Practice Set 20 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 20 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 20 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 20 Solutions Answers

Question 1.
Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 20 1
Solution:
∠APD and ∠APB are angles in a linear pair.
∴m∠APD + m∠APB = 180°
∴47 + m∠APB = 180
∴47 + m∠APB – 47 = 180 – 47 ….(Subtracting 47 from both sides)
∴m∠APB = 133°
m∠CPD = m∠APB = 133° … .(Vertically opposite angles)
m∠BPC = m∠APD = 47° … .(Vertically opposite angles)
∴The measures of ∠APB, ∠BPC and ∠CPD are 133°, 47° and 133° respectively.

Question 2.
Lines PQ and RS intersect at point M. m∠PMR = x°.What are the measures of ∠PMS, ∠SMQ and ∠QMR?Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 20 2
Solution:
∠PMR and ∠PMS are angles in a linear pair.
∴ m∠PMR + m∠PMS = 180°
∴ x + m∠PMS = 180
∴ m∠PMS = (180-x)°
m∠QMR = m∠PMS = (180 – x)° … .(Vertically opposite angles)
m∠SMQ = m∠PMR = x° …. (Vertically opposite angles)
∴The measures of ∠PMS, ∠SMQ and ∠QMR are (180 – x)°, x° and (180 – x)° respectively.

Class 7 Maths Solution Maharashtra Board

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

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

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.

Std 7 Maths Practice Set 19 Solutions Answers

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 1

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 2

iv.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 3

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

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)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 5

  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.

Class 7 Maths Solution Maharashtra Board

Practice Set 18 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 18 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 18 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 18 Solutions Answers

Question 1.
Name the pairs of opposite rays in the figure alongside.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 1
Solution:

  1. Ray PL and ray PM
  2. Ray PN and ray PT

Question 2.
Are the ray PM and PT opposite rays? Give reasons for your answer.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 2
Solution:
No.
Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 18 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 3

  1. Name the rays in the figure alongside.
  2. Name the origin of the rays
  3. Name the angle in the given figure

Solution:

  1. Ray BA and ray BC
  2. Point B
  3. ∠ABC or ∠CBA

Question 2.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 4

  1. Name the angle in the figure alongside.
  2. Name the rays whose origin is point B

Solution:

  1. ∠ABC or ∠CBA
  2. Ray BA and ray BC

Class 7 Maths Solution Maharashtra Board

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

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 17 Solutions Maharashtra Board

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.

Std 7 Maths Practice Set 17 Solutions Answers

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 17 1
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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 17 2

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

Solution:

  1. Straight angle
  2. 180°

Class 7 Maths Solution Maharashtra Board

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

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

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.

Std 7 Maths Practice Set 16 Solutions Answers

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)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 16 1

  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.

Class 7 Maths Solution Maharashtra Board

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

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

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

Std 7 Maths Practice Set 14 Solutions Answers

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
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 1
∴ LCM of 36 and 42 = 2 x 3 x 2 x 3 x 7
= 252

ii. 15, 25, 30
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 2
∴ LCM of 15, 25 and 30 = 5 x 3 x 5 x 2
= 150

iii. 18, 42, 48
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 3
∴ LCM of 18,42 and 48 = 2 x 3 x 2 x 2 x 3 x 7 x 2
= 1008

iv. 4, 12, 20
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 4
∴ LCM of 4, 12 and 20 = 2 x 2 x 3 x 5
= 60

v. 24, 40, 80, 120
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 5
∴ 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.
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 6

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

iii.
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 8

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.
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 14 9

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.

Class 7 Maths Solution Maharashtra Board

Practice Set 13 Class 7 Answers Chapter 3 HCF and LCM Maharashtra Board

HCF and LCM Class 7 Maths Chapter 3 Practice Set 13 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 13 Answers Solutions Chapter 3 HCF and LCM.

Std 7 Maths Practice Set 13 Solutions Answers

Question 1.
Find the LCM:
i. 12, 15
ii. 6, 8, 10
iii. 18, 32
iv. 10, 15, 20
v. 45, 86
vi. 15, 30, 90
vii. 105, 195
viii. 12,15,45
ix. 63,81
x. 18, 36, 27
Solution:
i. 12, 15
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 1
∴ LCM of 12 and 15 = 3 x 2 x 2 x 5
= 60

ii. 6, 8, 10
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 2
∴ LCM of 6, 8 and 10 = 2 x 2 x 3 x 2 x 5
= 120

iii. 18, 32
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 3
∴ LCM of 18 and 32 = 2 x 2 x 2 x 2 x 3 x 3 x 2
= 288

iv. 10, 15, 20
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 4
∴ LCM of 10, 15 and 20 = 5 x 2 x 3 x 2
= 60

v. 45, 86
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 5
∴ LCM of 45 and 86 = 2 x 3 x 3 x 5 x 43
= 3870

vi. 15, 30, 90
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 6
∴ LCM of 15,30 and 90 = 3 x 5 x 2 x 3
= 90

vii. 105, 195
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 7
∴ LCM of 105 and 195 = 5 x 3 x 7 x 13
= 1365

viii. 12, 15, 45
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 8
∴ LCM of 12, 15 and 45 = 3 x 3 x 2 x 5 x 2
= 180

ix. 63, 81
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 9
∴ LCM of 63 and 81 = 3 x 3 x 3 x 7 x 3
= 567

x. 18, 36, 27
Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 13 10
∴ LCM of 18, 36 and 27 = 3 x 3 x 2 x 2 x 3
= 108

Question 2.
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers:
i. 32, 37
ii. 46, 51
iii. 15, 60
iv. 18, 63
v. 78, 104
Solution:
i. 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 x 1
37 = 37 x 1
∴ HCF of 32 and 37 =1
LCM of 32 and 37 = 2 x 2 x 2 x 2 x 2 x 37
= 1184
HCF x LCM = 1 x 1184
= 1184
Product of the given numbers = 32 x 37
= 1184
∴ HCF x LCM = Product of the given numbers.

ii. 46 = 2 x 23 x 1
51 = 3 x 17 x 1
∴ HCF of 46 and 51 = 1
LCM of 46 and 51 = 2 x 23 x 3 x 17
= 2346
HCF x LCM = 1 x 2346
= 2346
Product of the given numbers = 46 x 51
= 2346
∴ HCF x LCM = Product of the given numbers.

iii. 15 = 3 x 5
60 = 2 x 30
= 2 x 2 x 15
= 2 x 2 x 3 x 5
∴ HCF of 15 and 60 = 3 x 5
= 15
LCM of 15 and 60 = 3 x 5 x 2 x 2
= 60
HCF x LCM = 15 x 60
= 900
Product of the given numbers = 15 x 60
= 900
∴ HCF x LCM = Product of the given numbers.

iv. 18 = 2 x 9
= 2 x 3 x 3
63 = 3 x 21
= 3 x 3 x 7
∴ HCF of 18 and 63 = 3 x 3
= 9
LCM of 18 and 63 = 3 x 3 x 2 x 7
= 126
HCF x LCM = 9 x 126
= 1134
Product of the given numbers = 18 x 63
= 1134
∴ HCF x LCM = Product of the given numbers.

v. 78 = 2 x 39
= 2 x 3 x 13
104 = 2 x 52
= 2 x 2 x 26
= 2 x 2 x 2 x 13
∴ HCF of 78 and 104 = 2 x 13
= 26
LCM of 78 and 104 = 2 x 13 x 3 x 2 x 2
= 312
HCF x LCM = 26 x 312
= 8112
Product of the given numbers = 78 x 104
= 8112
∴ HCF x LCM = Product of the given numbers.

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 13 Intext Questions and Activities

Question 1.
Write the tables of the given numbers and find their LCM. (Textbook pg. no. 19)
i. 6, 7
ii. 8, 12
iii. 5, 6, 15
Solution:
i. Multiples of 6 : 6, 12, 18, 24, 30, 36, 42
Multiples of 7 : 7, 14, 21, 28, 35, 42, 49
∴ LCM of 6 and 7 = 42

ii. Multiples of 8 : 8, 16, 24, 32, 40
Multiples of 12 : 12, 24, 36, 48
∴ LCM of 8 and 12 = 24

iii. Multiples of 5 : 5, 10, 15, 20, 25, 30, 35
Multiples of 6 : 6, 12, 18, 24, 30, 36
Multiples of 15 : 15, 30, 45, 60
∴ LCM of 5,6 and 15 = 30

Class 7 Maths Solution Maharashtra Board

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

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

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.

Std 7 Maths Practice Set 9 Solutions Answers

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
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 1

ii. 98 ÷ (-28)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 2

iii. (-51) ÷ 68
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 3

iv. 38 ÷ (-57)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 4

v. (-85) ÷ 20
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 5

vi. (-150) ÷ (-25)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 6

vii. 100 ÷ 60
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 7

viii. 9 ÷ (-54)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 8

ix. 78 ÷ 65
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 9

x. (-5) ÷ (-315)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 10

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 = \(\frac{-24}{9}=\frac{-8}{3}\)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 9 11
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.

Class 7 Maths Solution Maharashtra Board

Practice Set 8 Class 7 Answers Chapter 2 Multiplication and Division of Integers Maharashtra Board

Multiplication and Division of Integers Class 7 Maths Chapter 2 Practice Set 8 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 8 Answers Solutions Chapter 2 Multiplication and Division of Integers.

Std 7 Maths Practice Set 8 Solutions Answers

Question 1.
Multiply:

  1. (-5) × (-7)
  2. (-9) × (6)
  3. (9) × (-4)
  4. (8) × (-7)
  5. (-124) × (-1)
  6. (-12) × (-7)
  7. (-63) × (-7)
  8. (-7) × (15)

Solution:

  1. 35
  2. -54
  3. -36
  4. -56
  5. 124
  6. 84
  7. 441
  8. -105

Maharashtra Board Class 7 Maths Chapter 2 Multiplication and Division of Integers Practice Set 8 Intext Questions and Activities

Question 1.
In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below. (Textbook pg. no. 11)

  1. 5 + 7 = __
  2. 10 + (-5) = __
  3. -4 + 3 = __
  4. (-7) + (-2) = __
  5. (+8) – (+ 3) = __
  6. (+8) – (-3) = __

Solution:

  1. 12
  2. 5
  3. -1
  4. -9
  5. 5
  6. 11

Question 2.
Write a number in each bracket to obtain the answer ‘3’ in each operation. (Textbook pg. no. 11)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 8 1
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 8 2

Question 3.
Multiply the given integers and complete the table given below. (Textbook pg. no. 12)
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 8 3
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 2 Multiplication and Division of Integers Practice Set 8 4

Class 7 Maths Solution Maharashtra Board

Practice Set 7 Class 7 Answers Chapter 1 Geometrical Constructions Maharashtra Board

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 7 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 7 Answers Solutions Chapter 1 Geometrical Constructions.

Std 7 Maths Practice Set 7 Solutions Answers

Question 1.
Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 7 1
Solution:
i. ∠AOC ≅ ∠PQR
ii. ∠DOC ≅ ∠LMN
iii. ∠AOB ≅ ∠BOC ≅ ∠RST

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 7 Intext Questions and Activities

Question 1.
Observe the given angles and write the names of those having equal measures.
(Textbook pg. no. 8 and 9)
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 7 2
Solution:
i. ∠ABC and ∠SPM
ii. ∠NIT and ∠SRI
iii. ∠PTQ and ∠RTS

Question 2.
Observe the image shown in the adjacent figure and answer the following questions. (Textbook pg. no. 9)

  1. What time does this clock show?
  2. What is the measure of the angle between its two hands?
  3. At which other times is the angle between the hands congruent with this angle?

Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 7 3
Solution:

  1. 3 o’ clock.
  2. 90°.
  3. 9 o’ clock.

Question 3.
Get bangles of different sizes but equal thickness and find the congruent ones among them. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

Question 4.
Find congruent circles in your surroundings. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

Question 5.
Take some round bowls and plates. Place their edges one upon the other to find pairs of congruent edges. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

7th Std Maths Solution Maharashtra Board