CAT Algebra Questions & Solutions

A sample of real CAT Algebra past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 396 Algebra questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   A value of $c$ for which the minimum value of $f(x)=x^{2}-4cx+8c$ is greater than the maximum value of $g(x)=-x^{2}+3cx-2c$, is

   • $2$

   • $\dfrac{1}{2}$

   • $-\dfrac{1}{2}$

   • $-2$

   Show solution

   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  $\dfrac{1}{2}$

2. Q2.CAT 2024

   The sum of the infinite series $\cfrac{1}{5}\left(\cfrac{1}{5} - \cfrac{1}{7}\right) + \left(\cfrac{1}{5}\right)^2 \left(\left(\cfrac{1}{5}\right)^2 - \left(\cfrac{1}{7}\right)^2\right) + \left(\cfrac{1}{5}\right)^3 \left(\left(\cfrac{1}{5}\right)^3 - \left(\cfrac{1}{7}\right)^3\right) + ......$ is equal to

   • $\cfrac{7}{816}$

   • $\cfrac{5}{408}$

   • $\cfrac{7}{408}$

   • $\cfrac{5}{816}$

   Show solution

   Opening the brackets, we get the series as:  $\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^4-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2+\left(\dfrac{1}{5}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6+...$

   These are two infinite GPs when rearranged:
   $\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-...$

   The sum of the first series would be  $\dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}}=\dfrac{1}{24}$

   The sum of the second series would be  $\frac{\dfrac{1}{35}}{1-\dfrac{1}{35}}=\dfrac{1}{34}$

   The answer to the given series would then be  $\dfrac{1}{24}-\dfrac{1}{34}=\dfrac{10}{816}=\dfrac{5}{408}$

   Therefore, Option B is the correct answer. 

3. Q3.CAT 2023

   A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

   Answer: 340

   Show solution

   Let us assume the initial stock of all the fruits is S.

   Let us take we have 'b' and 'a' mangoes initially.

   Stock of Mangoes = 40% of S = 2S/5

   The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

   = 2S/10 + 96 + 4a/10 = S/2 (Given)

   => S/5 + 96 + 2a/5 = S/2

   => S = $\dfrac{\left(4a+960\right)}{3}$

   => $\dfrac{4a}{3}+320$

   'a' has to be a multiple of 3 for the above term to be an integer.

   But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

   => The smallest value of 'a' satisfying both conditions is 15.

   => $\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$ = 340

4. Q4.CAT 2022

   The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is

   Answer: 548

   Show solution

   General term = 38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

   Each term is in the form of 17k + 4

   Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106

   Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990

   106, 123, 140,..........., 990

   990 = 106 + 17(n-1)

   n = 53

   Sum =  $\frac{53}{2}\left(106+990\right)=53\times548$

   Average =  $53\times\frac{548}{53}=548$

5. Q5.CAT 2020

   The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals

   Answer: 1

   Show solution

   Let $a=x+\frac{1}{x}$
   So, the given equation is $a^2-3a+2=0$
   So, $a$ can be either 2 or 1.
   
   If $a=1$ , $x+\frac{1}{x}=1$ and it has no real roots. 
   If $a=2$ , $x+\frac{1}{x}=2$ and it has exactly one real root which is $x=1$

   So, the total number of distinct real roots of the given equation is 1

6. Q6.CAT 2019

   The number of common terms in the two sequences: 15, 19, 23, 27, . . . . , 415 and 14, 19, 24, 29, . . . , 464 is

   • 21

   • 20

   • 18

   • 19

   Show solution

   A: 15, 19, 23, 27, . . . . , 415

   B: 14, 19, 24, 29, . . . , 464

   Here the first common term = 19

   Common difference = LCM of 5, 4=20

   19+(n-1)20  \le\ 415

   (n-1)20  \le\ 396

   (n-1) \le\ 19.8

   n=20

7. Q7.CAT 2017

   If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b - 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is

   • 9:4

   • 81:4

   • 1:4

   • 25:4

   Show solution

   Since the square root can be positive or negative we will get two cases for each of the equation.

   For the first one,

   a + 3 = 3b .. i

   a + 3 = -3b ... ii

   For the second one,
   

   a - 1 = 2(b -1) ... iii

   a - 1 = 2 (1 - b) ... iv

   we have to solve i and iii, i and iv, ii and iii, ii and iv.

   Solving i and iii,

   a + 3 = 3b and a = 2b  - 1, solving, we get a = 3 and b = 2, which is not what we want.

   Solving i and iv

   a + 3 = 3b and a = 3 - 2b, solving, we get b = 1.2, which is not possible.

   Solving ii and iii

   a + 3 = -3b and a = 2b - 1, solving, we get b = 0.4, which is not possible.

   Solving ii and iv,

   a + 3 = -3b and a = 3 - 2b, solving, we get a = 15 and b = -6 which is what we want.

   Thus, $\frac{a^2}{b^2} = \frac{25}{4}$

8. Q8.CAT 2007

   A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?

   • -119

   • -159

   • -110

   • -180

   • -105

   Show solution

   Let the function be $ax^2 + bx + c$ .

   We know that x=0 value is 1 so c=1.

   So equation is $ax^2 + bx + 1$ .

   Now max value is 3 at x = 1.

   So after substituting we get a + b = 2.

   If f(x) attains a maximum at 'a' then the differential of f(x) at x=a, that is, f'(a)=0.

   So in this question f'(1)=0

   => 2*(1)*a+b = 0
   

   => 2a+b = 0.

   Solving the equations we get a=-2 and b=4.

   $-2x^2 + 4x + 1$ is the equation and on substituting x=10, we get -159.

9. Q9.CAT 2004

   Let $u = ({\log_2 x})^2 - 6 {\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has

   • no solution for x

   • exactly one solution for x

   • exactly two distinct solutions for x

   • exactly three distinct solutions for x

   Show solution

   $x^u = 256$

   Taking log to the base 2 on both the sides, 

   $u * \log_{2}{x} = \log_{2}{256}$

   => $[({\log_2 x})^2 - 6 {\log_2 x} + 12] * \log_{2}{x} = 8$

   $(log_2 x)^3 - 6(log_2 x)^2 + 12log_2 x = 8$

   Let $log_2 x = t$

   $t^3 - 6t^2 +12t - 8 = 0$

   $(t-2)^3 = 0$

   Therefore, $log_2 x = 2$  

   => $x = 4$ is the only solution
   

   Hence, option B is the correct answer.

10. Q10.CAT 2002

    Let S denotes the infinite sum $2 + 5x + 9x^2 + 14x^3 + 20x^4 + ...$ , where |x| < 1 and the coefficient of $x^{n - 1}$ is n( n + 3 )/2 , ( n = 1, 2 , . . . ) . Then S equals:

    • $(2-x)/(1-x)^3$

    • $(2-x)/(1+x)^3$

    • $(2+x)/(1-x)^3$

    • $(2+x)/(1+x)^3$

    Show solution

    Let  $S = 2+5x+9x^2+....$
    $S*x = 2x+5x^2+9x^3+...$
    $S(1-x) = 2+3x+4x^2+...$
    $S(1-x)*x = 2x+3x^2+4x^3+...$
    $S(1-x)(1-x) = 2+x+x^2+x^3+... = 2+x/(1-x)$
    So, S = [2(1-x) + x]/(1-x)^3 =&gt; S = (2-x)/(1-x)^3

11. Passage

    Directions for the next 2 questions:

    For real numbers x, y, let

    f(x, y) = Positive square-root of (x + y), if $(x + y)^{0.5}$ is real
    f(x, y) = $(x + y)^2$; otherwise

    g(x, y) = $(x + y)^2$, if $\sqrt{(x + y)}$ is real
    g(x, y) = $- (x + y)$ otherwise

    Q11.CAT 2000

    Under which of the following conditions is f(x, y) necessarily greater than g(x, y)?

    • Both x and y are less than -1

    • Both x and y are positive

    • Both x and y are negative

    • y>x

    Show solution

    When both x and y are less than -1, g(x,y) > 2 and f(x,y) > 4 and $f(x,y) = g(x,y)^2$
    So, f(x,y) > g(x,y)
    

12. Q12.CAT 1996

    Given the quadratic equation $x^2 - (A - 3)x - (A - 2)$, for what value of $A$ will the sum of the squares of the roots be zero?

    • -2

    • 3

    • 6

    • None of these

    Show solution

    For summation of square of roots to be zero, individual roots should be zero.
    Hence summation should be zero i.e.  A-3=0 ; A = 3
    And product of roots will also be zero i.e. A-2 = 0 ; A =2
    So there is no unique value of A which can satisfy above equation.

CAT Algebra previous year questions with solutions — AthenaPrep