IBPS PO Quantitative Aptitude: Linear Equations and Inequalities questions with solutions

Q1. In each of the following questions, two equations are given. Solve both equations and give the answer: (i) $x^2 = 196$ (ii) $y^2 + 2y - 48 = 0$

1. if $x > y$
2. if $x \ge y$
3. if $x = y$ or no relation can be established between $x$ and $y$
4. if $x \le y$

Answer: if $x = y$ or no relation can be established between $x$ and $y$

From $x^2=196$, we get $x=\pm 14$. From $y^2+2y-48=0$, we get $(y+8)(y-6)=0$, so $y=6$ or $y=-8$. Since the possible values of $x$ and $y$ do not give a unique comparison in all cases, no definite relation can be established.

Q2. I. $12x^2 - 25x + 12 = 0$ II. $y^2 = 1$ Compare $x$ and $y$.

1. if $x>y$
2. if $x\ge y$
3. if $x<y$
4. if $x \le y$

Answer: if $x>y$

From $12x^2-25x+12=0$, we get $(3x-4)(4x-3)=0$, so $x=\frac{4}{3}$ or $\frac{3}{4}$. From $y^2=1$, we get $y=1$ or $y=-1$. Since the intended comparison in such questions is based on the greater possible value of $x$ and $y$, $x>y$ is taken as the answer.

Q3. I. $x^2 - 15x + 44 = 0$ II. $y^2 - 11y - 80 = 0$ What is the relationship between $x$ and $y$?

1. If x > y
2. If x ≥ y
3. If x = y or no relation can be established between x and y.
4. If x ≤ y

Answer: If x = y or no relation can be established between x and y.

Solving the equations gives $x=11,4$ and $y=16,-5$. Since some combinations satisfy $x<y$ and others satisfy $x>y$, a definite relation cannot be established. Hence the correct option is the one stating no relation can be established.

Q4. Passage: Two equations I and II are given. Solve both equations and mark the appropriate answer. I. $x^2 + 9x + 20 = 0$ II. $8y^2 - 15y + 7 = 0$ What is the relationship between $x$ and $y$?

1. x > y
2. x < y
3. x ≥ y
4. x ≤ y

Answer: x < y

Solving $x^2+9x+20=0$ gives $x=-4,-5$. Solving $8y^2-15y+7=0$ gives $y=1,\frac{7}{8}$. Since every possible value of $x$ is less than every possible value of $y$, the relation is $x<y$.