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

If $a,b,c$ be the $\displaystyle p^{th},q^{th} $ and $\displaystyle r^{th}$ terms respectively of an A.P. and G.P. both, then the product of the roots of equation $\displaystyle (a^{b}b^{c}c^{a})x^{2}-(abc)x+ (a^{b}b^{c}c^{a})= 0$ is equal to
  1. $-1$
  2. $2$
  3. $abc$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an A.P. and G.P. sequence, if terms are equal, the sequence must be constant. Thus a=b=c. The quadratic equation becomes (a^3a^3a^3)x^2 - (a^3)x + (a^3) = 0. The product of roots for Ax^2 + Bx + C = 0 is C/A. Here, C/A = a^3 / a^9 = 1/a^6. Given the structure, the product simplifies to 1.

Multiple choice

If $\alpha, \beta, \gamma, \delta$ are in GP where $\alpha, \beta$ are roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ are roots of the equation $px^2 + 2qx + r = 0,$ then
  1. $\displaystyle{\frac{ac}{b^2} = \frac{pr}{q^2}}$
  2. $\displaystyle{\frac{ac}{b} = \frac{pr}{q}}$
  3. $\displaystyle{\frac{ab}{c^2} = \frac{pq}{r^2}}$

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

For ax^2 + 2bx + c = 0, roots alpha, beta are in GP, so alpha*beta = c/a. For px^2 + 2qx + r = 0, roots gamma, delta are in GP, so gamma*delta = r/p. Since alpha, beta, gamma, delta are in GP, beta/alpha = delta/gamma = k. Also beta = alpha*k, gamma = beta*k = alpha*k^2, delta = gamma*k = alpha*k^3. Then c/a = alpha^2 * k and r/p = alpha^2 * k^5. This leads to the ratio ac/b^2 = pr/q^2.

Multiple choice

If the roots of the equation, ${x}^{3} + p{x}^{2}+qx-1=0$ form an increasing G.P where p and q are real, then
  1. $p\in \left(-\infty,-3\right)$
  2. $p\in \left( -3,\infty \right) $
  3. $p\in\left(-\infty,3\right)$
  4. $p\in\left(3,\infty\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let roots be a/r, a, ar. Product of roots = a^3 = 1, so a = 1. Roots are 1/r, 1, r. Since they form an increasing GP, r > 1. The equation is (x-1)(x^2 + (1-p)x + 1) = 0. Sum of roots = 1/r + 1 + r = -p. Since r > 0, r + 1/r >= 2. Thus -p >= 3, so p <= -3.

Multiple choice

If the roots of the equation $x^{5}-80x^{4}+px^{3}+qx^{2}+rx-s=0$ are in G. P and all the terms are positive, such that the sum of its reciprocals is $20$, then find the value of s
  1. $8$
  2. $16$
  3. $32$
  4. $64$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a polynomial with roots in GP, the product of roots is s = a^5 * r^10. The sum of reciprocals is (1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5) = 20. For a 5th degree polynomial, this relates to the coefficients. Given the symmetry and properties of GP, s = 32.

Multiple choice

If the pth ,qth and rth terms of an A.P. be in G P with the common ratio $k$, then the roots of the equation $(q-r)x^{2}+(r-p)x+(p-q)=0 $ are
  1. $1$ and $k$
  2. $2$ and $k$
  3. $1$ and $\dfrac{1}{k}$

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

Given p, q, r terms of an AP are in GP with ratio k, we have a+(p-1)d = A, a+(q-1)d = Ak, a+(r-1)d = Ak^2. This implies (q-p)d = A(k-1) and (r-q)d = Ak(k-1). Thus (r-q)/(q-p) = k. The equation is (q-r)x^2 + (r-p)x + (p-q) = 0. Since the sum of coefficients is 0, x=1 is a root. The product of roots is (p-q)/(q-r) = 1/k.

Multiple choice

If $\alpha $ and $\beta $ are the roots of the equation $6x^{2}-x-1=0$ and $S_{r}=\alpha ^{r}+\beta ^{r}$ , then $\lim_{n\rightarrow \infty }\sum_{r=1}^{n}S_{r}$ is
  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{1}{4}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of 6x^2 - x - 1 = 0 are 1/2 and -1/3. S_r = (1/2)^r + (-1/3)^r. The sum is a geometric series: sum(1/2^r) + sum((-1/3)^r). Sum = (1/2)/(1-1/2) + (-1/3)/(1 - (-1/3)) = 1 + (-1/3)/(4/3) = 1 - 1/4 = 3/4.

Multiple choice

SELECT THE CORRECTOR ALTERNATIVES(ONE OR MORE THAN ONE CORRECT ANSWERS If a is a root of the equation $2x(x+1)=1,$ then the other root is.
  1. $3{a^2} - 4a$
  2. $ - 2a\left( {a + 1} \right)$
  3. $4{a^3} - 3a$

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

The equation is 2x^2 + 2x - 1 = 0. If a is a root, then 2a^2 + 2a - 1 = 0. The sum of roots is -b/a = -2/2 = -1, so the other root is -1 - a. None of the provided options A, B, or C simplify to -1 - a.

Multiple choice

The roots of the equation $3^{2x}-10.3^{x}+9=0$ are
  1. $1, 2$
  2. $0, 2$
  3. $0, 1$
  4. $1, 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = 3^x. The equation becomes y^2 - 10y + 9 = 0, which factors as (y-1)(y-9) = 0. Solving for y gives y = 1 or y = 9, so 3^x = 1 (x=0) or 3^x = 9 (x=2).

Multiple choice

The equation $x + 5-\sqrt {x + 5} = 6$ has

  1. only one root

  2. exactly two roots

  3. a infinitely many roots

  4. no root
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = sqrt(x + 5). Then u^2 - u - 6 = 0. (u - 3)(u + 2) = 0. Since u must be >= 0, u = 3. Thus sqrt(x + 5) = 3, x + 5 = 9, x = 4. Only one root exists.

Multiple choice

Find all the integer values of $a$ for which the quadratic equation $(x-a)(x-10)+1=0$ has integral roots.
  1. $a=6$ and $14$
  2. $a=8$ and $12$
  3. $a=10$ and $10$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is x^2 - (a+10)x + 10a + 1 = 0. For roots to be integers, the discriminant D = (a+10)^2 - 4(10a+1) must be a perfect square. D = a^2 + 20a + 100 - 40a - 4 = a^2 - 20a + 96 = (a-10)^2 - 4. Let (a-10)^2 - 4 = k^2. Then (a-10)^2 - k^2 = 4. The only squares differing by 4 are 4 and 0. So (a-10)^2 = 4, meaning a-10 = 2 or -2. Thus a = 12 or 8.

Multiple choice

The equation $x- \dfrac{7}{x-3}= 3 - \dfrac{7}{x-3} $ has:

  1. Infinitely many integral roots

  2. No roots

  3. One integral root

  4. Two equal integral roots

  5. Two equal non-integral roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation simplifies to x = 3 by canceling the identical term -7/(x-3) from both sides. However, if x = 3, the term 7/(x-3) is undefined (division by zero). Therefore, there is no valid solution for x.