NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 is covered in this article with a step by step explanation. Chapter 1 Real Numbers Exercise 1.1 covers basic concepts of divisibility of integers using Euclid’s division algorithm. Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b in a way that the remainder will be smaller than b.

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CBSE X Related Questions

1.
Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.
View Solution
2.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
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, \(AB \parallel EF\). 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
5.
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
6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

CBSE X Related Questions

1.
Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.

2.
The natural number 1 is :

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, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.
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 :

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

CBSE X Related Questions

1.Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.

2.The natural number 1 is :

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, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.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 :

6.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.

2.
The natural number 1 is :

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, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.
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 :

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is