CBSE Class 10 Sample Paper 2025 Science Standard Set 1 with Solutions

Mastering CBSE Class 10 Science requires structured practice with the latest board patterns. The CBSE Class 10 Sample Paper 2025 Science Standard Set 1 with Solutions serves as an essential resource for students targeting a 95+ score in the 2025–2026 board exam 10. This comprehensive guide breaks down the standard question paper pattern, key concepts, marking schemes, and step-by-step NCERT-aligned solutions for high-yield competency-based questions.

CBSE Class 10 Science Exam Pattern & Blueprint (2025–2026)

The annual CBSE Science theory examination carries 80 marks with a duration of 3 hours, complemented by 20 marks of internal assessment. The paper is structured into five distinct sections designed to test conceptual understanding, critical reasoning, and practical application:

Key Concepts and Chapter-Wise Breakdown

To solve CBSE Class 10 Sample Paper 2025 Science Standard Set 1 effectively, students must focus on fundamental laws, chemical reactions, ray optics, and biological mechanisms across the three core domains.

1. Chemical Substances – Nature and Behaviour (Chemistry)

Chemistry accounts for 25 marks. Key areas include:

• Types of Chemical Reactions: Balancing chemical equations, distinguishing combination, decomposition, displacement, double displacement, and redox reactions. In redox reactions, remember that the substance gaining oxygen or losing electrons undergoes oxidation, whereas the substance losing oxygen or gaining electrons undergoes reduction.
• Acids, Bases, and Salts: Understanding pH scale applications in daily life (e.g., tooth decay prevention at pH < 5.5, soil treatment), and preparation of industrial salts like Bleaching Powder (CaOCl₂), Baking Soda (NaHCO₃), Washing Soda (Na₂CO₃·10H₂O), and Plaster of Paris (CaSO₄·½H₂O).
• Metals and Non-Metals: Reactivity series, extraction of metals (roasting vs. calcination), electrolytic refining, and formation of ionic compounds with electron dot structures.
• Carbon and its Compounds: Covalent bonding, tetravalency and catenation of carbon, functional groups (alcohols, aldehydes, ketones, carboxylic acids), combustion, addition, substitution reactions, and the mechanism of soap micelle formation.

2. World of Living (Biology)

Biology carries 25 marks with heavy emphasis on physiological processes and genetics:

• Life Processes: Stomatal mechanism, human digestive enzyme actions, double circulation in the human heart, pathway of breakdown of glucose (aerobic respiration in mitochondria vs. anaerobic respiration in yeast and muscle cells), and nephron filtration in kidneys.
• Control and Coordination: Structure and transmission of nerve impulses in a neuron, reflex arc pathway, human brain lobes, and plant phytohormones (Auxin, Gibberellin, Cytokinin, Abscisic acid).
• Reproduction: Asexual reproduction modes (binary fission, budding, spore formation, vegetative propagation) and sexual reproduction in flowering plants (pollination and double fertilisation) and human reproductive systems.
• Heredity: Gregor Mendel’s experiments with garden pea (Pisum sativum) — Monohybrid cross (3:1 phenotypic, 1:2:1 genotypic ratio) and Dihybrid cross (9:3:3:1 phenotypic ratio) demonstrating the Law of Independent Assortment.

3. Natural Phenomena & Effects of Current (Physics)

Physics carries 25 marks and involves rigorous numerical problem solving:

• Light – Reflection and Refraction:
  • Mirror Formula: 1/f = 1/v + 1/u with magnification m = -v/u = h'/h
  • Lens Formula: 1/f = 1/v - 1/u with magnification m = v/u = h'/h
  • Power of a Lens: P = 1/f(in metres), measured in Dioptres (D). Convex lenses have positive power (+D); concave lenses have negative power (−D).
• Electricity:
  • Ohm’s Law: V = I × R
  • Resistance and Resistivity: R = ρ × (l / A) where ρ is electrical resistivity of the material.
  • Series Combination: Rs = R1 + R2 + R3
  • Parallel Combination: 1/Rp = 1/R1 + 1/R2 + 1/R3
  • Joule’s Law of Heating: H = I2Rt = VIt = (V2/R)t
  • Electric Power: P = VI = I2R = V2/R (1 kWh = 3.6 × 10⁶ J)
• Magnetic Effects of Electric Current: Magnetic field lines around straight conductors, circular loops, and solenoids; Fleming’s Left-Hand Rule for motor force; and the functioning of domestic electric circuits (earthing wire, fuse rating, short-circuiting).

Important CBSE Questions with Step-by-Step Solutions

Below are representative standard questions from the CBSE Class 10 Sample Paper 2025 Science Standard Set 1 with Solutions illustrating step-by-step marking criteria.

Section A (1 Mark Multiple Choice & Assertion-Reason)

Question 1: In the reaction: MnO2 + 4HCl → MnCl2 + 2H2O + Cl2, identify the substance oxidized and the reducing agent.
(a) MnO₂ is oxidized, HCl is the reducing agent.
(b) HCl is oxidized, MnO₂ is the reducing agent.
(c) HCl is oxidized, HCl is the reducing agent.
(d) MnO₂ is oxidized, MnO₂ is the reducing agent.

Solution: (c) HCl is oxidized, HCl is the reducing agent.
Explanation: Removal of hydrogen from HCl yields Cl₂ (oxidation). Since HCl undergoes oxidation by donating electrons/hydrogen to MnO₂, it acts as the reducing agent. MnO₂ loses oxygen (reduction) and acts as the oxidizing agent.

Question 2:
Assertion (A): The left ventricle of the human heart has a thicker muscular wall than the right ventricle.
Reason (R): The left ventricle has to pump oxygenated blood to all distant body organs against higher systemic resistance.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Solution: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Explanation: The right ventricle only pumps deoxygenated blood to the nearby lungs through the pulmonary artery under low pressure. The left ventricle pumps oxygenated blood through the aorta across the entire systemic circulation, requiring stronger muscular walls to withstand and generate greater pressure.

Section B (2 Marks Short Answer Type I)

Question 3: An electric lamp of resistance 20 Ω and a conductor of 4 Ω resistance are connected in series to a 6 V battery. Calculate: (a) Total resistance of the circuit, and (b) The current flowing through the circuit.

Solution:
Given:
Resistance of lamp, R1 = 20 Ω
Resistance of conductor, R2 = 4 Ω
Potential difference, V = 6 V

Step 1: Total Resistance (Series Connection)
Rs = R1 + R2 = 20 Ω + 4 Ω = 24 Ω (1 mark)

Step 2: Electric Current (Ohm’s Law)
I = V / Rs = 6 V / 24 Ω = 0.25 A (1 mark)
Final Answer: Total resistance is 24 Ω and current is 0.25 A.

Section C (3 Marks Short Answer Type II)

Question 4: Differentiate between roasting and calcination in metallurgy. Write balanced chemical equations for each using zinc ores.

Solution:

• Roasting: The process of strongly heating sulphide ores below their melting points in the presence of excess air to convert them into metal oxides.
  Equation (Zinc Blende ore):
  2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) ↑ (1.5 marks)
• Calcination: The process of heating carbonate ores strongly in limited air or in the absence of air to convert them into metal oxides.
  Equation (Calamine ore):
  ZnCO3(s) → ZnO(s) + CO2(g) ↑ (1.5 marks)

Section D (5 Marks Long Answer Type)

Question 5:
(a) An object 4 cm high is placed at a distance of 15 cm in front of a concave mirror of focal length 10 cm. Find the position, nature, and size of the image formed.
(b) Draw a neat labeled ray diagram to illustrate the image formation in this case.

Solution:
(a) Numerical Calculation:
Given:
Height of object, h = +4 cm
Object distance, u = -15 cm (Sign convention)
Focal length of concave mirror, f = -10 cm

Using Mirror Formula:
1/f = 1/v + 1/u ⇒ 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/(-15) = -1/10 + 1/15
1/v = (-3 + 2) / 30 = -1/30
v = -30 cm (1.5 marks)

Position of image: 30 cm in front of the mirror (on the same side as the object).

Using Magnification Formula:
m = -v / u = -(-30) / (-15) = -2 (1 mark)
m = h' / h ⇒ h' = m × h = -2 × 4 cm = -8 cm (1 mark)

Nature of Image: Real, inverted, and magnified (2 times). (0.5 mark)

(b) Ray Diagram Description:
When the object is placed between the Focus (F) and Centre of Curvature (C):
1. A ray parallel to the principal axis passes through F after reflection.
2. A ray passing through F emerges parallel to the principal axis after reflection.
The reflected rays intersect beyond C, forming a real, inverted, and enlarged image. (1 mark)

Section E (4 Marks Case-Based Study)

Question 6: Read the following passage and answer the questions that follow:
Gregor Johann Mendel crossed pure-breeding pea plants having round yellow seeds (RRYY) with pure-breeding pea plants having wrinkled green seeds (rryy). The F₁ generation plants were all round and yellow. On self-pollinating the F₁ plants, a total of 1600 seeds were harvested in the F₂ generation.

Sub-questions:
(i) What was the genotype of the F₁ progeny? (1 Mark)
(ii) State the phenotypic ratio of round-yellow, round-green, wrinkled-yellow, and wrinkled-green seeds observed in the F₂ generation. (1 Mark)
(iii) Calculate the expected number of seeds having a recombinant phenotype (wrinkled yellow and round green combined) in the F₂ generation. (2 Marks)

Solution:
(i) The genotype of the F₁ progeny is heterozygous round yellow: RrYy. (1 mark)
(ii) The phenotypic dihybrid ratio in F₂ is 9 : 3 : 3 : 1 (9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green). (1 mark)
(iii) Recombinant phenotypes are Round Green (3/16) and Wrinkled Yellow (3/16).
Total recombinant fraction = (3/16) + (3/16) = 6/16 = 3/8
Expected number of recombinant seeds = (6/16) × 1600 = 600 seeds. (2 marks)

How to Prepare for This Topic and Score 95+ in Board Exam 10

Achieving top scores in CBSE Class 10 Science requires deliberate revision and structured answer writing:

1. Maintain Sign Conventions in Numerical Problems: Always write down the given values with standard Cartesian sign conventions (e.g., object distance u is always negative, focal length of concave mirror/lens is negative, convex is positive). State explicit SI units in final answers.
2. Write Complete Balanced Chemical Equations: For every chemical reaction mentioned in Section C and Section D, provide the balanced chemical equation with physical state symbols (s, l, g, aq) and reaction conditions (temperature, catalyst, sunlight) over the arrow.
3. Practice NCERT Exemplar Diagrams: Perfect diagrams for human heart, nephron structure, reflex arc, female reproductive system, ray diagrams for spherical lenses/mirrors, and magnetic field lines around a solenoid. Always use a sharp pencil and write labels horizontally.
4. Structure Case-Based Questions: Read Section E extracts carefully, identify the core biological or physical law involved, and answer each sub-part with direct point-wise statements and calculations.
5. Time Management Strategy: Allocate 30 minutes for Section A (20 MCQs), 25 minutes for Section B, 45 minutes for Section C, 45 minutes for Section D, 25 minutes for Section E, and preserve 10 minutes strictly for final revision and unit checks.

Where to Practice More

To master the full spectrum of competency-based questions, download authentic model tests, chapter-wise mock tests, and standard solved question sets on qptool.theorify.in. Practice under strict 3-hour timed conditions to boost speed, accuracy, and confidence for your board exam.