NCERT Solutions for Class 9 Maths Chapter 11 Construction


A firm grip over mathematics from the school level helps a lot in the future. There are some complicated concepts in the NCERT Class 9 Maths syllabus that an average student may not understand in the first go. But going through CoolGyan PDF material once or twice shall clear all your doubts, and you will develop a thorough understanding of even the complex concepts. Solutions to NCERT Exercise questions will guide you through the process of Construction step by step. After you are done with NCERT questions, you can practice different questions from CoolGyan pdf material. 

We provide CBSE Solutions for each question of Class 9 Maths Chapter 11 in pdf format. You can download the solutions from our website. So, enrol in CoolGyan today and take advantage of quality study with thousands of other students. Download NCERT Solutions for class 9 Maths and NCERT Solutions for Class 9 Science from CoolGyan, which are curated by master teachers.


NCERT Solutions for Class 9 Maths Chapter 11 Construction - PDF Download

1.Construct an angle of $\mathbf{90}{}^\circ $ at the initial point of a given ray and justify the construction.

Ans: The steps given below will be followed to construct an angle of\[90{}^\circ \].

(i) Take the given ray $PQ$. Draw an arc of some radius by taking point $P$ as its centre, which intersects $PQ$ at $R$.

(ii) Take $R$ as centre & with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

(iii) Now take $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$(see figure).

(iv) Take $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

(v) Join $PU$, which is the required ray making \[90{}^\circ \] with the given ray $PQ$.

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Justification

We can justify the construction, if we can prove \[\angle UPQ=90{}^\circ .\] For this, join \[~PS\] and \[PT.\]

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We have, $\angle SPQ=\angle TPS=60{}^\circ .$In (iii) and (iv) steps of this construction, $PU$was drawn as the bisector of $\angle TPS.$

$\angle UPS=\frac{1}{2}\angle TPS $

 $ =\frac{1}{2}\times {{60}^{o}} $

 $ ={{30}^{o}}  $

Also, 

$ \angle UPQ=\angle SPQ+\angle UPS $

 $ ={{60}^{o}}+{{30}^{o}} $

 $ ={{90}^{o}}$  


2. Construct an angle of $\mathbf{45}{}^\circ $at the initial point of a given ray and justify the construction.

Ans: The steps given below will be followed to construct an angle of $45{}^\circ $.

Take the given ray $PQ$. Draw an arc of some radius taking point $P$as its centre, which intersects $PQ$ at $R$.

Take $R$ as centre & with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

Take $S$ as centre & with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

Take $S$ and $T$ as centre, draw an arc of the same radius to intersect each other at $U$.

Join $PU$. Let it intersect the arc at point $V$.

From $R$ and $V$, draw arcs with radius more than $\frac{1}{2}RV$to intersect each other at $W$. Join $PW$. $PW$ is the required ray making $45{}^\circ $with $PQ.$

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Justification of Construction:

We can justify the construction if we can prove ∠WPQ = 45°. For this, join PS and PT.

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 We have,$\angle SPQ=\angle TPS=60{}^\circ $. In (iii) and (iv) steps of this construction, $PU$was drawn as the bisector of \[\angle TPS.\]

  $ \angle UPS=\frac{1}{2}\angle TPS $

 $ =\frac{1}{2}\times {{60}^{o}} $

 $ ={{30}^{o}}  $

Also,

  $\angle UPQ=\angle SPQ+\angle UPS $

 $ ={{60}^{o}}+{{30}^{o}} $

 $ ={{90}^{o}} $

\[\]In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.

$ \angle WPQ=\frac{1}{2}\angle UPQ $

 $=\frac{1}{2}\times {{90}^{o}}$

 $={{45}^{o}}  $


3. Construct the angles of the following measurements:

(i) $30{}^\circ $  

Ans: The steps given below will be followed to construct an angle of $30{}^\circ $.

(1) Draw the given ray $PQ$. Taking $P$ as centre and with some radius, draw an arc of a circle that intersects $PQ$ at $R$.

(2) Take $R$ as centre and with the same radius as before & draw an arc intersecting the previously drawn arc at point $S$.

(3) Take $R$ and $S$ as centre and with radius more than $\frac{1}{2}RS$, draw arcs to intersect each other at $T$. Join $PT$which is the required ray making $30{}^\circ $ with the given ray $PQ$.

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(ii) ${{22}^{o}}\frac{1}{2}$

Ans: The steps given below will be followed to construct an angle of ${{22}^{o}}\frac{1}{2}$.

Take the given ray $PQ$. Draw an arc of some radius, taking point $P$ as its centre, which intersects $PQ$ at $R$.

Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

Join$PU$. Let it intersect the arc at point $V$.

From $R$ and $V$, draw arcs with radius more than $\frac{1}{2}RV$ to intersect each other at $W$. Join $PW$.

Let it intersect the arc at $X$. Taking $X$ and $R$as centre and radius more than half $RX$, draw arcs to intersect each other at $Y$.

Join $PY$ which is the required ray making ${{22}^{o}}\frac{1}{2}$ with the given ray $PQ$.

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(iii) $15{}^\circ $

Ans: The below given steps will be followed to construct an angle of $15{}^\circ $.

(1) Draw the given ray $PQ$. Taking $P$ as centre and with some radius, draw an arc of a circle which intersects $PQ$ at $R$.

(2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $S$.

(3) Taking $R$ and $S$ as centre and with radius more than $\frac{1}{2}RS$, draw arcs to intersect each other at $T$. Join $PT$.

(4) Let it intersect the arc at $U$. Taking $U$ and $R$ as centre and with radius more than $\frac{1}{2}RU$, draw an arc to intersect each other at $V$. Join $PV$ which is the required ray making $15{}^\circ $ with the given ray $PQ$.

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4. Construct the following angles and verify by measuring them by a protractor: 

(i) $75{}^\circ $

Ans: The below given steps will be followed to construct an angle of $75{}^\circ $.

Take the given ray $PQ$. Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R$.

Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

Join $PU$. Let it intersect the arc at $V$. Taking $S$ and $V$ as centre, draw arcs with radius more than $\frac{1}{2}SV$. Let those intersect each other at $W$. Join $PW$ which is the required ray making $75{}^\circ $ with the given ray $PQ$ .

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The angle so formed can be measured with the help of a protractor. It comes to be $75{}^\circ $.


(ii) $105{}^\circ $

Ans: The below given steps will be followed to construct an angle of $105{}^\circ $

Take the given ray $PQ$. Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R$.

Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

Taking $S$ and $T$as centre, draw an arc of same radius to intersect each other at U.

Join $PU$. Let it intersect the arc at $V$. Taking $T$ and $V$ as centre, draw arcs with radius more than $\frac{1}{2}TV$. Let these arcs intersect each other at W. Join $PW$ which is the required ray making $105{}^\circ $ with the given ray $PQ$.

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The angle so formed can be measured with the help of a protractor. It comes to be $105{}^\circ $


 (iii) $135{}^\circ $

Ans: The steps given below will be followed to construct an angle of $135{}^\circ $.

 (1) Take the given ray $PQ$. Extend $PQ$ on the opposite side of $Q$. Draw a semi-circle of some radius taking point P as its centre, which intersects $PQ$ at $R$and$W$.

(2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$. 

(3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure). 

(4) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at U. 

(5) Join $PU$. Let it intersect the arc at $V$. Taking $V$and $W$as centre and with radius more than $\frac{1}{2}TV$, draw arcs to intersect each other at $X$. Join $PX,$which is the required ray making $135{}^\circ $ with the given line $PQ$.

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The angle so formed can be measured with the help of a protractor. It comes to be $135{}^\circ $.


5. Construct an equilateral triangle, given its side and justify the construction.

Ans: Let us draw an equilateral triangle of side $5cm$. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be $5cm$. We also know that each angle of an equilateral triangle is$60{}^\text{o}$. The steps given below will be followed to draw an equilateral triangle of $5cm$ side. 

Step I: Draw a line segment $AB$of $5cm$ length. Draw an arc of some radius, while taking $A$ as its centre. Let it intersect $AB$ at $P$. 

Step II: Take $P$ as centre & draw an arc intersecting the previous arc at $E$. Join $AE$. 

Step III: Take $A$ as centre & draw an arc of $5cm$ radius, which intersects extended line segment $AE$ at $C$. Join $AC$ and $BC$. $\Delta ABC$is the required equilateral triangle of side $5cm$. 

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Justification of Construction: 

We can justify the construction by showing $ABC$as an equilateral triangle i.e., 

$AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5cm$

and $\angle A\text{ }=\angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ .$

In $\Delta ABC$, we have $AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5cm$and $\angle A\text{ }=\text{ }60{}^\circ $. 

Since $AC\text{ }=\text{ }AB,\angle B\text{ }=\angle C$(Angles opposite to equal sides of a triangle) 

In $\Delta ABC$,

 $\angle A\text{ }+\angle B\text{ }+\angle C\text{ }=\text{ }180{}^\circ $ (Angle sum property of a triangle)

$~\Rightarrow 60{}^\circ \text{ }+\angle C\text{ }+\angle C\text{ }=\text{ }180{}^\circ $

$\Rightarrow 60{}^\circ \text{ }+\text{ }2\angle C\text{ }=\text{ }180{}^\circ $

$\Rightarrow 2\angle C\text{ }=\text{ }180{}^\circ \text{ }-\text{ }60{}^\circ \text{ }=\text{ }120{}^\circ $

$\Rightarrow \angle C\text{ }=\text{ }60{}^\circ $

$\Rightarrow \angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ $

We have,$\angle A\text{ }=\angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ $... (1) 

$\angle A\text{ }=\angle B$and $\angle A\text{ }=\angle C$

$BC\text{ }=\text{ }AC$and $BC\text{ }=\text{ }AB$(Sides opposite to equal angles of a triangle) 

$AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5\text{ }cm$... (2) 

From Equations (1) and (2), $\Delta ABC$is an equilateral triangle.


Exercise (11.2)

1. Construct a triangle $\mathbf{ABC}$ in which $\mathbf{BC}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\angle \mathbf{B}\text{ }=\text{ }\mathbf{75}{}^\circ $and $\mathbf{AB}\text{ }+\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{13}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw a line segment $BC$of$7\text{ }cm$. At point $B$, draw an angle of$75{}^\circ $, say$\angle XBC$.

 Step II: Cut a line segment $BD\text{ }=\text{ }13\text{ }cm$(that is equal to$AB\text{ }+\text{ }AC$) from the ray$BX$.

 Step III: Join $DC$and make an angle $DCY$equal to $\angle BDC$.

 Step IV: Let \[CY\]intersect \[BX\]at \[A\]. $\Delta ABC$is the required triangle.

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2. Construct a triangle \[\mathbf{ABC}\] in which $\mathbf{BC}\text{ }=\text{ }\mathbf{8}\text{ }\mathbf{cm},\angle \mathbf{B}\text{ }=\text{ }\mathbf{45}{}^\circ $ and \[\mathbf{AB}\text{ }-\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm}\]. 

Ans: The steps given below will be followed to draw the required triangle. 

Step I: Draw the line segment $BC\text{ }=\text{ }8\text{ }cm$and at point $B$, make an angle of$45{}^\circ $, say$\angle XBC$. 

Step II: Cut the line segment $BD\text{ }=\text{ }3.5\text{ }cm$(equal to$AB\text{ }-\text{ }AC$) on ray$BX$. 

Step III: Join $DC$and draw the perpendicular bisector $PQ$of$DC$. 

Step IV: Let it intersect $BX$ at point $A$. Join $AC$. $\Delta ABC$is the required triangle.

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3. Construct a triangle $\mathbf{PQR}$ in which $\mathbf{QR}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\angle \mathbf{Q}\text{ }=\text{ }\mathbf{60}{}^\circ $and $\mathbf{PR}\text{ }-\text{ }\mathbf{PQ}\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{cm}$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw line segment $QR$ of $6\text{ }cm$. At point $Q$, draw an angle of $60{}^\circ $, say $\angle XQR$. 

Step II: Cut a line segment $QS$ of $2\text{ }cm$from the line segment $QT$extended in the opposite side of line segment$XQ$. (As $PR\text{ }>\text{ }PQ$and$PR\text{ }-\text{ }PQ\text{ }=\text{ }2\text{ }cm$). Join $SR$. 

Step III: Draw perpendicular bisector $AB$of line segment $SR$. Let it intersect $QX$at point $P$. Join$PQ,\text{ }PR$. $\Delta PQR$is the required triangle.

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4. Construct a triangle $\mathbf{XYZ}$ in which $\angle \mathbf{Y}\text{ }=\text{ }\mathbf{30}{}^\circ ,\angle \mathbf{Z}\text{ }=\text{ }\mathbf{90}{}^\circ $and $\mathbf{XY}\text{ }+\text{ }\mathbf{YZ}\text{ }+\text{ }\mathbf{ZX}\text{ }=\text{ }\mathbf{11}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw a line segment $AB$ of $11cm$. (As $XY\text{ }+\text{ }YZ\text{ }+\text{ }ZX\text{ }=\text{ }11\text{ }cm$) 

Step II: Construct an angle, $\angle PAB$, of $30{}^\circ $ at point A and an angle, $\angle QBA$, of $90{}^\circ $ at point $B$. 

Step III: Bisect $\angle PAB$ and $\angle QBA$. Let these bisectors intersect each other at point $\angle QBA$. 

Step IV: Draw perpendicular bisector $ST$ of $AX$ and $UV$ of $BX$.

Step V: Let $ST$ intersect $AB$ at $Y$ and $UV$ intersect $AB$ at $Z$. Join $XY,\text{ }XZ$. $\Delta XYZ$is the required triangle.

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5. Construct a right triangle whose base is $\mathbf{12}\text{ }\mathbf{cm}$and sum of its hypotenuse and other side is $\mathbf{18}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw line segment $AB$ of $12cm$. Draw a ray $AX$ making $90{}^\circ $with $AB$

Step II: Cut a line segment $AD$ of $18cm$ (as the sum of the other two sides is $18$) from ray $AB$. 

Step III: Join $DB$ and make an angle $DBY$equal to $ADB$. 

Step IV: Let $BY$ intersect $AX$ at $C$. Join$AC,\text{ }BC$. \[\Delta ABC\]is the required triangle.

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NCERT Solutions for Class 9 Maths Chapter 11 Construction - PDF Download

If you are looking for online study resources for CBSE Class 9 Mathematics, CoolGyan can be the best e-platform for you. Solutions of all the lessons in the book have been structured in a concise, easy to understand manner by experts who have years of experience in their field of study. The questions asked in the CBSE Maths Class 9 Textbook are crucial from an exam point of view, and a student should understand the way of solution writing to achieve good grades. The answers to problems are presented in a step by step manner for students to grasp them quickly. 

Class 9 Maths syllabus includes chapters like Number Systems, Algebra, Coordinate Geometry, Geometry, Mensuration, Statistics, and Probability. To answer the textbook's questions, and in the exam, one needs to understand various complex concepts. The explanations to these complex questions are provided in the CoolGyan platform in a simple, lucid manner for the students to understand and produce in their own words in the examination. One can download exclusive and best in class study material including Maths Class 9 Chapter 11, Constructions in easy to read pdf format by merely visiting the CoolGyan official website, CoolGyan.org.

You can opt for Chapter 11 - Constructions NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.


NCERT Solutions for Class 9 Maths


Construction Class 9 NCERT Solutions

Class 9 Mathematics, NCERT book, Chapter 11 is 'Constructions.' In this chapter, students would learn about the methodology of precisely constructing angles and various 2D shapes of the required measurements. Students in the process would learn about the use of Compass, Protractor, Dividers, and Set-Squares. Although the chapter is regarded as reasonably comfortable in the students' point of view, a student should still adequately understand the methodologies and the logic involved in construction to score full marks in the examination. Proper understanding and practice of all the NCERT questions in a stepwise and thorough manner are essential. 

Our subject experts at CoolGyan have taken utmost care in preparing the solutions by going through years of answer keys from CBSE. A student who is well versed with these NCERT questions and our solutions can easily score full marks in the CBSE Board exams. Additionally, students would get themselves acquainted with the type and pattern of questions asked in the exams. The NCERT book itself has numerous questions from varying degrees of difficulty that access a student's understanding. Mastery over the NCERT book will help you breeze over your exams.


We Cover All Exercises in The Chapter Given Below

EXERCISE 11.1 - 5 Questions with Solutions in PDF

EXERCISE 11.2 - 5 Questions with Solutions in PDF


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The materials by CoolGyan are enough to score good marks. Regular as well as weekly practise tests help the students to get acquainted with the exam patterns. The students will also get an idea to analyze themselves from the performance. To the meticulous students, the scholarship is also offered by CoolGyan, which will encourage them to study harder and perform better.

FAQs (Frequently Asked Questions)

1. How can I find the Solutions of NCERT Chapter 11 Maths Class 9?

You can find the solutions of NCERT Chapter 11 Maths Class 9 on the Internet easily. But CoolGyan is the best platform for NCERT solutions. A student who is keen to study will find the CoolGyan platform as his best study resource. The CoolGyan e-platform consists of pinpoint, well-structured study material, which will help immensely in your preparation. Solutions to all your NCERT book questions including Class 9th Maths Chapter 9 are provided in a concise step by step manner so that you understand the process of solving problems. This will help in your concept building. Moreover, we conduct live interactive classes on our CoolGyan platform. Expert mentors will make you learn concepts in a fun and exciting way, unlike boring traditional methods used in schools.


2. How can I resolve my doubts regarding NCERT Class 9 Maths Chapter 11?

There are various methods in which you can resolve your doubts on the CoolGyan platform. You can ask your mentor all your doubts in the live classes. Additionally, you can post all your doubts in our doubt forum, where they would be cleared instantly by our expert teachers. You can also message your mentor and get the doubt resolved. There is enough material for you to revise your concepts before exams. We have chapter-wise tests, timely tests that would access your preparation level and generate a competitive environment among your peers. Additional summary, formula snippets, etc., will help you in your last-minute revision.

3. How many exercises are there in Class 9 Maths Chapter 11?


There are two exercises covered in Class 9 Maths Chapter 11: Exercise 11.1 and Exercise 11.2 both are of equal importance. On the CoolGyan website (CoolGyan.org), you can find the NCERT Solutions of Class 9 Maths Chapter 11 for both these exercises and you can download them for free. However, first, try to solve the sums by yourself, and in case you are stuck, refer to the solutions of the same. If you solve the questions this way, then it will prepare you for the exams.

4. How do I study for Chapter 11  of Class 9 CBSE Maths?

Ans: For studying Chapter 11 of Class 9 CBSE Maths, you have to make a proper schedule as in for each subtopic. Practice is the core element. Unlike English or History, storytelling doesn’t occur in Maths as it is not a theory subject. You have to apply your mind to understand the practical aspects. Make a separate notebook, and in one place write all the important theorems and formulas to study easily for Class 9 Chapter 11 CBSE Maths. You can also paste important theorems and formulas of Chapter 11 in your study area. For a PDF of solutions, visit the page NCERT Solutions of Class 9 Maths and download the PDF for free.


5. What is the best reference book for Chapter 11  of Class 9 Maths students?


Ans: You can get the best reference book for Class 9 Chapter 11 Maths on CoolGyan. It has mainly all the reference books, which are required by the students of Class 9. These reference books are really helpful for the students who want to prepare properly for their exams. Chapter 11 solutions are created by subject matter experts and are equipped with exercises and have easy language which is understood by the students. All the study material is available on the CoolGyan mobile app. 


6. How many chapters are there in NCERT Class 9 Maths apart from Chapter 11?


Ans: In NCERT Class 9 Maths, there are 15 Chapters that start from ‘Number Systems’ and end with ‘Probability’. Each chapter has different concepts and types of questions. Some are easy chapters while the others are a bit difficult. In the starting chapter, you will get easy questions and after that, the difficulty level will rise. Chapter 11 is moderate, neither too easy nor too difficult. To get Chapter-wise questions and answers, you can visit the page NCERT Solutions for Class 9 Maths.


7. How can I get more than 80% marks in Class 9 Chapter 11 Maths class tests?


Ans: To score more than 80% marks in Class 9 Chapter 11 Maths Exams, the first thing you will need is a focused mind. If in any case, you lose your focus, devise a method to bring it back to the chapter as concentration is the key here. The students who score higher are not extraordinary, but the effort they put in is extraordinary. Devote more time to the questions, which do not seem easy for you, but also do a revision daily of the ones which are easy for you.