NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3 Solutions are based on finding the median of a grouped data, which is a measure of Central Tendency.

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Check out NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3 Solutions

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

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Chapter Related Topics
Mean of Grouped Data	Statistics Important Question	Statistics Revision Notes
Cumulative Frequency Curve	Class Mark	Statistics for Class 10
Mean Formula	Confidence Interval	Confidence Interval Formula
Step Deviation Method	Differences between Correlation and Regression	Mode of Grouped Data

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Math Formula	NCERT Solutions for Class 10 Maths	Algebra

CBSE X Related Questions

1.
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
2.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
3.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution
4.
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
5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
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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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

CBSE X Related Questions

1.
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 :

2.
The HCF of 960 and 432 is :

3.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

4.
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.

5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

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

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Exercise Solutions of Class 10 Maths Chapter 14 Statistics

CBSE X Related Questions

1.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 :

2.The HCF of 960 and 432 is :

3.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

4.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.

5.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

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.
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 :

2.
The HCF of 960 and 432 is :

3.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

4.
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.

5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

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.