CBSE Class 10 Mathematics Exam 2020: Important 4 Marks Questions with Solutions
This article brings the CBSE Class 10 Mathematics: Important 4 Marks Questions to prepare for Board Exams 2020. All the questions are provided with appropriate solutions to give you an idea to present answers in the best way in exam.
In this article we are providing the Important 4 Marks Questions for the CBSE Class 10 Mathematics Exam 2020. All these questions are completely solved.
In CBSE Class 10 Mathematics Board Exam 2020, Section - D will be comprised of 6 questions of 4 marks each. Thus almost 3.0 percent marks in Class 10 Maths paper can be secured by attempting the the 4 marks questions correctly. So, if one wishes to excel in the examination, a good practice of 4 marks questions is essential.
Given below are some sample questions for CBSE Class 10 Mathematics: Important 4 Marks Questions:
Q. For any positive integer n, prove that n3 – n is divisible by 6.
n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.
Now three consecutive positive integers are n, n + 1, n + 2.
Case I. If n = 3q.
n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)
But we know that the product of two consecutive integers is an even integer.
∴ (3q + 1) (3q + 2) is an even integer, say 2r.
⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.
Case II. If n = 3n + 1.
∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6r (q + 1),
which is divisible by 6.
Case III. If n = 3q + 2.
∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.
Q. The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?
Given, first term, a = 10
Last term, al = 361
And, common difference, d = 9
al = a + (n −1)d
361 = 10 + (n − 1)9
361 = 10 + 9n − 9
361 = 9n + 1
9n = 360
n = 40
Therefore, total number of terms in AP = 40
Now, sum of total number of terms of an AP is given as:
Sn = n/2 [2a + (n − 1)d]
S40 = 40/2 [2 x 10 + (40 − 1)9]
= 20[20 + 39 x 9]
=20[20 + 351]
=20 x 371 = 7420
Thus, sum of all 40 terms of AP = 7420
To get the complete set of questions, click on the following link:
All these Class 10 Mathematics: Important 4 Marks Questions have been prepared after carrying the thorough analysis of the trend followed in previous years’ Class 10 Maths CBSE question papers and the latest syllabus. Questions have been picked from the most important topics of Class 10 Maths. Thus practicing these questions will help to cover the significant part of the curriculum thus making an effective preparation for the CBSE Class 10 Board Exam 2020. Moreover, the solutions provided here will give an idea to present your answers in the best way to grab maximum scores.
To get more of such useful articles for CBSE Class 10 Board Exam preparations, click on the following links:
- CBSE Class 10 Mathematics Previous Years' Solved Question Papers (2010-2019)
- NCERT Exemplar Problems and Solutions Class 10 Maths: All Chapters
- NCERT Solutions for CBSE Class 10 Maths: All Chapters
- CBSE Class 10 Maths Exam 2020: Important questions with solutions
- CBSE Class 10 Board Exam 2020: Sample Papers and Marking Schemes of all subjects