NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3 is provided in the article. Class 10 Maths Chapter 3 Exercise 3.3 has questions related to solving for x and y in a system of linear equations.

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Check out the solutions of Class 10 Maths NCERT solutions chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Check out other exercise solutions of Class 10 Maths Chapter 3

Class 10 Chapter 3 Topics:

Pair of Linear Equations in Two Variables Important Questions	Pair of Linear Equations in Two Variables Formula	Pair of Linear Equations in Two Variables Revision Notes
Difference between linear and non-linear equations	Substitution Method	Cross-multiplication of solving linear equations

CBSE Class 10 Maths Study Guides:

Difference between in Maths	NCERT Solutions for Class 10 Maths	Math MCQs
Trigonometry	Math Study Notes	Math Formula
Calculus	Mensuration	Arithmetic

CBSE X Related Questions

1.
Find the H.C.F. and L.C.M. of 1530 and 2040.
View Solution
2.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
View Solution
3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
4.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution
5.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.
View Solution
6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution

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Similar Mathematics Concepts

Pair of Linear Equations In Two Variables Important Questions

Pair of Linear Equations in Two Variables: Revision Notes

Pair of Linear Equations in Two Variables Formula: Intersecting and Coincident Lines

NCERT Solutions for class 10 Maths Chapter 3: Pair Of Linear Equations In Two Variables

NCERT Solutions for Class 10 Maths: Chapter-wise Solutions and Important Topics

Pair of Linear Equations in Two Variables MCQs

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.4

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5

CBSE X Syllabus

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Area of Circle: Area Formula, Derivation with Solved Examples

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Pythagoras Theorem: Formula, Theorem, Proof and Examples

Nature of Roots of Quadratic Equation

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Trigonometry: Ratios, Identities & Trigonometric Table

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Circles: Definition, Properties, Parts, Theorems

Constructions Revision Notes

Pair of Linear Equations In Two Variables Important Questions

Circles Revision Notes: Theorems of Tangent to the Circle

Statistics Important Questions

Mathematics NCERT Solutions

Euclid's Division Lemma: Proof, Finding HCF & Examples

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NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

CBSE X Related Questions

1.
Find the H.C.F. and L.C.M. of 1530 and 2040.

2.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

5.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

CBSE X Related Questions

1.Find the H.C.F. and L.C.M. of 1530 and 2040.

2.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

3.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

5.In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

6.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the H.C.F. and L.C.M. of 1530 and 2040.

2.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

3.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

4.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

5.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.