CAT Linear Equations Questions & Solutions

A sample of real CAT Linear Equations past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 65 Linear Equations questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is

   Answer: 15

   Show solution

   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.

2. Q2.CAT 2024

   For some constant real numbers p, k and a, consider the following system of linear equations in x and y:
   px - 4y = 2
   3x + ky= a
   A necessary condition for the system to have no solution for (x, y ), is

   • $ap + 6 = 0$

   • $2a + k \neq 0$

   • $ap - 6 = 0$

   • $kp + 12 \neq 0$

   Show solution

   Arranging the equation, we know that there are no solutions when the lines are parallel:
   for that the condition had to  $\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$

   Checking through options: 
   Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

   Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

   Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

   Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

   Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

   Therefore, Option B is the correct answer. 

3. Q3.CAT 2022

   For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is

   Answer: 34

   Show solution

   It is given, y(x + z) = 19

   y cannot be 19. 

   If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

   Therefore, y = 1 and x + z = 19

   It is given, z(x + y) = 51

   z can take values 3 and 17

   Case 1:

   If z = 3, y = 1 and x = 16

   xyz = 3*1*16 = 48

   Case 2:

   If z = 17, y = 1 and x = 2

   xyz = 17*1*2 = 34

   Minimum value xyz can take is 34.

4. Q4.CAT 2021

   If $3x+2\mid y\mid+y=7$ and $x+\mid x \mid+3y=1$ then $x+2y$ is:

   • $-\frac{4}{3}$

   • $\frac{8}{3}$

   • $0$

   • $1$

   Show solution

   We need to check for all regions:

   x >= 0, y >= 0

   x >= 0, y < 0

   x < 0, y >= 0

   x < 0, y < 0

   However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

   Let us start with x >= 0, y >= 0,

   3x + 3y = 7

   2x + 3y = 1

   Hence, x = 6 and y = -11/3

   Since y > = 0, this is not satisfying the set of rules.

   Next, let us test x >= 0, y < 0,

   3x - y = 7

   2x + 3y = 1

   Hence, y = -1

   x = 2.

   This satisfies both the conditions. Hence, this is the correct point.

   WE need the value of x + 2y

   x + 2y = 2 + 2(-1) = 2 - 2 = 0.

5. Q5.CAT 2020

   In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹ 150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹ 450 on rice in April, then how much did he spend on wheat in May?

   • Rs.560

   • Rs.570

   • Rs.590

   • Rs.580

   Show solution

   Let John buy "m" kg of rice and "p" kg of wheat.

   Now let the price of rice be "r" in April. Price in May will be "1.2(r)"

   Now let the price of wheat be "w" in April . Price in April will be "1.12(w)".

   Now he spent ₹150 more in May , so 0.2(rm)+0.12(wp)=150

   Its also given that he had spent ₹450 on rice in April. So (rm)=450 

   So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60/0.12 or (wp)=500

   Amount spent on wheat in May will be 1.12(500)=₹560

6. Q6.CAT 2017

   The number of solutions $(x, y, z)$ to the equation $x - y - z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

   • 101

   • 99

   • 87

   • 105

   Show solution

   x - y - z = 25 and  $x\leq40,y\leq12$ ,  $z\leq12$
   If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
   If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
   If x = 37 then y + z = 12 which will give 11 solutions.
   Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
   Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
   = 12*13/2 + 21
   78 + 21 = 99

7. Q7.CAT 2006

   When you reverse the digits of the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?

   • 5

   • 6

   • 7

   • 8

   • 10

   Show solution

   Let the number be xy
   10y + x = 10x + y + 18
   => 9y - 9x = 18
   => y - x = 2
   So, y can take values from 9 to 4 (since 3 is already counted in 13)
   Number of possible values = 6

8. Q8.CAT 2004

   In NutsAndBolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts.

   Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?

   • 130 minutes

   • 135 minutes

   • 170 minutes

   • 180 minutes

   Show solution

   Machine A takes 15 min to produce 1000 nuts with clean time. machine b takes 30 min to make 1500 nuts with clean time . So B is slower. So with B 900 nuts will be made in 180 mins but at last round cleaning time of 10 min no need to count hence 170 mins

9. Q9.CAT 2003

   A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong answer, and -1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

   • 6

   • 12

   • 3

   • 9

   Show solution

   Let the number of questions answered correctly be x and the number of questions answered wrongly be y.

   So, number of questions left unattempted = (50-x-y)

   So, x - y/3 - (50-x-y)/6 = 32

   => 6x - 2y - 50 + x + y = 192 => 7x - y = 242 => y = 7x - 242

   If x = 35, y = 3

   If x = 36, y = 10

   So, min. value of y is 3.

   The number of wrongly answered questions cannot be less than 3.

10. Q10.CAT 2002

    A car rental agency has the following terms. If a car is rented for 5 hr or less, then, the charge is Rs. 60 per hour or Rs. 12 per kilometre whichever is more. On the other hand, if the car is rented for more than 5 hr, the charge is Rs. 50 per hour or Rs. 7.50 per kilometre whichever is more. Akil rented a car from this agency, drove it for 30 km and ended up playing Rs. 300. For how many hours did he rent the car?

    • 4 hr

    • 5 hr

    • 6 hr

    • None of these

    Show solution

    Suppose Akil drove the car for less than 5 hrs. In this case, by distance basis, Rs 360 should be charged. This is not the case.

    So he dove for more than 5 hrs. Cost comes more using time basis; which is Rs 300, i.e. he used the car for 6 hours.

11. Q11.CAT 2001

    A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is

    • 100

    • 140

    • 60

    • 150

    Show solution

    Let the number of coins of the three denominations be x, y and z respectively.
    x+y+z = 300  
    x+2y+5z = 960
    2x+y+5z = 920
    => 3(x+y) + 10z = 1880
    => 3(300 - z) + 10z = 1880
    => 900 + 7z = 1880 => z = 980/7 = 140
    So, the number of 5 rupee coins is 140

12. Q12.CAT 1996

    The points of intersection of three lines $2x+3y-5=0, 5x-7y+2=0$ and $9x-5y-4=0$

    • form a triangle

    • are on lines perpendicular to each other

    • are on lines parallel to each other

    • are coincident

    Show solution

    For points to be coincident, value of determinant should not be equal zero, so that they have a unique value of system.
    

    Here value of determinant is not equal to zero, simultaneously not any two lines are parallel or perpendicular.
    

    So system has a unique value
    

    Hence points are coincident.

CAT Linear Equations previous year questions with solutions — AthenaPrep