Mathematics · Quantitative Aptitude
Algebra and Arithmetic
406 Questions
Algebra and arithmetic questions cover fundamental mathematical operations, inequalities, and binomial products. They assess core quantitative reasoning skills required for various aptitude tests. Solving these problems strengthens the understanding of number systems and algebraic identities.
Binomial productsLinear inequalitiesLeast common multipleQuadratic equationsInteger properties
Algebra and Arithmetic Questions
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x
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x2
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$\frac{1}{x}$
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$\frac{1}{x^2}$
C
Correct answer
Explanation
$cothx = \frac{cosh x}{sinh x}
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\text{as} |x| < < 1, coshx \approx 1 \hspace{0.2cm} and \hspace{0.2cm} sinh x \approx x
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\text{Thus} \hspace{0.2cm} coth x \approx \frac{1}{x}$
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0.5 - j0.25
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1/(0.5 + j0.25)
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1/(0.5 - j0.25)
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2 + j4
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(1) and (2)
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(1) and (3)
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(3) and (4)
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(4) and (1)
A
Correct answer
Explanation
$\text{If} \hspace{0.5cm} x > y
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\text{Then} \hspace{0.5cm} e^x > e^y
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\text{Similarly, for x > y > 1 log x is increasing function. Hence,}
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\text{In} x > In y$
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$\bigg[1, \frac{2}{3} \bigg]$
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$\bigg[\frac{-1}{2} , 1\bigg]$
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$\bigg[-1, \frac{1}{2} \bigg]$
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[2, –4]
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positive
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K = R
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K < R
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none of the above
C
Correct answer
Explanation
When NPV is positive at a given discount rate, the project's internal rate of return (R) exceeds the cost of capital (K). The relationship K < R indicates that the project generates returns higher than the required rate, making it acceptable. Positive NPV and IRR > cost of capital are equivalent decision criteria.
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(A + B) + C = A + (B + C)
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A(B + C) = AB + AC
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A + AB = A
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A + B = B + A
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(a) A + A = A
B
Correct answer
Explanation
This is the distributive law.
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decrease by the same quantity
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increase by same quantity
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remains the same
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doubles
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None of the above
A
Correct answer
Explanation
Overall sum will be original sum - n times the constant quantity. Hence it decreases by same quantity.
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(A + B + C)(A . B . C)
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(A + B) . C
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(A + B)' . C
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((A + B) . C)'
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(A + B) + C
B
Correct answer
Explanation
((A + B)' + C')' = ((A' . B' + C')' = ((A' . B')' . (C')') = (((A')' + (B')') . C) = (A + B) . C
So, this is the correct answer.
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Additive inverse of - 121/10 is greater than 10.
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Multiplicative inverse of 10/18 is less than 1.
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Sum of number and its additive inverse is 0.
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Product of a number and its multiplicative inverse is - 1.
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Both (1) and (3)
E
Correct answer
Explanation
Statement (1) is correct, as additive inverse of - 121/10 is 121/10, which is greater than 10.
Statement (2) is incorrect, as multiplicative inverse of 10/18 is 18/10, which is more than 1.
Statement (3) is correct, as sum of a number and its additive inverse is 0.
Statement (4) is incorrect, as product of a number and its multiplicative inverse is 1.
Therefore, both (1) and (3) are correct.
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(A + B) + C = A + (B + C)
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A + B = B + A
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A (B + C) = A B + A C
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A + A = A
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A + A B = A
B
Correct answer
Explanation
Commutative: A+ B = B + A
A
Correct answer
Explanation
In Boolean algebra, the expression x · 0 (x AND 0) always equals 0. This is known as the null law or annihilator property - anything AND 0 is 0.
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ab
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a2 - b2
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a2 + b2
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a3 + b3
D
Correct answer
Explanation
Let (a, b) be (-1, -2). Then (1) = 2, (2) = - 2, (3) = - 3, (4) = 5 and (5) = - 7. So (4) is greatest among all the choices, is the answer.
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3( a + b)
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ab2
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ab – 1
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a2 – b
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b2 – a
B
Correct answer
Explanation
Let a = 3 and b = 2. Then, starting from (5), the choices are (4 – 3 = 1), (9 – 2 = 7), (6 – 1 = 5), (3 x 22 = 12) and {3(3 + 2) = 15}. Of these, only the second choice is an even integer.
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I only
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III only
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I, II and III
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II and III only
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I and III only
B
Correct answer
Explanation
Consistent with the given condition (a, b, c) can be 1/2, 1/3 and 6. So, I need not be true. Similarly, (a, b, c) can be (1/2, 1/2, 2). So, II need not be true. If ac and bc are both integers, then (ac)2 + (bc)2 must also be an integer. So, III must always be true. So, (3) is the answer.
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k2
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k(k - 1)
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k - 1
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3k + 1
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4k + 3
B
Correct answer
Explanation
Since we have to test all the choices, Let us consider one odd value (1) for k. Then (1) = 1, (2) = 0 , (3) = 0, (4) = 4 and (5) = 7. Now let us consider one even value (2) for k. Then (1) = 4, (2) = 2, (3) = 1, (4) = 7 and (5) = 11.In both cases Choice (2) is even, is the answer.
