NCERT Class 10 Maths Chapter 14 Statistics

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NCERT Solutions for class 10 Maths Chapter 14 Statistics are provided in the article below. Some of the important topics of Statistics chapter include:

Expected no of questions: 2 to 3 questions of total 5 marks

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 14 pdf

NCERT Solutions for Class 10 Mathematics Chapter 14

NCERT Solutions for Class 10 Maths Chapter 14 Statistics is given below.

NCERT Solution class 10 mathematics chapter 14 statistics

Class 10 Mathematics Chapter 14 Statistics – Important Topics

Study of collection, presentation, organisation and analysis of data is known as Statistics.

Important formulas in Statistics include:

Mean x = ∑x / n

Median [if n is odd] = (n+1 /2)

Median [if n is even] = [(n/2)th + (n+1 /2)] / 2

Standard Deviation σ = √∑ni=1(xi-μμ)² /n

Variance σ² = ∑ni=1(xi-μμ)² /n

Average of a given set of numbers is known as the mean.
Middle Value in a given set of data is known as median.
Most Occurring item in a given set of data is known as mode.
A measure which gives an approximate idea about data is known as variance. It is used to calculate standard deviation of a data set.
We can find how far a value is from the mean by the help of Standard Deviation. It can be found by taking the square root of Variance.

NCERT Solutions for Class 10 Maths Chapter 14 Exercises

NCERT Solutions for Class 10 Mathematics Chapter 14 Statistics Exercises are given below.

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CBSE X Related Questions

1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
  \(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
  \(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

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2.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

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3.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them
Monthly consumption
(in units)
Number of consumers
65 - 85
4
85 - 105
5
105 - 125
13
125 - 145
20
145 - 165
14
165 - 185
8
185 - 205
4

Monthly consumption (in units)	Number of consumers
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

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4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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5.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.
(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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6.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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NCERT Solutions For Class 10 Maths Chapter 14: Statistics

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 14 pdf

NCERT Solutions for Class 10 Mathematics Chapter 14

Class 10 Mathematics Chapter 14 Statistics – Important Topics

NCERT Solutions for Class 10 Maths Chapter 14 Exercises

CBSE X Related Questions

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Similar Mathematics Concepts

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NCERT Solutions For Class 10 Maths Chapter 14: Statistics

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 14 pdf

NCERT Solutions for Class 10 Mathematics Chapter 14

Class 10 Mathematics Chapter 14 Statistics – Important Topics

NCERT Solutions for Class 10 Maths Chapter 14 Exercises

CBSE X Related Questions

3. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare themMonthly consumption (in units) Number of consumers65 - 85 485 - 1055105 - 12513125 - 14520145 - 16514165 - 1858185 - 2054

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.