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

Given $ax^2+bx+c$ is a quadratic polynomial in x and leaves remainders 6, 11 and 18 respectively when divided by $(x+1), (x+2) and (x+3)$. Find the value of $a+b+c.$
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Remainder Theorem, P(-1)=6, P(-2)=11, P(-3)=18. Let P(x) = ax^2 + bx + c. P(-1)=a-b+c=6, P(-2)=4a-2b+c=11, P(-3)=9a-3b+c=18. Subtracting equations: (4a-2b+c)-(a-b+c) = 3a-b = 5. (9a-3b+c)-(4a-2b+c) = 5a-b = 7. Subtracting these: 2a=2 => a=1. Then 3(1)-b=5 => b=-2. Then 1-(-2)+c=6 => c=3. a+b+c = 1-2+3 = 2.

Multiple choice

Let p(x) be a quadratic polynomial with constant term 1. Suppose $p(x)$ when divided by $x -1$ leaves remainder 2 and when divided by $x +1$ leaves remainder 4. Then the sum of the roots of $p(x) = 0$ is
  1. $-1$
  2. $1$
  3. $-\dfrac{1}{2}$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let p(x) = ax^2 + bx + 1. Remainder theorem: p(1) = a + b + 1 = 2 => a + b = 1. p(-1) = a - b + 1 = 4 => a - b = 3. Adding: 2a = 4 => a = 2. Subtracting: 2b = -2 => b = -1. Sum of roots = -b/a = -(-1)/2 = 1/2.

Multiple choice

In the following questions, two questions I and II are given. You have to solve both the equations and give answer. I. $2x-15y=5$ II. $6x-5y=-1$
  1. If X $>$ Y
  2. If X $> =$ Y
  3. If X $<$ Y
  4. If X $= < $ Y
  5. If X$=$Y or the relationship cannot be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve I: 2x - 15y = 5. Solve II: 6x - 5y = -1. Multiply I by 3: 6x - 45y = 15. Subtract from II: (-5y - (-45y)) = -1 - 15, 40y = -16, y = -0.4. Substitute y into I: 2x - 15(-0.4) = 5, 2x + 6 = 5, 2x = -1, x = -0.5. Since -0.5 < -0.4, X < Y.

Multiple choice

Solve the following equations: $\cfrac{7}{{3}^{x}}-\cfrac{6}{{2}^{y}}=15$; $\cfrac{8}{{3}^{x}}=\cfrac{9}{{2}^{y}}$
  1. $(-2 , -3)$
  2. $(-1 , 2)$
  3. $(-4 , 5)$
  4. $(0 , 9)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = 1/3^x and v = 1/2^y. Equations: 7u - 6v = 15 and 8u = 9v (so v = 8u/9). Substitute: 7u - 6(8u/9) = 15 => 7u - 16u/3 = 15 => 5u/3 = 15 => u = 9. Since 1/3^x = 9 = 3^2, x = -2. Then 8(9) = 9v, so v = 8. Since 1/2^y = 8 = 2^3, y = -3. Solution is (-2, -3).

Multiple choice

Solve the following pair of simultaneous equations: $3\sqrt {2}x-5\sqrt {3}y+\sqrt {5}=0$ and $2\sqrt {3}x+7\sqrt {2}y-2\sqrt {5}=0$
  1. $x=\cfrac { 18\sqrt { 1 } +6\sqrt { 4 } }{ 56 } ;\quad y=\cfrac { 12\sqrt { 6 } +2\sqrt { 2 } }{ 52 } $
  2. $x=\cfrac { 11\sqrt { 15 } +7\sqrt { 16 } }{ 109 } ;\quad y=\cfrac { 1\sqrt { 13 } -7\sqrt { 11 } }{ 2 } $
  3. $x=\cfrac { 3\sqrt { 12 } -5\sqrt { 14 } }{ 23 } ;\quad y=\cfrac { 8\sqrt { 13 } +6\sqrt { 11 } }{ 56 } $
  4. $x=\cfrac { 10\sqrt { 15 } -7\sqrt { 10 } }{ 72 } ;\quad y=\cfrac { 2\sqrt { 15 } +6\sqrt { 10 } }{ 72 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can solve this system of linear equations using the elimination method. Multiplying the first equation by 2*sqrt(3) and the second by 3*sqrt(2) allows us to eliminate the x-term, yielding y = (2*sqrt(15) + 6*sqrt(10)) / 72. Similarly, eliminating the y-term by multiplying the equations by 7*sqrt(2) and 5*sqrt(3) respectively gives x = (10*sqrt(15) - 7*sqrt(10)) / 72.