Algebra
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📘 Comprehensive Syllabus & Examination Guide
Algebra
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Mapped Subjects & Topic Question Distribution
Total Question Pool
100%
29 MCQs
Combined Active Syllabus
Algebra
29 MCQs
Topic Pool
📊 Question Pool Structure
29 MCQs across fundamental, intermediate, and advanced concept tiers.
⚡ Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
⚖️ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Algebra, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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Solved Blueprint Examples
Showing 10 solved representative questions
📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Algebra
Hard • Quantitative Aptitude Test
If y−7=12, then y+19= ——?
A
19
B
28
C
38
D
42
E
50
✓ Correct Answer: C - 38
📖 Step-by-Step Solution & Conceptual Rationale:
y=12+7=19. Therefore, y+19=19+19=38.
Sample Question 2
Algebra
Hard • Quantitative Aptitude Test
How much is $\frac{1}{4} of \frac{x}{4}$?
A
4x
B
x
C
$\frac{1}{x}$
D
$\frac{x}{16}$
✓ Correct Answer: D - $\frac{x}{16}$
📖 Step-by-Step Solution & Conceptual Rationale:
$\frac{4}{4}×\frac{x}{4}=\frac{x}{16}$.
Sample Question 3
Algebra
Hard • Quantitative Aptitude Test
What is $\frac{1}{5}$ of $\frac{x}{5}$?
A
25x
B
$\frac{1}{25}$
C
$\frac{1}{x}$
D
$\frac{x}{25}$
✓ Correct Answer: D - $\frac{x}{25}$
📖 Step-by-Step Solution & Conceptual Rationale:
$\left(\frac{1}{5}\right) \times \left(\frac{x}{5}\right) = \frac{x}{25}$
Sample Question 4
Algebra
Hard • Quantitative Aptitude Test
Тo a certain number 8 is added. The sum is multiplied by 3, then the product is divided by 2 and 7 is subtracted from the quotient. The remainder left is 50. The number is:
A
30
B
40
C
50
D
60
✓ Correct Answer: A - 30
📖 Step-by-Step Solution & Conceptual Rationale:
We can solve this problem by working backward from the final remainder of 50:
Undo the subtraction:
Before subtracting 7, the number was:
\(50+7=57\)
Undo the division:
Before dividing by 2, the product was:
\(57\times 2=114\)
Undo the multiplication:
Before multiplying by 3, the sum was:
\(114\div 3=38\)
Undo the addition:
Before adding 8, the original number was:
\(38-8=\mathbf{30}\)
Undo the subtraction:
Before subtracting 7, the number was:
\(50+7=57\)
Undo the division:
Before dividing by 2, the product was:
\(57\times 2=114\)
Undo the multiplication:
Before multiplying by 3, the sum was:
\(114\div 3=38\)
Undo the addition:
Before adding 8, the original number was:
\(38-8=\mathbf{30}\)
Sample Question 5
Algebra
Hard • Quantitative Aptitude Test
Minimum value of $x^{3}+10x+7$?
A
7
B
-18
C
Zero
D
None of these
✓ Correct Answer: D - None of these
📖 Step-by-Step Solution & Conceptual Rationale:
To find the minimum value of the cubic function \(f(x) = x^3 + 10x + 7\), we look at its first derivative to check for critical points (peaks or valleys):
Find the derivative:
\(f^{\prime }(x)=3x^{2}+10\)
Analyze the derivative:
Since any real number squared (\(x^{2}\)) is always zero or positive, \(3x^2 \ge 0\). Adding \(10\) means that:
\(f^{\prime }(x)\ge 10\quad \text{for all real values of }x.\)
Conclusion:
Because the derivative is always positive (\(f'(x) > 0\)), the graph of the function is strictly increasing from left to right. It has no local minimum or local maximum points.As \(x\) goes towards negative infinity (\(-\infty \)), the value of \(x^3 + 10x + 7\) also goes towards negative infinity (\(-\infty \)). Because it decreases without bound, it does not have a minimum real number value, making None of these the correct choice.
Find the derivative:
\(f^{\prime }(x)=3x^{2}+10\)
Analyze the derivative:
Since any real number squared (\(x^{2}\)) is always zero or positive, \(3x^2 \ge 0\). Adding \(10\) means that:
\(f^{\prime }(x)\ge 10\quad \text{for all real values of }x.\)
Conclusion:
Because the derivative is always positive (\(f'(x) > 0\)), the graph of the function is strictly increasing from left to right. It has no local minimum or local maximum points.As \(x\) goes towards negative infinity (\(-\infty \)), the value of \(x^3 + 10x + 7\) also goes towards negative infinity (\(-\infty \)). Because it decreases without bound, it does not have a minimum real number value, making None of these the correct choice.
Sample Question 6
Algebra
Hard • Quantitative Aptitude Test
Expression \(x^{2}−3x+5\) has a minimum value for x equal to:
A
3.2
B
Zero
C
1.5
D
2.4
✓ Correct Answer: C - 1.5
📖 Step-by-Step Solution & Conceptual Rationale:
The expression \(f(x) = x^2 - 3x + 5\) is a quadratic equation where the coefficient of \(x^{2}\) is positive (\(a = 1\)). This means its graph is a parabola that opens upward, and its lowest point (minimum value) occurs at its vertex.
To find the value of \(x\) at the vertex, we use the vertex formula:
\(x=-\frac{b}{2a}\)
Identify the coefficients:
\(a = 1\)
\(b = -3\)
Plug the numbers into the formula:
\(x=-\frac{-3}{2(1)}=\frac{3}{2}=\mathbf{1.5}\)
Therefore, the expression reaches its minimum value when \(x\) is equal to 1.5.
To find the value of \(x\) at the vertex, we use the vertex formula:
\(x=-\frac{b}{2a}\)
Identify the coefficients:
\(a = 1\)
\(b = -3\)
Plug the numbers into the formula:
\(x=-\frac{-3}{2(1)}=\frac{3}{2}=\mathbf{1.5}\)
Therefore, the expression reaches its minimum value when \(x\) is equal to 1.5.
Sample Question 7
Algebra
Hard • Quantitative Aptitude Test
Value of $64×128×512×2^{-3}$ is?
A
$2^{20}$
B
$2^{21}$
C
$2^{19}$
D
$2^{17}$
✓ Correct Answer: C - $2^{19}$
📖 Step-by-Step Solution & Conceptual Rationale:
To solve this, express each number as a power of base 2:
\(64 = 2^6\)
\(128 = 2^7\)
\(512 = 2^9\)
Now, substitute these back into the expression:
\(2^{6}\times 2^{7}\times 2^{9}\times 2^{-3}\)
When multiplying terms with identical bases, you add the exponents together:
\(2^{6+7+9+(-3)}\)
\(2^{22-3}=\mathbf{2}^{\mathbf{19}}\)
\(64 = 2^6\)
\(128 = 2^7\)
\(512 = 2^9\)
Now, substitute these back into the expression:
\(2^{6}\times 2^{7}\times 2^{9}\times 2^{-3}\)
When multiplying terms with identical bases, you add the exponents together:
\(2^{6+7+9+(-3)}\)
\(2^{22-3}=\mathbf{2}^{\mathbf{19}}\)
Sample Question 8
Algebra
Hard • Quantitative Aptitude Test
Geometric mean of $a^{2n}×b^{2n}$?
A
$\frac{(a^{2n}b^{2n})}{2}$
B
$a^nb^n$
C
$\frac{(a^{2n}+b^{2n})}{2}$
D
$(ab)^{2n}$
✓ Correct Answer: B - $a^nb^n$
📖 Step-by-Step Solution & Conceptual Rationale:
The Geometric Mean (GM) of two quantities \(x\) and \(y\) is found by taking the square root of their product:
\(\text{GM}=\sqrt{x\times y}\)
Substitute the given expressions:
Here, \(x = a^{2n}\) and \(y = b^{2n}\).
\(\text{GM}=\sqrt{a^{2n}\times b^{2n}}\)
Combine the bases under the exponent:
\(\text{GM}=\sqrt{(ab)^{2n}}\)
Simplify the square root:
Taking the square root of an exponential expression is the same as dividing its exponent by 2:
\(\text{GM}=\left((ab)^{2n}\right)^{\frac{1}{2}}=(ab)^{n}=\mathbf{a}^{\mathbf{n}}\mathbf{b}^{\mathbf{n}}\)
\(\text{GM}=\sqrt{x\times y}\)
Substitute the given expressions:
Here, \(x = a^{2n}\) and \(y = b^{2n}\).
\(\text{GM}=\sqrt{a^{2n}\times b^{2n}}\)
Combine the bases under the exponent:
\(\text{GM}=\sqrt{(ab)^{2n}}\)
Simplify the square root:
Taking the square root of an exponential expression is the same as dividing its exponent by 2:
\(\text{GM}=\left((ab)^{2n}\right)^{\frac{1}{2}}=(ab)^{n}=\mathbf{a}^{\mathbf{n}}\mathbf{b}^{\mathbf{n}}\)
Sample Question 9
Algebra
Hard • Quantitative Aptitude Test
Value of $(2/7)^{−2}$?
A
$\frac{-2}{7}$
B
$(\frac{7}{2})^2$
C
$(\frac{2}{7})^{-6}$
D
$(\frac{2}{7})^{6}$
✓ Correct Answer: B - $(\frac{7}{2})^2$
📖 Step-by-Step Solution & Conceptual Rationale:
Apply the negative exponent rule:
A negative exponent tells you to invert the fraction (find its reciprocal) and make the exponent positive:
\(\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^{n}\)
\(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
A negative exponent tells you to invert the fraction (find its reciprocal) and make the exponent positive:
\(\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^{n}\)
\(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)
Sample Question 10
Algebra
Hard • Quantitative Aptitude Test
Expansion of $(X+3)^{19}$ will contain __ terms?
A
19
B
21
C
20
D
18
✓ Correct Answer: C - 20
📖 Step-by-Step Solution & Conceptual Rationale:
In binomial expansion (a+b)n, the number of terms is always n+1.
According to the Binomial Theorem, when you expand a binomial expression raised to a positive integer power \(n\) (in the form \((a+b)^n\)), the total number of terms in the expanded form is always equal to \(n + 1\).
Calculation
Given exponent (\(n\)) = \(19\)
Total terms = \(19 + 1 = \mathbf{20}\)
According to the Binomial Theorem, when you expand a binomial expression raised to a positive integer power \(n\) (in the form \((a+b)^n\)), the total number of terms in the expanded form is always equal to \(n + 1\).
Calculation
Given exponent (\(n\)) = \(19\)
Total terms = \(19 + 1 = \mathbf{20}\)
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