Here you will get the CBSE Class 10 Mathematics Practice Paper 2017-2018: Set-I, designed as per the latest examination pattern to help you prepare for the CBSE class 10 Board Exam 2018.
All the questions in CBSE Class 10 Mathematics Solved Practice Paper : Set-I have been asked from the most important topics of class 10 Mathematics. These questions will not only help students to prepare for the exam in a better manner but will also help them in understanding the depth with which a topic should be studied.
Salient features of CBSE Class 10 Mathematics Solved Practice Paper : Set-II, are:
- Includes questions asked from the complete syllabus of CBSE class 10 Maths
- Consists of 30 questions divided into four sections A, B, C and D
- Section A contains 6 questions of 1 mark each
- Section B contains 6 questions of 2 marks each
- Section C contains 10 questions of 3 marks each
- Section D contains 8 questions of 4 marks each
- Total marks: 80
- Maximum time: 3 Hours
Solving this practice paper will help you fine tune your preparation for the final exam by letting you know your weak areas which you can recover with more practice and revision. This way you may access your preparedness for the exam.
CBSE Class 10 Mathematics Board Exam 2018: Latest Exam Pattern
Some of the sample questions from CBSE Class 10 Mathematics Solved Practice Paper : Set-I are given below:
Q. If the sum of first m terms of an AP is 2m^{2} + 3m, then what is its second term?
Sol.
S_{m }= 2m^{2} + 3m
Putting m = 1
S_{1 }= 2 + 3 = 5 = a_{1}
S_{2} = 8 + 6 = 14
Hence, a_{1} = 5
And a_{1} + a_{2} = 14
⟹ a_{2} = 14 – 5
⟹ a_{2} = 9
Q. Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y = 5.
Sol.
Since point (3, a) lies on line 2x – 3y = 5.
Then, 2 x 3 – 3 x a = 5
⟹ 6 – 5 = 3a
⟹ a = 1/3
Q. Find the value of tan^{2}10^{o} − cot^{2}80^{o}.
Sol.
We have,
tan^{2}10^{o} − cot^{2}80^{o} = tan^{2}10^{o} − cot^{2}(90^{o} – 10^{o})
= tan^{2}10^{o} − tan^{2}10^{o}
Q. Show that any positive odd integer can be written in the form 6m + 1, 6m + 3 or 6m + 5 for some integer m.
Sol.
Let a be any positive integer and b = 6.
Then, by Euclid’s lemma a = 6m + r, 0 ≤ r < 6.
So the possible remainders are 0, 1, 2, 3, 4 and 5.
The value of a can be 6m or 6m + 1 or 6m + 2 or 6m + 3 or 6m + 4 or 6m + 5, where m is the quotient and also a positive integer.
An odd integer can be written in the form of 2n + 1, where n ϵ N.
Now,
If a = 6m + 1
⟹ a = 2(3m) + 1
⟹ a = 2n + 1 (where, n = 2m)
Hence, 6m + 1 is an odd integer.
If a = 6m + 3
⟹ a = 2(3m) + 2 + 1
⟹ a = 2(3m + 1) + 1
⟹ a = 2n + 1 (where, 3m + 1)
Hence, 6m + 3 is an odd integer.
If a = 6m + 5
⟹ a = 2(3m) + 4 + 1
⟹ a = 2(3m + 2) + 1
⟹ a = 2n + 1 (where, n = 3m + 2)
Hence, 6m + 5 is an odd integer.
Q. Find the median using an empirical relation, when it is given that mode and mean are 8 and 9 respectively.
Sol.
The relation between Mean, Median and Mode of the given data is:
Mode = 3Median − 2Mean
⟹ 8 = 3Median − 2 × 9
⟹ 3Median = 8 + 18
⟹ Median = 26/3
⟹ Median = 8.67
You may get the complete practice paper by clicking on the following link:
CBSE Class 10 Mathematics Solved Practice Paper 2017-2018: Set-I |