Skip to content

Quant · free practice

Linear and Quadratic Equations questions for IPMAT, with solutions

10 free Linear and Quadratic Equations practice questions for IPMAT Indore, Rohtak and JIPMAT, 2 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Linear and Quadratic Equations on Learn.

  1. Question 1Type the answerWritten by IPMATwayExponential Equations

    If 3x−1+3x+1=903^{x-1}+3^{x+1}=90, then xx equals

    Show the answer and solution

    Answer: 3

    Factor 3x−13^{x-1}: 3x−1(1+32)=903^{x-1}(1+3^{2})=90, so 3x−1=9=323^{x-1}=9=3^{2} and x=3x=3.

  2. Question 2Type the answerWritten by IPMATwayExponential Equations

    If 5x+5x+1=7505^{x}+5^{x+1}=750, then xx equals

    Show the answer and solution

    Answer: 3

    Factor 5x5^{x}: 5x(1+5)=7505^{x}(1+5)=750, so 5x=125=535^{x}=125=5^{3} and x=3x=3.

  3. Question 3Type the answerWritten by IPMATwayHigher-Degree Equations

    The product of all real roots of x4−7x2+12=0x^{4}-7x^{2}+12=0 is

    Show the answer and solution

    Answer: 12

    (x2−3)(x2−4)=0(x^{2}-3)(x^{2}-4)=0 gives x=±3x=\pm\sqrt{3} and x=±2x=\pm2; the product is (−3)(−4)=12(-3)(-4)=12.

  4. Question 4Multiple choiceWritten by IPMATwayLinear Equations

    The system x+2y=3x+2y=3 and 2x+4y=72x+4y=7 has

    1. A.exactly two solutions
    2. B.a unique solution
    3. C.no solution
    4. D.infinitely many solutions
    Show the answer and solution

    Answer: C. no solution

    2x+4y=2(x+2y)=62x+4y=2(x+2y)=6 for every solution of the first equation, which contradicts 2x+4y=72x+4y=7. No solution.

  5. Question 5Type the answerWritten by IPMATwayLinear Equations

    A two-digit number is such that six times the tens digit added to ten times the units digit equals the number itself. The number is

    Show the answer and solution

    Answer: 94

    4t=9u4t=9u with digits forces u=4u=4, t=9t=9 (the next multiple would need t=18t=18). The number is 9494.

  6. Question 6Multiple choiceAsked in JIPMAT 2023Quadratic Equations

    Statement (I): (x2+3x+1)=(x−2)2(x^{2}+3x+1)=(x-2)^{2} is not a quadratic equation

    Statement (II): The nature of roots of quadric equations x2+2x3+3=0x^{2}+2x\sqrt{3}+3=0 are real and equal.

    In light of the above statements, choose the most appropriate answer from the options given below.

    1. A.Both Statement (I) and Statement (II) are true.
    2. B.Both Statement (I) and Statement (II) are false.
    3. C.Statement (I) is true but Statement (II) is false.
    4. D.Statement (I) is false but Statement (II) is true.
    Show the answer and solution

    Answer: A. Both Statement (I) and Statement (II) are true.

    Expanding, x2+3x+1=x2−4x+4x^{2}+3x+1=x^{2}-4x+4 reduces to 7x=37x=3, which is linear, so Statement I is true. For Statement II the discriminant is 12−12=012-12=0: real and equal roots. Both true.

  7. Question 7Multiple choiceAsked in JIPMAT 2024Quadratic Equations

    If α\alpha and β\beta are the roots of the equation ax2+bx+c=0ax^{2}+bx+c=0, then value of 1aα+b+1aβ+b\frac{1}{a\alpha+b}+\frac{1}{a\beta+b} is:

    1. A.abc\frac{a}{bc}
    2. B.bac\frac{b}{ac}
    3. C.cab\frac{c}{ab}
    4. D.1abc\frac{1}{abc}
    Show the answer and solution

    Answer: B. bac\frac{b}{ac}

    From α+β=−ba\alpha+\beta=-\frac{b}{a}, aα+b=−aβa\alpha+b=-a\beta and aβ+b=−aαa\beta+b=-a\alpha. The expression is −1a(1β+1α)=−1a⋅α+βαβ=−1a⋅−b/ac/a=bac-\frac{1}{a}\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)=-\frac{1}{a}\cdot\frac{\alpha+\beta}{\alpha\beta}=-\frac{1}{a}\cdot\frac{-b/a}{c/a}=\frac{b}{ac}. Check with x2−5x+6x^{2}-5x+6: 12−5+13−5=−56=−56\frac{1}{2-5}+\frac{1}{3-5}=-\frac{5}{6}=\frac{-5}{6}.

  8. Question 8Type the answerWritten by IPMATwayQuadratic Equations

    The value of pp for which the sum of the squares of the roots of x2−(p−3)x+(p−4)=0x^{2}-(p-3)x+(p-4)=0 is least is

    Show the answer and solution

    Answer: 4

    α2+β2=S2−2P=(p−3)2−2(p−4)=(p−4)2+1\alpha^{2}+\beta^{2}=S^{2}-2P=(p-3)^{2}-2(p-4)=(p-4)^{2}+1, least at p=4p=4. The roots are real for every pp since D=(p−5)2≥0D=(p-5)^{2}\ge0.

  9. Question 9Multiple choiceWritten by IPMATwayLinear Equations

    For which value of kk does the system (k+1)x+8y=4k(k+1)x+8y=4k,   kx+(k+3)y=3k−1\;kx+(k+3)y=3k-1 have no solution?

    1. A.k=1k=1
    2. B.k=1k=1 or k=3k=3
    3. C.k=−3k=-3
    4. D.k=3k=3
    Show the answer and solution

    Answer: D. k=3k=3

    The system has no solution when a1a2=b1b2≠c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}. From the first equality, (k+1)(k+3)=8k⇒k2−4k+3=0⇒k=1(k+1)(k+3)=8k\Rightarrow k^2-4k+3=0\Rightarrow k=1 or 33. For k=1k=1: 2x+8y=42x+8y=4 and x+4y=2x+4y=2 are identical (ratio 22 for all three), so infinitely many solutions. For k=3k=3: 4x+8y=124x+8y=12 and 3x+6y=83x+6y=8; ratios 43,43,128=32\frac43,\frac43,\frac{12}{8}=\frac32, so the constants break the pattern: parallel lines, no solution. Answer k=3k=3.

  10. Question 10Multiple choiceWritten by IPMATwayLinear Equations

    A new temperature scale XX marks 20∘X20^\circ X at the freezing point of water and 140∘X140^\circ X at its boiling point. At what Celsius temperature does the XX scale show the same number as the Fahrenheit scale?

    1. A.−40∘-40^\circC
    2. B.−20∘-20^\circC
    3. C.20∘20^\circC
    4. D.−4∘-4^\circC
    Show the answer and solution

    Answer: B. −20∘-20^\circC

    The XX scale spans 120120 units over 100∘100^\circC, so X=20+1.2CX=20+1.2C. Fahrenheit is F=1.8C+32F=1.8C+32. Setting X=FX=F: 20+1.2C=1.8C+32⇒−12=0.6C⇒C=−2020+1.2C=1.8C+32\Rightarrow-12=0.6C\Rightarrow C=-20. Check: X=20−24=−4X=20-24=-4 and F=−36+32=−4F=-36+32=-4. The question asks for the Celsius temperature, −20∘-20^\circC, not the shared reading −4-4.