In this article we are providing important questions for CBSE class 10 Mathematics exam 2018 to help you prepare easily and effectively for the final exam.
The collection of important questions given here has been prepared after carrying a thorough analysis of the past year exam trends and the latest syllabus. All these questions provided with appropriate solutions will come in handy to revise the significant part of class 10 Maths syllabus. Moreover the solutions given here will help students frame a perfect solution for each question in the Class 10 Maths Exam 2018.
Find below the important questions for CBSE Class 10 Mathematics Board Exam 2018:
Important 1 Mark Questions
Here we provide important questions which should be prepared for the 1 mark questions set to be asked in class 10 Mathematics exam.
Some sample questions from the set of Important 1 Mark Questions for CBSE Class 10 Mathematics Exam, are given below:
Q. Find the length of the tangent from a point M which is at a distance of 17 cm from the centre O of the circle of radius 8 cm.
Sol.
Consider the figure:
Since, MN is the tangent of the circle,
∠MNO = 90⁰
⟹ MO^{2} = MN^{2} + ON^{2}
⟹ 17^{2} = MN^{2} + 8^{2}
⟹ 289 = MN^{2} + 64
⟹ 289 – 64 = MN^{2}
⟹ MN^{2} = 225
⟹ MN = 15
Thus, the length of the tangent is 15 cm.
Q. If the common difference of an A.P. is 3, then find a_{20 }– a_{15}.
Sol.
Let the first term of the AP be a.
a_{n }= a(n − 1)d
a_{20 }– a_{15} = [a + (20 – 1)d] – [a + (15 – 1)d]
= 19d – 14d
= 5d
= 5 × 3
Get here the complete set of CBSE Class 10 Mathematics: Important 1 Mark Questions |
Important 2 Marks Questions
Here you will get important 2 marks questions which might be asked in CBSE class 10 Mathematics paper.
Some sample questions from the set of Important 2 Marks Questions for CBSE Class 10 Mathematics Exam, are given below:
Q. Solve the following system of linear equations by substitution method:
2x – y = 2
x + 3y =15
Sol.
Here, 2x – y = 2
⟹ y = 2x – 2
⟹ x + 3y = 15
Substituting the value of y from (i) in (ii), we get
x + 6x – 6 = 15
⟹ 7x = 21
⟹ x = 3
From (i), y = 2 × 3 – 2 = 4
∴ x = 3 and y = 4
Q. From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semi-circular portion with BC as diameteris cut off. Find the area of remining paper (use π = 22/7)
Sol.
Given situation can be represented as the following diagram:
Length of paper, AB = l = 40cm
Width of paper, AD = b = 40cm
Area of paper = l × b = 40 × 28 = 1120 cm^{2}
Diameter of semi-circle = 28cm
∴ Radius of semi-circle, r = 14cm
Thus, area of semi-circle = 1/2. πr^{2}
= 1/2 × 22/7 × 14 × 14
= 308cm^{2}
∴ Area of remaining paper = 1120 – 308 = 812 cm^{2}
Get here the complete set of CBSE Class 10 Mathematics: Important 2 Marks Questions |
Important 3 Marks Questions
Get here a collection of important 3 marks questions which students must practice to make an effective preparation for the CBSE Class 10 Mathematics Paper 2018.
Some sample questions from the set of Important 3 Marks Questions for CBSE Class 10 Mathematics Exam, are given below:
Q. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Sol.
Given that sum of the 4th and 8th terms of an AP is 24.
⟹ a + 3d + a + 7d = 24
⟹ 2a + 10d = 24 ...(i)
Also the sum of the 6th and 10th term is 44.
⟹ a + 5d + a + 9d = 44
⟹ 2a + 14d = 44 ...(ii)
Subtracting equation (i) from equation (ii), we get:
4d = 20
⟹ d = 5
Substituting d = 5 in equation (i), we have:
2a + 10d = 24
⟹ 2a + 10 (5) = 24
⟹ 2a + 50 = 24
⟹ 2a = −26
⟹ a = −13
Hence first term of given A.P. is −13 and common difference is 5.
Q. The taxi charges in a city comprise of a fixed charge together with the charges for the distance covered. For a journey of 10 km the charge paid is Rs. 75 and for a journey of 15 km the charge paid is Rs. 110.
(i) What will a person have to pay for travelling a distance of 25 km?
(ii)Which mathematical concept is used in this question?
(iii) What is its value?
Sol.
Let the fixed charge of taxi be Rs. x per km and the running charge be ` y per km.
According to the question,
x + 10y = 75
x + 15y = 110
Subtracting equation (ii) from equation (i), we get
– 5y = – 35
⟹ y = 7
Putting y = 7 in equation (i), we get x = 5
∴Total charges for travelling a distance of 25 km = x + 25y
= (5 + 25 × 7)
= (5 + 175)
= 180
Get here the complete set of CBSE Class 10 Mathematics: Important 3 Marks Questions |
Important 4 Marks Questions
Here you will get a set of important 4 marks questions to prepare for the CBSE Class 10 Mathematics Paper 2018.
Some sample questions from the set of Important 4 Marks Questions for CBSE Class 10 Mathematics Exam, are given below:
Q. A motor boat whose speed is 24 km/hr in still water takes 1 hr more to go 32km upstream than to return downstream to the same spot. Find the speed of the stream.
Sol.
Let the speed of stream be x.
Then,
Speed of boat in upstream is 24 ‒ x
In downstream, speed of boat is 24 + x
According to question,
Time taken in the upstream journey ‒ Time taken in the downstream journey = 1 hour
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
Now a_{l} =a + (n −1)
⟹ 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:
S_{n} = n/2 [2a + (n − 1)d]
⟹ S_{40} = 40/2 [2 × 10 + (40 − 1)9]
= 20|20 + 39 x 9]
=20[20 + 351]
=20 × 371 = 7420
Thus, sum of all 40 terms of AP = 7420
Get here the complete set of CBSE Class 10 Mathematics: Important 4 Marks Questions |
Maths is one such subject which can be learned by intensive practice and having a good hold on basic concepts. Therefore, students are suggested to solve the class 10 Mathematics important questions given here in order to assess their preparation and excel the efforts made to perform well in the board exams.