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NCERT Most Important question & Solutions for class-9 Chapter-1 Matter in Our Surrounding (Chemistry)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1. Arrange the three states of matter in the increasing order of: (i) rate of diffusion(ii) particle motion.
Answer
(i) Rate of diffusion: solid < liquid < gas.

(ii) Particle motion: solid < liquid < gas

2.Explain why heat energy is needed to melt a solid. Define latent heat of fusion.
Answer
Heat energy is needed to melt a solid because there exist forces of attraction between the particles of solid. The heat energy helps in overcoming the forces of attraction between the particles and thus helps in melting a solid.Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid at atmospheric pressure at its melting point.

3.Why does the temperature remain constant during the change of state of matter? Explain it on the basis of change of solid state into liquid state.

Answer

The temperature remains constant as the heat gets used up in changing the state by overcoming the forces of attraction between the particles.For example, a solid melts on heating. Its temperature does not rise until the entire solid is converted into liquid. This heat energy gets hidden into the contents and is known as the latent heat.

4.List any four factors on which evaporation depends. Explain in short any three factors.

11. Convert the following temperature to Celsius scale:

(a) 300 K
► 300 K = (300 – 273)°C
= 27°C
(b) 573 K► 573 K = (573 – 273)°C
= 300°C

NCERT Quick Revision Notes For Chapter-1 Matter in Our Surrounding

NCERT Solution For Chapter-1 Matter in Our Surrounding

NCERT MCQs For Chapter-1 Matter in Our Surrounding

NCERT MCQs For Class-9 Chapter-15 Probability

1) The probability of each event, when a coin is tossed for 1000 times with frequencies: Head:455 & Tail: 545 is:

a. 0.455 & 0.545

b. 0.5 & 0.5

c. 0.45 & 0.55

d. 455 & 545

Answer: a

Explanation: Let E and F are the event of the occurrence of Head and Tail, respectively.

Probability of Occurrence of Head P(E) = No. of heads/total number of trials

P(E) = 455/1000 = 0.455

Similarly,

P(F) = No. of tails/total number of trials

P(F) = 545/1000 = 0.545

2) The sum of all probabilities equal to:

a. 4

b. 1

c. 3

d. 2

Answer: b

3) The probability of each event lies between:

a. 1 & 2

b. 1 & 10

c. 0 & 1

d. 0 & 5

Answer: c

4) If P(E) = 0.44, then P(not E) will be:

a. 0.44

b. 0.55

c. 0.50

d. 0.56

Answer: d

Explanation: We know;

P(E) + P(not E) = 1

0.44 + P(not E) = 1

P(not E) = 1 – 0.44 = 0.56

5) If P(E) = 0.38, then probability of event E, not occurring is:

a. 0.62

b. 0.38

c. 0.48

d. 1

Answer: a

Explanation: P(not E) = 1 – P(E) = 1-0.38 = 0.62

6) The probability of drawing an ace card from a deck of cards is:

a. 1/52

b. 1/26

c. 4/13

d. 1/13

Answer: d

Explanation: There are 4 aces in a deck of card.

Hence, the probability of taking one ace out of 52 card = 4/52 = 1/13

7) If the probability of an event to happen is 0.3 and the probability of the event not happening is:

a. 0.7

b. 0.6

c. 0.5

d. None of the above

Answer: a

Explanation: Probability of an event not happening = 1 – P(E)

P(not E) = 1 – 0.3 = 0.7

8) A dice is thrown. The probability of getting 1 and 5 is:

a. ⅙

b. ⅔

c. ⅓

d. ½

Answer: c

Explanation: The probability of getting 1 and 5 = 2/6 = ⅓

9) A batsman hits boundaries for 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

a. ⅕

b. ⅖

c. ⅗

d. ⅘

Answer: d

Explanation: No. of boundaries = 6

No. of balls = 30

No. of balls without boundaries = 30 – 6 =24

Probability of no boundary = 24/30 = ⅘

10. Three coins were tossed 200 times. The number of times 2 heads came up is 72. Then the probability of 2 heads coming up is:

a. 1/25

b. 2/25

c. 7/25

d. 9/25

Answer: d

Explanation: Probability = 72/200 = 9/25

11.Which of the following cannot be the empirical probability of an event?
(a) 2/3
(b) 3/2
(c) 0
(d) 1

Answers: (b)

12.The sum of the probabilities of all events of a trial is
(a) 1
(b) greater than 1
(c) less than 1
(d) between 0 and 1

Answers: (A)

13.Two coins are tossed 1000 times and the outcomes are recorded as below:

The probability of getting at the most one head is:
(a) 1/5
(b) 1/4
(c) 4/5
(d) 3/4

Answers: (b)

14.A card is selected at random from a deck of 52 cards. The probability of its being a red face card is
(a) 3/13
(b) 1/2
(c) 2/13
(d) 3/26

Answers: (D)

15.In a series of 6 cricket matches, the number of runs scored by the captain of a team are 54, 32, 48, 55, 29, 35. So in the next match, the probability that he will cross the half century is
(a) 0.33
(b) 0.24
(c) 0.35
(d) 0.48

Answers: (a)

NCERT Quick revision Notes For Chapter-15 Probability

NCERT Solutions For Chapter-15 Probability

NCERT Most Important Questions For Chapter-15 Probability

NCERT MCQs For Class-9 Chapter-14 Statistics

1) The ratio of the sum of observations and the total number of observations is called:

a. Mean

b. Median

c. Mode

d. Central tendency

Answer: a

2) The mean of x+2, x+3, x+4 and x-2 is:

a. (x+7)/4

b. (2x+7)/4

c. (3x+7)/4

d. (4x+7)/4

Answer: d

Explanation: Mean = (x+2+x+3+x+4+x-2)/4 = (4x+7)/4

3) The median of the data: 4, 6, 8, 9, 11 is

a. 6

b. 8

c. 9

d. 11

Answer: b

4) The median of the data: 155 160 145 149 150 147 152 144 148 is

a. 149

b. 150

c. 147

d. 144

Answer: a

Explanation: First arrange the data in ascending order.

144 145 147 148 149 150 152 155 160

Since, the number of observations here is odd, therefore,

Median = (n+1)/2 th = (9+1)/2 = 10/2 = 5th number = 149

5) The median of the data: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 is:

a. 10

b. 24

c. 12

d. 8

Answer: c

Explanation: Arrange the given data in ascending order:

2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48

Since, the number of observations givere here is even, hence,

Median will be average of two middle terms.

n/2th = 16/2 = 8th term

(n/2 +1)th = (16/2 + 1)th = 9th term

Therefore,

Median = (10+14)/2 = 12

6) The mode of the given data: 4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9 is;

a. 7

b. 9

c. 10

d. 6

Answer: b

Explanation: First arrange the data in order:

2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10

Hence, mode = 9

7) The value which appears very frequently in a data is called:

a. Mean

b. Median

c. Mode

d. Central tendency

Answer: c

8) The collection of information, collected for a purpose is called:

a. Mean

b. Median

c. Mode

d. Data

Answer: d

9) The mean of the data 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 is

a. 2

b. 2.2

c. 2.4

d. 2.8

Answer: d

Explanation: Mean = (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8

10) Which of the following is not a measure of central tendency?

a. Standard deviation

b. Mean

c. Median

d. Mode

Answer: a

11)The class mark of the class 90-130 is:
(a) 90
(b) 105
(c) 115
(d) 110

Answers: (d)

12.The range of the data:
25, 81, 20, 22, 16, 6, 17,15,12, 30, 32, 10, 91, 8, 11, 20 is
(a) 10
(b) 75
(c) 85
(d) 26

Answers: (c)

13.In the class intervals 10-20, 20-30, the number 20 is included in:
(a) 10-20
(b) 20-30
(c) both the intervals
(d) none of these intervals

Answers: (b)

14.A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304,402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 370-390 is:
(a) 0
(b) 1
(c) 3
(d) 5

Answers: (a)

15.If x1¯, x2¯, x3¯, …….., xn¯ are the means of n groups with n₁, n₂, ……. n_n number of observations respectively, then the mean x of all the groups taken together is given by:

Answer: (c)

NCERT Quick revision Notes Chapter-14 statistics

NCERT Solutions Chapter-14 statistics

NCERT Most Important Questions Chapter-14 statistics

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NCERT MCQs For Class-9 Chapter-3 Coordinate Geometry

1. The points (–4,–8) lies in:

a) First quadrant

b) Second quadrant

c) Third quadrant

d) Fourth quadrant

Answer: (d)

2. The point (0, –5) lies:

a) On the x-axis

b) On the y-axis

c) In the first quadrant

d) None of the above

Answer: (b)

3. Ordinate of all the points in the x-axis is:

a) 0

b) 1

c) –1

d) Any natural number

Answer: (a)

4. Points (1, -2), (1, -3), (-4, 5), (0, 0), (3, -3)

a) Lie in III quadrant

b) Lie in II quadrant

c) Lie in IV quadrant

d) Do not lie in the same quadrant

Answer: (d)

5. If the x-coordinate of a point is zero, then this point lies:

a) In II quadrant

b) In I quadrant

c) On x-axis

d) On y-axis

Answer: (d)

6. If the perpendicular distance of a point P from the x-axis is 7 units and the foot of the perpendicular lies on the negative direction of x-axis, then the point P has:

a) y-coordinate = 7 or –7 only

b) y-coordinate = 7 only

c) y-coordinate = –7 only

d) x-coordinate = –7

Answer: (a)

7. On plotting P (–3, 8), Q (7, –5), R (–3, –8) and T (–7, 9) are plotted on the graph paper, then point(s) in the third quadrant are:

a) P and T

b) Q and R

c) Only R

d) P and R

Answer: (c)

8. If the coordinates of the two points are P (–7, 5) and Q (–6, 9), then (abscissa of P) – (abscissa of Q) is

a) –3

b) 1

c) –2

d) –1

Answer: (d)

9. Abscissa of a point is positive in:

a) I and II quadrants

b) I and IV quadrants

c) I quadrant only

d) II quadrant only

Answer: (b)

10. The point whose ordinate is 8 and lies on y-axis:

a) (0, 8)

b) (8, 0)

c) (5, 8)

d) (8, 5)

Answer: (a)

11. The coordinates of any point on the y-axis are of the form (0, k), where |k| is the distance of the point from the:

a) y-axis

b) x-axis

c) (0, 1)

d) (1, 0)

Answer: (b)

12. The mirror of a point (3, 4) on y-axis is:

a) (3, 4)

b) (–3, 4)

c) (3, –4)

d) (–3, –4)

Answer: (b)

13. The distance of the points (5, 0) and (–3, 0) from x-axis is:

a) –3

b) 5

c) 0

d) 2

Answer: (c)

14. The perpendicular distance of a point P (5, 8) from the y-axis is:

a) 5

b) 8

c) 3

d) 13

Answer: (a)

15. A point (x + 2, x + 4) lies in the first quadrant, the mirror image for which for x-axis is (5, –7). What is the value of x?

a) (–5, –7)

b) (–5, 7)

c) (5, –7)

d) (5, 7)

Answer: (d)

NCERT Quick revision Notes Of Chapter-3 Coordinate Geometry

NCERT Solutions Of Chapter-3 Coordinate Geometry

NCERT Most Important Questions Of Chapter-3 Coordinate Geometry

NCERT MCQs For Class-9 Chapter-4 Linear Equation in Two Variables

1. The linear equation 4x – 10y = 14 has:

a) A unique solution

b) Two solutions

c) Infinitely many solutions

d) No solutions

Answer: (c)

2. Find the number of solutions of the following pair of linear equations. x + 2y – 8 = 0 and 2x + 4y = 16:

a) 0

b) 1

c) 2

d) Infinite

Answer: (d)

3. If (2, 0) is a solution of the linear equation 2x +3y = k, then the value of k is:

a) 4

b) 6

c) 5

d) 2

Answer: (a)

4. The graph of the linear equation 2x +3y = 6 cuts the y-axis at the point:

a) (2, 0)

b) (0, 3)

c) (3, 0)

d) (0, 2)

Answer: (d)

5. The equation y = 5, in two variables, can be written as:

a) 1 .x + 1 .y = 5

b) 0 .x + 0 .y = 5

c) 1 .x + 0 .y = 5

d)0 .x + 1 .y = 5

Answer: (d)

6. Any point on the line y = x is of the form:

a) (a, –a)

b) (0, a)

c) (a, 0)

d) (a, a)

Answer: (d)

7. The graph of x = 5 is a line:

a) Parallel to x-axis at a distance 5 units from the origin

b) Parallel to y-axis at a distance 5 units from the origin

c) Making an intercept 5 on the x-axis

d) Making an intercept 5 on the y-axis

Answer: (b)

8. x = 9, y = 4 is a solution of the linear equation:

a) 2x + y = 17

b) x + y = 17

c) x + 2y = 17

d) 3x – 2y = 17

Answer: (c)

9. Any point on the x-axis is of the form:

a) (0, y)

b) (x, 0)

c) (x, x)

d) (x, y)

Answer: (b)

10. If a linear equation has solutions (–3, 3), (0, 0) and (3, –3), then it is of the form:

a)y – x = 0

b)x + y = 0

c) –2x + y = 0

d) –x + 2y = 0

Answer: (b)

11. The positive solutions of the equation ax + by + c = 0 always lie in the:

a) 1st quadrant

b) 2nd quadrant

c) 3rd quadrant

d) 4th quadrant

Answer: (a)

12. The graph of the linear equation 5x + 3y = 10 is a line which meets the x-axis at the point:

a) (0, 3)

b) (3, 0)

c) (2, 0)

d) (0, 2)

Answer: (c)

13. The point of the form (a, –a) always lies on the line:

a) x = a

b) y = –a

c) y = x

d) x + y = 0

Answer: (d)

14. The graph of x = 9 is a straight line:

a) Intersecting both the axes

b) parallel to y-axis

c) parallel to x-axis

d) Passing through the origin

Answer: (b)

15. Equation of the line parallel to x-axis and 6 units above the origin is:

a) x = 6

b) x = –6

c)y = 6

d)y = –6

Answer: (c)

16.The value of y at x = -1 in the equation 5y = 2 is
(a) 5/2
(b) 2/5
(c) 10
(d) 0

Answers: (2/5)

17. To which equation does the graph represent?

(a) 3x – 7y = 10
(b) y – 2x = 3
(c) 8y – 6x = 4
(d) 5x +35/2 y = 25

Answers: (C)

18. Cost of book (x) exceeds twice the cost of pen (y) by Rs 10. This statement can be expressed as linear equation.
(a) x – 2y – 10 = 0
(b) 2x – y – 10 = 0
(c) 2x + y – 10 = 0
(d) x – 2y + 10 = 0

Answers: (a)

19.Equation of a line passing through origin is
(a) x + y = 1
(b) x = 2y – 4
(c) x + y = 0
(d) y = x – 1

Answers: (c)

20.The condition that the equation ax + by + c = 0 represents a linear equation in two variables is
(a) a ≠ 0, b = 0
(b) b ≠ 0, a = 0
(c) a = 0, b = 0
(d) a ≠ 0, b ≠ 0

Answers: (d)

NCERT Quick revision Notes Of Chapter-4 Linear Equation In two Variable

NCERT Solutions Of Chapter-4 Linear Equation In two Variable

NCERT Most Important Questions Of Chapter-4 Linear Equation In two Variable

NCERT MCQs For Class-9 Chapter-2 Polynomial

1.  Which one is not a polynomial     

 (a)   4x² + 2x – 1 (b)    (c)   x³ – 1 (d)   y² + 5y + 1

Answers:-(b)

2.  The polynomial px² + qx + rx⁴ + 5 is of tyep     

 (a)   linear (b)   quadratic (c)   cubic (d)   Biquadratic

Answers:-(d)

3.  Identify the polynomial      

(a)   x^–2 + x^–1 + 5 (b)    c)    (d)   3x² + 7

Answers:-(d)

4.  The zero of the polynomial p(x) = 2x + 5 is  

    (a)   2 (b)   5 (c)    (d)   

Answers:-(d)

5.  The number of zeros of x² + 4x + 2     

 (a)   1 (b)   2 (c)   3 (d)   none of these

Answers:-(b)

6.  The polynomial of type ax² + bx + c, a = 0 is of type     

 (a)   linear (b)   quadratic (c)   cubic (d)   Biquadratic

Answers:- (a)

7.  The value of k, if (x – 1) is a factor of 4x³ + 3x² – 4x + k, is 

    (a)   1 (b)   2 (c)   –3 (d)   3

Answers:-(c)

8.  The degree of polynomial  is  

    (a)   0 (b)   2 (c)   1 (d)   3

Answers:-(c)

9.  If 3 + 5 – 8 = 0, then the value of (3)³ + (5)³ – (8)³ is    

  (a)   260 (b)   –360 (c)   –160 (d)   160

Answers:-(b)

10.  If value of 104 × 96 is    

  (a)   9984 (b)   9469 (c)   10234 (d)   11324

Answers:-(a)

11.  The value of 5.63 × 5.63 + 11.26 × 2.37 + 2.37 × 2.37 is    

  (a)   237 (b)   126 (c)   56 (d)   64

Answers:-(d)

12.  The value of      

 (a)   300 (b)   500 (c)   400 (d)   600

Answers:-(b)

13.  If x + y = 3, x² + y² = 5 then xy is    

  (a)   1 (b)   3 (c)   2 (d)   5

Answers:-(c)

14.  If x + 2 is a factor of x³ – 2ax² + 16, then value of a is  

    (a)   3 (b)   1 (c)   4 (d)   2

Answers:-(b)

15.  If one of the factor of x² + x – 20 is (x + 5). Find the other   

   (a)   x – 4 (b)   x + 2 (c)   x + 4 (d)   x – 5

Answers:-(A)

NCERT Quick revision Notes Of Chapter-2 Polynomials

NCERT Solutions Of Chapter-2 Polynomials

NCERT Most Important Questions Of Chapter-2 Polynomials

NCERT Important Questions & Solutions for Class 9 Maths Chapter 15 Probability

You can find Chapter 2 Probability Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions for Chapter 15 Probability Ex 15.1

1.In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Here, the total number of trials = 30
∵ Number of times, the ball touched the boundary =6
∴ Number of times, the ball missed the boundary = 30 – 6 = 24
Let the event of not hitting the boundary be represented by E, then

P(E)= 24/30= 4/5

Thus, the required probability =4/5

2.In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained.

Find the probability that a student of the class was born in August.
Solution:
From the graph, we have
Total number of students born in various months = 40
Number of students born in August = 6
∴ Probability of a student of the Class-IX who was born in August = 6 / 40 = 3/20

3.An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below.

Suppose a family is chosen. Find the probability that the family has chosen is
(i) earning ₹ 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹13000-16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Here, the total number of families = 2400
(i) ∵ Number of families earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = 29
∴ Probability of a family earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles =29 / 2400

(ii) ∵ Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ Probability of a family earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579 /2400

(iii) ∵ Number of families earning less than Rs. 7000 per month and do not own any vehicle = 10
∴ Probability of a family earning less than Rs. 7000 per month and does not own any vehicle = 10 / 2400 =1 /240

(iv) ∵ Number of families earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25
∴ Probability of a family earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25/2400 = 1/96

(v) ∵ Number of families owning not more than 1 vehicle
= [Number of families having no vehicle] + [Number of families having only 1 vehicle]
= [10 + 1 + 2 + 1] + [160 + 305 + 535 + 469 + 579] = 14 + 2048 = 2062
∴ Probability of a family owning not more than 1 vehicle = 2062 /2400 = 1031/ 1200

4.The distance (in km) of 40 engineers from their residence to their place of work were found as follows

What is the empirical probability that an engineer lives
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within 1/2 km from her place of work?
Solution:
Here, total number of engineers = 40

(i) ∵ Number of engineers who are living less than 7 km from their work place = 9
∴ Probability of an engineer who is living less than 7 km from her place of work = 9/40

(ii) ∵ Number of engineers living at a distance more than or equal to 7 km from their work place = 31
∴ Probability of an engineer who is living at distance more than or equal to 7 km from her place of work = 31/40

(iii) ∵ The number of engineers living within 1/2km from their work place = 0
∴ Probability of an engineer who is living within 1/2km from her place of work = 0/40 = 0

5.Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg)
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags,chosen at random contains more than 5 kg of flour.
Solution:
Here, total number of bags = 11
∵ Number of bags having more than 5 kg of flour = 7
∴ Probability of a bag having more than 5 kg of flour = 7/11

NCERT Quick revision Notes of Chapter-15 Probability

NCERT Solutions of Chapter-15 Probability

NCERT MCQs of Chapter-15 Probability

NCERT Important Questions & Solutions for Class 9 Maths Chapter 14 Statistics

You can find Chapter 2 Statistics Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.2

1.The blood groups of 30 students of class VIII are recorded as follows
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Solution:
The required frequency distribution table is

From the above table, we have The most common blood group is O. The rarest blood group is AB.

2.The relative humidity (in %) of a certain city for a month of 30 days was as follows

(i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Here, the lowest value of observation = 84.9
The highest value of observation = 99.2
So, class intervals are 84 – 86, 86 – 88, 88 – 90, ……. , 98 – 100

(i) Thus, the required frequency distribution table is

(ii) Since, the relative humidity is high during the rainy season, so, the data appears to be taken in the rainy season.
(iii) Range = (Highest observation) – (Lowest observation) = 99.2 – 84.9 = 14.3

3.A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08 and so on.
(ii) For how many day’s was the the concentration of sulphur dioxide more than 0.11 parts per million ?
Solution:
(i) Here, the lowest value of the observation = 0.01
The highest value of the observation = 0.22
∴ Class intervals are 0.00 – 0.04, 0.04 – 0.08,……., 0.20 – 0.24
The required frequency distribution table is

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

4.The value of π upto 50 decimal places is given below
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) The required frequency distribution table

(ii) The most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

1.A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) The required graphical representation is shown as follows:

(ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) Two factors may be un education and poor background.

2.Given below are the seats won by different political parties in the polling outcome of a state assembly elections

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i) The required bar graph is shown below:

(ii) The political party A won the maximum number of seats.

3.The length of 40 leaves of a plant measured correct to one millimetre and the obtained data is represented in the following table

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves 153 mm long and Why?
Solution:
(i) The given frequency distribution table is not continuous. Therefore, first we have to modify it to be continuous distribution.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex. 126-127= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 118-0.5=117.5 & 126+0.5=126.5 so the new modified row is (117.5 – 126.5 ) so continue this in all the row of table.

Thus, the modified frequency distribution table is:

Now, the required histogram of the frequency distribution is shown below :

(ii) Yes, other suitable graphical representation is a ‘frequency polygon’.
(iii) No, it is not a correct statement. The maximum number of leaves lie in the class interval 145 – 153.

4.The runs scored by two teams A and B on the first 60 balls in a cricket match are given below

Represent the data of both the teams on the same graph by frequency polygons.
Solution:
The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex.6-7= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 1-0.5=0.5 & 6+0.5=6.5 so the new modified row is (0.5 – 6.5 ) so continue this in all the row of table.

Now also Find Class Mark for the given Number of balls. Class Mark= Upper limit+Lower limit/2 ( Ex. 1+6/2=7/2=3.5 )

Now, the required frequency polygons are as shown below:

5.100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class size = 6 – 4 = 2.
Therefore, we have the following table for length of rectangles.

The required histogram is shown below:

(ii) The maximum frequency is 44, which is corresponding to the class interval 6 – 8.
∴ Maximum number of surnames lie in the class interval 6 – 8.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

1.The following number of goals were scored by a team in a series of 10 matches
2, 3, 4, 5, 0, 1, 3, 3, 4, 3.
Find the mean, median and mode of these scores.
Solution:
To find the mean :
Here, n = 10

Thus, mean = 2.8

To find median:
Now arranging the given data in ascending order,
we have 0,1, 2, 3, 3, 3, 3, 4, 4, 5
∵ n = 10, an even number

Thus, median = 3

To find mode:
In the given data, the observation 3 occurs 4 times,
i.e., maximum number of times.
Thus, mode = 3

2.The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Here, the given observations are in ascending order.
Since, n = 10 (an even number)

Since, median = 63 [Given]
∵ x + 1 = 63 ⇒ x = 63 – 1 = 62
Thus, the required value of x is 62.

3.Find the mean salary of 60 workers of a factory from the following table

Solution:

Thus, the required mean salary = Rs. 5083.33

NCERT Quick revision Notes For Chapter-4 Statistics

NCERT Solutions For Chapter-4 Statistics

NCERT MCQs For Chapter-4 Statistics

NCERT Important questions & Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

You can find Chapter 2 Linear Equations in Two Variables Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.1

1. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35     (ii) 5 = 2x

Solutions

(i) We have 2x + 3y = 9.35
or (2)x + (3)y + (−9.35 ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35

(ii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.2

1.Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9

Solutions:-

(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

2.Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.3

1.Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2

Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the ime is the required graph of x – y = 2

2.Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
(2, 14) means x = 2 and y = 14
Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0
There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

3.The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.

Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

4.In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is a linear equation that converts Fahrenheit to Celsius:
F = (9/5 )C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.
Solution:
(i) We have
F = (9/5 )C + 32
When C = 0 , F = (9/5 ) x 0 + 32 = 32
When C = 15, F = (9/5 )(-15) + 32= -27 + 32 = 5
When C = -10, F = 9/5 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (9/5)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (9/5)C + 32 ⇒ −32×59/9 = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = 9/5F + 32 ⇒ F – 9/5F = 32
⇒ −4/5F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.4

1.Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,
i.e. y = 3.

NCERT Quick revision Notes For Chapter-4 Linear Equation in Two Variables

NCERT Solutions For Chapter-4 Linear Equation in Two Variables

NCERT MCQs For Chapter-4 Linear Equation in Two Variables

NCERT Important Questions & Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

You can find Chapter 2 Coordinate Geometry Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

1. Write the answer of each of the following questions:
   (i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
   (ii) What is the name of each part of the plane formed by these two lines?
   (iii) Write the name of the point where these two lines intersect.
   Solution:
   (i) The horizontal line: x – axis and the vertical line: y – axis.
   (ii) Each part is called “Quadrant”.
   (iii) Origin

2.See the given figure and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3,-5).
(iv) The point identified by the coordinates (2,-4).
(v) The abscissa of point D.
(vi) The ordinate of the point H.
(vii) The coordinates of the point L.
(viii) The coordinates of the point M.

NCERT Quick revision Notes Chapter-3 Coordinate Geometry

NCERT Solutions Chapter-3 Coordinate Geometry

NCERT MCQ Chapter-3 Coordinate Geometry