Here you will get the CBSE Class 10 Mathematics Solved Guess Paper to prepare for the board exam to be held on 28 March, 2018.
Mathematics is one such subject in which students can score full marks by following the right preparatrion technique and exam writing techniques. You cannot learn Maths by just memorizing a set of formulae given in the book. You need to understand how to use those formulae, for which you would require more and more practice. So, it’s very important to solve practice papers, guess papers, sample papers. This will help you track your preparedness for the exam, letting you know your weak areas which you may improve later.
CBSE Class 10 Mathematics Solved Guess Paper 2018, has been specially prepared by the subject experts after the brief analysis of previous year question papers and the latest examination format. The paper contains entirely fresh questions picked from the most important topics of class 10 Mathematics. Practice these questions will definitely help to fine tune your preparations for the board exam 2018.
Some significant features of this paper are:
- Covers the whole syllabus of CBSE class 10 Mathematics.
- Follows the latest examination pattern for class 10 Mathematics board exam 2018.
- Each question has been provided with an apt and easy solution.
- Questions have been asked from only those topics which are important from exam point of view.
Practicing this paper will help the students in understanding the depth with which a topic should be studied in order to prepare in a more effective way to get the desired results.
CBSE Class 10 Science Board Exam 2018: Latest Exam Pattern
Structure of this paper is as below:
- The question paper 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 markseach.
- Section C contains 10 questions of 3 marks each.
- Section D contains 8 questions of 4 marks each.
- Covers the whole syllabus of CBSE class 10 Mathematics.
- Total Marks: 80
- Maximum Time: 3 hours
This CBSE Class 10 Mathematics Guess Paper will give you a preview of the question paper which you will be solving in the class 10 Mathematics board exam 2018.
Some sample questions and their solutions from the CBSE Class 10 Mathematics Solved Guess Paper 2018 are given below:
Q. If a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5 then calculate the least prime factor of (a + b).
Sol.
a and b are two positive integers such that the least prime factor of a is 3 and least prime factor of b is 5.
Then least prime factor of (a + b) is 2.
Q. If the sum of two natural numbers is 8 and their product is 15, find the numbers.
Sol.
Let the numbers be x and (8 – x)
According to the question:
x(8 – x) = 15
⟹ 8 x – x^{2} = 15
⟹ 0 = x^{2} − 8x + 15
⟹ x^{2} − 5x − 3x + 15 = 0
⟹ x(x – 5) – 3 (x – 5) = 0
⟹ (x – 5) (x – 3) = 0
⟹ x – 5 = 0 Or x – 3 = 0
⟹ x = 3 Or x = 5
When x = 3, numbers are 3 and 5.
When x = 5, numbers are 5 and 3.
NCERT Solutions for CBSE Class 10 Science
Q. If two positive integers p and q are written as are prime numbers, then verify:
LCM (p, q) × HCF (p, q) = pq
Sol.
Here, LCM (p, q) = a^{3}b^{3}
And LCM (p, q) a^{2}b
⟹ LCM (p, q) × LCM (p, q) = a^{5}b^{4}
= (a^{3}b^{3}) (a^{3}b) = pq
Q. In the following figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is I Q cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then calculate the length of CD.
Construction: Join OR
Here, ∠1 = ∠2 = 90^{o}
[Tangent is perpendicular to the radius through the point of contact]
And ∠3 = 90^{o }[Given]
Thus, ORDS is a square.
DR = OS = 10 cm ….(i)
BP = BQ = 27 cm
[Tangents drawn from an external point]
Therefore, CQ = 38 – 27 = 11 cm
RC = CQ = 11 cm ….(ii)
[Tangents drawn from an external point]
Thus, from (i) and (ii), we have:
DC = DR + RC = 10 + 11 = 21cm
Q. Two cubes each of volume 27 cm^{3} are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Sol.
Volume of a cube = 27 cm^{3}
⟹ (Side)^{3} = (3)^{3}
⟹ Side = 3
Length of resulting cuboid, l = 2 × 3 = 6 cm
Breadth of resulting cuboid, b = 3 cm
Height of resulting cuboid, h = 3 cm
Therefore, surface area of resulting cuboid = 2 (lb + bh + hl)
= 2 (6 × 3 + 3 × 3 + 3 × 6)
= 2 (18 + 9 + 18)
= 2 (45) = 90 cm^{2}
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