CAT Algebra Marathon Questions 2026 PDF with Video Solutions

Formula used: Properties of Modulus and Inequalities

$$\log_2\left|x-2\right|\le3\ \ $$
$$\left|x-2\right|\le2^3\ $$
$$-2^3\le x-2\le2^3\ $$
$$-6\le x\le10\ $$   -----> Eq(1)

$$\left|2x-5\right|+\left|x+3\right|>15$$
The critical points occur when the modulus terms are equal to 0.
$$2x-5=0$$ and $$x+3=0$$

$$x=\dfrac{5}{2}$$ and $$x=-3$$

There are three regions in which the inequality can be evaluated:
$$x<-3$$,    $$-3\le x<\dfrac{5}{2}$$,    $$x\ge\dfrac{5}{2}\ $$

Case 1:
$$x<-3$$
$$-\left(2x-5\right)-\left(x+3\right)>15$$

$$x<\dfrac{-13}{3}$$

Using Eq(1) and inequality above, two values of $$x$$ possible: $$x=-5,-6$$

Case 2:
$$-3\le x<\dfrac{5}{2}$$

$$-\left(2x-5\right)+x+3>15$$
$$-x+8>15$$
$$x<-7$$

No solution.

Case 3:
$$x\ge\dfrac{5}{2}\ $$

$$2x-5+x+3>15$$

$$x>\dfrac{17}{3}$$

Using Eq(1) and inequality above, values of $$x$$ possible: $$x=6,7,8,9,10$$

Values of $$x$$ possible: $$-5,-6,6,7,8,9,10$$

Sum of the values$$=-5-6+6+7+8+9+10=29$$

The answer is 29.

We have the following four cases:

Case A.

$$x+y\geq 0$$ and $$x-y<0$$ which give $$x\geq -y$$ and $$x<y$$

Combined with $$(x+y)+(y-x)=16$$ or $$y=8$$, we get the values of $$x$$ in the interval $$[-8,8)$$

Case B.

$$x+y\geq 0$$ and $$x-y\geq 0$$ which give $$x\geq -y$$ and $$x\geq y$$

Combined with $$(x+y)+(x-y)=16$$ or $$x=8$$, we get the only value of $$x$$ satisfying the equation as $$8$$.

Case C.

$$x+y<0$$ and $$x-y<0$$ which give $$x<-y$$ and $$x<y$$

Combined with $$(-x-y)+(y-x)=16$$ or $$x=-8$$, we get the only value of $$x$$ satisfying the equation as $$-8$$.

Case D.

$$x+y<0$$ and $$x-y\geq0$$ which give $$x<-y$$ and $$x\geq y$$

Combined with $$(-x-y)+(x-y)=16$$ or $$y=-8$$, we get the values of $$x$$ satisfying the interval $$[-8,8)$$

Based on the four intervals, we get all the possible values of $$x$$ lying in the interval $$[-8,8]$$

Therefore, the least positive integral value of $$x$$ that does not satisfy any value of $$y$$ would be $$\boxed{9}$$.

We are given the expression that $$||x-1|+x-2|<3$$

First, let's look at the term $$|x-1|$$, from this term we have 2 cases:

Case 1: $$x<1$$

Case 2: $$x \geq 1$$

Now, let's proceed case-wise to get all solutions.

Case 1 : $$x<1$$

$$||x-1|+x-2|<3$$

or $$|-(x-1)+x-2|<3$$

or $$|-x+1+x-2|<3$$

or $$|-1|<3$$

or $$1<3$$ ( Always true)

So, this means that $$||x-1|+x-2|<3$$ whenever $$x<1$$.

Therefore, $$x\in (-\infty,1)$$

Case 2: $$x\geq1$$

$$||x-1|+x-2|<3$$

or $$|x-1+x-2|<3$$

or $$|2x-3|<3$$

Now, let's look at the term $$|2x-3|$$, from this term we again have 2 cases and we will have to proceed casewise to get all solutions.

Case 2.1 : $$1 \leq x<1.5$$

$$|2x-3|<3$$

or $$-(2x-3)<3$$

or $$-2x+3<3$$

Subtracting 3 from both sides of the inequality

$$-2x+3-3<3-3$$

or $$-2x<0$$

Dividing both sides of the inequality by $$-2$$

$$\dfrac{-2x}{-2}>\dfrac{0}{-2}$$

or $$x>0$$

Therefore, from here we are getting that $$x>0$$ when $$1 \leq x<1.5$$.

This means $$||x-1|+x-2|<3$$ is true when $$1 \leq x<1.5$$.

Therefore, $$x\in [1,1.5)$$

Case 2.2 : $$x \geq 1.5$$

$$|2x-3|<3$$

or $$(2x-3)<3$$

Adding 3 both sides of the inequality

$$2x<6$$

or $$x<3$$

Therefore, from here we get that $$x<3$$ when $$x\geq1.5$$.

Therefore, $$x\in [1.5,3)$$

Now, combining all solutions from all cases, we get:

$$x\in (-\infty,1)\cup[1,1.5)\cup[1.5,3)$$

or $$ x \in (-\infty,3)$$

Hence, Option C is correct.

We are given the following inequalities $$y-2x<7$$, $$y+2x<15$$

Now, we can rewrite them as follows:

$$y<2x+7$$,

$$y<-2x+15$$

Now, let's also find the intersection point of the two lines $$y-2x=7$$ and $$y+2x=15$$

$$y-2x=7$$

$$y+2x=15$$

Adding the above equations

$$2y=22$$

or $$y=11$$

Now, $$y+2x=15$$

$$11+2x=15$$

or $$x=2$$

Hence, the coordinates of the intersection point of these two lines ,let's call this point A, are $$=(2,11)$$.

Now, let's plot these inequalities and find the enclosed area

Now, the green region represents the inequality: $$y<2x+7$$

And the blue region represents the inequality $$y<-2x+15$$

The triangle ABC is the enclosed region between $$y-2x<7$$, $$y+2x<15$$ and the $$x$$ axis

Now, the height of triangle ABC is equal to the $$y$$ coordinate of point A and the corresponding base BC is equal to $$3.5+7.5=11$$ units

Hence, the area of triangle is $$=\dfrac{11*11}{2}=60.5$$ square units

Hence, Option A is correct.

$$\text{Win}\ \text{ratio}=$$ $$\dfrac{\text{Number}\ \text{of}\ \text{matches}\ \text{won}}{\text{Total}\ \text{number}\ \text{of}\ \text{matches}\ \text{played}}$$

Let the total number of matches played by the team be $$T$$ and the total number of matches won by the team be $$W$$

Since, the win ratio before the world cup was $$0.8$$,

$$\dfrac{W}{T}=0.8$$

$$W=\dfrac{4}{5}T$$            $$\longrightarrow\ i$$

Now, in the world cup, India played a total of $$10$$ matches and suffered only one defeat.

Total matches played by India = $$(T+10)$$

Total matches won by India = $$(W+9)$$

Win ratio after the world cup = $$\dfrac{W+9}{T+10}$$

Since, the win ratio of India after the world cup was greater than $$0.84$$,

$$\dfrac{W+9}{T+10}>0.84$$

$$100W+900>84T+840$$            $$\longrightarrow\ ii$$

From equation $$i$$ and $$ii$$, we get,

$$80T+900>84T+840$$

$$4T<60$$

$$T<15$$            $$\longrightarrow\ iii$$

Now, from equation $$i$$ and $$iii$$, we can say that, $$T$$ should be less than $$15$$ and a multiple of $$5$$.

Possible values of $$T$$ = $$10,5$$

Possible values of $$W$$ = $$8,4$$ (from equation $$i$$)

Sum of all the possible values of $$W$$ =  $$8+4=12$$

Hence, the sum of all the possible matches won by India before the start of the world cup is 12.

$$\therefore\ $$ The required answer is B.

The sum of the first n odd numbers is $$n^2$$.
The highest square number less than 666 is 625.
So, the number double counted is 666-625 = 41.

The nth term of the series is $$\frac{n(n+1)}{2}$$.
So, sum of the first n terms is $$\frac{n(n+1)(2n+1)}{12}$$ +$$\frac{n(n+1)}{4}$$.
So, sum of the first 50 terms is 22100

Given initially

$$M+O+G = 145$$

After selling the fruits

$$\dfrac{4}{5}M+\dfrac{1}{2}O+\dfrac{1}{3}G = 66$$

Since we can only sell a natural number of fruits, $$G$$ must be a multiple of $$3$$ as well, and $$O$$ must be a multiple of $$2$$, and M must be a multiple of 5.

Thus, $$G$$ must be a multiple of $$75$$ and $$O$$ must be a multiple of $$50$$ ( given that they are multiples of 25 in the question)

Hence, G can be $$75, 150,..$$ while O can be $$50,100,..$$

From the original equation, we can see that $$O+G$$ must be less than $$145$$. Thus the only values of $$O$$ and $$G$$ that satisfy this are $$(50,75)$$

Thus, $$ M = 145 - (75+50) = 20$$

We are given that the pair of linear equations $$ax+6y=13$$ and $$7x+by=7$$ has no solution.

We know that this pair of linear equation has no solutions if:

$$\dfrac{a}{7}=\dfrac{6}{b}\neq\dfrac{13}{7}$$

So $$\dfrac{a}{7}=\dfrac{6}{b}$$

or $$ab=7*6$$

or $$ab=2*3*7$$

Therefore, the number of factors of the product $$ab$$ is$$=(1+1)(1+1)(1+1)=8$$

Therefore, $$k=8$$

Hence, Option C is correct.

The given equation is $$9x + 2y = 9-3xy$$

$$3xy+9x+2y=9$$

In this equation, if we look at the first two terms, we can take $$3x$$ common and we will get $$(y+3)$$.

So, if we can write one more term as $$(y+3)$$, then, we can take $$(y+3)$$ as common from the LHS.

Add $$6$$ on both sides.

$$3xy+9x+2y+6=9+6$$

Take $$3x$$ common from the first two terms and $$2$$ common from the last two terms.

$$3x\left(y+3\right)+2\left(y+3\right)=15$$

$$\left(3x+2\right)\times\left(y+3\right)=15$$

15 on the RHS has to be written in the form of product of two numbers (as we have a product of two numbers on LHS)

$$15=1\times15, 3\times5, -1\times-15, -3\times-5\ \ $$

Now, $$(3x+2)$$ can be either one of $$5$$, $$-5$$, $$3$$, $$-3$$, $$15$$, $$-15$$, $$1$$, $$-1$$.

So, we get the value of $$x$$ as $$1$$, $$\dfrac{-7}{3}$$, $$\dfrac{1}{3}$$, $$\dfrac{-5}{3}$$, $$\dfrac{13}{3}$$, $$\dfrac{-17}{3}$$, $$\dfrac{-1}{3}$$, $$-1$$.

Since, it is given that the solution is a non-negative integer,

We get the value of $$x$$ as $$1$$.

Hence, $$(3x+2)$$ = $$\left(3\times1\right)+2$$ = 5

So, we get the value of $$(y+3)$$ as 3.

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

Hence, $$x=1$$ and $$y=0$$.

$$(x\times y)=(1\times0)=0$$

Option B.

5a + 4b + 6c = 100 --> (1)
and 10a+6b+9c = 180 --> (2)
(2) - 1.5*(1) gives a = 12

The given equation is $$4xy+4x+y=26$$

In this equation, if we look at the first two terms, we can take $$4x$$ common and we will get $$(y+1)$$.

So, if we can write one more term as $$(y+1)$$, then, we can take $$(y+1)$$ as common.

Add $$1$$ on both sides.

$$4xy+4x+y+1=26+1$$

Take $$4x$$ common from the first two terms and 1 common from the next two terms.

$$4x\left(y+1\right)+1\left(y+1\right)=27$$

$$\left(4x+1\right)\times\left(y+1\right)=27$$

27 on the RHS has to be written in the form of product of two numbers (as we have a product of two numbers on LHS)

$$27=1\times27, 3\times9, -1\times-27, -3\times-9\ \ $$

Now, $$(4x+1)$$ can be either one of $$9$$, $$-9$$, $$3$$, $$-3$$, $$27$$, $$-27$$, $$1$$, $$-1$$.

So, we get the value of $$x$$ as $$2$$, $$\dfrac{-10}{4}$$, $$\dfrac{2}{4}$$, $$-1$$, $$\dfrac{26}{4}$$, $$-7$$, $$0$$, $$\dfrac{-2}{4}$$.

Since, it is given that the solution is a positive integer,

We get the value of $$x$$ as $$2$$.

Hence, $$(4x+1)$$ = $$\left(4\times2\right)+1$$ = 9

So, we get the value of $$(y+1)$$ as 3.

$$y=(3-1)=2$$

Hence, $$x=2$$ and $$y=2$$.

$$(x+y)=(2+2)=4$$

Let the fixed part for a group be $$F$$ and the varying constant be $$k$$ for a particular number of tourists $$n$$ in a group, such that the total price for the group becomes $$F+kn^2$$

We have,

$$F+64k = 516000$$ ----------(1)  and

$$(F+100k) + (F+4k) = 876000$$  or  $$2F+104k = 876000$$ --------------(2)

The first equation can also be written as $$2F+128k = 1032000$$ ------------(3)

Subtracting equation (2) from equation (3), we get

$$24k = 156000$$ or $$k= 6500$$. Which gives $$F= 100000$$

Thus, for the total 10 friends which we have to divide into two groups, we have,

$$100000 + 6500n^2 + 100000 + 6500 (12-n)^2$$ as the total amount, this gives

$$200000 + 6500(2n^2+144-24n)$$ or $$200000 + 13000(n^2-12n+72)$$

We need to minimise $$n^2-12n+72$$

The minimum value of this quadratic will occur at $$-\dfrac{(-12)}{2} = 6$$, where $$n=6$$ gives the minimum cost as;

$$200000+13000(36-72+72) = 668000$$

The correct answer is Rs. 6,68,000.

We have $$x^2-x-1=0$$, which gives $$x^2=x+1$$ and $$x^4 = x^2+2x+1$$

Again substituting $$x^2=x+1$$, we get $$x^4=3x+2$$

We have $${\alpha}^4=3\alpha+ 2$$, and $${\beta}^4=3\beta +2$$

Giving $${\alpha}^{4}+{\beta}^{4} = 3(\alpha + \beta)+4 = 3(1)+4 = 7$$

Alternate Solution:

We know that $${\alpha}^2+{\beta}^2 = (\alpha + \beta)^2 - 2\alpha \beta = \dfrac{b^2-2ac}{a^2} = 3$$

Further, $${\alpha}^4 + {\beta}^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2$$

Therefore, $${\alpha}^4 + {\beta}^4 = 9 - 2(-1)^2 = 7$$

$$\dfrac{(x^2+1)^2 +9x^2}{x(x^2+1)}  = \dfrac{x^2 + 1}{x} + \dfrac{9x}{x^2 +1}$$

Let's assume $$  \dfrac{x^2 + 1}{x} = t$$

Now the equation reduces to the form

$$ t+\dfrac{9}{t} $$

To find the minima, we can use the AM-GM inequality. We know that the lowest value AM takes is when the terms are equal.

Thus,

$$ t = \dfrac{9}{t} $$ or $$ t = \pm 3$$

We know that $$ x>1$$ hence $$t$$ cannot be negative and we must solve for

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

$$ x^2 - 3x + 1 =0 $$

Solving using the quadratic formula, we end up with two values of x

$$ x = \dfrac{3\pm\sqrt{5}}{2}$$

We reject $$ x = \dfrac{3-\sqrt{5}}{2}$$ as it's mentioned in the question that x>1.

Thus, $$ x = \dfrac{3+\sqrt{5}}{2}$$

We are given the equation: $$(2-\sqrt3)^{x^2-2x-2}+(2+\sqrt3)^{x^2-2x-2}=4$$

Let us assume that $$a=(2-\sqrt3)^{x^2-2x-2}$$

Then, $$\dfrac{1}{a}=\left(\dfrac{1}{2-\sqrt3}\right)^{x^2-2x-2}=(2+\sqrt3)^{x^2-2x-2}$$

Now, we can rewrite the given equation as:
$$a+\dfrac{1}{a}=4$$

or, $$a^2-4a+1=0$$

or, $$a=\dfrac{4\pm \sqrt{16-4}}{2}$$

or, $$a=2\pm\sqrt3$$

Now, we get $$2$$ cases from here:

CASE 1: $$a=2-\sqrt3$$

or, $$(2-\sqrt3)^{x^2-2x-2}=(2-\sqrt3)^1$$

or, $$x^2-2x-2=1$$

or, $$x^2-2x-3=0$$

From here, we get: $$x=-1$$ or $$x=3$$

CASE 2: $$a=2+\sqrt3$$

or, $$(2-\sqrt3)^{x^2-2x-2}=(2+\sqrt3)^1=(2-\sqrt3)^{-1}$$

or, $$x^2-2x-2=-1$$

or, $$x^2-2x-1=0$$

From here, we get: $$x=1+\sqrt2$$ or $$x=1-\sqrt2$$

Hence, the sum of all possible real values of $$x$$ is $$=-1+3+1+\sqrt2+1-\sqrt{2}=4$$

$$f(x)=kx\ ⇒\ x^2+px+q=kx\ ⇒\ x^2+(p−k)x+q=0$$

Now, for the real roots, $$D\ge0$$

So, $$(p−k)^2−4q\ge0$$

So, this must be true for every real value of k.

So, $$(p−k)^2−4q\ge0\ \ ∀\ \ \ k∈R$$

$$(p−k)^2\ge4q\ \ ∀\ \ k$$

$$(p−k)^2$$ is minimum when $$k=p$$

So, the minimum value = 0

So,$$0\ge4q⇒q\le0$$

Hence, Option A is correct.

Alternative Method:

$$x^2 + (p-k)x + q = 0$$

$$D_x = (p-k)^2 - 4q \ge 0$$

$$(p^2 - 2pk + k^2) - 4q \ge 0$$

$$k^2 - 2pk + (p^2 - 4q) \ge 0$$

For a quadratic $$Ak^2 + Bk + C$$ to be $$\ge 0$$ for all real $$k$$:

The leading coefficient must be positive, and the quadratic must have at most one real root. This means its discriminant ($$D_k$$) must be less than or equal to zero.

Here: $$A=1, B=-2p, C=p^2-4q$$:

$$D_k = B^2 - 4AC \le 0$$

$$(-2p)^2 - 4(1)(p^2 - 4q) \le 0$$

$$4p^2 - 4p^2 + 16q \le 0$$

$$16q \le 0$$

$$q \le 0$$

$$\log_2\left(x-1\right)+\log_4\left(x+3\right)=\dfrac{3}{2}+\log_43$$

$$\log_2\left(x-1\right)+\dfrac{\log_2\left(x+3\right)}{2}=\dfrac{3}{2}+\dfrac{\log_23}{2}$$

$$2\log_2\left(x-1\right)+\log_2\left(x+3\right)=3\log_22+\log_23$$

$$\log_2\left(x-1\right)^2\left(x+3\right)=\log_224$$

$$(x^2-2x+1)(x+3)=24$$

$$x^3+x^2-5x-21=0$$

$$x=3$$ satisfies the above equation.

$$(x-3)(x^2+4x+7)=0$$

$$(x-3)[(x+2)^2+3]=0$$

Thus, the values of x that satisfies the above equation is $$x=3$$.

For a number N, the number of digits $$=\left[\log_{10}N\right]+1$$

$$A=\left[\log_{10}12^{12}+1\right]=\left[12\left(2\log_{10}2+\log_{10}3\right)+1\right]=\left[12\times1.08+1\right]=13$$

$$B=\left[\log_{10}18^{12}+1\right]=\left[12\left(\log_{10}2+2\log_{10}3\right)+1\right]=\left[12\times1.26+1\right]=16$$

Therefore, $$A+B=13+16=29$$

We are given that $$log_3(x^5)+log_{(x^3)}(27)=6$$

or $$log_3(x^5)+log_{(x^3)}(27)=6$$

or $$log_3(x^5)+log_{(x^3)}(3^3)=6$$

or 5$$log_3x+log_{x}3=6$$

or 5$$log_3x+\dfrac{1}{log_{3}x}=6$$

Let $$a=log_3x$$.

so,  $$5a+\dfrac{1}{a}=6$$

or $$5a^2-6a+1=0$$

On solving, we get $$a=\dfrac{1}{5}$$ or $$a=1$$

$$log_3x=\dfrac{1}{5}$$ or $$log_3x=1$$

$$x=3^{-5}$$ or $$x=3$$ and in both cases $$x$$ is positive and not equal to $$1$$.

Therefore, we have $$2$$ solutions for $$x$$

Hence, Option B is correct.

Let us assume that $$a=2-\sqrt3$$

Then, $$\dfrac{1}{a}=\dfrac{1}{2-\sqrt3}=2+\sqrt3$$

So, the given equation can be rewritten as:

$$log_{a}(x+9)+log_{\left(\dfrac{1}{a}\right)}(2x+7)=0$$

or, $$log_{a}(x+9)-log_{a}(2x+7)=0$$

or, $$log_{a}\dfrac{x+9}{2x+7}=0$$

or, $$x+9=2x+7$$

$$x=2$$

Hence, Option A is correct.