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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)

Solution:

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

Solution:

Class 7 Maths Solution Maharashtra Board

• Multiplication and Division of Integers Practice Set 8 Class 7 Maths Solution
• Multiplication and Division of Integers Practice Set 9 Class 7 Maths Solution
• HCF and LCM Practice Set 10 Class 7 Maths Solution
• HCF and LCM Practice Set 11 Class 7 Maths Solution
• HCF and LCM Practice Set 12 Class 7 Maths Solution
• HCF and LCM Practice Set 13 Class 7 Maths Solution
• HCF and LCM Practice Set 14 Class 7 Maths Solution

Rational and Irrational Numbers Class 8 Maths Chapter 1 Practice Set 1.3 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.3 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Std 8 Maths Practice Set 1.3 Chapter 1 Solutions Answers

Question 1.
Write the following rational numbers in decimal form.
i. \(\frac { 9 }{ 37 }\)
ii. \(\frac { 18 }{ 42 }\)
iii. \(\frac { 9 }{ 14 }\)
iv. \(-\frac { 103 }{ 5 }\)
v. \(-\frac { 11 }{ 13 }\)
Solution:
i. \(\frac { 9 }{ 37 }\)

ii. \(\frac { 18 }{ 42 }\)

iii. \(\frac { 9 }{ 14 }\)

iv. \(-\frac { 103 }{ 5 }\)

v. \(-\frac { 11 }{ 13 }\)

Maharashtra Board Class 8 Maths Solutions

• Rational and Irrational Numbers Practice Set 1.1 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.2 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.3 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.4 Class 8 Maths Solutions

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.

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)

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?

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

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 6 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 6 Answers Solutions Chapter 1 Geometrical Constructions.

Std 7 Maths Practice Set 6 Solutions Answers

Question 1.
Write the names of pairs of congruent line segments. (Use a divider to find them.)

i. ___
ii. ___
iii. ___
iv. ___
Solution:
i. seg BG ≅ seg CG
ii. seg NG ≅ seg MG ≅ seg EG ≅ seg RG

Question 2.
On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks.

i. seg AB ≅ seg ___
ii. seg AP ≅ seg ___
iii. seg AC ≅ seg ___
iv. seg ___ ≅ seg BY
v. seg __ ≅ seg YQ
vi. seg BW ≅ seg ___
Solution:
i. BC
ii. QW
iii. QZ
iv. AZ
v. AY
vi. AC

Note: The above problem has many solutions. Students may write solutions other than the ones given.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 6 Intext Questions and Activities

Question 1.
Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so. (Textbook pg. no. 7)
i. ∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.
Solution:
m∠A + m∠B = 85° + 115°
= 200°>180°
But the sum of the measures of the angles of a triangle is 180°
Hence, ∆ABC cannot be drawn.

ii. ∆PQR in which l(QR) = 2cm, l(PQ) = 4cm, l(PR) = 2cm.
Solution:
l(QR) + l(PR) = 2 cm + 2cm
= 4 cm
= l(PQ)
But in a triangle, the sum of the length of any two sides of a triangle is always greater than the length of the third side.
Hence, ∆PQR cannot be drawn.

Question 2.
Draw ∆ABC such that l(BC) = 8 cm, l(CA) = 6 cm, m∠ABC = 40°.
Draw a ray to make an angle of 40° with the base BC, l(BC) = 8 cm. We have to obtain point ‘A’ on the ray. With ‘C’ as the centre, draw an arc of radius 6 cm to do so. What do we observe? The arc intersects the ray in two different points. Thus, we get two triangles of two different shapes having the given measures. (Textbook pg. no. 7)
Solution:

Here ∠B is an acute angle. ∠C can be an acute angle or an obtuse angle.
Hence we get two triangles of two different shapes.

Question 3.
Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn? (Textbook pg. no. 7)
Solution:
Yes a triangle can be drawn.
Since the length of side is not given, any length of side can be selected and then triangle can be constructed. We will get different triangles for different length of sides.

Question 4.
Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q. (Textbook pg. no. 7)

l(AB) = ___
l(PQ) = ___
Solution:
l(AB) = 4 cm
l(PQ) = 4 cm
Since the length of two segments is the same, if placed on one another, they will coincide.

Question 5.
From the shape shown below, write the names of the pairs of congruent line segments. (Textbook pg. no. 8)
i. seg AB ≅ seg DC
ii. seg AE ≅ seg BH
iii. seg EF ≅ seg ___
iv. seg DF ≅ seg ___

Solution:
seg EF ≅ seg AD ≅ seg BC ≅ seg HG
seg DF ≅ seg CG ≅ seg AE ≅ seg BH

Question 6.
Take a rectangular paper. Place two opposite sides upon each Other. What do you observe? (Textbook pg. no. 7)
Solution:
Opposite sides of the rectangular paper coincide and hence are congruent.

7th Std Maths Solution Maharashtra Board

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

Rational and Irrational Numbers Class 8 Maths Chapter 1 Practice Set 1.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Std 8 Maths Practice Set 1.2 Chapter 1 Solutions Answers

Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, \(\frac { -9 }{ 5 }\)
iii. \(\frac { 8 }{ 7 }\), 0
iv. \(-\frac{5}{4}, \frac{1}{4}\)
v. \(\frac{40}{29}, \frac{141}{29}\)
vi. \(-\frac{17}{20},-\frac{13}{20}\)
vii. \(\frac{15}{12}, \frac{7}{16}\)
viii. \(-\frac{25}{8},-\frac{9}{4}\)
ix. \(\frac{12}{15}, \frac{3}{5}\)
x. \(-\frac{7}{11},-\frac{3}{4}\)
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

ii. 0, \(\frac { -9 }{ 5 }\)
On a number line, \(\frac { -9 }{ 5 }\) is to the left of zero.
∴ 0 > \(\frac { -9 }{ 5 }\)

iii. \(\frac { 8 }{ 7 }\), 0
On a number line, zero is to the left of \(\frac { 8 }{ 7 }\) .
∴ \(\frac { 8 }{ 7 }\) > 0

iv. \(-\frac{5}{4}, \frac{1}{4}\)
We know that, a negative number is always less than a positive number.
∴ \(-\frac{5}{4}<\frac{1}{4}\)

v. \(\frac{40}{29}, \frac{141}{29}\)
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ \(\frac{40}{29}<\frac{141}{29}\)

vi. \(-\frac{17}{20},-\frac{13}{20}\)
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ \(-\frac{17}{20}<-\frac{13}{20}\)

vii. \(\frac{15}{12}, \frac{7}{16}\)
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7

viii. \(-\frac{25}{8},-\frac{9}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

ix. \(\frac{12}{15}, \frac{3}{5}\)
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

x. \(-\frac{7}{11},-\frac{3}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities

Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. \(\frac{5}{4}<\frac{7}{4}\) but \(\frac{-5}{4}<\frac{-7}{4}\)
Solution:

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3

Maharashtra Board Class 8 Maths Solutions

• Rational and Irrational Numbers Practice Set 1.1 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.2 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.3 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.4 Class 8 Maths Solutions

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

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

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.

7th Std Maths Solution Maharashtra Board

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

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

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

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°

7th Std Maths Solution Maharashtra Board

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

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

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

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:

7th Std Maths Solution Maharashtra Board

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 2 Answers Solutions Chapter 1 Geometrical Constructions.

Std 7 Maths Practice Set 2 Solutions Answers

Question 1.
Draw triangles with the measures given below:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.
ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.
iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.
Solution:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.

ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.

iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.

Question 2.
Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.
Solution:

Question 3.
Draw an equilateral triangle with side 6.5 cm.
Solution:

Question 4.
Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene triangle.
Solution:
i. Equilateral triangle LMN, l(LM) = l(MN) = l(LN) = 4 cm.

ii. Isosceles triangle STU, l(ST) = l(TU) = 4cm, l(SU) = 6 cm

iii. Scalene triangle XYZ, l(XY) = 4.5 cm, l(XY) = 6.5 cm, l(XZ) = 5.5 cm

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 2 Intext Questions and Activities

Question 1.
Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm. (Textbook pg. no. 3)

1. Can this triangle be drawn?
2. A number of triangles can be drawn to fulfill these conditions. Try it out.
3. Which further condition must be placed if we are to draw a unique triangle using the above information?

Solution:

1. ∆ABC triangle cannot be drawn as length of third side is not given.
2. For ∆ABC to draw l(AC) > l(AB) + l(BC)
   i.e., l(AC) > 4 + 3
   i.e., l(AC) > 7 cm
   ∴ number of triangles can be drawn if l(AC) > 7 cm
3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

7th Std Maths Solution Maharashtra Board

• Geometrical Constructions Practice Set 1 Class 7 Maths Solution
• Geometrical Constructions Practice Set 2 Class 7 Maths Solution
• Geometrical Constructions Practice Set 3 Class 7 Maths Solution
• Geometrical Constructions Practice Set 4 Class 7 Maths Solution
• Geometrical Constructions Practice Set 5 Class 7 Maths Solution
• Geometrical Constructions Practice Set 6 Class 7 Maths Solution
• Geometrical Constructions Practice Set 7 Class 7 Maths Solution

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

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

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

Class 7 Maths Solution Maharashtra Board

• Multiplication and Division of Integers Practice Set 8 Class 7 Maths Solution
• Multiplication and Division of Integers Practice Set 9 Class 7 Maths Solution
• HCF and LCM Practice Set 10 Class 7 Maths Solution
• HCF and LCM Practice Set 11 Class 7 Maths Solution
• HCF and LCM Practice Set 12 Class 7 Maths Solution
• HCF and LCM Practice Set 13 Class 7 Maths Solution
• HCF and LCM Practice Set 14 Class 7 Maths Solution