PracticeVideosBlogFlashcardsEspañol
All Practice Exams

Free Practice Questions for NSTSE

Exam-style questions and explanations by OpenExamPrep.

✓ No registration✓ No credit card
Exam pass rate: Competitive ranking system; top 1-2% receive top distinction awards.
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More Indian School Olympiads

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

SOF IEO — International English Olympiad

Practice Questions

SOF IMO — International Mathematics Olympiad

Practice Questions

SOF NSO — National Science Olympiad

Practice Questions
Exam Review

Key Facts: NSTSE Exam

100

Questions

Unified Council

90 mins

Exam Time

Unified Council

₹354

Individual Fee

Unified Council 2026

0

Negative Marks

Unified Council

Class 1-12

Eligibility

Unified Council

SPR

Feedback Report

Unified Council Diagnostic

The NSTSE is a national-level diagnostic exam for school students from Class 1 to 12, administered by the Unified Council. For Class 10, it consists of 100 multiple-choice questions across Mathematics, Physics, Chemistry, Biology, and Critical Thinking, with a 90-minute time limit and no negative marking. Registration fees are ₹354 for individual online applicants and ₹150 for school registrations. The exam focuses on assessing conceptual understanding rather than rote recall.

Sample NSTSE Practice Questions

Try these sample questions to review concepts for the NSTSE exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x$ and $y$ are prime numbers, then what is the HCF ($a, b$) and LCM ($a, b$)?
A.HCF = x y^2, LCM = x^3 y^3
B.HCF = x^3 y^3, LCM = x y^2
C.HCF = x^2 y^2, LCM = x^4 y^5
D.HCF = x y, LCM = x^3 y^3
Explanation: To find the Highest Common Factor (HCF) of two numbers expressed in terms of their prime factors, we take the product of the terms with the lowest power of each common prime factor. For $a = x^3 y^2$ and $b = x y^3$, the lowest power of $x$ is $x^1$ and the lowest power of $y$ is $y^2$, giving an HCF of $x y^2$. For the Least Common Multiple (LCM), we take the highest power of each prime factor, giving $x^3 y^3$.
2If one zero of the quadratic polynomial $f(x) = 4x^2 - 8kx - 9$ is the negative of the other, then what is the value of $k$?
A.0
B.1
C.-1
D.9/4
Explanation: Let the zeroes of the quadratic polynomial be $\alpha$ and $-\alpha$. According to the relationship between the zeroes and the coefficients of a quadratic polynomial $ax^2 + bx + c$, the sum of the zeroes is given by $-b/a$. Thus, $\alpha + (-\alpha) = -(-8k)/4 \implies 0 = 2k \implies k = 0$.
3For what value of $k$ do the equations $3x - y + 8 = 0$ and $6x - ky = -16$ represent coincident lines?
A.2
B.1/2
C.-2
D.-1/2
Explanation: For two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to represent coincident lines, the condition of consistency is $a_1/a_2 = b_1/b_2 = c_1/c_2$. Rearranging the second equation gives $6x - ky + 16 = 0$. Comparing coefficients: $3/6 = -1/(-k) = 8/16 \implies 1/2 = 1/k \implies k = 2$.
4If the quadratic equation $(1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0$ has real and equal roots, which of the following is the correct relationship between $a$, $c$, and $m$?
A.c^2 = a^2(1 + m^2)
B.a^2 = c^2(1 + m^2)
C.c^2 = a^2(1 - m^2)
D.c^2 = a^2 + m^2
Explanation: For equal roots, the discriminant $D = B^2 - 4AC$ must be zero. Here, $A = 1 + m^2$, $B = 2mc$, and $C = c^2 - a^2$. Setting $D = 0 \implies (2mc)^2 - 4(1+m^2)(c^2-a^2) = 0 \implies 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \implies m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \implies -c^2 + a^2(1 + m^2) = 0 \implies c^2 = a^2(1 + m^2)$.
5If the sum of the first $n$ terms of an Arithmetic Progression is given by $S_n = 3n^2 + 5n$, what is its common difference ($d$) and its 12th term ($a_{12}$)?
A.d = 6, a_12 = 74
B.d = 3, a_12 = 41
C.d = 6, a_12 = 77
D.d = 5, a_12 = 65
Explanation: The $n$-th term is given by $a_n = S_n - S_{n-1}$. For $n=1$, $S_1 = 3(1)^2 + 5(1) = 8$, so the first term $a_1 = 8$. For $n=2$, $S_2 = 3(2)^2 + 5(2) = 22$. The second term is $a_2 = S_2 - S_1 = 22 - 8 = 14$. The common difference is $d = a_2 - a_1 = 14 - 8 = 6$. The 12th term is $a_{12} = a_1 + 11d = 8 + 11(6) = 74$.
6In $\Delta ABC$, $DE \parallel BC$ such that $AD/DB = 3/5$. If $AC = 5.6 \text{ cm}$, then what is the length of $AE$?
A.2.1 cm
B.3.5 cm
C.1.8 cm
D.2.4 cm
Explanation: By the Basic Proportionality Theorem (Thales Theorem), if $DE \parallel BC$, then $AD/DB = AE/EC$. Let $AE = x$. Since $AC = 5.6 \text{ cm}$, $EC = 5.6 - x$. Thus, $3/5 = x/(5.6 - x) \implies 3(5.6 - x) = 5x \implies 16.8 - 3x = 5x \implies 8x = 16.8 \implies x = 2.1 \text{ cm}$.
7In what ratio does the y-axis divide the line segment joining the points $A(5, -6)$ and $B(-1, -4)$?
A.5:1
B.1:5
C.2:3
D.3:4
Explanation: Let the y-axis divide the line segment joining $A(5, -6)$ and $B(-1, -4)$ in the ratio $k:1$. The coordinates of the point of division are given by the section formula: $(\frac{-k + 5}{k + 1}, \frac{-4k - 6}{k + 1})$. Since the point lies on the y-axis, its x-coordinate is 0. So, $\frac{-k + 5}{k + 1} = 0 \implies -k + 5 = 0 \implies k = 5$. Thus, the ratio is $5:1$.
8If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, then what is the value of $\cos \theta - \sin \theta$?
A.\sqrt{2} \sin \theta
B.\sqrt{2} \cos \theta
C.1/\sqrt{2}
D.\sin \theta
Explanation: Given $\sin \theta + \cos \theta = \sqrt{2} \cos \theta \implies \sin \theta = (\sqrt{2} - 1) \cos \theta$. Multiplying both sides by $(\sqrt{2} + 1)$, we get $(\sqrt{2} + 1) \sin \theta = (\sqrt{2} - 1)(\sqrt{2} + 1) \cos \theta \implies \sqrt{2} \sin \theta + \sin \theta = \cos \theta \implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta$.
9From a point $P$ which is at a distance of $13 \text{ cm}$ from the center $O$ of a circle of radius $5 \text{ cm}$, a pair of tangents $PQ$ and $PR$ are drawn to the circle. What is the area of the quadrilateral $PQOR$?
A.60 sq cm
B.30 sq cm
C.120 sq cm
D.65 sq cm
Explanation: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Thus, $\angle PQO = 90^\circ$. In the right-angled triangle $PQO$, using the Pythagoras theorem: $PQ = \sqrt{OP^2 - OQ^2} = \sqrt{13^2 - 5^2} = 12 \text{ cm}$. The area of $\Delta PQO = (1/2) \times \text{base} \times \text{height} = (1/2) \times 12 \times 5 = 30 \text{ sq cm}$. Since the quadrilateral $PQOR$ is composed of two such congruent triangles, its area is $2 \times 30 = 60 \text{ sq cm}$.
10A metallic solid sphere of radius $10.5 \text{ cm}$ is melted and recast into smaller solid cones, each of radius $3.5 \text{ cm}$ and height $3 \text{ cm}$. How many such cones can be formed?
A.126
B.21
C.63
D.252
Explanation: Let $n$ be the number of cones. The volume of the sphere is equal to the total volume of $n$ cones. Volume of sphere = $(4/3) \pi R^3$, and volume of each cone = $(1/3) \pi r^2 h$. Thus, $(4/3) \pi R^3 = n \times (1/3) \pi r^2 h \implies n = (4 R^3) / (r^2 h) = (4 \times 10.5 \times 10.5 \times 10.5) / (3.5 \times 3.5 \times 3) = (4 \times 3 \times 3 \times 10.5) / 3 = 4 \times 3 \times 10.5 = 126$ cones.

About the NSTSE Exam

The National Level Science Talent Search Examination (NSTSE) is a diagnostic test conducted by the Unified Council for students in Classes 1 to 12. It measures conceptual understanding and analytical thinking, moving away from rote memorization. The exam covers Mathematics, Physics, Chemistry, Biology, and Critical Thinking, providing students with a detailed Student's Performance Report (SPR) to identify their strengths and weaknesses.

Exam sponsor: Unified Council. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

100 multiple-choice questions covering Mathematics, Physics, Chemistry, Biology, and General Questions (Critical Thinking)

Time Limit

90 minutes

Passing Score

Percentile-based ranking across national and state levels

Exam / Certification Fees

₹354 (Individual) / ₹150 (School)

Exam sponsor website

Reported exam pass rate: Competitive ranking system; top 1-2% receive top distinction awards.. based on national performance curves This describes exam candidates, not OpenExamPrep users or results from using our resources. Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

25%

Mathematics

Number Systems, Polynomials, Linear and Quadratic Equations, Progressions, Geometry, Trigonometry, Coordinate Geometry, Mensuration, and Probability.

25%

Physics

Light, Human Eye and Colorful World, Electricity, Magnetic Effects of Electric Current, Motion, Forces, and Gravitation.

20%

Chemistry

Chemical Reactions and Equations, Acids, Bases and Salts, Metals and Non-metals, Carbon and its Compounds, and Atomic Structure.

20%

Biology

Life Processes, Control and Coordination, Organism Reproduction, Heredity and Evolution, Cell and Tissues, and Environmental Management.

10%

Critical Thinking

Logical deduction, Syllogisms, Pattern recognition, Coding-Decoding, blood relations, and logical decision-making scenarios.

Preparing for the NSTSE Exam

What You Need to Know

  • Passing score: Percentile-based ranking across national and state levels
  • Assessment: 100 multiple-choice questions covering Mathematics, Physics, Chemistry, Biology, and General Questions (Critical Thinking)
  • Time limit: 90 minutes
  • Exam / certification fees: ₹354 (Individual) / ₹150 (School) Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

NSTSE: Suggested Study Strategy

1Focus on the 'why' behind scientific phenomena rather than memorizing definitions.
2Understand the derivation and logical application of mathematical formulas in real-world contexts.
3Practice solving logical puzzles and reasoning patterns to score high on the Critical Thinking section.
4Use NCERT and standard boards' textbooks as the baseline, and supplement with past papers from the Unified Council.
5Learn to manage your time: for Class 10, you have 90 minutes for 100 questions, meaning less than a minute per question.

Frequently Asked Questions

What is the primary objective of the NSTSE?

Unlike traditional school exams which often test memory recall, the NSTSE is a diagnostic test designed to assess a student's actual understanding of concepts in science and mathematics, as well as their analytical and critical thinking skills.

Is there any negative marking in the NSTSE?

No, there is no negative marking in the NSTSE. Students are encouraged to answer all questions.

Which syllabus is the NSTSE based on?

The exam is based on the CBSE syllabus. However, it is designed in a way that is highly suitable for ICSE, ISC, and various State Board students as well.

How much does it cost to register for the NSTSE?

The registration fee is ₹354 (including GST) for individual candidates registering online. For students registering through their respective schools, the fee is typically around ₹150.

What is the Student's Performance Report (SPR)?

The SPR is a highly detailed report provided by the Unified Council to each participant. It provides a question-by-question analysis, subject-wise marks, skill-wise analysis, and national/state-level rank comparisons, helping students identify their exact learning gaps.

Can I take the NSTSE online?

Yes, individual registered students can take the exam in an online proctored format. School-registered students typically write the exam offline in their respective schools.