4-36=-32 <0), so always positive. But in this case (x<-9), so inequality is satisfied. Thus (x < -9) also works.

$$x < -3  \text{or}  x>3$$

So the solution set is $${(-\infty,-3) \cup (3,\infty)}$$

Let the marked price be M and the initial discount rate be d. The cost price is 1650. A profit of 20% means the selling price is

$$SP = 1650 \times 1.2 = 1980$$

With discount d,

$$M(1-d) = 1980$$

If the discount is doubled, the selling price becomes (M(1-2d)), and the profit is 110, so

$$M(1-2d) - 1650 = 110 \Rightarrow M(1-2d) = 1760$$

Subtracting the two equations:

$$M(1-d) - M(1-2d) = 1980 - 1760 \Rightarrow Md = 220 \Rightarrow M = \frac{220}{d}$$

Plug into the first equation:

$$ \frac{220}{d}(1-d) = 1980 \Rightarrow \frac{1-d}{d} = 9 \Rightarrow 1-d = 9d \Rightarrow d = 0.1 $$

So the initial discount rate is 10% and the marked price is

$$M = \frac{220}{0.1} = 2200 $$

Now, let the discount rate be (r) such that the profit percentage equals the discount percentage. Then

$$\frac{2200(1-r) - 1650}{1650} = r$$

Simplify:

$$2200(1-r) - 1650 = 1650 r \Rightarrow 2200 - 2200r - 1650 = 1650r \Rightarrow 550 = 3850 r \Rightarrow r = \frac{550}{3850} \approx 0.142857 $$

Thus, the required discount rate is approximately 14%

Sure! We can solve it without explicitly introducing new variables for substitution.

We are given $$(m+2n)(2m+n) = 27$$ and we want to maximize $$2m-3n$$

Because m and n are integers, both (m+2n) and (2m+n) are going to be integers.

Factor pairs of 27 (including negatives) are: $$(1,27),(3,9),(9,3),(27,1),(-1,-27),(-3,-9),(-9,-3),(-27,-1)$$

For each factor pair (a,b), take $$a=m+2n$$, $$b=2m+n$$ and solve for integers (m,n).

If we notice $$a+b=\left(m+2n\right)+\left(2m+n\right)=3\left(m+n\right)$$

So, basically the sum of the two numbers has to be a multiple of $$3$$.

So, the ordered pairs we will consider are $$\left(3,9\right),\left(9,3\right),\left(-3,-9\right),\left(-9,-3\right)$$

For $$(a,b)=(9,3)$$:

$$ m+2n=9,\quad 2m+n=3 $$

Upon solving, we get $$n=5$$ and $$m=-1$$

For $$(a,b)=(3,9)$$:

$$ m+2n=3,\quad 2m+n=9 $$

Upon solving, we get $$n=-1$$ and $$m=5$$

Then $$2m-3n=2\cdot5-3(-1)=10+3=13$$ 

For negative factor pairs, we similarly get integer solutions $$(-5,1)$$ and $$(1,-5)$$, giving $$2m-3n=-13  \text{and}  17$$, respectively.

Thus, the maximum value of $$(2m-3n)$$ among all solutions is 17

$$625=5^4$$ and $$128=2^7$$ 

So, $$(625)^{65},(128)^{36}=5^{260},2^{252}=(2^{252}5^{252})\cdot5^8=10^{252}\cdot5^8$$

Thus, the number equals $$5^8$$ followed by 252 zeros. Now (5^8=390625), and the sum of the digits of (390625) is

$$ 3+9+0+6+2+5=25.$$

Zeros add nothing, so the required sum of digits is 25.

Let's assume that the common root is r. 

The sum of the roots of the first equation is 5/3 and that of the second equation is 1.

We want the sum of the other two roots:

$$ \text{Sum} = \left(\frac{5}{3}-r\right) + (1-r) = \frac{8}{3}-2r$$

We now need to express r in terms of p and q.

Since r is a common root, it satisfies:

$$3r^2 - 5r + p = 0 \quad (1)$$

$$ 2r^2 - 2r + q = 0 \quad (2)$$

Eliminate $$ r^2$$ .

Multiply (2) by 3:

$$ 6r^2 - 6r + 3q = 0$$

Multiply (1) by 2:

$$ 6r^2 - 10r + 2p = 0$$

Subtract:

$$ (6r^2 - 6r + 3q) - (6r^2 - 10r + 2p) = 0$$

$$ 4r + 3q - 2p = 0$$

$$  r = \frac{2p - 3q}{4}$$

Now substitute into $$ \frac{8}{3} - 2r$$ :

$$ \frac{8}{3} - 2\left(\frac{2p - 3q}{4}\right)$$

$$ = \frac{8}{3} - \frac{2p - 3q}{2}$$

$$ = \frac{8}{3} - p + \frac{3}{2}q$$

$$64 = 8^2  \text{and}  512 = 8^3$$

$$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$$,

Property of log: $$\log_{b^m}\ a^{n\ }=\frac{n}{m}\ \log_ba$$

$$\log_{8^2}{x^{2}+\log_{8}{\sqrt{y}+3\log_{8^3}{(\sqrt{y}z)}}}=4$$

Using the above-mentioned property, the expression becomes $$\log_{8}{x}+\log_{8}{\sqrt{y}+\log_{8}{(\sqrt{y}z)}}=4$$

$$\log_8x\sqrt{y}\cdot(\sqrt{y}z)=4$$

$$\log_8xyz=4$$

$$xyz =8^4=2^{12}$$

Using AM-GM inequality

$$\frac{\left(x+y+z\right)}{3}\ge\sqrt[\ 3]{xyz}$$

$$\frac{\left(x+y+z\right)}{3}\ge2^4$$

$$\left(x+y+z\right)\ge48$$

Let the river speed be v km/h. For Sneha still-water speed = 6 km/h:

Upstream speed = (6-v), downstream speed = (6+v).

Given $$\dfrac{14}{(6-v)}-\dfrac{14}{(6+v)}=48$$minutes =0.8 hours. 

So, $$\dfrac{14(6+v-6+v)}{36-v^2}=0.8 \Rightarrow \dfrac{28v}{36-v^2}=0.8$$

$$28v=0.8(36-v^2)=28.8-0.8v^2 \Rightarrow 0.8v^2+28v-28.8=0$$

Multiply by 5: $$4v^2+140v-144=0$$

$$v^2+35v-36=0$$

We get v = 1

Thus, the river speed is 1 km/h. For Rita, the still water speed is 5 km/h:
Upstream speed = 5-1=4 km/h, Downstream Speed = (5+1=6) km/h.

If she rows d km upstream and returns d km downstream, the total time

$$ \dfrac{d}{4}+\dfrac{d}{6}=d\Big(\dfrac{1}{4}+\dfrac{1}{6}\Big)=d\cdot\dfrac{5}{12}\ \text{hours}$$

This equals 100 minutes $$=100/60=\dfrac{5}{3}$$ hours. 

So, $$d\cdot\dfrac{5}{12}=\dfrac{5}{3}\ \Rightarrow\ d=4\ \text{km}$$

Total distance covered 2d=8 km.

Our task is to minimise $$3a+2b+c$$.

Here, the coefficient for $$c$$ is the minimum. 

$$3ac=8(a+b)$$

We know that a, b, and c are natural numbers. So, the product $$ac$$ should definately be a multiple of 8.

Case 1: a = 1, c = 8 and b = 2 $$\Rightarrow$$ 3a+2b+c = 15

Case 2: a = 2, c = 4 and b = 1 $$\Rightarrow$$ 3a+2b+c = 12

So, 12 is the correct answer.

We check where $$f(x)=\frac{x}{2x-1}$$ or $$g(x)=\frac{x}{x-1}$$, or their compositions, become undefined.

First, f(x) is undefined at $$x=\tfrac12$$ and g(x) is undefined at $$x=1$$.

Next, for $$f(g(x)) = \frac{x}{x+1}$$

Since, the denominator can't be zero, x=-1 must also be excluded.

For $$g(f(x)) = \frac{x}{1-x}$$

$$\Rightarrow x=1$$ is not possible, which is already excluded.

So the values at which $$h(x)=f(g(x))+g(f(x))$$ is undefined are $${-1,\ \tfrac12,\ 1}$$

Let the annual interest rate be (r) (in decimal). Discount the two instalments to present value:

$$\dfrac{530}{1+r}+\dfrac{594}{(1+r)^2}=1000$$

Set $$x=\dfrac{1}{1+r}$$. Then

$$594x^2+530x-1000=0$$

Discriminant = $$530^2+4\cdot594\cdot1000=280900+2376000=2656900=1630^2$$.

$$x=\dfrac{\ -b\pm\sqrt{b^2-4ac}}{2a}$$

$$x=\dfrac{-530+1630}{2\cdot594}=\frac{1100}{1188}=\dfrac{275}{297}$$

So $$1+r=\dfrac{297}{275}$$

$$r=\dfrac{297-275}{275}=\dfrac{22}{275}=\dfrac{2}{25}=0.08=8\%$$

We can draw the following diagram as per the given question :

Now, taking $$\triangle AOT$$ and $$\triangle AOQ$$

AO is the common side

AT=AQ (Tangents drawn from an external point are equal in length)

OT=OQ (radius of circle)

So, by S.S.S., $$\triangle\ AOT$$ is congruent to $$\triangle\ AOQ$$

So, by C.P.C.T., $$\angle\ AOT=\angle\ AOQ$$

Similarly, for $$\triangle\ BOT\ $$ and $$\triangle\ BOR\ $$, we can do the same thing,

So, we can say $$\triangle\ BOT$$ and $$\triangle\ BOR$$ will also be congruent

Now, $$\angle\ QOR=\angle\ AOQ+\angle\ AOB+\angle\ ROB=\angle\ AOT+\angle\ AOB+\angle\ BOT=2\angle\ AOB=2\times\ 50^{\circ\ }=100^{\circ\ }$$

Now, in quadrilateral OQPR,

$$\angle\ OQP=\angle\ ORP=90^{\circ\ }$$ (Tangents drawn to a point is perpendicular to the radius drawn at the same point)

Now, sum of angles of a quadrilateral is $$360^{\circ\ }$$

So, $$\angle\ APB+\angle\ QOR=360^{\circ\ }-\left(90^{\circ\ }+90^{\circ\ }\right)=180^{\circ\ }$$

So, $$\angle\ APB=180^{\circ\ }-100^{\circ\ }=80^{\circ\ }$$

The divisors of the given number will have the form $$2^a*3^b*5^c*7^d$$ with $$0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$$

Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0). 

Reduce modulo 3: $$2  \text{mod}  3 = 2, 5  \text{mod}  3 = 2, 7  \text{mod}  3 = 1$$ 

Hence,  $$2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$$

2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even

$$a\in{0,\dots,6}$$ has 4 even, 3 odd values 

$$c\in{0,\dots,3}$$ has 2 even, 2 odd

Number of (a,c) with (a+c) even is $$4\cdot2 + 3\cdot2 = 8+6=14$$

For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $$14\times3=42$$.

Let the side of the hexagon be a. 

The area of the whole hexagon is going to be $$\frac{3\sqrt{3}a^2}{2}$$

We know that the longest diagonal of a hexagon is 2 times the side of the hexagon.

PQ = $$\frac{\left(BC+AD\right)}{2}= \frac{\left(a+2a\right)}{2} =1.5a$$

The area of the trapezium =  Average of parllel sides * Height = $$\frac{\left(BC+PQ\right)}{2}\cdot h=\frac{5}{4}a\cdot h$$

The distance between BC and AD will be $$\sqrt{3}a$$, that is, the length of diagonal BF.

The distance between BC and AD will be $$\frac{\sqrt{3}a}{2}$$ 

And the height of the trapezium BCQP will be  $$\frac{\sqrt{3}a}{4}$$

Area of trapezium = $$\frac{5}{4}a\cdot \frac{\sqrt{3}a}{4}$$ = $$\frac{5\sqrt{3}a^2}{16}$$ 

Area of hexagon ABCDEF = $$\frac{3\sqrt{3}}{2}a^2$$

We have to find the ratio  = $$\frac{5\sqrt{3}a^2}{16}$$ : $$\frac{3\sqrt{3}}{2}a^2$$ = 5:24

Given expression: $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$

The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive.

If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal. 

So, the nearest four values are 12, 11, 11, 12.

With this, the minimum value is $$(12-11)^2+(12-11)^2+(12-12)^2 = 2$$

Let Lakshmi's income = A and expenditure = B.

Let Meenakshi's income = C and expenditure = D.

From B:D = 2:3, take B = 2k and D = 3k.

From A:D = 6:7, $$A = \tfrac{6}{7} D = \tfrac{6}{7}\cdot 3k = \tfrac{18}{7}k$$.

Lakshmi's saving = $$A - B = \tfrac{18}{7}k - 2k = \tfrac{4}{7}k$$.

Let Meenakshi's saving $$= C - D = C - 3k$$. Given the savings ratio $$\frac{\frac{4k}{7}}{C-3k} = \tfrac{4}{9}$$.

Solve: $$\dfrac{4}{7}k\cdot\dfrac{9}{4} = C - 3k \Rightarrow \dfrac{9}{7}k = C - 3k \Rightarrow C = \dfrac{30}{7}k$$

So incomes ratio $$A:C = \dfrac{18}{7}k : \dfrac{30}{7}k = 18:30 = 3:5.$$

Let the first term be a and the common ratio be r.

Given

$$a+a r + a r^2 = 52 \quad\text{and}\quad a^2 r + a^2 r^2 + a^2 r^3 = 624.$$

From the first equation $$a(1+r+r^2)=52$$, so $$a=\dfrac{52}{1+r+r^2}$$. 

Substitute into the second:

$$\frac{52^2}{(1+r+r^2)^2}*(r+r^2+r^3)=624$$

$$\frac{r}{(1+r+r^2)}=\frac{3}{13}$$

$$3r^2-10r+3=0$$

$$(3r-1)(r-3)=0$$

Simplify to get an equation in (r); its real solutions are $$r=\tfrac{1}{3}$$ and $$r=3$$. For an infinite geometric progression, we must have (|r|<1), so $$r=\dfrac{1}{3}$$.

Now
$$a=\frac{52}{1+\tfrac{1}{3}+\tfrac{1}{9}}=\frac{52}{13/9}=36$$

The sum of the infinite progression is

$$ S=\frac{a}{1-r}=\frac{36}{1-\tfrac{1}{3}}=\frac{36}{2/3}=54.$$

Let coffee price = C Rs/kg and cocoa price = K Rs/kg.

From the two given mixtures:

$$0.16C + 0.84K = 240$$

$$0.36C + 0.64K = 320$$

Multiply both equations by 100 to remove decimals:

$$16C + 84K = 24000$$ 

$$36C + 64K = 32000$$

Subtract the first from the second:

$$(36C+64K)-(16C+84K)=32000-24000$$

$$20C - 20K = 8000 \implies C - K = 400$$

Put $$C=K+400$$ into $$16C+84K=24000$$:

$$16(K+400)+84K=24000 $$

$$16K+6400+84K=24000 $$

$$100K = 17600 \implies K = 176$$

So $$C = 176 + 400 = 576$$ (Rs/kg).

For the new mixture priced at Rs 376/kg, let the coffee fraction be p. Then

$$ p\cdot 576 + (1-p)\cdot 176 = 376$$

Upon solving $$p = \tfrac{1}{2}$$

Thus, coffee is (50%) of the new mixture. In 10 kg of this mixture, coffee = $$10\times 0.5 = 5\text{kg}$$

In the question, it is given $$AD:AT=4:3$$

So,$$TD:AT=1:3$$

Let, $$TD=x,\ AT=3x$$

Also, $$BE:BT=5:4$$

So, $$BT:TE=4:1$$

Let, $$BT=4y,\ TE=y$$

Given, DF is parallel to BE

So, DF is parallel to TE

So, $$\triangle\ ATE~\triangle\ ADF$$ (by A.A., as $$\angle\ ATE=\angle\ ADF$$ and $$\angle\ AET=\angle\ AFD$$)

So, $$\dfrac{AT}{AD}=\dfrac{TE}{DF}$$

So, $$\dfrac{3x}{x+3x}=\dfrac{y}{DF}$$

So, $$DF=\dfrac{4y}{3}$$

Also, since DF is parallel to BE,

So, $$\triangle\ CFD~\triangle\ CBE$$ (by A.A., as $$\angle\ CFD=\angle\ CEB$$ and $$\angle\ CDF=\angle\ CBE$$)

So, $$\dfrac{CD}{BC}=\dfrac{DF}{BE}$$

So, $$\dfrac{CD}{BC}=\dfrac{\dfrac{4y}{3}}{5y}=\dfrac{4y}{15y}=\dfrac{4}{15}$$

So, $$\dfrac{CD}{BD}=\dfrac{4}{11}$$

So, $$BD:CD=11:4$$

Let Chandan’s one-day work = x.
Then Bipin = 2x and Ankita = 4x.

All three work together for the first 20 days. Their combined daily work = $$x + 2x + 4x = 7x$$.
Work done in 20 days = $$20\times7x = 140x$$.

Bipin leaves after 20 days; Ankita and Chandan continue for the remaining (60-20=40) days.
Their combined daily work = $$4x + x = 5x$$. Work done in 40 days = $$40\times5x = 200x$$.

Total work = $$140x+200x=340x$$. 

We know that Chandan does x work in a day. So, the number of days it takes him to finish the work is 340.

Answer: 340 days.

Let the total amount be 100 units (taking 100 makes percentage calculations easy).
Pinu receives 20% of the total, so Pinu gets 20 units.
The amount left after giving Pinu his share is 100 - 20 = 80 units.

Meena receives 40% of this remaining amount, so Meena gets 40% of 80 = 32 units.
Now, the amount left for Rinu and Seema together is 80 - 32 = 48 units.

Seema receives 20% less than Pinu. Since Pinu gets 20 units, Seema gets
20 - 20% of 20 = 20 - 4 = 16 units.

Out of the 48 units left for Rinu and Seema, Seema’s share is 16 units, so Rinu gets
48 - 16 = 32 units.

Therefore, Pinu receives 20 units, and Rinu gets 32 units.
The required ratio of the amounts received by Pinu : Rinu = 20 : 32 = 5 : 8.