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CBSE Class 10 Pre Board 2026 : Biology (Ryan International School, Rundh, Surat)

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Garvita Nandwana
Ryan International School, Rundh, Surat

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Marking Scheme Strictly Confidential (For Internal and Restricted use only) Secondary School Examination, 2026 (Xth) SUBJECT NAME SCIENCE (Q.P. CODE /Set No 31/1/1) General Instructions: 1 You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and understand the spot evaluation guidelines carefully. 2 Evaluation policy is a confidential policy as it is related to the confidentiality of the examinations conducted, evaluation done and several other aspects. Its leakage to public in any manner could lead to derailment of the examination system and affect the life and future of millions of candidates. Sharing this policy/document to anyone, publishing in any magazine and printing in Newspaper/Website, etc. may invite action under various rules of the Board and IPC. 3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one s own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed. However, while evaluating, answers which are based on latest information or knowledge and/or are innovative, they may be assessed for their correctness otherwise and due marks be awarded to them. In Class-X, while evaluating two competency-based questions, please try to understand given answer and even if reply is not from marking scheme but correct competency is enumerated by the candidate, due marks should be awarded. 4 The Marking scheme carries only suggested value points for the answers. These are in the nature of Guidelines only and do not constitute the complete answer. The students can have their own expression and if the expression is correct, the due marks should be awarded accordingly. 5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. If there is any variation, the same should be zero after deliberation and discussion. The remaining answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the marking of individual evaluators. 6 Evaluators will mark ( ) wherever answer is correct. For wrong answer CROSS X be marked. Evaluators will not put right ( ) while evaluating which gives an impression that answer is correct and no marks are awarded. This is most common mistake which evaluators are committing. 7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of the question should then be totaled up and written in the left-hand margin and encircled. This may be followed strictly. 8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be followed strictly. 9 If a student has attempted an extra question, answer of the question deserving more marks should be retained and the other answer scored out with a note Extra Question . 10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once. PAGE 1 {31/1/1} 11 A full scale of marks 80 (example 0 to 80/70/60/50/40/30 marks as given in Question Paper) has to be used. Please do not hesitate to award full marks if the answer deserves it. 12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions in question paper. 13 Ensure that you do not make the following common types of errors committed by the Examiner in the past : Leaving answer or part thereof unassessed in an answer book. Giving more marks for an answer than assigned to it. Wrong totaling of marks awarded on an answer. Wrong transfer of marks from the inside pages of the answer book to the title page. Wrong question wise totaling on the title page. Wrong totaling of marks of the two columns on the title page. Wrong grand total. Marks in words and figures not tallying/not same. Wrong transfer of marks from the answer book to online award list. Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.) Half or a part of answer marked correct and the rest as wrong, but no marks awarded. 14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and awarded zero (0) Marks. 15 Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously. 16 The Examiners should acquaint themselves with the guidelines given in the Guidelines for Spot Evaluation before starting the actual evaluation. 17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly totaled and written in figures and words. 18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. PAGE 2 {31/1/1} MARKING SCHEME SCIENCE (Subject Code-086) (PAPER CODE: 31/1/1) (10-01-86K) EXPECTED OUTCOMES/VALUE POINTS Q.No. Marks Total Marks SECTION A (Biology) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. (B) / Stomata (D) / Binary Fission (D) / they reproduce asexually (B) / Cerebellum (B) / Trypsin digests proteins and lipase digests emulsified fats. (A) / (ii) and (iii) (C) / Polythene bag, rubber band, ball pen (A) / Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (C) / Assertion (A) is true, but Reason (R) is false. 11. The main function of the diaphragm is to flatten during inhalation, which expands the chest cavity and draws the air into the lungs / helps in breathing. It is located at the base of chest cavity. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 (a) Chewing of food i) Salivation on sight of food It is a voluntary action. ii) It is controlled by forebrain. i) It is a reflex/involuntary action. 1 ii) It is controlled by medulla in the hind brain. 1 (any other suitable difference) OR (b) Pollination i) It is the transfer of pollen grains from anther to the suitable stigma. ii) It occurs in plants. Fertilization i) It is the fusion of male gamete with the female gamete. 1 ii) It occurs in both plants and animals. 1 (any other suitable difference) 2 PAGE 3 {31/1/1} 12. Diagram (a) Stigma (b) Pollen Tube 13. (a) (b) (c) 14. 1 (i) Grass (ii) Deer, Rabbit (iii) Snake, Lion (iv) Lion Primary consumers feed on green plants which have large amount of energy. Only 10% of its energy is available/passed for the next secondary consumer /trophic level. The base is broad as the number/energy/mass of producers is usually the highest in comparison to other trophic levels of the pyramid. 2 1 1 1 3 (a) Nephron (i) (ii) Neuron Filtration/Structural/ (i) Functional unit of the kidney. Filters nitrogenous wastes (ii) from the blood. Structural/ Functional unit of the nervous system. Transmits information from one part of the body to another. 1 (any one, any other suitable difference) (b) Sensory Nerve (i) Carries impulse from receptors to CNS/ Brain and Spinal cord. Motor Nerve (i) Carries impulse from CNS/Brain and Spinal cord to the motor area/ effector organ. 1 (any other suitable difference) PAGE 4 {31/1/1} (c) Consumers (i) (a) Organisms that breakdown dead organic matter into simpler inorganic substances. Transfer energy through the (ii) Recycle nutrients back into food chain. the environment. (any one, any other suitable difference) In F1 progeny, pea plants have Tt where T is dominant over t (b) (c) so all the plants of F1 progeny were tall. / Tall height is dominant trait over short height. Self-pollination (i) (ii) 15. Organisms that feed on (i) producers and other consumers. Dominant trait i) ii) (c) 16. Decomposers Expresses itself over recessive trait. Expresses in both conditions-TT and Tt. Recessive trait i) Unable to express itself in presence of a dominant trait. ii) Expresses itself only when it is tt or in pure condition. 1 3 1 1 2 (any one, any other suitable difference) OR (ii) Mendel s observations: All plants of F1 progeny were tall. No medium/ no short height plants observed in F1 progeny. F1 progeny resembled one parent only. (any two observations) 1+1 4 (a) (i) Most of these bacteria would die, but the few variants resistant to heat would survive and grow further. 1 (ii) Fertilization occurs to form a zygote. 1 / No 1 (iv) If the egg is not fertilised, the thick and spongy lining of the uterus breaks and comes out through vagina as blood and mucus, known as menstruation. / Menstruation will take place. 1 (v) The seed will develop into a seedling. / Germination will take place. OR 1 (iii) Cross pollination may occur leading to fruit formation. fertilization. / No fruit formation. (b) (i) When spores land on a substance and get adequate moisture and temperature, it will develop into new Rhizopus. 1 PAGE 5 {31/1/1} (ii) 17. 18. 19. 20. 21. 22. 23. 24. 25. New plants grow from the buds located in the notches of the leaf. 1 (iii) The pollen tube will not be formed. / No fertilisation will take place. 1 (iv) Fertilization /Pregnancy will be prevented. 1 (v) Each fragment or piece grows into a new individual organism. 1 SECTION B (Chemistry) (C) / Both, (i) and (ii) are double displacement reactions and precipitation reactions. (B) / Vanilla essence (D) / NO2 and O2 (B) / -CHO (A) / Pb (A) / tomato, curd, ant-sting (B) / Calcium (A) / Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (a) Metal oxides which can react with both acids as well as bases to produce salt and water. (b) ZnO Amphoteric oxide 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Na2O Basic oxide 1 CO2 Acidic oxide 26. 5 2 (a) (b) Substance oxidised - C Substance reduced - ZnO Pb(NO3)2 + 2 KI PbI2 + 2 KNO3 1 (c) 27. 2H2O 2AgCl electricity sunlight 2H2 + O2 2Ag + Cl2 (any other example in each case) 3 (a) When electricity is passed through brine, it decomposes to form sodium hydroxide (alkali) and chlorine, hence this process is called chlor-alkali process. 2NaCl(aq) + 2H2 O(l) Electricity At anode: Cl2 At cathode: H2 2NaOH(aq) + H2 (g) + Cl2 (g) 1 1 OR PAGE 6 {31/1/1} (b) (i) NaCl + H2 O + NH3 + CO2 NaHCO3 + NH4 Cl (ii) Ca(OH)2 + Cl2 CaOCl2 + 2 1 1 / 2Ca(OH)2 + 2Cl2 Ca(ClO)2 + CaCl2 + 2H2O (iii) 28. CaSO4 .2H2 O 373K 1 1 CaSO4 . 2 H2 O + 1 2 H2 O (deduct mark for no / incorrect balancing) (a) Because it is easier to obtain metal from its oxide. / 1 3 1 Because it is easier to reduce metal oxide to metal (b) Fe2 O3 (s)+ 2Al(s) 2Fe(l) + Al2 O3 (s) + Heat 3MnO2 (s) + 4Al(s) 3Mn(l) + 2Al2 O3 (s) + Heat (balancing is optional) (c) (i) 2Cu2S + 3O2(g) 2Cu2O + Cu2S Heat Heat 1 (any one equation) 1 2Cu2O(s) + 2SO2 (g) 1 6Cu(s) + SO2(g) OR (c)(ii) (I) 1 Because highly reactive metals have more affinity for oxygen than carbon. 1 (II) Because of its low melting point. 29. 4 (a) (i) (I) They do not give rise to charged particles/ ions. 1 (II) Soap reacts with calcium and magnesium salts present in 1 hard water and forms insoluble substances called Scum. (III) C-C bonds are strong and stable whereas Si-Si bonds are 1 relatively weak. (ii) (I) (II) CH3 CH2 OH acidified K2 Cr2 O7 +heat CH2 =CH2 + H2 Ni CH3 COOH CH3 -CH3 1 1 OR (b) (i) X - CH3COOH/ethanoic acid /acetic acid Y - CH3COOC2H5/CH3COOCH2CH3/ ester/ ethyl ethanoate Z - CH3COONa/sodium ethanoate/sodium acetate (ii) Catalyst/ dehydrating agent PAGE 7 {31/1/1} (iii) 1 Esterification reaction CH3 COOC2 H5 + NaOH CH3 COONa +C2 H5 OH Saponification reaction/ De-esterification reaction 1 5 30. SECTION C (Physics) (C) / - 30 cm and + 30 cm from lens 1 1 31. (A) / Ciliary muscles of your eye contract and the eye lens become thick 1 1 32. (C) / Assertion (A) is true, but Reason (R) is false. 1 1 33. (a) Convex lens / Converging lens (b) m = - 2, v= 30 cm 1 m =u u=m 30 u = 2 u = -15cm / The object was placed at 15 cm in front of the lens. 34. (a) 2 r = 0.01 cm=1 x 10-4 m = 1 cm = 0.01 m R = = = RA A = x 2 7 x 22 x 10 8 1 7 x 0.01 = 22 x 10-8 x 102 = 22 x 10-6 m =2.2 x 10-5 m OR (b) Resistance of electric heater R = V I 220 R = 11 R = 20 V2 P =R P = 20 P = 2000W/ 2 kW 200 x 200 2 PAGE 8 {31/1/1} 35. 1 Hypermetropia/ Far sightedness 1 1 3 (deduct mark for not showing the direction of ray of light) 36. (a) Procedure Take a small aluminium rod AB and using two connecting wires suspend it horizontally from a stand. Place a strong horse-shoe magnet in such a way that the rod lies between the two poles perpendicularly. 1 Connect the aluminium rod in series with a battery and a key. Now pass a current through the aluminium rod from one end to another. / (Procedure can also be explained with a Diagram) Observation It is observed that the rod is displaced on passing current though it. (b) Magnetic field will be vertically downwards. 1 1 3 PAGE 9 {31/1/1} 37. (a) 1+1 (any two cases) / (b) 38. Magnetic field pattern for a current carrying straight conductorconcentric circles. Magnetic field pattern for a current carrying solenoid- magnetic field lines similar to that of a bar magnet. Magnetic field pattern for a current carrying circular loop- a pair of concentric circles with parallel straight lines at the centre. (any two cases) At X. Magnetic field decreases as the distance from the conducting wire increases. Position - Image will form at 40 cm / 2F /C Nature Real and inverted 3 (a) / Alternate answer f = +20 cm, u = -40 cm Using Lens Formula, 1 1 1 = v u f 1 1 1 = v 40 20 1 1 1 = + v 20 40 v = + 40 cm Position - 40 cm on the other side of the lens Nature - Real and inverted (b) 1 PAGE 10 {31/1/1} (c) (i) f1 = 30 cm = 0.3 m, f 2 = -15cm = -0.15 m 1 P=f +1 1 P1= 0.3 D ; P2= 0.15 D Equivalent power, P = P1+P2 P = - 3.33D 1 Equivalent focal length, f = P 1 f = 3.33 = - 0.3 m = - 30 cm OR (c) (ii) Combination Lens will behave like convex lens f 1 = - 2 m, f 2 = 1.5 m 1 1 P=f P1= 1 2 +1 D , P2= 1.5 D P = P1+P2 1 P=6 f = + 6m The focal length of combination is positive. 1 / 39. Alternate answer Combination Lens will behave like convex lens. Convex lens - Less f , More P1 Concave lens - More f , Less P2 Combined Power = P1+P2, which will be positive. 4 (a) As Resistance, R = A , it changes with change in length and area of cross section of conductor. But resistivity of conductor is the characteristic property of material and hence it does not change. (ii) The resistivity of an alloy is generally higher than that of its (i) constituent metals. / Alloys do not oxidise (burn) readily at high temperatures. (iii) 1 Ampere is constituted by the flow of 1 Coulomb of charge per second. / 1 1 2 1 1A= 1C/1s OR PAGE 11 {31/1/1} (b) (i) V= 4V, I= 2A Resistance of circuit R = V I 4 R=2 R = 2 Let n be the number of bulbs 1 R 1 R 1 2 1 1 1 1 8 n 8 8 8 = + + + + n = = 8 n 8 n=4 Therefore, 4 bulbs of resistance 8 should be connected in parallel. (ii) 1 Ammeter In series 1 (iii) Heat generated through a current carrying conductor is directly proportional to square of current, resistance of conductor and time for which current flows in conductor. / 1 2 H=I Rt 5 PAGE 12 {31/1/1}

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