Class 7

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 Maths Chapter 3 Practice Set 14 Solutions Maharashtra Board

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

∴ 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.

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

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 Maths Chapter 3 Practice Set 13 Solutions Maharashtra Board

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

∴ LCM of 12 and 15 = 3 x 2 x 2 x 5
= 60

ii. 6, 8, 10

∴ LCM of 6, 8 and 10 = 2 x 2 x 3 x 2 x 5
= 120

iii. 18, 32

∴ LCM of 18 and 32 = 2 x 2 x 2 x 2 x 3 x 3 x 2
= 288

iv. 10, 15, 20

∴ LCM of 10, 15 and 20 = 5 x 2 x 3 x 2
= 60

v. 45, 86

∴ LCM of 45 and 86 = 2 x 3 x 3 x 5 x 43
= 3870

vi. 15, 30, 90

∴ LCM of 15,30 and 90 = 3 x 5 x 2 x 3
= 90

vii. 105, 195

∴ LCM of 105 and 195 = 5 x 3 x 7 x 13
= 1365

viii. 12, 15, 45

∴ LCM of 12, 15 and 45 = 3 x 3 x 2 x 5 x 2
= 180

ix. 63, 81

∴ LCM of 63 and 81 = 3 x 3 x 3 x 7 x 3
= 567

x. 18, 36, 27

∴ 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

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

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 Maths Chapter 3 Practice Set 11 Solutions Maharashtra Board

Std 7 Maths Practice Set 11 Solutions Answers

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

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 10 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Maths Chapter 3 Practice Set 10 Solutions Maharashtra Board

Std 7 Maths Practice Set 10 Solutions Answers

Question 1.
Which number is neither a prime number nor a composite number?
Solution:
1

Question 2.
Which of the following are pairs of co-primes?
i. 8,14
ii. 4,5
iii. 17,19
iv. 27,15
Solution:
i. Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
∴ Common factors of 8 and 14: 1,2
∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4
Factors of 5: 1, 5
∴ Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17: 1, 17
Factors of 19: 1, 19
∴ Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.

iv. Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15 .
∴ Common factors of 27 and 15 : 1,3
∴ 27 and 15 are not a pair of co-prime numbers.

Question 3.
List the prime numbers from 25 to 100 and say how many they are.
Solution:
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are 16 prime numbers from 25 to 100.

Question 4.
Write all the twin prime numbers from 51 to 100.
Solution:

1. 59 and 61
2. 71 and 73

Question 5.
Write 5 pairs of twin prime numbers from 1 to 50.
Solution:

1. 3,5
2. 5,7
3. 11,13
4. 17,19
5. 29,31
6. 41,43

Question 6.
Which are the even prime numbers?
Solution:
2

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 10 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 15)
i. Which is the smallest prime number?
ii. List the prime numbers from 1 to 50. How many are they?
iii. Identify the prime numbers in the list below.
17, 15 ,4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97.
Solution:
i. 2 is the smallest prime number.
ii. There are 15 prime numbers from 1 to 50.
They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
iii. [17], 15 ,4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97].

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

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 Maths Chapter 2 Practice Set 9 Solutions Maharashtra Board

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

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.

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

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.

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

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)

Solution:

Question 3.
Multiply the given integers and complete the table given below. (Textbook pg. no. 12)

Solution:

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 5 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 5 Solutions Maharashtra Board

Std 7 Maths Practice Set 5 Solutions Answers

Construct triangles of the measures given below:

Question 1.
In ∆MAN, m∠MAN = 90°, l(AN) = 8 cm, l(MN) = 10 cm.
Solution:

Question 2.
In the right-angled ∆STU, hypotenuse SU = 5cm and l(ST) = 4cm.
Solution:

Question 3.
In ∆ABC, l(AC) = 7.5 cm, m∠ABC = 90°, l(BC) = 5.5cm.
Solution:

Question 4.
In ∆PQR, l(PQ) = 4.5 cm, l(PR) = 11.7cm, m∠PQR = 90°.
Solution:

Question 5.
Students should take examples of their own and practice construction of triangles.
i. In ∆PQR, l(PQ) = 5 cm, l(QR) = 6.8 cm, l(PR) = 5.5 cm.
ii. In ∆XYZ, l(XY) = 5.7 cm, m∠Y = 120°, l(YZ) = 7 cm.
iii. In ∆RST, l(ST) = 6.7 cm, m∠S = 60°, m∠T = 40°.
iv. In ∆UVW, m∠U = 90°, l(UV) = 5 cm, l(VW) = 6 cm.
Solution:
i. In ∆PQR, l(PQ) = 5 cm, l(QR) = 6.8 cm, l(PR) = 5.5 cm.

ii. In ∆XYZ, l(XY) = 5.7 cm, m∠Y = 120°, l(YZ) = 7 cm.

iii. In ∆RST, l(ST) = 6.7 cm, m∠S = 60°, m∠T = 40°.

iv. In ∆UVW, m∠U = 90°, l(UV) = 5 cm, l(VW) = 6 cm.

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 4 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 4 Solutions Maharashtra Board

Std 7 Maths Practice Set 4 Solutions Answers

Construct triangles of the measures given below:

Question 1.
In ∆SAT, l(AT) = 6.4 cm, m∠A = 45°, m∠T = 105°.
Solution:

Question 2.
In ∆MNP, l(NP) = 5.2 cm, m∠N = 70°, m∠P = 40°
Solution:

Question 3.
In ∆EFG, l(EG) = 6 cm, m∠F = 65°, m∠G = 45°.
Solution:
In ∆EFG,
m∠E + m∠F + m∠G = 180° …(sum of measures of angles of a triangle)
m∠E + 65° + 45° = 180°
m∠E + 110° = 180°
m∠E = 180° – 110°
m∠E = 70°

Question 4.
In ∆XYZ, l(XY) = 7.3 cm, m∠X = 34°, m∠Y = 95°.
Solution:

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 4 Intext Questions and Activities

Question 1.
In ∆ABC, m∠A = 60°, m∠B = 40°, l(AC) = 6 cm. (Textbook pg. no. 5)
1. Can you draw ∆ABC?
2. What further information is required before it can be drawn?
3. Which property can be used to get it?
4. Draw the rough figure to find out.
Solution:
1. ∆ABC cannot be drawn using the given information.
Seg AC is included inside the angles ∠A and ∠C. Since measure of ∠C is not known, the triangle cannot be drawn.
2. To draw the triangle, measure of ∠C is required.
3. The property of sum of the measures of the angles of a triangle can be used to find out m∠C.
4. In ∆ABC,
m∠A + m∠B + m∠C = 180°
∴ 60° + 40° + m∠C = 180°
∴ 100° + m∠C = 180°
m∠C = 180°- 100°
∴ m∠C = 80°

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 3 Answers Solutions Chapter 1 Geometrical Constructions.

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

Std 7 Maths Practice Set 3 Solutions Answers

Draw triangles with the measures given below:

Question 1.
In ∆MAT, l(MA) = 5.2 cm, m∠A = 80°, l(AT) = 6 cm.
Solution:

Question 2.
In ∆NTS, m∠T = 40°, l(NT) = l(TS) = 5 cm.
Solution:

Question 3.
In ∆FUN, l(FU) = 5 cm, l(UN) = 4.6 cm, m∠U = 110°.
Solution:

Question 4.
In ∆PRS, l(RS) = 5.5 cm, l(RP) = 4.2 cm, m∠R = 90°.
Solution:

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 12 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Maths Chapter 3 Practice Set 12 Solutions Maharashtra Board

Std 7 Maths Practice Set 12 Solutions Answers

Question 1.
i. 25, 40
ii. 56, 32
iii. 40, 60, 75
iv. 16, 27
v. 18, 32,48
vi. 105, 154
vii. 42, 45, 48
viii. 57, 75, 102
ix. 56, 57
x. 777, 315, 588
Solution:
i. 25, 40

∴ 25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2.
Find the HCF by the division method and reduce to the simplest form:
i. \(\frac { 275 }{ 525 }\)
ii. \(\frac { 76 }{ 133 }\)
iii. \(\frac { 161 }{ 69 }\)
Solution:
i. \(\frac { 275 }{ 525 }\)

ii. \(\frac { 76 }{ 133 }\)

iii. \(\frac { 161 }{ 69 }\)

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1.
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17)
i. 28, 42
ii. 51, 27
iii. 25, 15, 35
Solution:
i. Factors of 28 = 1,2,4, 7, 14, 28
Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42
∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5

Std 7 Maths Digest

• Practice Set 8 Class 7 Answers
• Practice Set 9 Class 7 Answers
• Practice Set 10 Class 7 Answers
• Practice Set 11 Class 7 Answers
• Practice Set 12 Class 7 Answers
• Practice Set 13 Class 7 Answers
• Practice Set 14 Class 7 Answers