Design Mathematics Model Test Papers Class 8 (2024-25) PDF

## DESIGN MATHEMATICS ### I. Weightage to Learning Objectives : | Objective | Remembering and Understanding | Application | Skills | Total | |---|---|---|---|---| | Suggestive marks | 44 | 28 | 8 | 80 | ### II. Weightage to form of questions | Form of Ques. | MCQ/(AR) | CBQ | SA-1 | SA-2 | LA |...

## DESIGN MATHEMATICS ### I. Weightage to Learning Objectives : | Objective | Remembering and Understanding | Application | Skills | Total | |---|---|---|---|---| | Suggestive marks | 44 | 28 | 8 | 80 | ### II. Weightage to form of questions | Form of Ques. | MCQ/(AR) | CBQ | SA-1 | SA-2 | LA | Total | |---|---|---|---|---|---|---| | No. of Ques. | 18(MCQ) + 2 (AR) | 3 | 5 | 6 | 4 | 38 | | Marks Allotted | 20 | 12 | 10 | 18 | 20 | 80 | ### III. Scheme of Sections : * Section I (MCQ+AR) : 18 (MCQ) + 2 (AR) Questions : 1 Mark each * Section II (SA I) : 5 Questions : 2 Marks each * Section III(SA II) : 6 Questions : 3 Marks each * Section IV (LA) : 4 Questions : 5 Marks each * Section V (Case Study Based Questions) : 3 Questions : 4 Marks each (Each having 3 sub parts as (1+1+2)* ### IV. Scheme of Options : * No overall option, only internal choice in 9 questions. * In each case study based question there is a choice in the two-mark question. * 2 questions in Section II * 2 questions in Section III * 2 questions in Section IV ### V. Weightage to Difficulty Level : | | Difficult Questions | Average Questions | Easy Questions | |---|---|---|---| | | 16 marks | 48 marks | 16 marks | | | 20% | 60% | 20% | ### VI. Abbreviations used : * HOTS: Higher Order Thinking Skills * CBQ: Case Based Questions * MCQ: Multiple Choice Questions * LA: Long Answer Type * AR: Assertion-Reasoning * SA 1: Short Answer Type 1 * SA 2: Short Answer Type 2 ## MODEL TEST PAPER-I CLASS-VIII (2024-25) MATHEMATICS Time allowed : 3 hours Maximum Marks : 80 General Instructions : 1. The question paper consists of five sections : * Section I: Question No. 1 to 20 are of 1 mark each (18 are MCQ Type and 2 are Assertion-Reasoning type questions). * Section II : Question No. 21 to 25 are Short Answer Type-1 questions of 2 marks each. * Section III : Question No. 26 to 31 are Short Answer Type-2 questions of 3 marks each. * Section IV : Question No. 32 to 35 are Long Answer Type questions of 5 marks each. * Section V : Question No. 36 to 38 are of Case Based questions. Each case study has 3 case based sub parts, two are of 1 marks each and third sub part is a short answer type (2 marks) having internal choice. 2. Please write the serial number of the question before attempting it. 3. In questions of constructions/graph, the drawing should be neat, clean and exactly as per given measurements. Use ruler and compass only. 4. All questions are compulsory. However, internal choices have been given in some questions. ## MODEL TEST PAPER-II CLASS-VIII (2024-25) MATHEMATICS Time allowed : 3 hours Maximum Marks : 80 General Instructions : 1. The question paper consists of five sections : * Section I: Question No. 1 to 20 are of 1 mark each (18 are MCQ Type and 2 are Assertion-Reasoning type questions). * Section II : Question No. 21 to 25 are Short Answer Type-1 questions of 2 marks each. * Section III : Question No. 26 to 31 are Short Answer Type-2 questions of 3 marks each. * Section IV : Question No. 32 to 35 are Long Answer Type questions of 5 marks each. * Section V : Question No. 36 to 38 are of Case Based questions. Each case study has 3 case based sub parts, two are of 1 mark each and third sub part is a short answer type (2 marks) having internal choice. 2. Please write the serial number of the question before attempting it. 3. In questions of constructions/graph, the drawing should be neat, clean and exactly as per given measurements. Use ruler and compass only. 4. All questions are compulsory. However, internal choices have been given in some questions. ## SAMPLE PAPER-2 MATHEMATICS Marking Scheme/Hints to Solutions Note: Any other relevant answer, not given here in but given by the candidates, be suitably awarded. | Q.No. | Value points / key points | Marks alloted to each key point/Value point | Total Marks | |---|---|---|---| | | **(Section-I)** | | | | 1 | (d) 10 | 1 | 1 | | 2 | (a) 4 | 1 | 1 | | 3 | (b) 256 | 1 | 1 | | 4 | (a) Distance travelled and the taxi fare | 1 | 1 | | 5 | (c) Avi earns less profit than Bobby. | 1 | 1 | | 6 | (d) 50% | 1 | 1 | | 7 | (d) a compound interest of 10% compounded quarterly. | 1 | 1 | | 8 | (b) (-x+2y+3z)² | 1 | 1 | | 9 | (b) 48 | 1 | 1 | | 10 | (c) 10° | 1 | 1 | | 11 | (b) 230° | 1 | 1 | | 12 | (b) 15 | 1 | 1 | | 13 | (c) y-axis | 1 | 1 | | 14 | (b) a line parallel to the x-axis | 1 | 1 | | 15 | (b) 12 m | 1 | 1 | | 16 | (d) 625 cm³ | 1 | 1 | | 17 | (c) 3/26 | 1 | 1 | | 18 | (b) 16.5 | 1 | 1 | | 19 | (b) Both Assertion & Reason are true but Reason is not the correct explanation of assertion. | 1 | 1 | | 20 | (d) Assertion (A) s false but Reason (R) is true. | 1 | 1 | | | **(Section-II)** | | | | 21 | $3√x-12=9$ $x-12=9³$ $x-12=729$ $x=729+12$ $x=741$ OR $3√(17/27)$ $3√(81-17/27)$ $3√(64/27)$ $4/3$ | 1/2, 1/2, 1/2, 1/2, 1 | 2 | | 22 | S.P. = `54 Let C.P. = x x Loss = 1/10 C.P. - Loss = SP x x = 54 10 10x-x=54 10 9x = 54 10 54×10 x = 9 x = 60 C.P. of pen = `60 2 | 1/2, 1/2, 1/2, 1/2 | 2 | | 23 | $(a-1/α)²=a²+2×a×1/α$ $(a-1/α)²=a²+1/2$ $(a-1/α)²=18-2$ $(a-1/α)²=16$ $(a-1/α)=√16$ $a-1/α=4$ $a=4$ OR (10.1)² = (10+0.1)² By using the identity (a + b)² = a² + 2ab + b² (10+0.1)² = (10)² + 2 × 10 × 0.1 + (0.1)² = 100 + 2 + 0.01 = 102.01 | 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2 | 2 | | 24 | ↑P 130° 1 2 l α m n b ∠1 + ∠2 = 180° (linear pair) 130° + ∠2 = 180° ∠2 = 50° ∠a = ∠2 (corresponding angles) ..∠a = 50° ∠b = ∠1 (alternate exterior ∠s) ..∠b = 130° a:b = 50° : 130° a:b = 5:13 | 1/2, 1/2, 1/2, 1/2 | 2 | | 25 | Let total no. of employees = x 15% of x = 75 15 x x = 75 100 75×100 x = 15 x = 500 | 1/2, 1/2, 1/2 | 2 | | | **(Section-III)** | | 26 | $2x-1 + 2x+1= 320$ $2x * (1/2 + 2) = 320$ $2x * (5/2) = 320$ $2x = 320 * 2 / 5$ $2x = 64 * 2$ $2x = 128$ $2x = 27$ $x = 7$ OR $3-4 × 216-1/3 × 25½$ $1251/3 × 16 × 3-6$ $3^4/36/16/1251/3 * 2161/3$ $3^4/36/16/1251/3 * 2161/3$ $36-4 * (52)1/3$ $(24)1/4 * (53)1/3 * (63) 1/3$ $32 * 5$ $2×5×6$ $45/3/60/4$ | 1/2, 1/2, 1/2, 1/2, 1/2, 1, 1, 1/2 | 3 | | 27 | No. of students in the beginning = 500 No. of students after 12 days = 500 + 300 = 800 For 500 students food will remain for 48 days. Let for 800 students food will remain = x days No. of students 500 800 No. of days 48 x It is a case of inverse variation 500 × 48 = 800 × x 500 × 48 x = 800 x = 30 The food will last for 30 days. OR Let length of the tunnel = xm Length of the train = 270 m Total distance = (270 + x)m | 1/2, 1/2, 1/2, 1/2, 1/2, 1 | 3 | | 28 | $ y²-2y-35$ $y-7$ $ y²-7y+5y-35$ $y- 7$ $y(y-7)+ 5(y-7)$ $y-7$ $(y)(y+5)$ $(y-7)$ = y + 5 | 1/2, 1, 1/2, 1/2, 1/2 | 3 | | 29 | Correct axes Plotting of each point Joining of points | 1/2, 1/2×4=2, 1/2 | 3 | | | **Alternative question for visually challenged students in lieu of Q. 29** Let Ravi's present age be 5x years and Hema's present age be 7x years 4 years later Ravi's age = (5x + 4) years, Hema's age = (7x + 4) years ATQ 5x + 4 3 = 7x + 4 4 => 4 (5x + 4) = 3 (7x + 4) => 20x + 16 = 21x + 12 => 20x - 21x = 12 - 16 => -x = -4 => x = 4 .. Ravi's age = 5 x 4 = 20 years Hema's age = 7 x 4 = 28 years | 1/2, 1/2, 1/2, 1, 1/2 | 3 | | 30 | Let the lengths of the parallel sides be 4x and 5x height of the trapezium = 18 cm area of the trapezium = 405 cm² 1 2 xh × h × sum of parallel sides = Area of trapezium 1 × 18 × (4x + 5x) = 405 2 1 9 × 18 × 9x = 405 2 81x = 405 405 x = 81 x = 5 Length of parallel sides are 20 cm and 25 cm. | 1/2, 1/2, 1/2, 1, 1/2 | 3 | | 31 | | Height (in cm) | Tally marks | Number of students | |---|---|---| | 125-130 | III | 5 | | 130-135 | III II | 7 | | 135-140 | III I | 6 | | 140-145 | III | 4 | | | **Total | 22 |** | | 3 | 3 | | | **(Section-IV)** | | | | 32 | Area of square lawn = (Total cost)² Cost per m² = 10,935² 15 = 729 m² Also area of square lawn = (side)² ..(side)²= 729 m² side = √729 m² = 27 m Perimeter of square lawn = 4 × side = 4 × 27 m = 108 m Cost of fencing @ 25 per m = 108 × 25 = 2700 OR Let no. of students in the class = x Amount of money donated by each student = x Money donated by whole class = money donated by class teacher .. Money donated by whole class 6272 2 = 3136 .. x × x = 3136 x² = 3136 x = √3136 x = 56 No. of students in the class = 56 Money donated by the class teacher = 3136 | 1/2, 1, 1, 1, 1/2, 1/2, 1, 1/2, 1/2 | 5 | | 33 | Standard form of polynomial 3x (5x² + 3x³ + 2) - (2x² + 8 - x) 15x³ + 9x⁴ + 6x - 2x² - 8 + x 9x⁴ + 15x³ -2x² + 7x - 8 Standard form of divisor -2 + 3x = 3x - 2 3x-2 9x⁴ + 15x³ -2x² + 7x - 8 9x⁴ + 6x³ 21x³ -2x² + 7x - 8 21x³ + 14x² - 12x² + 7x - 8 -12x² + 8x 15x - 8 15x - 10 2 Q = 3x³ + 7x² + 4x + 5, Check: Dividend = Divisor × Quotient + Remainder LHS = 9x⁴ + 15x³ -2x² + 7x - 8 RHS = (3x - 2)(3x³ + 7x² + 4x + 5) + 2 = 3x (3x³ + 7x² + 4x + 5) - 2 (3x³ + 7x² + 4x + 5) + 2 = 9x⁴ + 21x³ + 12x² + 15x - 6x³ - 14x² - 8x - 10 + 2 = 9x⁴ + 15x³ - 2x² + 7x - 8 LHS = RHS | 1/2, 1/2, 2, 1/2, 11/2 | 5 | | 34 | | S x R 75°. y 4 cm 75° 105° P 5 cm Q Draw PQ = 5 cm Constructing ∠P = 75° Constructing ∠Q = 105° Finding point R such that QR = 4 cm Constructing ∠R = 75° Finding point S PQRS is a parallelogram Alternative question for visually challenged students in lieu of Q. 34 Let the principal be X S.I. = P × RxT = x × 10 × 2/100 = x/5 C.I. = P (1 + R/100)ⁿ - 1 = x (1 + 18/100)² - 1 = x (11/10)² - 1 = x (121/100)-1 = x [121-100]/100 = 21x/100 CI - SI = 65 21x 100 x = 65 21x - 20x 100 x = 65 x = 6500 Principal = 6500 35. Length of wall = 15 m = 1500 cm Breadth of wall = 30 cm Height of wall = 4 m = 400 cm Volume of wall = LxBxH = 1500 × 30 × 400 = 1,80,00,000 cm³ | 1/2, 1, 1, 1/2, 1/2, 2, 1, 1 | 5 | | 36 | (1) Rate of growth per annum = 60% 60% Rate of growth semi annually = 2 = 30% (2) Let two years ago population of rabbits = P Rate of growth per annum = 60% Let A be the present population = 5120 2 5120 = P (1 + 60/100)² 5120 = P(16/10)² 5120 = P(16/10)² 5120 = P(64/25) P = 5120 x 25/64 P = 2000 OR Let present population be P = 5120 Let A be the population after 2 years A = P( 1 + R/100)ⁿ = 5120 (1 + 60/100)² = 5120 (16/10)² = 5120 x (16/10)² = 5120 × 16 × 16 10 × 10 = 13107.2 = 13107 (3) No. of rabbits in the beginning of 2020 = 5120 No. of rabbits die = 120 No. of rabbits left = 5120 - 120 = 5000 No. of rabbits increased in the next year = 60% of 5000 = 3000 | 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2 | 4 | | 37 | 1. (4x+20) 2. Neena's expenditure on day 1 = Neena's expenditure on day 2 = (6x+20) ATQ 4x+20 = 11 6x+20 16 16 (4x + 20) = 11 (6x + 20) 64x + 320 = 66x + 220 64x - 66x = 220 - 320 -2x = -100 100 x = -2=50 .. Cost of Neena's ride is `50 OR Given that cost of Neena's ride is x 3 Cost of Arun's ride is -x 2 Neena's expenditure on day 1 = (4x + 20) Arun's expenditure on day 1 = ATQ (4x + 20) + (15x + 20) = 615 2 => 4x + 15 x + 40 = 615 2 => 8x + 21x = 615 - 40 2 => 23x = 575 2 x = 575 x 2 23 => x = 50 Cost of Neena's ride = `50 25 Cost of Arun's ride = ×50 3 2 = `75 3. Cost of each ride on day 3 = `50 Expenditure of Neena on day 3 = 5 × 50 + 20 = 250 + 20 = 270 Expenditure of Arun on day 3 = 5 × 50 + 20 = 270 Total expenditure = 270 + 270 = 540 | 1/2, 1, 1, 1/2, 1/2, 1/2, 1/2, 1/2 | 4 | | 38 | (1) Since opposite sides are equal and parallel .. It is a parallelogram. (2) A parallelogram having a pair of adjacent sides equal, is called a rhombus. .. a + b = c (3) a // b ..∠1=∠2 (alternate interior angles) ..∠2 = 40° (given) ∠3 = 2∠2 ..∠3 = 80° ∠1 + ∠3 + ∠4 = 180° (By angle sum property of a Δ) 40° + 80° + ∠4 = 180° ∠4 = 180° - 120° ∠4 = 60° OR Let a = 2x, b = 3x a + b + 2c = Perimeter of trapezium 2x + 3x + 2(10) = 40 mm 2x + 3x = 40 - 20 5x = 20 x = 4 a = 8 mm, b = 12 mm | 1, 1/2, 1/2, 1/2, 1/2, 1/2, 1, 1/2, 1/2 | 4 |