NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5 is given in the article. Class 10 Maths Chapter 3 Exercise 3.5 has questions related to solving a pair of linear equations using various methods.

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Check out the solutions of Class 10 Maths NCERT solutions chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Check out other exercise solutions of Class 10 Maths Chapter 3

Class 10 Chapter 3 Topics:

Difference between linear and non-linear equations	Graph of a Linear Equation in two variables	Consistent System of Linear Equations
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

CBSE Class 10 Maths Study Guides:

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

CBSE X Related Questions

1.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
- \(2.5\) cm
- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
View Solution
2.
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
3.
Find the H.C.F. and L.C.M. of 1530 and 2040.
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.
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
6.
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

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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 3.3

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.4

CBSE X Syllabus

Mathematics Preparation Guides

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Distance Formula and Derivation of Coordinate Geometry

Probability- A Theoretical Approach: An Overview and Examples

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

Probability: Concepts, Problems and Solved Examples

Trigonometry: Ratios, Identities & Trigonometric Table

Heights and Distances: Applications in Trigonometry

Real Numbers: Euclid’s Division Lemma, Properties & Sets

Surface Areas and Volumes Revision Notes

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

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Real Numbers: Definition, Properties and Examples

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Relation Between HCF and LCM: Definition, Methods and Formulas

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NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5

CBSE X Related Questions

1.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.
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.

3.
Find the H.C.F. and L.C.M. of 1530 and 2040.

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.
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.

6.
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.

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5

CBSE X Related Questions

1.In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.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.

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

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.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.

6.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.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

2.
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.

3.
Find the H.C.F. and L.C.M. of 1530 and 2040.

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.
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.

6.
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.