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

In the following questions two equations (I) and (II) are given. You have to solve both the equation and give answer- ( (I) ) ( y^2 - x^2 = 32 ) ( (II) ) ( y - x = 2 )

  1. If x > y

  2. If x < y

  3. If x ≥ y

  4. If x ≤ y

  5. If x = y or the relationship can not be established.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From (II): y = x + 2. Substitute into (I): (x+2)² - x² = 32 gives x² + 4x + 4 - x² = 32, so 4x = 28 and x = 7. Then y = 7 + 2 = 9. Since 7 < 9, x < y.

Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - (I.) x^2 – 23x + 132 = 0 (II.) y^2 – 15y + 56 = 0

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or the relationship cannot be established.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factor both equations: x²-23x+132=(x-11)(x-12) giving x=11,12. y²-15y+56=(y-7)(y-8) giving y=7,8. Since 11>7, 11>8, 12>7, 12>8, all x values exceed all y values.

Multiple choice

In the following question two equation numbered I and II are given .You have to solve both the equations and give answer if - I. (x^2 + 17x + 72 = 0) II. (y^2 + 12y + 32 = 0)

  1. x > y

  2. x ≥ y

  3. x < y

  4. x ≤ y

  5. x = y or the relationship cannot be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation I: x²+17x+72=(x+8)(x+9) gives x=-8,-9. Equation II: y²+12y+32=(y+4)(y+8) gives y=-4,-8. x values are -8,-9. When x=-8, y can be -8 (equal) or -4 (greater). When x=-9, it's less than both y values. So x≤y holds.

Multiple choice

(\text{In each of the following questions, one or two equations are given. You have to solve them and}\ (\text{(I)}\quad p^2 + 16 = 8p\ (\text{(II)}\quad 4q^2 + 64 = 32q)

  1. If p>q

  2. If p≥q

  3. If p<q

  4. If p≤q

  5. If p=q or relationship can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation (I) simplifies to p² - 8p + 16 = 0, which is (p-4)² = 0, giving p = 4. Equation (II) simplifies to 4q² - 32q + 64 = 0, or 4(q-4)² = 0, giving q = 4. Since both p and q are exactly 4, p = q is the correct relationship.

Multiple choice

In each of the following questions, one or two equations are given. You have to solve them and ( (I) ) ( 2p^2 + 12p + 16 = 0 ) ( (II) ) ( 2q^2 +14q + 24 = 0 )

  1. If p>q

  2. If p≥q

  3. If p<q

  4. If p≤q

  5. If p = q or relationship can not be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation (I) factors to 2(p+4)(p+2) = 0, giving p = -4 or p = -2. Equation (II) factors to 2(q+4)(q+3) = 0, giving q = -4 or q = -3. Since p can be greater than, less than, or equal to q depending on which roots are chosen, the relationship cannot be established.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and Give answer- ( (I) ) ( a^2 + 3a + 2 = 0 ) ( (II) ) ( b^2 + 5b + 6 = 0 )

  1. If a > b

  2. If a ≥ b

  3. If a < b

  4. If a ≤ b

  5. If a = b or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: a² + 3a + 2 = 0 factors to (a+1)(a+2) = 0, giving a = -1, -2. Equation II: b² + 5b + 6 = 0 factors to (b+2)(b+3) = 0, giving b = -2, -3. Since a can be -1 (which is > -3) and a can be -2 (which equals b's -2), the relationship is a ≥ b. Option A is wrong (a > b is not always true; when a=-2 and b=-2, they're equal). Option C and D are wrong because a is not always less than b.

Multiple choice

In the following questions two equation I and II are given you have to solve both the equation and give answer- I. (x^{2}–7x+12=0) II. (y^{2}–12y+32=0)

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

  5. If x=y or the relationship can not be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solve equation I: x²-7x+12=0 factors to (x-3)(x-4)=0, so x=3 or x=4. Solve equation II: y²-12y+32=0 factors to (y-4)(y-8)=0, so y=4 or y=8. Comparing values: x can be 3 or 4, while y can be 4 or 8. In all cases x≤y (when x=3: 3≤4, 3≤8; when x=4: 4≤4, 4≤8). Therefore option D (x≤y) is correct.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and give answers- (I) (P^2 + 4 = -4P) ()\n(II) (Q^2 + 7Q + 12 = 0)

  1. If P < Q

  2. If P ≤ Q

  3. If P > Q

  4. If P ≥ Q

  5. If P = Q or relationship cannot be established.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For equation I: P^2 + 4 = -4P gives P^2 + 4P + 4 = 0, so (P+2)^2 = 0, hence P = -2. For equation II: Q^2 + 7Q + 12 = 0 factors to (Q+3)(Q+4) = 0, so Q = -3 or Q = -4. Since P = -2 and Q = -3 or -4, we have P > Q.

Multiple choice

For what value of k are the roots of the equadratic equation (k+1)x2-2(k-1)x+1=0 real and equal? k के किस मान के लिए, द्विघात समीकरण (k+1)x2-2(k-1)x+1=0 के मूल वास्तविक और समान होंगें?

  1. k=0 only/केवल

  2. k=-3 only/केवल

  3. k=0 or/या k = 3

  4. k=0 or/;k k=-3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real and equal roots, discriminant D = 0. Given equation: (k+1)x² - 2(k-1)x + 1 = 0. D = 4(k-1)² - 4(k+1)(1) = 0. Simplifying: 4[(k-1)² - (k+1)] = 0 → (k² - 2k + 1) - k - 1 = 0 → k² - 3k = 0 → k(k-3) = 0. Thus k = 0 or k = 3. We must also ensure k+1 ≠ 0 (coefficient of x²), so k ≠ -1. Both k=0 and k=3 satisfy this. Option C correctly states k=0 or k=3. Option D incorrectly suggests k=-3 as an alternative, which doesn't satisfy the equation.

Multiple choice

If one of the roots of the equation ax2+x-3=0 is -1.5, then what is the value of a ? यदि समीकरण ax2+x-3=0 के मूलों में एक मूल -1.5 है तो a का मान क्या है?

  1. 4

  2. 3

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If x = -1.5 is a root of ax² + x - 3 = 0, then substituting gives a(-1.5)² + (-1.5) - 3 = 0. This simplifies to 2.25a - 4.5 = 0, giving 2.25a = 4.5, so a = 2. Verification: 2(2.25) - 4.5 = 4.5 - 4.5 = 0. The distractor 3 would result from incorrectly calculating (-1.5)² as 1.5 instead of 2.25.

Multiple choice

For what value of k, will the roots of the equation kx2-5x+6=0 be in the ratio of 2 : 3? समीकरण kx2-5x+6=0 में k के किस मान के लिए उसके मूल 2 : 3 अनुपात में होंगे?

  1. 0

  2. 1

  3. -1

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the equation kx² - 5x + 6 = 0 to have roots in ratio 2:3, let roots be 2α and 3α. By Vieta's formulas: Sum = 2α + 3α = 5α = 5/k, and Product = 2α × 3α = 6α² = 6/k. From sum: α = 1/k. Substituting in product: 6(1/k)² = 6/k, which gives k = 1. This satisfies the condition.

Multiple choice

If α and β are the roots of the equation x2 + px +q = 0, then – α–1, –β–1 are the roots of which one of the following equations? यदि समीकरण x2 + px +q = 0 के मूल α और β हैं, तो निम्नलिखित समीकरणों में से किस समीकरण के मूल्य – α–1, –β–1 हैं?

  1. qx2 – px + 1 = 0

  2. qx2 + px + 1 = 0

  3. x2 + px – q = 0

  4. x2 – px + q = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If α and β are roots of x^2 + px + q = 0, then sum of roots = α + β = -p and product = αβ = q. For the transformed roots -1/α and -1/β, the new sum = -(1/α + 1/β) = -(α+β)/(αβ) = p/q and new product = 1/(αβ) = 1/q. The required equation is x^2 - (sum)x + (product) = 0, which gives x^2 - (p/q)x + (1/q) = 0. Multiplying by q: qx^2 - px + 1 = 0. Option B has the wrong sign for the px term.

Multiple choice

In each of the following questions, two equations are given. You have to solve them and\ I. ((8 + x)^2 = x^2 + 12x)\ II. (7y^2 + 15y + 4 = 6y^2 + 20)

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solving Equation I: (8 + x)^2 = x^2 + 12x. Expanding: 64 + 16x + x^2 = x^2 + 12x, which simplifies to 64 + 4x = 0, giving x = -16. Solving Equation II: 7y^2 + 15y + 4 = 6y^2 + 20, which simplifies to y^2 + 15y - 16 = 0. Factoring: (y - 1)(y + 16) = 0, so y = 1 or y = -16. Comparing x = -16 with y values: x = y when y = -16, and x < y when y = 1. Therefore x ≤ y in all cases.

Multiple choice

In each of the following question two equations (I) and (II) are given. You have to solve them and ( (I) ) ( x^2 + x = 56 ) ( (II) ) ( y^2 - 17y + 72 = 0 )

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. If x = y or, relation cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solve equation (I): x²+x-56=0 gives x=(−1±√(1+224))/2=(−1±15)/2, so x=7 or x=−8. Solve equation (II): y²−17y+72=0 gives y=(17±√(289−288))/2=(17±1)/2, so y=9 or y=8. Comparing the positive roots: x=7 is less than both y=8 and y=9. So x