Prasanna

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 21 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Practice Set 21 Answers Solutions Chapter 4

Question 1.
∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

Solution:
Let the measures of ∠A be x°.
m∠A = m∠B = x°
∠ACD is the exterior angle of ∆ABC
∴ m∠ACD = m∠A + m∠B
∴ 140 = x + x
∴ 140 = 2x
∴ 2x = 140
∴ x = \(\frac { 140 }{ 2 }\)
= 70
∴ The measures of the angles ∠A and ∠B is 70° each.

Question 2.
Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Solution:
m∠EOD = m∠AOB = 8y ….(vertically opposite angles)
∠FOL, ∠EOD and ∠COD form a straight angle.
∴ m∠FOE + m∠EOD + m∠COD = 180°
∴ 4y + 8y + 6y = 180
∴ 18y = 180
∴ y = \(\frac { 180 }{ 18 }\)
∴ y = 10
m∠EOD = 8y = 8 x 10 = 80°
m∠AOF = m∠COD ….(Vertically opposite angles)
= 6y = 6 x 10 = 60°
m∠BOC = m∠FOE ….(Vertically opposite angles)
= 4y = 4 x 10 = 40°
∴ The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

Question 3.
In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

Solution:
Let the measure of ∠A be y°. A
∴ m∠A = m∠B = y°
∠ACB and ∠ACD form a pair of linear angles.
∴ m∠ACB + m∠ACD = 180°
∴ (3x – 17) + (8x + 10) = 180
∴ 3x + 8x – 17 + 10 = 180
∴ 11x – 7 = 180
∴ 11x – 7 + 7 = 180 + 7 …(Adding 7 on both sides.)
∴ 11x = 187
∴ x = \(\frac { 187 }{ 11 }\) = 17
m∠ACB = 3x – 17 = (3 x 17) – 17 = 51 – 17 = 34°
m∠ACD = 8x + 10 = 8 x 17 + 10 = 136 + 10 = 146°
Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.
∴ m∠ACD = m∠A + m∠B
∴ 146 = y + y
∴ 146 = 2y
∴ 2y = 146
∴ y = \(\frac { 146 }{ 2 }\) = 73
∴ The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 21 Intext Questions and Activities

Question 1.
Use straws or sticks to make all the kinds of angles that you have learnt about. (Textbook pg. no. 29)
Solution:
(Student should attempt the activity on their own)

Question 2.
Observe the table given below and draw your conclusions (Textbook pg. no. 31)

Solution:
i. 180°
ii. 360°
iii. 540°
iv. 720°
v. 180° x 5 = 900°
vi.  , 180° x 6 = 1080°

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 22 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Practice Set 22 Answers Solutions Chapter 5

Question 1.
Carry out the following additions of rational numbers:
i. \(\frac{5}{36}+\frac{6}{42}\)
ii. \(1 \frac{2}{3}+2 \frac{4}{5}\)
iii. \(\frac{11}{17}+\frac{13}{19}\)
iv. \(2 \frac{3}{11}+1 \frac{3}{77}\)
Solution:
i. \(\frac{5}{36}+\frac{6}{42}\)

ii. \(1 \frac{2}{3}+2 \frac{4}{5}\)

iii. \(\frac{11}{17}+\frac{13}{19}\)

iv. \(2 \frac{3}{11}+1 \frac{3}{77}\)

Question 2.
Carry out the following subtractions involving rational numbers.
i. \(\frac{7}{11}-\frac{3}{7}\)
ii. \(\frac{13}{36}-\frac{2}{40}\)
iii. \(1 \frac{2}{3}-3 \frac{5}{6}\)
iv. \(4 \frac{1}{2}-3 \frac{1}{3}\)
Solution:
i. \(\frac{7}{11}-\frac{3}{7}\)

ii. \(\frac{13}{36}-\frac{2}{40}\)

iii. \(1 \frac{2}{3}-3 \frac{5}{6}\)

iv. \(4 \frac{1}{2}-3 \frac{1}{3}\)

Question 3.
Multiply the following rational numbers.
i. \(\frac{3}{11} \times \frac{2}{5}\)
ii. \(\frac{12}{5} \times \frac{4}{15}\)
iii. \(\frac{(-8)}{9} \times \frac{3}{4}\)
iv. \(\frac{0}{6} \times \frac{3}{4}\)
Solution:
i. \(\frac{3}{11} \times \frac{2}{5}\)
\(=\frac{3 \times 2}{11 \times 5}=\frac{6}{55}\)

ii. \(\frac{12}{5} \times \frac{4}{15}\)
\(=\frac{4}{5} \times \frac{4}{5}=\frac{4 \times 4}{5 \times 5}=\frac{16}{25}\)

iii. \(\frac{(-8)}{9} \times \frac{3}{4}\)
\(=\frac{(-2)}{3} \times \frac{1}{1}=\frac{-2}{3}\)

iv. \(\frac{0}{6} \times \frac{3}{4}\)
\(=0 \times \frac{3}{4}=0\)

Question 4.
Write the multiplicative inverse of.
i. \(\frac{2}{5}\)
ii. \(\frac{-3}{8}\)
iii. \(\frac{-17}{39}\)
iv. 7
v. \(-7 \frac{1}{3}\)
Solution:
i. \(\frac{5}{2}\)
ii. \(\frac{-8}{3}\)
iii. \(\frac{-39}{17}\)
iv. \(\frac {1}{7}\)
v. \(\frac {-3}{22}\)

Question 5.
Carry out the divisions of rational numbers:
i. \(\frac{40}{12} \div \frac{10}{4}\)
ii. \(\frac{-10}{11} \div \frac{-11}{10}\)
iii. \(\frac{-7}{8} \div \frac{-3}{6}\)
iv. \(\frac{2}{3} \div(-4)\)
v. \(2 \frac{1}{5} \div 5 \frac{3}{6}\)
vi. \(\frac{-5}{13} \div \frac{7}{26}\)
vii. \(\frac{9}{11} \div(-8)\)
viii. \(5 \div \frac{2}{5}\)
Solution:
i. \(\frac{40}{12} \div \frac{10}{4}\)
\(=\frac{40}{12} \times \frac{4}{10}=\frac{4}{3}\)

ii. \(\frac{-10}{11} \div \frac{-11}{10}\)
\(=\frac{-10}{11} \times \frac{-10}{11}=\frac{100}{121}\)

iii. \(\frac{-7}{8} \div \frac{-3}{6}\)
\(=\frac{-7}{8} \times \frac{-6}{3}=\frac{-7}{4} \times \frac{-3}{3}=\frac{7}{4}\)

iv. \(\frac{2}{3} \div(-4)\)
\(=\frac{2}{3} \times \frac{-1}{4}=\frac{1}{3} \times \frac{-1}{2}=\frac{-1}{6}\)

v. \(2 \frac{1}{5} \div 5 \frac{3}{6}\)
\(=\frac{11}{5} \div \frac{33}{6}=\frac{11}{5} \times \frac{6}{33}=\frac{1}{5} \times \frac{6}{3}=\frac{2}{5}\)

vi. \(\frac{-5}{13} \div \frac{7}{26}\)
\(=\frac{-5}{13} \times \frac{26}{7}=\frac{-10}{7}\)

vii. \(\frac{9}{11} \div(-8)\)
\(=\frac{9}{11} \times \frac{-1}{8}=\frac{-9}{88}\)

viii. \(5 \div \frac{2}{5}\)
\(=\frac{5}{1} \times \frac{5}{2}=\frac{25}{2}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 22 Intext Questions and Activities

Question 1.
Complete the table given below. (Textbook pg. no. 34)

	-3	\(\frac {3}{5}\)	-17	\(\frac { -5 }{ 11 }\)	5
Natural Numbers	x				✓
Integers	✓
Rational Numbers	✓

Solution:

	-3	\(\frac {3}{5}\)	-17	\(\frac { -5 }{ 11 }\)	5
Natural Numbers	x	x	x	x	✓
Integers	✓	x	✓	x	✓
Rational Numbers	✓	✓	✓	✓	✓

Question 2.
Discuss the characteristics of various groups of numbers in class and complete the table below. In front of each group, write the inference you make after carrying out the operations of addition, subtraction, multiplication and division, using a (✓) or a (x).
Remember that you cannot divide by zero. (Textbook pg. no. 35)

Group of Numbers	Addition	Subtraction	Multiplication	Division
Natural Numbers	✓	x (7- 10 =-3)	✓	x (3÷5=\(\frac { 3 }{ 5 }\))
Integers
Rational Numbers

Solution:

Group of Numbers	Addition	Subtraction	Multiplication	Division
Natural Numbers	✓	x (7- 10 =-3)	✓	x (3÷5=\(\frac { 3 }{ 5 }\))
Integers	✓	✓	✓	x (4÷9=\(\frac { 4 }{ 9 }\))
Rational Numbers	✓	✓	✓	✓

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 17 Answers Solutions.

6th Standard Maths Practice Set 17 Answers Chapter 5 Decimal Fractions

Question 1.
Carry out the following divisions.
i. 4.8÷2
ii. 17.5÷5
iii. 20.6÷2
iv. 32.5÷25
Solution:
i. 4.8÷2

ii. 17.5÷5

iii. 20.6÷2

iv. 32.5÷25

Question 2.
A road is 4 km 800 m long. If trees are planted on both its sides at intervals of 9.6 m, how many trees were planted?
Solution:
Length of road = 4 km 800 m
= 4 × 1000 m + 800 m
= 4000 m + 800 m
= 4800 m
Number of trees on one side = 4800 ÷ 9.6

= 500
∴ Number of trees on both sides = 2 x number of trees on one side
= 2 x 500 = 1000
If the trees are planted at the beginning of the road, then
Total number of trees = 1000 + 2 = 1002
∴ Total number of trees planted is 1000 or 1002.

Question 3.
Pradnya exercises regularly by walking along a circular path on a field. If she walks a distance of 3.825 km in 9 rounds of the path, how much does she walk in one round?
Solution:
Total distance walked in 9 rounds = 3.825 km
∴Distance walked in 1 round = 3.825 4 ÷ 9

= 0.425 km
∴ Total distance walked in 1 round is 0.425 km.

Question 4.
A pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for Rs 9500. What is the cost per quintal of hirada? (1 quintal = 100 kg)
Solution:
Cost of 0.25 quintal of hirada = Rs 9500
∴ Cost of 1 quintal of hirada = 9500 ÷ 0.25

= Rs 38,000
∴ Cost per quintal of hirada is Rs 38,000.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 17 Intext Questions and Activities

Question 1.
Maths is fun! (Textbook pg. no. 34)

1. Consider any three-digit number (say 527).
2. Multiply the number by 7. Then multiply the product obtained by 13, and this product by 11.
3. The found product is 5,27,527.

Take two or three other numbers. Do the same multiplication and find out how it is done.
Solution:
7 × 13 × 11 = 1001
∴ 527 × 1001 = 527 × (1000+ 1)
= (527 × 1000) + (527 × 1)
= 527000 + 527 = 527527
Thus, when any three-digit number is multiplied with 1001, the product obtained is a six-digit number in which the original three-digit number is written back to back twice.
(Students may consider any other three-digit numbers and verify the property.)

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.

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

Question 1.
Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.

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?
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.

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.

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

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.

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.

iv.

v.

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)

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.

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 Practice Set 11 Answers Solutions Chapter 3

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

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 Practice Set 10 Answers Solutions Chapter 3

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].

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 55 Answers Solutions Chapter 15 Statistics.

Statistics Class 7 Practice Set 55 Answers Solutions Chapter 15

Question 1.
The height of 30 children in a class is given in centimeters. Draw up a frequency table of this data.
131, 135, 140, 138, 132, 133, 135, 133, 134, 135, 132, 133, 140, 139, 132, 131, 134, 133, 140, 140, 139, 136, 137, 136, 139, 137, 133, 134, 131, 140
Solution:

Question 2.
In a certain colony, there are 50 families. The number of people in every family is given below. Draw up the frequency table.
5, 4, 5, 4, 5, 3, 3, 3, 4, 3, 4, 2, 3, 4, 2, 2, 2, 2, 4, 5, 1, 3, 2, 4, 5, 3, 3, 2, 4, 4, 2, 3, 4, 3, 4, 2, 3, 4, 5, 3, 2, 3, 2, 3, 4, 5, 3, 2, 3, 2
Solution:

Question 3.
A dice was cast 40 times and each score noted is given below. Draw up a frequency table for this data.
3, 2, 5, 6, 4, 2, 3, 1, 6, 6, 2, 3, 5, 3, 5, 3, 4, 2, 4, 5, 4, 2, 6, 3, 3, 2, 4, 3, 3, 4, 1, 4, 3, 3, 2, 2, 5, 3, 3, 4.
Solution:

Question 4.
The number of chapatis that 30 children in a hostel need at every meal is given below. Make a frequency table for these scores.
3, 2, 2, 3, 4, 5, 4, 3, 4, 5, 2, 3, 4, 3, 2, 5, 4, 4, 4, 3, 3, 2, 2, 2, 3, 4, 3, 2, 3, 2.
Solution:

Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 55 Intext Questions and Activities

Question 1.
Make groups of 10 children in your class. Find the average height of the children in each group. (Textbook pg. no. 96)
Solution:
(Students should attempt the above activities on their own.)

Question 2.
With the help of your class teacher, note the daily attendance for a week and find the average attendance. (Textbook pg. no. 96)
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 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.

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.

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

Solution:

1. Straight angle
2. 180°

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 16 Answers Solutions.

6th Standard Maths Practice Set 16 Answers Chapter 5 Decimal Fractions

Question 1.
If, 317 × 45 = 14265, then 3.17 × 4.5 = ?
Solution:
3.17 × 4.5
= 14.265

Question 2.
If, 503 × 217 = 109151, then 5.03 × 2.17 = ?
Solution:
5.03 x 2.17
= 10.9151

Question 3.
i. 2.7 × 1.4
ii. 6.17 × 3.9
iii. 0.57 × 2
iv. 5.04 × 0.7
Solution:
i. 2.7 × 1.4

= 3.75

ii. 6.17 × 3.9

= 24.063

iii. 0.57 × 2

= 1.14

iv. 5.04 × 0.7

= 3.528

Question 4.
Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs Rs 42 per kg, how much did he pay for it?
Solution:
Weight of one bag of rice = 5.250 kg
Number of bags of rice = 18
∴ Total Weight = 18 × 5.250

Cost of 1 kg of rice = Rs 42
∴ Cost of 94.5 kg of rice = 42 × 94.5

∴ Total rice bought by Virendra is 94.5 kg, and the amount paid for it is Rs 3969.

Question 5.
Vedika has 23.5 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?
Solution:
We know, that 1 m = 100 cm
Cloth required to make 1 curtain = 4 m 25 cm
= 4 m + \(\frac { 25 }{ 100 }\) m
= 4 m + 0.25 m

= 4.25 m
∴ Cloth required to make 5 curtains = 5 × 4.25

= 21.25 m
Cloth remaining with Vedika = Total cloth with Vedika – Cloth used
= 23.5 m – 21.25 m

= 2.25 m
∴ The length of cloth remaining with Vedika is 2.25 m.