Maharashtra Board Practice Set 19 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 19 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 19 Answers Solutions Chapter 4

Question 1.
Draw the pairs of angles as described below. If that is not possible, say why.
i. Complementary angles that are not adjacent.
ii. Angles in a linear pair which are not supplementary.
iii. Complementary angles that do not form a linear pair.
iv. Adjacent angles which are not in linear pair.
v. Angles which are neither complementary nor adjacent.
vi. Angles in a linear pair which are complementary.
Solution:
i.

ii. Sum of angles in a linear pair is 180°.
i.e. they are supplementary .
∴ Angles in a linear pair which are not supplementary cannot be drawn.

iii.

iv.

v.

vi. Angles in linear pair have their sum as 180° But, complementary angles have their sum as 90°.
∴ Angles in a linear pair which are complementary cannot be drawn.

Note: Problem No. i, iii, iv, and v have more than one answers students may draw angles other than the once given.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 19 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions: (Textbook pg. no. 29)

1. Write the names of the angles in the figure alongside.
2. What type of a pair of angles is it?
3. Which arms of the angles are not the common arms?
4. m∠PQR = __.
5. m∠RQS = __.

Solution:

1. ∠PQR and ∠RQS
2. Angles in a linear pair
3. Ray QP and ray QS
4. 125
5. 55
Here, m∠PQR + m∠RQS = 125° + 55°
= 180°
∴The adjacent angles ∠PQR and ∠RQS are supplementary.

Maharashtra Board Practice Set 11 Class 7 Maths Solutions Chapter 3 HCF and LCM

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 11 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 11 Answers Solutions Chapter 3

Question 1.
Factorize the following numbers into primes:
i. 32
ii. 57
iii. 23
iv. 150
v. 216
vi. 208
vii. 765
viii. 342
ix. 377
x. 559
Solution:
i. 32

∴ 32 = 2 × 2 × 2 × 2 × 2

ii. 57

∴ 57 = 3 × 19

iii. 23

∴ 23 = 23 × 1

iv. 150

∴ 150 = 2 × 3 × 5 × 5

v. 216

∴ 216 = 2 × 2 × 2 × 3 × 3 × 3

vi. 208

∴ 208 = 2 × 2 × 2 × 2 × 13

vii. 765

∴ 765 = 3 × 3 × 5 × 17

viii. 342

∴ 342 = 2 × 3 × 3 × 19

ix. 377

∴ 377 = 13 × 29

x. 559

∴ 559 = 13 × 43

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:

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:

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:

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 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 17 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 17 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 17 Answers Solutions Chapter 4

Question 1.
Write the measures of the supplements of the angles given below:
i. 15°
ii. 85°
iii. 120°
iv. 37°
v. 108°
vi. 0°
vii. a°
Solution:
i. Let the measure of the supplementary angle be x°.
∴ 15 + x = 180
∴ 15 + x – 15 = 180 – 15
….(Subtracting 15 from both sides)
∴ x = 165
∴ The measures of the supplement of an angle of 15° is 165°.

ii. Let the measure of the supplementary angle be x°.
∴ 85 + x = 180
∴ 85 + x – 85 = 180 – 85
….(Subtracting 85 from both sides)
∴ x = 95
∴ The measures of the supplement of an angle of 85° is 95°.

iii. Let the measure of the supplementary angle be x°.
∴ 120 + x = 180
∴ 120 + x – 120 = 180 – 120
….(Subtracting 120 from both sides)
∴ x = 60
∴ The measures of the supplement of an angle of 120° is 60°.

iv. Let the measure of the supplementary angle be x°.
∴ 37 + x = 180
∴ 37 + x – 37 = 180 – 37
….(Subtracting 37 from both sides)
∴ x = 143
∴ The measures of the supplement of an angle of 37° is 143°.

v. Let the measure of the supplementary angle be x°.
∴ 108 + x = 180
∴ 108 + x – 108 = 180 – 108
….(Subtracting 108 from both sides)
∴ x = 72
∴ The measures of the supplement of an angle of 108° is 72°.

vi. Let the measure of the supplementary angle be x°.
∴0 + x = 180
∴ x = 180
∴ The measures of the supplement of an angle of 0° is 180°.

vii. Let the measure of the supplementary angle be x°.
∴ a + x = 180
∴ a + x – a = 180 – a
….(Subtracting a from both sides) x = (180 – a)
∴ The measures of the supplement of an angle of a° is (180 – a)°.

Question 2.
The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.
m∠B = 60°
m∠N = 30°
m∠Y = 90°
m∠J = 150°
m∠D = 75°
m∠E = 0°
m∠F = 15°
m∠G = 120°
Solution:
i. m∠B + m∠N = 60° + 30°
= 90°
∴∠B and ∠N are a pair of complementary angles.

ii. m∠Y + m∠E = 90° + 0°
= 90°
∴∠Y and ∠E are a pair of complementary angles.

iii. m∠D + m∠F = 75° + 15°
= 90°
∴∠D and ∠F are a pair of complementary angles.

iv. m∠B + m∠G = 60° + 120°
= 180°
∴∠B and ∠G are a pair of supplementary angles.

v. m∠N + m∠J = 30° + 150°
= 180°
∴∠N and ∠J are a pair of supplementary angles.

Question 3.
In ΔXYZ, m∠Y = 90°. What kind of a pair do ∠X and ∠Z make?
Solution:
In ΔXYZ,
m∠X + m∠Y + m∠Z = 180° ….(Sum of the measure of the angles of a triangle is 180°)
∴m∠X + 90 + m∠Z = 180
∴m∠X + 90 + m∠Z – 90 = 180 – 90 ….(Subtracting 90 from both sides)
∴m∠X + m∠Z = 90°
∴∠X and ∠Z make a pair of complementary angles.

Question 4.
The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles.
Solution:
Let the measure of one angle be x°.
∴Measure of other angle = (x + 40)°
x + (x + 40) = 90 …(Since, the two angles are complementary)
∴ 2x + 40 – 40 = 90 – 40 ….(Subtracting 40 from both sides)
∴2x = 50
∴x = $$\frac { 50 }{ 2 }$$
∴x = 25
∴x + 40 = 25 + 40
= 65
∴The measures of the two angles is 25° and 65°.

Question 5.
₹PTNM is a rectangle. Write the names of the pairs of supplementary angles.

Solution:
Since, each angle of the rectangle is 90°.
∴ Pairs of supplementary angles are:
i. ∠P and ∠M
ii. ∠P and ∠N
iii. ∠P and ∠T
iv. ∠M and ∠N
v. ∠M and ∠T
vi. ∠N and ∠T

Question 6.
If m∠A = 70°, what is the measure of the supplement of the complement of ∠A?
Solution:
Let the measure of the complement of ∠A be x° and the measure of its supplementary angle be y°.
m∠A + x = 90°
∴70 + x = 90
∴70 + x – 70 = 90 – 70 ….(Subtracting 70 from both sides)
∴x = 20
Since, x and y are supplementary angles.
∴x + y = 180
∴20 + y = 180
∴20 + y – 20 = 180 – 20 ….(Subtracting 20 from both sides)
∴y = 160
∴The measure of supplement of the complement of ∠A is 160°.

Question 7.
If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A?
Solution:
Since, ∠A and ∠B are supplementary angles.
∴m∠A + m∠B = 180
∴m∠A + x + 20 = 180
∴m∠A + x + 20 – 20 = 180 – 20 ….(Subtracting 20 from both sides)
∴m∠A + x = 160
∴m∠A + x – x = 160 – x ….(Subtracting x from both sides)
∴m∠A = (160 – x)°
∴The measure of ∠A is (160 – x)°.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 17 Intext Questions and Activities

Question 1.
Observe the figure and answer the following questions. (Textbook pg. no. 26)
T is a point on line AB.

1. What kind of angle is ∠ATB?
2. What is its measure?

Solution:

1. Straight angle
2. 180°

Maharashtra Board Practice Set 16 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 16 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 16 Answers Solutions Chapter 4

Question 1.
The measures of some angles are given below. Write the measures of their complementary angles.
i. 40°
ii. 63°
iii. 45°
iv. 55°
v. 20°
vi. 90°
vii. x°
Solution:
i. Let the measure of the complementary angle be x°.
∴ 40 + x = 90
∴ 40 + x – 40 = 90 – 40
….(Subtracting 40 from both sides)
∴ x = 50
∴ The measure of the complement of an angle of measure 40° is 50°.

ii. Let the measure of the complementary angle be x°.
∴ 63 + x = 90
∴ 63+x-63 = 90-63
….(Subtracting 63 from both sides)
∴ x = 27
∴ The measure of the complement of an angle of measure 63° is 27°.

iii. Let the measure of the complementary angle be x°.
∴ 45 + x = 90
∴ 45+x-45 = 90-45
….(Subtracting 45 from both sides)
∴ x = 45
∴ The measure of the complement of an angle of measure 45° is 45°.

iv. Let the measure of the complementary angle be x°.
∴ 55 + x = 90
∴ 55 + x-55 = 90-55
….(Subtracting 55 from both sides)
∴ x = 35
∴ The measure of the complement of an angle of measure 55° is 35°.

v. Let the measure of the complementary angle be x°.
∴ 20 + x = 90
∴ 20 + x – 20 = 90 – 20
….(Subtracting 20 from both sides)
∴ x = 70
∴ The measure of the complement of an angle of measure 20° is 70°.

vi. Let the measure of the complementary angle be x°.
∴ 90 + x = 90
∴ 90 + x – 90 = 90 – 90
….(Subtracting 90 from both sides)
∴ x = 0
∴ The measure of the complement of an angle of measure 90° is 0°.

vii. Let the measure of the complementary angle be a°.
∴ x + a = 90
∴ x + a – x = 90 – x
….(Subtracting x from both sides)
∴ a = (90 – x)
∴ The measure of the complement of an angle of measure x° is (90 – x)°.

Question 2.
(y – 20)° and (y + 30)° are the measures of complementary angles. Find the measure of each angle.
Solution:
(y – 20)° and (y + 30)° are the measures of complementary angles.
∴ (y – 20) + (y + 30) = 90
∴ y + y + 30 – 20 = 90
∴ 2y+10 = 90
∴ 2y = 90 – 10
∴ 2y = 80
∴ $$y=\frac { 80 }{ 2 }$$
= 40
Measure of first angle = (y – 20)° = (40 – 20)° = 20°
Measure of second angle = (y + 30)° = (40 + 30)° = 70°
∴ The measure of the two angles is 20° and 70°.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 16 Intext Questions and Activities

Question 1.
Observe the angles in the figure and enter the proper number in the empty place. (Textbook pg. no. 26)

1. m∠ABC = ___°.
2. m∠PQR = ___°.
3. m∠ABC + m∠PQR = ___°.

Solution:

1. 40
2. 50
3. 90

Note: Here, the sum of the measures of ∠ABC and ∠PQR is 90 °. Therefore, they are complementary angles.

Maharashtra Board Practice Set 15 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 15 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 15 Answers Solutions Chapter 4

Question 1.
Observe the figure and complete the table for ∠AWB.

 Points in the interior Points in the exterior Points on the arms of the angles

Solution:

 Points in the interior point C, point R, point N, point X Points in the exterior point T, point U, point Q, point V, point Y Points on the arms of the angles point A, point W, point G, point B

Question 2.
Name the pairs of adjacent angles in the figures below.

Solution:
i. ∠ANB and ∠ANC
∠BNA and ∠BNC
∠ANC and ∠BNC

ii. ∠PQR and ∠PQT

Question 3.
Are the following pairs adjacent angles? If not, state the reason.

1. ∠PMQ and ∠RMQ
2. ∠RMQ and ∠SMR
3. ∠RMS and ∠RMT
4. ∠SMT and ∠RMS

Solution:

1. ∠PMQ and ∠RMQ are adjacent angles.
2. ∠RMQ and ∠SMR not adjacent angles since they do not have separate interiors.
3. ∠RMS and ∠RMT not adjacent angles since they do not have separate interiors.
4. ∠SMT and ∠RMS are adjacent angles.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 15 Intext Questions and Activities

Question 1.
Observe the figure alongside and write the answers. (Textbook pg. no. 24)

1. Write the name of the angle shown alongside___.
2. Write the name of its vertex___.
3. Write the names of its arms___.
4. Write the names of the points marked on its arms___.

Solution:

1. ∠ABC
2. Point B
3. Ray BA, ray BC
4. Points A, B, C

Maharashtra Board Practice Set 14 Class 7 Maths Solutions Chapter 3 HCF and LCM

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 14 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 14 Answers Solutions Chapter 3

Question 1.
Choose the right option.
i. The HCF of 120 and 150 is __
(A) 30
(B) 45
(C) 20
(D) 120
Solution:
(A) 30

Hint:
120 = 2 x 2 x 2 x 3 x 5
150 = 2 x 3 x 5 x 5
∴ HCF of 120 and 150 = 2 x 3 x 5 = 30

ii. The HCF of this pair of numbers is not 1.
(A) 13,17
(B) 29,20
(C) 40, 20
(D) 14, 15
Solution:
(C) 40, 20

Hint:
40 = 2 x 2 x 2 x 5
20 = 2 x 2 x 5
∴ HCF of 40 and 20 = 2 x 5 = 10

Question 2.
Find the HCF and LCM.
i. 14,28
ii. 32,16
iii. 17,102,170
iv. 23,69
v. 21,49,84
Solution:
i. 14 = 2 x 7
28 = 2 x 14
= 2 x 2 x 7
∴ HCF of 14 and 28 = 2 x 7
= 14
LCM of 14 and 28 = 2 x 2 x 7
= 28

ii. 32 = 2 x 16
= 2 x 2 x 8
= 2 x 2 x 2 x 4
= 2 x 2 x 2 x 2 x 2
16 = 2 x 8
= 2 x 2 x 4
= 2 x 2 x 2 x 2
∴ HCF of 32 and 16 = 2 x 2 x 2 x 2
= 16
∴ LCM of 32 and 16 = 2 x 2 x 2 x 2 x 2
= 32

iii. 17 = 17 x 1
102 = 2 x 51
= 2 x 3 x 17
170 = 2 x 85
= 2 x 5 x 17
∴ HCF of 17, 102 and 170 = 17
∴ LCM of 17, 102 and 170 = 17 x 2 x 3 x 5
= 510

iv. 23 = 23 x 1
69 = 3 x 23
∴ HCF of 23 and 69 = 23
∴ LCM of 23 and 69 = 23 x 3
= 69

v. 21 = 3 x 7
49 = 7 x 7
84 = 2 x 42
= 2 x 2 x 21
= 2 x 2 x 3 x 7
∴ HCF of 21, 49 and 84 = 7
∴ LCM of 21, 49 and 84 = 7 x 3 x 7 x 2 x 2
= 588

Question 3.
Find the LCM.
i. 36, 42
ii. 15, 25, 30
iii. 18, 42, 48
iv. 4, 12, 20
v. 24, 40, 80, 120
Solution:
i. 36, 42

∴ LCM of 36 and 42 = 2 x 3 x 2 x 3 x 7
= 252

ii. 15, 25, 30

∴ LCM of 15, 25 and 30 = 5 x 3 x 5 x 2
= 150

iii. 18, 42, 48

∴ LCM of 18,42 and 48 = 2 x 3 x 2 x 2 x 3 x 7 x 2
= 1008

iv. 4, 12, 20

∴ LCM of 4, 12 and 20 = 2 x 2 x 3 x 5
= 60

v. 24, 40, 80, 120

∴ LCM of 24, 40, 80 and 120 = 2 x 2 x 2 x 5 x 3 x 2
= 240

Question 4.
Find the smallest number which when divided by 8,9,10,15,20 gives a remainder of 5 every time.
Solution:
Here, the smallest number for division is LCM of 8, 9, 10,15 and 20.
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
15 = 3 x 5
20 = 2 x 2 x 5
LCM of given numbers = 2 x 2 x 2 x 3 x 3 x 5 = 360
∴ Required, smallest number = LCM + Remainder
= 360 + 5
= 365
∴ The required smallest number is 365.

Question 5.
Reduce the fractions $$\frac{348}{319}, \frac{221}{247}, \frac{437}{551}$$ to the lowest terms.
Solution:
i.

ii.

iii.

Question 6.
The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other?
Solution:
Here, LCM = 432, HCF = 72, First number = 216
First number x Second number = LCM x HCF
∴ 216 x Second number = 432 x 72
∴ Second number = $$\frac{432 \times 72}{216}=432 \times \frac{72}{216}=432 \times \frac{1}{3}=144$$
∴ The other number is 144.

Question 7.
The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM?
Solution:
Here, HCF = 3, Product of the given numbers = 765
Now, HCF x LCM = Product of the given numbers
∴ 3 x LCM = 765
∴ LCM = $$\frac { 765 }{ 3 }$$ = 255
∴ The LCM of the two two-digit numbers is 255.

Question 8.
A trader has three bundles of string 392 m, 308 m and 490 m long. What is the greatest length of string that the bundles can be cut up into without any left over string?
Solution:
The required greatest length of the string is the highest common factor (HCF) of 392, 308 and 490.
∴ 392 = 2 x 2 x 2 x 7 x 7
308 = 2 x 2 x 7 x 11
490 = 2 x 7 x 7 x 5
∴ HCF of 392, 308 and 490 = 2 x 7
= 14
∴ The required greatest length of the string is 14 m.

Question 9.
Which two consecutive even numbers have an LCM of 180?
Solution:
LCM of two consecutive even numbers = 180
But, HCF of two consecutive even numbers = 2
Now, product of the given number = HCF x LCM
= 2 x 180
= 360
To find the two consecutive even numbers, we have to factorize 360.
360 = 2 x 2 x 2 x 3 x 3 x 5
360 = (2 x 3 x 3) x (2 x 2 x 5)
= 18 x 20
∴ The two consecutive even numbers whose LCM is 180 are 18 and 20.

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. $$\frac{1}{2} y^{2}-8 z^{2}$$

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

Maharashtra Board Practice Set 9 Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 9 Answers Solutions Chapter 2 Multiplication and Division of Integers.

Multiplication and Division of Integers Class 7 Practice Set 9 Answers Solutions Chapter 2

Question 1.
Solve
i. (-96) ÷ 16
ii. 98 ÷ (-28)
iii. (-51) ÷ 68
iv. 38 ÷ (-57)
v. (-85) ÷ 20
vi. (-150) ÷ (-25)
vii. 100 ÷ 60
viii. 9 ÷ (-54)
ix. 78 ÷ 65
x. (-5) ÷ (-315)
Solution:
i. (-96) ÷ 16

ii. 98 ÷ (-28)

iii. (-51) ÷ 68

iv. 38 ÷ (-57)

v. (-85) ÷ 20

vi. (-150) ÷ (-25)

vii. 100 ÷ 60

viii. 9 ÷ (-54)

ix. 78 ÷ 65

x. (-5) ÷ (-315)

Question 2.
Write three divisions of integers such that the fractional form of each will be $$\frac { 24 }{ 5 }$$.
Solution:

1. $$\frac{24}{5}=\frac{24 \times 1}{5 \times 1}=\frac{24}{5}=24 \div 5$$
2. $$\frac{24}{5}=\frac{24 \times 2}{5 \times 2}=\frac{48}{10}=48 \div 10$$
3. $$\frac{24}{5}=\frac{24 \times(-10)}{5 \times(-10)}=\frac{-240}{-50}=(-240) \div(-50)$$

Question 3.
Write three divisions of integers such that the fractional form of each will be $$\frac { -5 }{ 7 }$$.
Solution:

1. $$\frac{-5}{7}=\frac{-5 \times 2}{7 \times 2}=\frac{-10}{14}=(-10) \div 14$$
2. $$\frac{-5}{7}=\frac{-5 \times(-5)}{7 \times(-5)}=\frac{25}{-35}=25 \div(-35)$$
3. $$\frac{-5}{7}=\frac{-5 \times 7}{7 \times 7}=\frac{-35}{49}=(-35) \div 49$$

Question 4.
The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.
Examples:
i. (-13) × (-15) = 195
ii. (-24) ÷ 9 = $$\frac{-24}{9}=\frac{-8}{3}$$

Solution:

1. (-13) × 9 = -117
2. 12 × 13 = 156
3. 9 × (-37) = -333
4. (-15) × (-8) = 120
5. $$(-28) \div 12=\frac{-28}{12}=\frac{(-1) \times(28)}{12}=\frac{-7}{3}$$
6. $$12 \div 9=\frac{12}{9}=\frac{4}{3}$$
7. $$9 \div(-24)=\frac{9}{-24}=\frac{9}{(-1) \times 24}=\frac{-3}{8}$$
8. $$(-18) \div(-27)=\frac{-18}{-27}=\frac{(-1) \times 18}{(-1) \times 27}=\frac{2}{3}$$

Note: Problems 2, 3 and 4 have many answers. Students may write answers other than the ones given.