CBSE Class 10 Science Question Paper 2025 with Solutions PDF: Complete Paper Pattern, Chapter-Wise Solutions, and Score-Booster Guide

Mastering the CBSE Class 10 Science Question Paper 2025 with Solutions PDF is essential for every student aiming to score 95+ in their upcoming board exam 10 for the 2025–2026 academic sessions. The Central Board of Secondary Education has shifted its assessment framework toward competency-based questions, demanding conceptual clarity over rote memorization. This comprehensive guide dissects the 2025 question paper structure, provides step-by-step solutions aligned with the official CBSE marking scheme, and provides actionable preparation strategies to excel in CBSE Science.

CBSE Class 10 Science Question Paper Pattern and Blueprint

The annual CBSE Class 10 Science theoretical paper comprises 80 marks distributed across 39 mandatory questions to be completed within 3 hours. An internal assessment of 20 marks complements this theoretical evaluation.

Section	Question Types	Number of Questions	Marks per Question	Total Marks
Section A	Multiple Choice Questions (MCQs) & Assertion-Reasoning	20 (Q1 to Q20)	1 Mark	20 Marks
Section B	Very Short Answer Questions (VSA)	6 (Q21 to Q26)	2 Marks	12 Marks
Section C	Short Answer Questions (SA)	7 (Q27 to Q33)	3 Marks	21 Marks
Section D	Long Answer Questions (LA)	3 (Q34 to Q36)	5 Marks	15 Marks
Section E	Source-based / Case-based Assessments	3 (Q37 to Q39)	4 Marks	12 Marks
Total	Comprehensive Theory Paper	39 Questions	—	80 Marks

Core Scientific Concepts and High-Yield Formulas

A rigorous review of the CBSE Class 10 Science syllabus spans three core disciplines: Chemistry, Biology, and Physics. Below are foundational concepts, formulas, and balanced equations frequently evaluated in the 2025 examination.

1. Chemical Substances: Nature and Behaviour (Chemistry)

CBSE consistently tests balanced chemical equations with correct physical states, reaction mechanisms, and qualitative observations:

Redox Processes: Oxidation involves the addition of oxygen or removal of hydrogen; reduction involves the addition of hydrogen or removal of oxygen. In the industrial extraction of copper:
2Cu₂S(s) + 3O₂(g) → 2Cu₂O(s) + 2SO₂(g)
2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
Chlor-Alkali Process: Electrolysis of aqueous sodium chloride (brine):
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)
Applications: Chlorine at the anode for water treatment; hydrogen at the cathode for fuel/ammonia; sodium hydroxide for soap and paper manufacturing.
Esterification and Saponification:
Esterification: CH₃COOH + C₂H₅OH —(conc. H₂SO₄)→ CH₃COOC₂H₅ + H₂O
Saponification: CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH

2. World of Living (Biology)

High-scoring answers in Biology require precise terminology and clean flowchart representations:

Aerobic vs. Anaerobic Respiration: Aerobic breakdown of glucose occurs in the mitochondria yielding 6CO₂ + 6H₂O + 38 ATP, whereas anaerobic respiration in human muscle cells during vigorous exercise produces lactic acid (C₃H₆O₃ + 2 ATP), causing muscular cramps.
Excretion Mechanism in Nephrons: Involves three sequential steps:
1. Ultrafiltration: Blood is filtered under pressure in the glomerulus into Bowman's capsule.
2. Selective Reabsorption: Glucose, amino acids, salts, and major fractions of water are reabsorbed along the renal tubule.
3. Tubular Secretion: Excess ions and nitrogenous wastes (urea) are secreted into the collecting duct to form urine.
Mendelian Inheritance: A monohybrid cross of pure tall (TT) and pure dwarf (tt) pea plants generates a phenotypic ratio of 3:1 (Tall : Dwarf) and a genotypic ratio of 1:2:1 (1 TT : 2 Tt : 1 tt) in the F₂ generation.

3. Natural Phenomena and Effects of Current (Physics)

Physics numericals demand correct sign conventions (New Cartesian Sign Convention) and standard SI units:

Mirror Formula & Magnification:
1/f = 1/v + 1/u | m = -v/u = h'/h
Lens Formula & Lens Power:
1/f = 1/v - 1/u | m = v/u = h'/h | P = 1/f (in meters), measured in Dioptres (D).
Current Electricity & Joule's Heating Effect:
Ohm's Law: V = I · R | Resistance: R = ρ · (l / A)
Resistors in Series: R_s = R₁ + R₂ + R₃
Resistors in Parallel: 1/R_p = 1/R₁ + 1/R₂ + 1/R₃
Joule's Heat Energy: H = I²Rt = V · I · t = (V² / R) · t

Official-Pattern CBSE Class 10 Science Questions with Step-by-Step Solutions

Below are representative high-weightage questions from the 2025 CBSE Class 10 Science paper model, accompanied by official marking scheme breakdowns.

Section A (1 Mark - Objective / Competency)

Question 1: An aqueous solution of a salt turns red litmus paper blue and produces effervescence with dilute hydrochloric acid. Identify the salt and write the balanced chemical reaction.
(a) Calcium sulphate
(b) Sodium carbonate
(c) Potassium chloride
(d) Ammonium chloride

Answer: (b) Sodium carbonate (Na₂CO₃)
Explanation: Sodium carbonate is a basic salt formed from a strong base (NaOH) and a weak acid (H₂CO₃). Its solution is alkaline, turning red litmus blue. When treated with dilute HCl, it releases carbon dioxide gas with brisk effervescence:
Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g) [1 Mark]

Question 2 (Assertion-Reason):
Assertion (A): The magnetic field lines around a straight current-carrying conductor are concentric circles.
Reason (R): The direction of the magnetic field is given by Fleming's Left-Hand Rule.
(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.

Answer: (c) (A) is true, but (R) is false.
Explanation: The magnetic field lines are concentric circles as demonstrated by iron filings around a straight conductor. However, the direction of magnetic field around a current-carrying straight wire is determined by the Right-Hand Thumb Rule, not Fleming's Left-Hand Rule (which is used for finding magnetic force on a moving charge/conductor). [1 Mark]

Section B (2 Marks - Short Answer I)

Question 3: What is bile juice? Although it contains no digestive enzymes, why is it considered crucial for digestion in humans?

Solution:
Bile juice is a greenish-yellow digestive fluid secreted by the liver and stored in the gallbladder. [0.5 Mark]
It plays two vital roles in human digestion despite having no enzymes:

Alkalinization of Chyme: It converts the acidic food arriving from the stomach into an alkaline medium, which is mandatory for pancreatic enzymes (like trypsin and lipase) to act efficiently. [0.75 Mark]
Emulsification of Fats: Bile salts break down large, insoluble fat globules into smaller micro-droplets, vastly increasing the surface area for the enzyme lipase to digest lipids. [0.75 Mark]

Section C (3 Marks - Short Answer II)

Question 4: An electric heater of resistance 20 Ω draws a current of 15 A from the service mains. Calculate:
(i) The rate at which heat is developed in the heater.
(ii) The total heat produced in 2 hours in kilojoules (kJ).

Solution:
Given: Resistance R = 20 Ω, Current I = 15 A, Time t = 2 h = 2 × 3600 s = 7200 s.
(i) Rate of heat developed (Electric Power P):
P = I² × R
P = (15)² × 20 = 225 × 20 = 4500 W (or J/s) [1.5 Marks]
(ii) Total heat produced in 2 hours:
H = P × t = 4500 W × 7200 s = 32,400,000 J
In kilojoules: H = 32,400,000 / 1000 = 32,400 kJ [1.5 Marks]

Section D (5 Marks - Long Answer)

Question 5:
(a) State the reason why carbon forms covalent compounds rather than ionic compounds.
(b) Define structural isomerism. Draw the structures of two structural isomers of butane (C₄H₁₀).
(c) Explain the cleansing action of soaps with the help of a micelle diagram description.

Solution:
(a) Reason for Covalency: Carbon has atomic number 6 with electronic configuration (2, 4). To achieve inert gas configuration, it cannot gain 4 electrons (forming C^4-) because 6 protons in the nucleus cannot hold 10 electrons. It cannot lose 4 electrons (forming C⁴⁺) because removing 4 electrons requires massive ionization energy. Hence, carbon shares valence electrons to form covalent bonds. [1.5 Marks]
(b) Structural Isomers of Butane: Compounds having identical molecular formula but different structural arrangements are called structural isomers. [0.5 Mark]
Isomers of C₄H₁₀:

n-Butane (straight chain): CH₃—CH₂—CH₂—CH₃
Isobutane / 2-Methylpropane (branched): CH₃—CH(CH₃)—CH₃ [1 Mark]

(c) Cleansing Action of Soap: A soap molecule (e.g., sodium stearate) consists of two ends:

Hydrophilic head: Ionic end (—COO^-Na⁺) that dissolves in water.
Hydrophobic tail: Hydrocarbon chain that dissolves in oily dirt.

When soap is applied to dirty fabric in water, the hydrophobic tails align inward toward the oil droplet while the hydrophilic heads orient outwards facing water, forming spherical clusters called micelles. On mechanical agitation, the entrapped oily dirt is lifted and rinsed away with water. [2 Marks]

Section E (4 Marks - Case-Based Study)

Question 6 (Case Study - Light: Refraction and Lenses):
A student places an object 20 cm in front of a convex lens. A sharp real, inverted image is formed on a screen placed 30 cm on the other side of the lens.
(a) Determine the focal length of the convex lens. [2 Marks]
(b) Find the magnification and nature of the image formed. [1 Mark]
(c) If the object is now shifted to a distance of 10 cm from the lens, predict the nature and position of the new image formed. [1 Mark]

Solution:
(a) Calculation of Focal Length:
According to Cartesian sign conventions:
Object distance, u = -20 cm
Image distance, v = +30 cm
Using lens formula: 1/f = 1/v - 1/u
1/f = 1/30 - (1/-20) = 1/30 + 1/20 = (2 + 3) / 60 = 5 / 60 = 1 / 12
f = +12 cm (Positive focal length confirms a convex lens). [2 Marks]
(b) Magnification:
m = v / u = (+30) / (-20) = -1.5
The image is real, inverted, and magnified (1.5 times the size of the object). [1 Mark]
(c) Shifted Position:
When u = -10 cm, the object lies between the optical centre and principal focus (since u < f = 12 cm).
Result: The lens forms a virtual, erect, and magnified image on the same side of the lens as the object. [1 Mark]

How to Prepare for CBSE Class 10 Science in 2025 and 2026

Achieving top percentiles in CBSE Science requires a systematic approach to syllabus coverage and answer presentation:

Thorough NCERT Mastery: Every diagram (Nephron, Human Brain, Female Reproductive System, Ray Diagrams) and in-text activity box in NCERT textbooks must be practiced at least twice.
Adhere to CBSE Step-Marking Rules: Always write the relevant formula before numerical substitution, state the correct units (e.g., Ω, W, J, D), and include balanced state symbols in chemical equations ((s), (l), (g), (aq)).
Focus on Competency and Assertion-Reason Questions: At least 50% of the paper now evaluates conceptual synthesis rather than factual recall. Practice reasoning-based question banks regularly.
Time Allocation Strategy: Allocate 30 minutes for Section A, 25 minutes for Section B, 45 minutes for Section C, 50 minutes for Section D, and 20 minutes for Section E, reserving the final 10 minutes for thorough paper verification.

Where to Practice More and Download Practice Papers

Consistent practice with timed, syllabus-accurate mock exams is the most effective way to eliminate exam anxiety and build speed for your board exam 10. Visit Theorify QPTool (qptool.theorify.in) to generate customized chapter tests, download the latest CBSE Class 10 Science Question Paper 2025 with Solutions PDF, and access instant AI-powered evaluations with step-by-step marking rubrics.

CBSE Class 10 Science Question Paper 2025 with Solutions PDF