Quantitative Aptitude: Cubic Equations Set 1

With the changing pattern in exam, One should be ready for each type of questions. So we are providing you with questions on Cubic Equations.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly –

  1. I. x3 + 7x2 + 16x + 12 = 0,
    II. y3 – 2y2 – 5y + 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option D
    Solution:

    In these equations, you will have to first satisfy the equation with some root.
    Like start with 1: Put x = 1 in x3 + 7x2 + 16x + 12 = 0 and see if it equals to 0. It does not. Next move on to -1, again not, next put 2 and then -2
    -2 satisfies the equation: -8 + 28 – 32 + 12 = 0. Since -2 satisfies the equation so (x+2) is a factor of this equation.
    So x = -2, -3
    Next divide x3 + 7x2 + 16x + 12 = 0 by (x+2) . It gives x2 + 5x + 6 which is a quadratic equation. Its roots are -2 and -3
    Now same with y3 – 2y2 – 5y + 6 = 0
    Putting y = 1, satisfies the equation, so (y-1) is a factor of this equation> Now divide y3 – 2y2 – 5y + 6 = 0 by (y-1) gives y2 – y – 6 = 0 which is a quadratic equation with roots -2 and 3
    So y = 1, -2, 3
    Now put on number line
    -3……-2…..1……3
    so x ≤ y
  2. I. x3 – x2 – 2x = 0,
    II. y3 – 7y – 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x3 – x2 – 2x = 0 becomes
    x (x2 – x – 2) = 0
    So x = 0 and (x2 – x – 2) = 0
    Second is a quadratic equation with roots -1 and 2
    So x = 0, -1, 2
    Now y3 – 7y – 6 = 0
    Put y = 1 – Does not satisfy.
    Put y = -1 – Satisfies the equation so (y+1) is a factor
    Divide y3 – 7y – 6 = 0 by (y+1) gives y2 – y – 6 = 0
    So y = -1, -2, 3
    Put on number line
    -2……-1..…0……2……3 so no relation
  3. I. x3 – 6x2 + 11x – 6 = 0,
    II. y3 + 9y2 + 26y + 24 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option A
    Solution:

    For x3 – 6x2 + 11x – 6 = 0, x = 1 satisfies the equation so (x-1) is a factor
    Divide x3 – 6x2 + 11x – 6 = 0 by (x-1) => gives x2 – 5x + 6 = 0
    Gives x = 1, 2, 3
    Similarly for second equation, y = -2, -3, -4
    put on number line
    So x > y
  4. I. x3 + x2 – 4x – 4 = 0,
    II. y3 – 7y2 + 14y – 8 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -1 satisfies x3 + x2 – 4x – 4 = 0
    So (x+1) is a factor. On dividing gives x2 – 4 = 0
    So x = -1, 2, -2
    Similarly for second equation
    y = 1, 2, 4
  5. I. 3x3 – 13x2 + 18x – 8 = 0,
    II. 3y3 + 14y2 – 32 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = 1 satisfies 3x3 – 13x2 + 18x – 8 = 0 so (x-1) is a factor. On dividing gives 3x2 – 10x + 8 = 0 which is quadratic equation with roots x = 2, 4/3
    So x = 1, 2, 4/3
    Similarly with second equation, y = -2, -4, 4/3
  6. I. 2x3 + 15x2 + 13x – 30 = 0,
    II. 2y3 + 22y2 + 5y – 18 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -6, -5/2, 1
    y = -9/2, -2, 1
    Put on number line
    -6 … -9/2 …. -5/2 …. -2…. 1
  7. I. 3x3 + 25x2 + 68x + 60 = 0,
    II. 3y3 + 5y2 – 4y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option D
    Solution:

    x = -10/3, -3, -2
    y = -2, -2/3, 1
    Put on number line
    -10/3….-3…. -2…. -2/3….1
  8. I. x3 + 2x2 – 19x – 20 = 0,
    II. 2y3 + 15y2 + 28y + 15 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -5, -1, 4
    y = -5, -3/2, -1
    Put on number line
    -5…. -3/2…-1….. 4
  9. I. 5x3 – 12x2 + x + 6 = 0,
    II. 5y3 + 28y2 + 35y + 12 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option C
    Solution:

    x = -3/5, 1, 2
    y = -4, -1, -3/5
    Put on number line
    -4…..-1…. -3/5…1. 2
  10. I. 2x3 – 13x2 + 22x – 8 = 0,
    II. 2y3 + 9y2 + 3y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option C
    Solution:

    x = 2 satisfies 2x3 – 13x2 + 22x – 8 = 0. On dividing gives 2x2 – 9x + 4 = 0
    x = 4 , 1/2, 2
    y = -1 satisfies 2y3 + 9y2 + 3y – 4 = 0. On dividing gives 2y2 + 7y – 4 = 0
    y = -4, 1/2, -1
    Put on number line
    -4……-1….. 1/2……2…. 4

Related posts

15 Thoughts to “Quantitative Aptitude: Cubic Equations Set 1”

  1. Pawan Kandpal

    in question 10 equation 2nd it is 14 or 4 as a constant as i am solving a question i am getting here 14 instead of 4 please check

    1. 4 only

      Put y = -1
      -2 + 9 – 3 – 4 = 0

  2. •?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

    mam ek doubt hai mera…………..jab cubic equations ke root three hain to why we compare only 2 ???

    1. Nhi 2 nhi.
      All 3 compare kiye h

      1. •?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

        ok mam and jab 1 root given ho to bhi 3 ko hi compare karna hai na ???

        1. 1 root given means ?
          I dint get

          1. •?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

            mam ek type ke ques hote hain jinme one root given hota hai and humein remaining 2 find karne hote hain

          2. •?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

            jese mam x^3 – 5 x^2 – 2x + 24/ (x + 2) = 0

          3. Isme 2 lo. x+2 to cancel ho jaega
            And also -2 put karke denominator 0 a ra h to not defined ho jaega

          4. •?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

            ok mam

  3. Its such as you read my mind! You seem to know a lot about this, like you wrote the ebook in it or something. I think that you simply could do with a few to force the message home a bit, but instead of that, this is great blog. An excellent read. I’ll certainly be back.

  4. Usually I don’t read article on blogs, but I would like to say that this write-up very forced me to try and do so! Your writing style has been amazed me. Thanks, quite nice post.

  5. This is one awesome article.Much thanks again. Really Great.

  6. Aw, this was a very good post. Finding the time and actual effort to produce a really good article… but what can I say… I procrastinate a lot and don’t seem to get nearly anything done.

Leave a Comment