Maharashtra Board Practice Set 23 Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 23 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Practice Set 23 Answers Solutions Chapter 5

Question 1.
Write three rational numbers that lie between the two given numbers.
i. \(\frac{2}{7}, \frac{6}{7}\)
ii. \(\frac{4}{5}, \frac{2}{3}\)
iii. \(-\frac{2}{3}, \frac{4}{5}\)
iv. \(\frac{7}{9},-\frac{5}{9}\)
v. \(\frac{-3}{4}, \frac{+5}{4}\)
vi. \(\frac{7}{8}, \frac{-5}{3}\)
vii. \(\frac{5}{7}, \frac{11}{7}\)
viii. \(0, \frac{-3}{4}\)
Solution:
i. \(\frac{2}{7}, \frac{6}{7}\)
The three numbers lying between \(\frac { 2 }{ 7 }\) and \(\frac { 6 }{ 7 }\) are \(\frac{3}{7}, \frac{4}{7}, \frac{5}{7}\)

ii. \(\frac{4}{5}, \frac{2}{3}\)
\(\frac{4}{5}=\frac{24}{30}, \frac{2}{3}=\frac{20}{30}\)
The three numbers between \(\frac { 4 }{ 5 }\) and \(\frac { 2 }{ 3 }\) are \(\frac{21}{30}, \frac{22}{30}, \frac{23}{30}\)

iii. \(-\frac{2}{3}, \frac{4}{5}\)
\(\frac{-2}{3}=\frac{-10}{15}, \frac{4}{5}=\frac{12}{15}\)
The three numbers between \(\frac { -2 }{ 3 }\) and \(\frac { 4 }{ 5 }\) are \(\frac{-9}{15}, \frac{-7}{15}, \frac{4}{15}\)

iv. \(\frac{7}{9},-\frac{5}{9}\)
The three numbers between \(\frac { 7 }{ 9 }\) and \(\frac { -5 }{ 9 }\) are \(\frac{6}{9}, 0, \frac{-4}{9}\)

v. \(\frac{-3}{4}, \frac{+5}{4}\)
The three numbers between \(\frac { -3 }{ 4 }\) and \(\frac { +5 }{ 4 }\) are \(\frac{-2}{4}, \frac{-1}{4}, \frac{3}{4}\)

vi. \(\frac{7}{8}, \frac{-5}{3}\)
\(\frac{7}{8}=\frac{21}{24}, \frac{-5}{3}=\frac{-40}{24}\)
The three numbers between \(\frac { 7 }{ 8 }\) and \(\frac { -5 }{ 3 }\) are \(\frac{17}{24}, \frac{11}{24}, \frac{-13}{24}\)

vii. \(\frac{5}{7}, \frac{11}{7}\)
The three numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 11 }{ 7 }\) are \(\frac{6}{7}, \frac{8}{7}, \frac{9}{7}\)

viii. \(0, \frac{-3}{4}\)
The three numbers between 0 and \(\frac { -3 }{ 4 }\) are \(\frac{-1}{8}, \frac{-2}{8}, \frac{-5}{8}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 23 Intext Questions and Activities

Question 1.
Answer the following questions: (Textbook pg. no. 36)

  1. Write all the natural numbers between 2 and 9.
  2. Write all the integers between -4, and 5.
  3. Which rational numbers are there between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) ?

Solution:

  1. 3, 4, 5, 6, 7, 8
  2. -3, -2, -1, 0, 1, 2, 3, 4
  3. \(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}=\frac{2 \times 10}{4 \times 10}=\frac{20}{40}\)
    \(\frac{3}{4}=\frac{3 \times 10}{4 \times 10}=\frac{30}{40}\)
    ∴ The rational numbers between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) are \(\frac{21}{40}, \frac{22}{40}, \frac{25}{40}, \frac{27}{40}\) etc.

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

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

HCF and LCM Class 7 Practice Set 13 Answers Solutions Chapter 3

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

Maharashtra Board Practice Set 52 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 52 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 52 Answers Solutions Chapter 14

Algebraic Formulae Expansion Of Squares Class 7 Question 1.
Factorise the following expressions and write them in the product form.
i. 201a³b²
ii. 91xyt²
iii. 24a²b²
iv. tr²s³
i. 201a³b²
= 3 × 67 × a³ × b²
= 3 × 67 × a × a × a × b × b

ii. 91xyt²
= 7 × 13 × x × y × t²
= 7 × 13 × x × y × t × t

iii. 24a²b²
= 2 × 2 × 2 × 3 × a² × b²
= 2 × 2 × 2 × 3 × a × a × b × b

iv. tr²s³
= t × r² × s³
= t × r × r × s × s × s

Maharashtra Board Practice Set 35 Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 35 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 35 Answers Solutions Chapter 8

Question 1.
Multiply:
i. 16xy × 18xy
ii. 23xy² × 4yz²
iii. (12a + 17b) × 4c
iv. (4x + 5y) × (9x + 7y)
Solution:
i. 16xy × 18xy
= 16 × 18 × xy × xy
= 288x²y²

ii. 23xy² × 4yz²
= 23 × 4 × xy² × yz²
= 92xy³z²

iii. (12a + 17b) × 4c = 12a × 4c + 17b × 4c
= 48ac + 68bc

iv. (4x + 5y) × (9x + 7y)
= 4x × (9x + 7y) + 5y × (9x + 7y)
= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²

Question 2.
A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area. Solution:
Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
∴ Area of the rectangle = length × breadth
= (8x + 5) × (5x + 3)
= 8x × (5x + 3) + 5 × (5x + 3)
= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15
∴ The area of the rectangle is (40x² + 49x + 15) sq. cm.

Maharashtra Board Practice Set 32 Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 32  Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 32 Answers Solutions Chapter 8

Question 1.
Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
i. 7x
ii. 5y – 7z
iii. 3x³ – 5x² – 11
iv. 1 – 8a – 7a² – 7a³
v. 5m – 3
vi. a
vii. 4
viii. 3y² – 7y + 5
Solution:
i. Monomial
ii. Binomial
iii. Trinomial
iv. Polynomial
v. Binomial
vi. Monomial
vii. Monomial
viii. Trinomial

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.
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°

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)
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.

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 15 1

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

Solution:

Points in the interiorpoint C, point R, point N, point X
Points in the exteriorpoint T, point U, point Q, point V, point Y
Points on the arms of the anglespoint A, point W, point G, point B

Question 2.
Name the pairs of adjacent angles in the figures below.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 15 2
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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 15 3

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

  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
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.

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
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.