CBSE Maths Important Questions

Important Questions For Class 10 Maths

The question paper of 10th CBSE Maths will be for 80 marks. There will be four sections of the paper. Section-A a will have all the 1 mark questions and the number of questions in this section is 20. So, the total weightage of section-A is 20 marks. The marks weightage of Section-B is 12 marks where each question will be for 2 marks each. The number of questions in this section is 6. In Section-C, there will be 8 questions asked which carry 3 marks each and the total weightage of this section is 24. 

10th Maths Important Questions

There will be 40 questions in total asked in the question paper, The type of questions is different in each section. There will be objective-type questions from Section-A which will carry 1 mark. Section-B contains short-type questions which are for 2 marks each. The short type questions will be asked in section-C too which will be for 3 marks each. The section-D will have long-type questions which will carry 4 marks each.

The Important Question Of Maths Class 10 Short Type Questions

These questions will be mostly based on the formulae or describing the part of the formula. The questions will be helpful to find the surface area of the given shape, finding the decimal representation, completing the sequence of Arithmetic progression, the ratio of areas of two similar triangles when the ratio of sides are given, basic probability questions, finding if the pair of straight lines are parallel, nature of roots in a quadratic equation. 

CBSE Maths Important Questions of Section- C

The section-C will have short-type questions which will be mostly from the chapter Arithmetic progressions. These will include finding the sum to n terms. The next important topic is from applications of trigonometry like finding the height of the tree if they make a certain angle of elevation or deviation. Questions will also be asked from real numbers (which include finding) if the number is irrational, terminating or non-terminating, etc.

The statistics chapter will have good weightage and the questions can be related to mean median, mode. Coming to the chapter of coordinate geometry, the section formula, distance formula, midpoint of 2 points will be mostly asked in the exam.  

CBSE Maths Important Questions From Section D

There will be an internal choice in this section to answer the paper. In each question, two questions will be asked from which you can answer any one of your choices. 

The questions from quadratic equations will be to solve for the value of x. The practical questions from surface areas and volumes will be asked where the real-time examples will be asked and the area or volume should be calculated. The questions from statistics will be related to the mean, median, and mode of the given data. 

Solved Examples

1. If ΔAB\[\tilde{C}\] ΔPQR and the Ratio of AB and PQ is 4 : 5. Find the Ratio of their Areas and Perimeters.

Given: ΔAB\[\tilde{C}\] ΔPQR

AB : PQ = 4 : 5

\[\frac{AB}{PQ}\] = \[\frac{4}{5}\]

We know that the ratio of areas of 2 similar triangles is equal to the square of the ratio of their corresponding sides.

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{AB}{PQ}\])\[^{2}\]

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{4}{5}\])\[^{2}\]

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = \[\frac{16}{25}\]

Ar(\[\triangle\]ABC) : Ar(\[\triangle\]PQR) = 16 : 25  

Hence, the ratio of areas of two triangles is 16 : 25 .

We know that the ratio of areas of 2 similar triangles is equal to the square of their perimeters.

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{\text{Perimeter of \triangleABC}}{\text{Perimeter of \trianglePQR}}\])\[^{2}\]

\[\sqrt{\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\sqrt{\frac{16}{25}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\sqrt{(\frac{4}{5})^{2}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\frac{4}{5}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

Perimeter (\[\triangle\]ABC) : Perimeter (\[\triangle\]PQR) = 4 : 5

Hence, the ratio of the perimeters of two similar triangles is 4:5.

Q2. The Point P(x, y) Divides the Line Segment Joining the Points A(1, 4) and B(5, -1). If the x-coordinate of a Point P is 2, Find the Ratio in Which the Point P Divides Point A,B and also Find the y -coordinate of a Point P.

Given: A(1,4) and B(5,-1)

Let Point P(x,y) divides the points A, B in the ratio k : 1

x -coordinate is 2 ⇒ P(2,y) 

Section formula: If a point p(x,y) divides (x\[_{1}\],y\[_{1}\]) and (x\[_{2}\],y\[_{2}\]) in the ratio m : n, then

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p(x,y) = (\[\frac{mx_{2} + nx_{1}}{m + n}\] , \[\frac{my_{2} + ny_{1}}{m + n}\]

p(2,y) = (\[\frac{5k+1}{k+1}\] , \[\frac{-k+4}{k+1}\])

On comparing the x -coordinate, we get

2 = \[\frac{5k + 1}{k + 1}\]

2(k + 1) = 5k + 1

2k + 2 = 5k + 1

5k - 2k = 2 - 1

3k = 1

k = \[\frac{1}{3}\]

So, the ratio of k : 1 = \[\frac{1}{3}\] : 1 

On comparing the y -coordinates, we get

y = \[\frac{-k+4}{k+1}\]

⇒ y = \[\frac{-\frac{1}{3} + 4}{\frac{1}{3} + 1}\]

⇒ y = \[\frac{-\frac{-1+12}{3}}{\frac{1+3}{3}}\]

⇒ y = \[\frac{\frac{11}{3}}{\frac{4}{3}}\]

⇒ y = \[\frac{11}{4}\]

Hence, the ratio in which point P divides points A, B is \[\frac{1}{3}\] : 1 and the value of y -coordinate is \[\frac{11}{4}\].