Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

Simplify the quadratic fraction: $\dfrac{-9+x^2}{x^2+2x-15}$
  1. $\dfrac{-x-3}{x+5}$
  2. $\dfrac{-x+3}{x+5}$
  3. $\dfrac{x+3}{x+5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Factor the numerator: x^2 - 9 = (x-3)(x+3). Factor the denominator: x^2 + 2x - 15 = (x+5)(x-3). The fraction is ((x-3)(x+3)) / ((x+5)(x-3)). Canceling (x-3) leaves (x+3)/(x+5).

Multiple choice

State whether the following statement is true or false. If A and G are respectively arithmetic and geometric means between two positive quantities a and b, then the quadratic equation having a, b as its roots is $x^2-2Ax+G^2=0$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The arithmetic mean of two positive numbers a and b is A = (a + b) / 2, which means their sum is 2A. Their geometric mean is G = sqrt(a * b), which means their product is G^2. A quadratic equation with roots a and b is represented by x^2 - (sum of roots)x + (product of roots) = 0, which simplifies to x^2 - 2Ax + G^2 = 0.

Multiple choice

The quadratic equation whose roots are the $AM$ and $HM$ of the roots of the equation $\displaystyle x^{2}+7x-1=0$ is
  1. $\displaystyle 14x^{2}+14x-45=0$
  2. $\displaystyle 45x^{2}-14x+14=0$
  3. $\displaystyle 14x^{2}+45x-14=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x^2 + 7x - 1 = 0, let roots be a, b. a+b = -7, ab = -1. AM = (a+b)/2 = -3.5. HM = 2ab/(a+b) = 2(-1)/(-7) = 2/7. The equation is x^2 - (AM+HM)x + (AM*HM) = 0. AM+HM = -7/2 + 2/7 = (-49+4)/14 = -45/14. AM*HM = -7/2 * 2/7 = -1. Equation: x^2 + (45/14)x - 1 = 0, which is 14x^2 + 45x - 14 = 0.

Multiple choice

If $A, G, H$ be respectively A.M., GM., H.M. of three numbers (>0) then the equation whose roots are the numbers is given by
  1. $\displaystyle x^{3}+3Ax^{2}+3\frac{G}{H}x-G^{3}=0$
  2. $\displaystyle x^{3}-3Ax^{2}-3(\frac{G^3}{H})+G^{3}=0$
  3. $x^{3}-3Ax^{2}+G^{3}\left ( 3x-1 \right )=0$
  4. $\displaystyle x^{3}-3Ax^{2}+3(\frac{G^3}{H})x-G^{3}=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots a, b, c, the cubic is x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0. Given A = (a+b+c)/3, G^3 = abc, and H = 3/(1/a+1/b+1/c) = 3abc/(ab+bc+ca). Thus, ab+bc+ca = 3abc/H = 3G^3/H. Substituting gives x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.

Multiple choice

Solve the following quadratics $ x^{2}+x+1 = 0 $
  1. $\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2} $
  2. $-\dfrac{1}{2}\pm \dfrac{i}{2} $
  3. $-\dfrac{1}{2}\pm \dfrac{i\sqrt{3}}{2} $
  4. $\dfrac{1}{2}\pm \dfrac{i}{2} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, for x^2 + x + 1 = 0, we get x = (-1 +/- sqrt(1 - 4)) / 2 = (-1 +/- sqrt(-3)) / 2 = -1/2 +/- i*sqrt(3)/2.

Multiple choice

Solve the following quadratic $ 21x^{2}-28x+10 = 0 $
  1. $ \dfrac{2}{3}\pm \dfrac{\sqrt{15}}{21}i $
  2. $ \dfrac{2}{3}\pm \dfrac{\sqrt{16}}{21}i $
  3. $ \dfrac{2}{3}\pm \dfrac{\sqrt{14}}{21}i $
  4. $ \dfrac{-2}{3}\pm \dfrac{\sqrt{14}}{21}i $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, we have a=21, b=-28, c=10. The discriminant is (-28)^2 - 4(21)(10) = 784 - 840 = -56. The roots are (28 +/- sqrt(-56)) / 42 = (28 +/- i*sqrt(56)) / 42 = (28 +/- 2*i*sqrt(14)) / 42 = 2/3 +/- i*sqrt(14)/21.

Multiple choice

The real roots of the equation $5^{\log_5 (x^2-4x+5)}=x-1$ are
  1. $1$ and $2$
  2. $2$ and $3$
  3. $3$ and $4$
  4. $4$ and $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation 5^(log5(x^2-4x+5)) = x-1 simplifies to x^2-4x+5 = x-1, which rearranges to x^2-5x+6 = 0. Factoring gives (x-2)(x-3) = 0, so the roots are 2 and 3. Both roots satisfy the domain condition of the logarithm (x^2-4x+5 > 0).