Practice IBA Undergraduate Aptitude questions with answers and explanations.
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The sequence increases by 3 each time.
Adding 3 to 11 gives 14.
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The sum of the first n natural numbers is n(n + 1)/2.
For n = 10, the sum is 10 × 11 ÷ 2 = 55.
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The sum of the first n odd numbers equals n².
For n = 8, the sum is 8² = 64.
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The least common multiple of 3 and 4 is 12.
A number divisible by both must therefore be divisible by 12.
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The units digits of powers of 7 cycle as 7, 9, 3, 1.
The fourth power therefore ends in 1.
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The greatest common divisor of 18 and 24 is 6.
Dividing both terms by 6 gives 3/4.
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The multiplicative inverse of a nonzero fraction is its reciprocal.
Interchanging numerator and denominator gives 7/5.
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Multiplication is performed before addition.
Thus, 3 × 4 = 12 and 5 + 12 = 17.
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The operation inside parentheses is completed first.
Then 12 ÷ 3 = 4.
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The identity of the arresting authority is obvious or unimportant.
Omitting the agent keeps the passive sentence concise.
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The present perfect passive question uses 'have + subject + been + past participle.'
The indefinite agent 'anyone' does not need to be retained.
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'Slept' is intransitive and has no direct object to become a passive subject.
Only transitive constructions normally form the passive.
