The Central Board of Secondary Education (CBSE) introduced Mathematics Basic as an alternative to the standard Mathematics option for Class X students in the academic year 2019-20. This change was implemented to cater to students who might not want to pursue Mathematics in higher studies but still require a basic understanding of mathematical concepts for everyday life.
The Mathematics Basic paper is designed with a comparatively simpler approach, focusing on fundamental concepts and their applications. It aims to reduce mathematics anxiety among students while ensuring they develop essential mathematical skills.
The Class X Mathematics Basic Sample Question Paper 2019-20 serves as a valuable resource for students to understand the examination pattern, question types, and marking scheme. This sample paper reflects the changes introduced by CBSE to make mathematics assessment more student-friendly and application-oriented.
The Mathematics Basic examination follows a structured format to assess students on various mathematical competencies. The paper is divided into four sections (A, B, C, and D) with different question types to test different skills.
| Section | Question Type | Number of Questions | Marks per Question | Total Marks |
|---|---|---|---|---|
| A | Objective Type (MCQs/Very Short Answer) | 20 | 1 | 20 |
| B | Short Answer Type I | 6 | 2 | 12 |
| C | Short Answer Type II | 8 | 3 | 24 |
| D | Long Answer Type | 6 | 4 | 24 |
The examination has a total duration of 3 hours and carries 80 marks. The remaining 20 marks are allocated for internal assessment, which includes periodic tests, notebook submission, and lab activities.
The Mathematics Basic sample paper follows a balanced distribution of questions across various chapters, with emphasis given to certain key topics. Understanding the weightage helps students prioritize their preparation accordingly.
| Unit/Chapter | Marks |
|---|---|
| Number Systems | 6 |
| Algebra | 20 |
| Coordinate Geometry | 6 |
| Geometry | 15 |
| Trigonometry | 12 |
| Mensuration | 10 |
| Statistics and Probability | 11 |
| Total | 80 |
Q1: The decimal expansion of the rational number 87/2500 will terminate after how many decimal places?
Answer: (d) 4
Q2: If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is:
Answer: (c) 15/4
Q3: Find the discriminant of the quadratic equation 2x - 5x + 3 = 0 and hence find the nature of its roots.
Solution: The discriminant D = b - 4ac = (-5) - 423 = 25 - 24 = 1. Since D > 0, the equation has two distinct real roots.
Q4: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution: Let O be the common centre. Let AB be the chord of the larger circle touching the smaller circle at C. Then OC is the radius of the smaller circle. By the property of tangent to a circle, OC AB. Hence, triangle OCB is a right-angled triangle with right angle at C. Using Pythagoras theorem, BC = OB - OC = 5 - 3 = 25 - 9 = 16. Hence, BC = 4 cm. Therefore, AB = 2BC = 8 cm.
Q5: If tan(A + B) = 3 and tan(A - B) = 1/3, where 0 < A + B 90 and A > B, find A and B.
Solution: Since tan(A + B) = 3 = tan 60, we have A + B = 60. Since tan(A - B) = 1/3 = tan 30, we have A - B = 30. Adding these two equations: 2A = 90, hence A = 45. Substituting in A + B = 60, we get 45 + B = 60, hence B = 15.
Q6: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution: Let AB be the tree and CD be the broken part. Let BD = 8 m be the distance from the foot of the tree to the point where the top touches the ground. Let CDB = 30. In right triangle CDB, we have tan 30 = CB/BD = CB/8. Therefore, CB = 8 tan 30 = 8/3 = 83/3 m. Also, cos 30 = BD/CD = 8/CD. Therefore, CD = 8/cos 30 = 8/(3/2) = 163/3 m. Height of the tree = CB + CD = 83/3 + 163/3 = 243/3 = 83 m 13.86 m.
Q7: The sum of first nine terms of an AP is 171 and the sum of its first twenty four terms is 996. Find the first term and common difference of the AP.
Solution: Let the first term be a and common difference be d. The sum of first n terms of an AP is given by Sn = n/2[2a + (n-1)d]. Given S9 = 171 = 9/2[2a + 8d], which gives 2a + 8d = 38, or a + 4d = 19 (Equation 1). Given S24 = 996 = 24/2[2a + 23d], which gives 2a + 23d = 83 (Equation 2). Subtracting Equation 1 from Equation 2, we get 19d = 64, hence d = 64/19. Substituting in Equation 1, a + 4(64/19) = 19, which gives a = 19 - 256/19 = (361 - 256)/19 = 105/19.
Q8: The following distribution gives the daily income of 50 workers of a factory:
Daily Income (in Rs): 100-120, 120-140, 140-160, 160-180, 180-200
Number of workers: 12, 14, 8, 6, 10
Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.
Solution: The less than type cumulative frequency distribution is as follows:
Daily Income (less than) (in Rs): 120, 140, 160, 180, 200
Cumulative frequency: 12, 26, 34, 40, 50
To draw the ogive, we plot the points (120,12), (140,26), (160,34), (180,40), and (200,50) on a graph with daily income on x-axis and cumulative frequency on y-axis. These points are then joined by a free hand smooth curve to obtain the less than ogive.
Students often wonder about the differences between Mathematics Basic and Standard Mathematics. Here are some key distinctions:
Success in the Mathematics Basic examination requires strategic preparation and a thorough understanding of fundamental concepts. Here are some effective preparation tips:
The Class X Mathematics Basic Sample Question Paper 2019-20 represents a thoughtful approach by CBSE to make mathematics assessment more accessible and less intimidating for students. While simpler than the Standard Mathematics option, it still covers essential mathematical concepts that are crucial for everyday life applications.
Students preparing for this examination should focus on understanding the fundamental principles, practicing regularly with sample papers, and approaching the subject with a positive mindset. With consistent effort and strategic preparation, students can perform well in this examination regardless of their previous mathematical background.
The Basic Mathematics option opens doors for students who may not wish to pursue Mathematics at the higher secondary level but still value the importance of mathematical literacy in their academic and everyday lives. This curriculum design reflects a progressive educational approach that recognizes diverse student needs and aspirations.
