Maharashtra Board Practice Set 55 Class 7 Maths Solutions Chapter 15 Statistics

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 55 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 55 Answers Solutions Chapter 15

Question 1.
The height of 30 children in a class is given in centimeters. Draw up a frequency table of this data.
131, 135, 140, 138, 132, 133, 135, 133, 134, 135, 132, 133, 140, 139, 132, 131, 134, 133, 140, 140, 139, 136, 137, 136, 139, 137, 133, 134, 131, 140
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 55 1

Question 2.
In a certain colony, there are 50 families. The number of people in every family is given below. Draw up the frequency table.
5, 4, 5, 4, 5, 3, 3, 3, 4, 3, 4, 2, 3, 4, 2, 2, 2, 2, 4, 5, 1, 3, 2, 4, 5, 3, 3, 2, 4, 4, 2, 3, 4, 3, 4, 2, 3, 4, 5, 3, 2, 3, 2, 3, 4, 5, 3, 2, 3, 2
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 55 2

Question 3.
A dice was cast 40 times and each score noted is given below. Draw up a frequency table for this data.
3, 2, 5, 6, 4, 2, 3, 1, 6, 6, 2, 3, 5, 3, 5, 3, 4, 2, 4, 5, 4, 2, 6, 3, 3, 2, 4, 3, 3, 4, 1, 4, 3, 3, 2, 2, 5, 3, 3, 4.
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 55 3

Question 4.
The number of chapatis that 30 children in a hostel need at every meal is given below. Make a frequency table for these scores.
3, 2, 2, 3, 4, 5, 4, 3, 4, 5, 2, 3, 4, 3, 2, 5, 4, 4, 4, 3, 3, 2, 2, 2, 3, 4, 3, 2, 3, 2.
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 55 4

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 55 Intext Questions and Activities

Question 1.
Make groups of 10 children in your class. Find the average height of the children in each group. (Textbook pg. no. 96)
Solution:
(Students should attempt the above activities on their own.)

Question 2.
With the help of your class teacher, note the daily attendance for a week and find the average attendance. (Textbook pg. no. 96)
Solution:
(Students should attempt the above activities on their own.)

Maharashtra Board Practice Set 54 Class 7 Maths Solutions Chapter 15 Statistics

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 54 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 54 Answers Solutions Chapter 15

Question 1.
The daily rainfall for each day of a week in a certain city is given in millimeters. Find the average rainfall during the week.
9, 11, 8, 20, 10, 16, 12
Solution:
\(\text { Average rainfall during the week }=\frac{\text { sum of rainfall for each day of the week }}{\text { number of days }}\)
= \(\frac{9+11+8+20+10+16+12}{7}\)
= \(\frac { 86 }{ 7 }\)
= 12.285 ≈ 12.29
∴ The average rainfall during the week is 12.29 mm.

Question 2.
During the annual function of a school, a Women’s Self-help Group had set up a snacks stall. Their sales every hour were worth Rs 960, Rs 830, Rs 945, Rs 800, Rs 847, Rs 970 respectively. What was the average of the hourly sales?
Solution:
\(\text { Average hourly sales }=\frac{\text { sum of sales every hour }}{\text { number of hours }}\)
= \(\frac{960+830+945+800+847+970}{6}\)
= \(\frac { 5352 }{ 6 }\)
= Rs 892
∴ The average of the hourly sales was Rs 892.

Question 3.
The annual rainfall in Vidarbha in five years is given below. What is the average rainfall for those 5 years?
900 mm, 650 mm, 450 mm, 733 mm, 400 mm.
Solution:
\(\text { Average rainfall for } 5 \text { years }=\frac{\text { sum of annual rainfall in five years }}{\text { number of years }}\)
= \(\frac{900+650+450+733+400}{5}\)
= \(\frac { 3133 }{ 5 }\)
= 626.6
∴ The average rainfall in Vidarbha for 5 years was 626.6 mm.

Question 4.
A farmer bought some sacks of animal feed. The weights of the sacks are given below in kilograms. What is the average weight of the sacks?
49.8, 49.7, 49.5, 49.3, 50,48.9, 49.2, 48.8.
Solution:
\(\text { Average weight of the sacks }=\frac{\text { sum of weight of each sack }}{\text { number of sacks }}\)
= \(\frac{49.8+49.7+49.5+49.3+50+48.9+49.2+48.8}{8}\)
= \(\frac { 395.2 }{ 8 }\)
= \(\frac { 3952 }{ 80 }\)
= 49.4
∴ The average weight of the sacks is 49.4 kg.

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 54 Intext Questions and Activities

Question 1.
Rutuja practised skipping with a rope all seven days of a week. The number of times she jumped the rope in one minute every day is given below. Find the average number of jumps per minute.
60, 62, 61, 60, 59, 63, 58. (Textbook pg. no. 96)
Solution:
\(\text { Average }=\frac{\text { Sum of the number of jumps ons even days }}{\text { Total number of days }}\)
= \(\frac{[60]+[62]+[61]+[60]+[59]+[63]+[58]}{7}\)
= \(\frac { 423 }{ 7 }\)
= 60.42
∴ Average number of jumps per minute = 60.4

Maharashtra Board Practice Set 53 Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 53 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 53 Answers Solutions Chapter 14

Question 1.
Factorize the following expressions:
i. p² – q²
ii. 4x² – 25y²
iii. y² – 4
iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
v. \(9 x^{2}-\frac{1}{16} y^{2}\)
vi. \(x^{2}-\frac{1}{x^{2}}\)
vii. a²b – ab
viii. 4x²y – 6x²
ix. \(\frac{1}{2} y^{2}-8 z^{2}\)
x. 2x² – 8y²
Solution:
i. p² – q²
Here, a = p, b = q
∴ p² – q² = (p + q)(p – q)
….[(a² – b²) = (a + b)(a – b)]

ii. 4x² – 25y²
= (2x)² – (5y)²
Here, a = 2x, b = 5y
∴ (2x)² – (5y)² = (2x + 5y)(2x – 5y)
….[(a² – b²) = (a + b)(a – b)]

iii. y² – 4
= y² – 2²
Here, a = y, b = 2
∴ y² – 2² = (y + 2)(y – 2)
….[(a² – b²) = (a + b)(a – b)]

iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
Here a = \(\frac { 1 }{ 25 }\), b = \(\frac { 1 }{ 5 }\)
\(p^{2}-\left(\frac{1}{5}\right)^{2}=\left(p+\frac{1}{5}\right)\left(p-\frac{1}{5}\right)\)
….[(a² – b²) = (a + b)(a – b)]

v. \(9 x^{2}-\frac{1}{16} y^{2}\)
Here a = 3x, b = \(\frac { 1 }{ 4 }y\)
∴\((3 x)^{2}-\left(\frac{1}{4} y\right)^{2}=\left(3 x+\frac{1}{4} y\right)\left(3 x-\frac{1}{4} y\right)\)
….[(a² – b²) = (a + b)(a – b)]

vi. \(x^{2}-\frac{1}{x^{2}}\)
Here a = x, b = \(\frac { 1 }{ x }\)
\(x^{2}-\left(\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
….[(a² – b²) = (a + b)(a – b)]

vii. a²b – ab
= a (ab – b)
= ab (a – 1)

viii. 4x²y – 6x²
= 2 (2x²y – 3x²)
= 2x² (2y – 3)

ix. \(\frac{1}{2} y^{2}-8 z^{2}\)
Maharashtra Board Class 7 Maths Solutions Chapter 14 Algebraic Formulae - Expansion of Squares Practice Set 53 1

x. 2x² – 8y²
= 2 (x² – 4y²)
= 2 [x² – (2y)²]
= 2(x + 2y)(x – 2y)
….[(a² – b²) = (a + b)(a – b)]