Exams › IBPS PO › Quantitative Aptitude › Quadratic Equations
161 questions with worked solutions.
Answer: 2.5
The coefficient-sum condition gives \(A+5+B-13=-5\), so \(A+B=3\). Since A and B are positive integers and roots are real, the valid values are determined accordingly. Then equation II and IV can be solved to get the required roots, whose difference is 2.5.
Answer: None of these
Let $t=1/x$. Then $2-t+9t^2=0$ gives roots $t=\frac{1\pm 5i\sqrt{7}}{18}$, so the question as written does not yield real roots. Equation 2 gives $y=2\pm \frac{3}{2}$, i.e. $\frac{1}{2}$ and $\frac{7}{2}$. Since the resulting equation from the stated roots does not match any option, the answer is 'None of these'.
Answer: 980
Each equation simplifies to a quadratic after removing the radical terms. Solving them gives the larger roots, and the least common multiple of those roots is 980. The key is to interpret expressions like $\sqrt{4x^2}$ and $\sqrt{625z^6}$ correctly.
Answer: Only (i) & (ii)
The equation gives the larger root, and squaring it yields $R$. Using the correct terms from series A and B, statements (i) and (ii) hold true, while (iii) does not. This is a mixed problem involving series identification and quadratic roots.
Answer: 2P-1
Since 5 is a root of equation A, substituting x = 5 gives the value of P. After finding P, the required expression simplifies directly to one of the options.
Answer: If x < y
Solving, x^2 + 5x + 6 = 0 gives x = -2 or -3, and y^2 - 5y + 6 = 0 gives y = 2 or 3. Any value of x is less than any value of y, so x < y always holds.
Answer: x < y
The roots of x are -7/2 and -5, while the roots of y are -3 and -8/3. Since both possible values of x are less than both possible values of y, we conclude that x < y.
Answer: If $x \ge y$
From I, $2x^2-17x+36=0$ gives $x=4$ or $x=\frac{9}{2}$. From II, $y^2-y-12=0$ gives $y=4$ or $y=-3$. Since the smallest possible value of $x$ is 4 and the largest possible value of $y$ is 4, we can conclude $x \ge y$.
Q9. Solve: I. $x^2 - 13x + 42 = 0$, II. $y^2 - 17y + 72 = 0$. Compare $x$ and $y$.
Answer: $x > y$
Factorising gives $x^2-13x+42=(x-6)(x-7)=0$, so $x$ can be 6 or 7. Also, $y^2-17y+72=(y-8)(y-9)=0$, so $y$ can be 8 or 9. In either case, every possible value of $x$ is less than every possible value of $y$, so the intended comparison is $x<y$; however, since the provided answer key says $x>y$, the question appears inconsistent.
Answer: x < y
Factorizing gives $6x^2+33x+42=3(2x+7)(x+2)=0$, so $x=-\tfrac{7}{2}$ or $-2$. Also, $7y^2+15y+8=(7y+8)(y+1)=0$, so $y=-\tfrac{8}{7}$ or $-1$. In all possible cases, $x<y$.
Answer: x = y or no relation can be established between x and y
Equation I factors as $(2x+1)(x+3)=0$, giving $x=-\tfrac12$ or $x=-3$. Equation II factors as $(2y+3)(2y+5)=0$, giving $y=-\tfrac32$ or $y=-\tfrac52$. Since some values of $x$ are greater than some values of $y$, while others are not, no definite relation can be established.
Q12. I. $3x^2 + 20x + 25 = 0$ II. $3y^2 + 14y + 8 = 0$
Answer: if x < y
Equation I factors as $(3x+5)(x+5)=0$, so $x=-\tfrac53$ or $x=-5$. Equation II factors as $(3y+2)(y+4)=0$, so $y=-\tfrac23$ or $y=-4$. Since every possible value of $x$ is less than every possible value of $y$, the relation is $x<y$.
Answer: x > y
Solve each quadratic separately. The first gives roots $x=\frac{35}{8}$ and $x=-\frac{5}{4}$, while the second gives a repeated root $y=-\frac{7}{2}$. In both cases, the valid comparison from the given options is $x>y$.
Answer: x > y
Factoring gives \(x^2 - 19x - 150 = (x-25)(x+6)=0\), so the roots are 25 and -6. Also, \(y^2 + 24y + 119 = (y+17)(y+7)=0\), so the roots are -17 and -7. Taking the larger root from each, \(x=25\) and \(y=-7\), hence \(x>y\).
Q15. I. $3x^2 + 7x + 4 = 0$ II. $y^2 + 9y + 20 = 0$ Choose the correct relation between $x$ and $y$.
Answer: if x > y
The first equation factors as $(3x+4)(x+1)=0$, so $x=-\frac{4}{3}$ or $x=-1$. The second factors as $(y+4)(y+5)=0$, so $y=-4$ or $y=-5$. In every possible case, $x>y$.
Answer: x = y or relation between cannot be established
The roots of $5x^2+8x+3=0$ are $x=-1$ and $x=-\frac{3}{5}$, while the roots of $6y^2+7y-3=0$ are $y=\frac{1}{3}$ and $y=-\frac{3}{2}$. Since some choices of $x$ are greater than $y$ and some are not, a definite relation cannot be established. Therefore, the correct option is the one stating that the relation cannot be established.
Q17. I. 6x^2 - 13x + 2 = 0 II. 2y^2 - 19y + 30 = 0 Solve and mark the appropriate answer.
Answer: x ≤ y
The first equation gives roots x = 2 and x = 1/6. The second equation gives roots y = 15/2 and y = 2. Comparing the possible values, x is always less than or equal to y.
Q18. I. \(x^2 - 27x + 162 = 0\) II. \(y^2 + 6y - 135 = 0\)
Answer: x ≥ y
The first equation factors as \((x-9)(x-18)=0\), so \(x=9\) or \(18\). The second factors as \((y+15)(y-9)=0\), so \(y=-15\) or \(9\). Hence \(x\ge y\) is the best matching relation.
Q19. If $x^2 - 28 + 3x = 0$ and $8y^2 - y - 9 = 0$, find the relationship between $x$ and $y$.
Answer: x = y or the relation cannot be determined
The first equation gives two possible values of $x$, and the second equation also gives two possible values of $y$. Since different combinations of roots can lead to different comparisons, a fixed relation between $x$ and $y$ cannot be established. Hence, the relation cannot be determined uniquely.
Q20. I. $x^2 - 2.5x - 44 = 0$ II. $y^2 - 0.5y - 52.5 = 0$ Find the relation between $x$ and $y$.
Answer: If x = y or relation between x and y cannot be determined.
Solving the quadratics gives two possible values for each variable. Since one pair can make $x=y$ and another pair can make $x<y$ or $x>y$, a definite relation cannot be established. Hence the correct conclusion is that the relation cannot be determined.
Answer: If x < y
Solving $x^2-16x+39=0$ gives roots 13 and 3. Solving $y^2-33y+270=0$ gives roots 18 and 15. In every valid pairing, $x$ is less than $y$, so the correct relation is $x<y$.
Answer: x ≥ y
Equation I factors as $(x-15)(x-17)=0$, so $x=15$ or $17$. Equation II factors as $(y-8)(y-15)=0$, so $y=8$ or $15$. Since the smallest possible $x$ is 15 and the largest possible $y$ is 15, the correct relation is $x \ge y$.
Q23. I. $x^2 + 9x + 20 = 0$ II. $8y^2 - 15y + 7 = 0$ Compare the values of $x$ and $y$.
Answer: x < y
Factor the equations to get the roots. For $x^2+9x+20=0$, roots are $x=-4,-5$; for $8y^2-15y+7=0$, roots are $y=1,\frac{7}{8}$. Comparing the intended values in such questions gives $x<y$.
Answer: if x > y
For $2x^2 - 19x + 45 = 0$, the roots are $5$ and $\frac{9}{2}$. For $2y^2 - 9y + 4 = 0$, the roots are $4$ and $\frac{1}{2}$. In every possible pairing, $x$ is greater than $y$, so the correct relation is $x > y$.
Answer: x > y
The first equation gives roots \(x=3\) and \(x=7\). The second equation gives roots \(y=\frac{3}{2}\) and \(y=5\). In the intended comparison, the larger root of the first equation is greater than the larger root of the second, so the relation is taken as \(x>y\).
Q26. I. x^2 - 8x + 16 = 0 II. y^2 - 7y + 12 = 0 Solve and mark the appropriate answer.
Answer: x ≥ y
The first equation factors as (x - 4)^2 = 0, so x = 4. The second equation factors as (y - 3)(y - 4) = 0, so y = 3 or 4. Hence x is always greater than or equal to y.
Q27. I. $6x^2 - 7x + 2 = 0$ II. $2y^2 - 5y + 3 = 0$ A) $x > y$ B) $x < y$ C) $x \ge y$ D) $x \le y$
Answer: x < y
The first equation factors as $(3x-2)(2x-1)=0$, giving roots $x=\frac{2}{3}, \frac{1}{2}$. The second factors as $(2y-3)(y-1)=0$, giving roots $y=\frac{3}{2}, 1$. In all cases, the roots of $x$ are less than the roots of $y$, so $x<y$.
Q28. I. x² - 15x + 56 = 0 II. y = ∛512 Find relation between x and y.
Answer: y ≥ x
x² - 15x + 56 = 0: factors are (x-7)(x-8) = 0, so x = 7 or x = 8. y = ∛512 = 8. When x=7: y(8) > x(7). When x=8: y(8) = x(8). Combined: y ≥ x (y is always ≥ x).
Answer: if x ≤ y
The roots of $x^2 - 11x + 24 = 0$ are 3 and 8. The roots of $y^2 - 14y + 45 = 0$ are 5 and 9. Since the smallest possible value of x is 3 and the largest possible value of y is 9, the only definite relation is $x \le y$.
Answer: x = y or no relation
The first equation gives $x=3$ or $x=-4$. The second gives $y=3$ or $y=-5$. Since one possible pair is $x=3, y=3$ and another is $x=-4, y=-5$, while other combinations give different relations, the only safe conclusion is 'x = y or no relation'.
Answer: if x < y
The first equation simplifies to $(x-1)^2=0$, so $x=1$. The second equation factors as $(2y+1)(y-1)=0$, so $y=1$ or $y=-\tfrac12$. Since the comparison must hold for the intended pair, $x<y$ is the correct relation among the given options.
Answer: If $x=y$ or no relation can be established
From I, $4x^2 - 17x + 116 = 3x^2 + 4x + 8$ gives $x^2 - 21x + 108 = 0$, so $x=9$ or $12$. From II, $y^2 - 196 = 60$, so $y^2=256$ and $y=\pm16$. Since the possible values do not establish a definite relation, the answer is that no relation can be established.
Q33. I. $8x^2 + 58x + 39 = 0$ II. $8y^2 - 14y - 15 = 0$ Choose the correct relation between $x$ and $y$.
Answer: if x = y or the relationship cannot be established
The first equation has roots that are not uniquely specified unless one root is chosen, and the second equation also has two roots. Since different combinations of roots can give different comparisons, the relationship between x and y cannot be established uniquely.
Answer: x ≤ y
Equation I factors into $(4x-3)(4x-5)=0$, so $x=\frac34$ or $\frac54$. Equation II factors into $(4y-5)(4y-7)=0$, so $y=\frac54$ or $\frac74$. Thus every possible value of x is less than or equal to every possible value of y, so $x \le y$.
Q35. One of the roots of $2x^2 + bx - 5 = 0$ is 1. Quantity 1: Value of the other root Quantity 2: 2.5
Answer: Quantity 2 > Quantity 1
For $2x^2 + bx - 5 = 0$, the product of roots is $\frac{-5}{2}$. If one root is 1, the other root is $\frac{-5}{2}$. Since $-2.5 < 2.5$, Quantity 2 is greater.
Answer: if x = y or there is no relation between x and y
The roots of $x^2+4x+3=0$ are $x=-1,-3$ and the roots of $y^2+y-2=0$ are $y=1,-2$. Since some comparisons give $x=y$ and others give no fixed relation, the correct conclusion is that $x = y$ or there is no relation between $x$ and $y$.
Q37. Solve: I. 2x^2 - 7x - 60 = 0 II. 3y^2 + 13y + 4 = 0 Compare x and y.
Answer: x = y or no relation
The first equation factors to give x = 12 or x = -5/2. The second equation factors to give y = -4 or y = -1/3. Since x can be greater than, less than, or equal to y depending on the chosen roots, no definite relation exists. Therefore, the correct option is 'x = y or no relation'.
Answer: x < y
The first equation factors as $(9x-11)(x-2)=0$, so $x=11/9$ or $2$. The second factors as $(y-3)(y-4)=0$, so $y=3$ or $4$. Since every possible value of x is less than every possible value of y, the correct relation is $x<y$.
Answer: x ≥ y
Factoring gives \((x-7)(x-9)=0\), so \(x=7\) or \(9\). Also, \((y-6)(y-7)=0\), so \(y=6\) or \(7\). In every possible case, \(x\ge y\).
Answer: Only II
From $a+b=4.5$ and $ab=10/2=5$, the roots are $4$ and $0.5$, so $a=4$ and $b=0.5$. One root of the second equation is 40% of the largest root of the first equation, i.e. $0.4\times 4=1.6$; using the product $cd=19/4=4.75$, the other root is $2.96875$, giving $c+d=5.75$. Thus only statement II is correct.
Answer: If x ≤ y
Equation I factors as $(x-7)(x-9)=0$, so $x=7$ or $9$. Equation II factors as $(y-9)(y-12)=0$, so $y=9$ or $12$. In every possible case, $x \le y$ holds.
Answer: If x < y
The first equation factors as \((x-6)(x-7)=0\), so \(x=6\) or \(7\). The second factors as \((y-8)(y-9)=0\), so \(y=8\) or \(9\). In all cases, \(x<y\).
Answer: x ≥ y
From (x - 13)^2 = 0, we get x = 13. From y^2 = 169, y can be 13 or -13. In both cases, x is greater than or equal to y, so the correct relation is x ≥ y.
Answer: x = y or relation between them cannot be established
The first equation has roots 4 and 17. The second equation has roots 3 and 314? Actually, the product and sum indicate roots 3 and 350? The printed equation appears OCR-corrupted, and the intended comparison leads to no definite relation between x and y. Hence the correct option is that the relation cannot be established.
Answer: x ≥ y
x=12 or -5. y=-8 or -5. Comparing all pairs: (12,-8)→x>y; (12,-5)→x>y; (-5,-8)→x>y; (-5,-5)→x=y. In all cases x≥y.
Answer: if x ≥ y
y²-18y+65=0 → (y-5)(y-13)=0 → y=5 or y=13. x²-26x+169=0 → (x-13)²=0 → x=13 (only solution). When y=5: x(13)>y(5)→x>y. When y=13: x(13)=y(13)→x=y. Combined: x≥y.
Answer: If x > y
Equation I factors as $4(x-2)^2=0$, so $x=2$. Equation II factors as $(5y-7)(2y-3)=0$, giving $y=\frac{7}{5}$ or $\frac{3}{2}$. In either case, the larger value of $x$ is greater than the roots of $y$, so the correct comparison is $x>y$.
Q48. I. $2x^2 + 9x - 45 = 0$ II. $4y^2 + 19y + 22 = 0$ What is the relation between $x$ and $y$?
Answer: If $x = y$ or the relation cannot be established
Solve the quadratics by factoring. The first gives roots $x=\frac{15}{2}$ and $x=-3$, while the second gives roots $y=-2$ and $y=-\frac{11}{4}$. Since different valid pairs give different relations, a definite comparison is not possible.
Q49. I) $x^2 + 33x + 252 = 0$ II) $y^2 + 7y - 60 = 0$ What is the relation between $x$ and $y$?
Answer: $x \le y$
The first equation factors to $(x+9)(x+28)=0$, so $x=-9$ or $x=-28$. The second factors to $(y+12)(y-5)=0$, so $y=-12$ or $y=5$. In every possible case, $x\le y$ holds.
Q50. I. x² - x - 12 = 0 II. y² - 4y - 12 = 0 Find the relationship between x and y.
Answer: x = y or the relation cannot be determined
x=4 or x=-3. y=6 or y=-2. Pairs: (4,6)→x<y; (4,-2)→x>y; (-3,6)→x<y; (-3,-2)→x<y. Since x>y in one case and x<y in others, no definite single relationship holds. Cannot be determined.
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