Category: Class 7

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 54 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 54 Answers Solutions Chapter 15

Question 1.
The daily rainfall for each day of a week in a certain city is given in millimeters. Find the average rainfall during the week.
9, 11, 8, 20, 10, 16, 12
Solution:
\(\text { Average rainfall during the week }=\frac{\text { sum of rainfall for each day of the week }}{\text { number of days }}\)
= \(\frac{9+11+8+20+10+16+12}{7}\)
= \(\frac { 86 }{ 7 }\)
= 12.285 ≈ 12.29
∴ The average rainfall during the week is 12.29 mm.

Question 2.
During the annual function of a school, a Women’s Self-help Group had set up a snacks stall. Their sales every hour were worth Rs 960, Rs 830, Rs 945, Rs 800, Rs 847, Rs 970 respectively. What was the average of the hourly sales?
Solution:
\(\text { Average hourly sales }=\frac{\text { sum of sales every hour }}{\text { number of hours }}\)
= \(\frac{960+830+945+800+847+970}{6}\)
= \(\frac { 5352 }{ 6 }\)
= Rs 892
∴ The average of the hourly sales was Rs 892.

Question 3.
The annual rainfall in Vidarbha in five years is given below. What is the average rainfall for those 5 years?
900 mm, 650 mm, 450 mm, 733 mm, 400 mm.
Solution:
\(\text { Average rainfall for } 5 \text { years }=\frac{\text { sum of annual rainfall in five years }}{\text { number of years }}\)
= \(\frac{900+650+450+733+400}{5}\)
= \(\frac { 3133 }{ 5 }\)
= 626.6
∴ The average rainfall in Vidarbha for 5 years was 626.6 mm.

Question 4.
A farmer bought some sacks of animal feed. The weights of the sacks are given below in kilograms. What is the average weight of the sacks?
49.8, 49.7, 49.5, 49.3, 50,48.9, 49.2, 48.8.
Solution:
\(\text { Average weight of the sacks }=\frac{\text { sum of weight of each sack }}{\text { number of sacks }}\)
= \(\frac{49.8+49.7+49.5+49.3+50+48.9+49.2+48.8}{8}\)
= \(\frac { 395.2 }{ 8 }\)
= \(\frac { 3952 }{ 80 }\)
= 49.4
∴ The average weight of the sacks is 49.4 kg.

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 54 Intext Questions and Activities

Question 1.
Rutuja practised skipping with a rope all seven days of a week. The number of times she jumped the rope in one minute every day is given below. Find the average number of jumps per minute.
60, 62, 61, 60, 59, 63, 58. (Textbook pg. no. 96)
Solution:
\(\text { Average }=\frac{\text { Sum of the number of jumps ons even days }}{\text { Total number of days }}\)
= \(\frac{[60]+[62]+[61]+[60]+[59]+[63]+[58]}{7}\)
= \(\frac { 423 }{ 7 }\)
= 60.42
∴ Average number of jumps per minute = 60.4

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 53 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 53 Answers Solutions Chapter 14

Question 1.
Factorize the following expressions:
i. p² – q²
ii. 4x² – 25y²
iii. y² – 4
iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
v. \(9 x^{2}-\frac{1}{16} y^{2}\)
vi. \(x^{2}-\frac{1}{x^{2}}\)
vii. a²b – ab
viii. 4x²y – 6x²
ix. \(\frac{1}{2} y^{2}-8 z^{2}\)
x. 2x² – 8y²
Solution:
i. p² – q²
Here, a = p, b = q
∴ p² – q² = (p + q)(p – q)
….[(a² – b²) = (a + b)(a – b)]

ii. 4x² – 25y²
= (2x)² – (5y)²
Here, a = 2x, b = 5y
∴ (2x)² – (5y)² = (2x + 5y)(2x – 5y)
….[(a² – b²) = (a + b)(a – b)]

iii. y² – 4
= y² – 2²
Here, a = y, b = 2
∴ y² – 2² = (y + 2)(y – 2)
….[(a² – b²) = (a + b)(a – b)]

iv. \(\mathrm{p}^{2}-\frac{1}{25}\)
Here a = \(\frac { 1 }{ 25 }\), b = \(\frac { 1 }{ 5 }\)
\(p^{2}-\left(\frac{1}{5}\right)^{2}=\left(p+\frac{1}{5}\right)\left(p-\frac{1}{5}\right)\)
….[(a² – b²) = (a + b)(a – b)]

v. \(9 x^{2}-\frac{1}{16} y^{2}\)
Here a = 3x, b = \(\frac { 1 }{ 4 }y\)
∴\((3 x)^{2}-\left(\frac{1}{4} y\right)^{2}=\left(3 x+\frac{1}{4} y\right)\left(3 x-\frac{1}{4} y\right)\)
….[(a² – b²) = (a + b)(a – b)]

vi. \(x^{2}-\frac{1}{x^{2}}\)
Here a = x, b = \(\frac { 1 }{ x }\)
\(x^{2}-\left(\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
….[(a² – b²) = (a + b)(a – b)]

vii. a²b – ab
= a (ab – b)
= ab (a – 1)

viii. 4x²y – 6x²
= 2 (2x²y – 3x²)
= 2x² (2y – 3)

ix. \(\frac{1}{2} y^{2}-8 z^{2}\)

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

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

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

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

ii. 98 ÷ (-28)

iii. (-51) ÷ 68

iv. 38 ÷ (-57)

v. (-85) ÷ 20

vi. (-150) ÷ (-25)

vii. 100 ÷ 60

viii. 9 ÷ (-54)

ix. 78 ÷ 65

x. (-5) ÷ (-315)

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

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

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

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

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

Solution:

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

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

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 Practice Set 8 Answers Solutions Chapter 2

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:

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

Geometrical Constructions Class 7 Practice Set 7 Answers Solutions Chapter 1

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.]

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

Geometrical Constructions Class 7 Practice Set 6 Answers Solutions Chapter 1

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.

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 Practice Set 5 Answers Solutions Chapter 1

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.

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 Practice Set 4 Answers Solutions Chapter 1

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°

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 Practice Set 3 Answers Solutions Chapter 1

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:

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

Geometrical Constructions Class 7 Practice Set 2 Answers Solutions Chapter 1

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.