In this article we are providing the CBSE Class 10 Mathematics: Important 4 Marks Questions to prepare for the board exam 2019. All these questions are completely solved.

In CBSE Class 10 Mathematics Board Exam 2019, Section - D will comprise 8 questions of 4 marks each. Thus almost 40 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 3 marks question is essential.

**CBSE Class 10 Mathematics Board Exam 2018: Latest Exam Pattern**

**Given below are some sample questions for CBSE Class 10 Mathematics: Important 4 Marks Questions:**

**Q.** For any positive integer n, prove that n^{3} – n is divisible by 6.

**Solution.**

n^{3} – 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 3*q*, 3*q* + 1 or 3*q* + 2 for some positive integer *q*.

Now three consecutive positive integers are *n, n* + 1, *n* + 2.

**Case I.** If *n *= 3*q*.

*n*(*n* + 1) (*n* + 2) = 3*q*(3*q* + 1) (3*q* + 2)

But we know that the product of two consecutive integers is an even integer.

∴ (3*q* + 1) (3*q* + 2) is an even integer, say 2*r*.

⟹ *n*(*n* + 1) (*n* + 2) = 3*q* × 2*r* = 6*qr*, which is divisible by 6.

**Case II. **If *n *= 3*n* + 1.

∴ *n*(*n* + 1) (*n* + 2) = (3*q* + 1) (3*q* + 2) (3*q* + 3)

= (even number say 2*r*) (3) (*q* + 1)

= 6*r* (*q* +* *1),

which is divisible by 6.

**Case III. **If *n *= 3*q* + 2.

∴ *n*(*n* + 1) (*n* + 2) = (3*q* + 2) (3*q* + 3) (3*q* + 4)

= multiple of 6 for every *q*

= 6*r* (say),

which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6.

**CBSE Class 10 Mathematics Exam 2018: Important 3 Marks Questions**

**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?

**Sol.**

Given, first term, *a* = 10

Last term, *a _{l}* = 361

And, common difference, *d* = 9

*a _{l} = a *+

*(n*

*−1)d*

361 = 10 + (

*n*− 1)9

361 = 10 + 9

*n*− 9

361 = 9

*n*+ 1

9

*n*= 360

*n*= 40

Therefore, total number of terms in AP = 40

Now, sum of total number of terms of an AP is given as:

S* _{n}* =

*n*/2 [2

*a*+ (

*n*− 1)

*d*]

S

_{40}= 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 2019. 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 (2010-2018)****NCERT Exemplar Problems and Solutions Class 10 Maths: All Chapters****NCERT Solutions for CBSE Class 10 Maths: All Chapters****CBSE Class 10 Maths Exam 2019: Important questions with solutions****CBSE Class 10 Board Exam 2019: Sample Papers and Marking Schemes of all subjects**