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

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 Maths Chapter 5 Practice Set 22 Solutions Maharashtra Board

Std 7 Maths Practice Set 22 Solutions Answers

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. \(\frac{5}{36}+\frac{6}{42}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 1

ii. \(1 \frac{2}{3}+2 \frac{4}{5}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 2

iii. \(\frac{11}{17}+\frac{13}{19}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 3

iv. \(2 \frac{3}{11}+1 \frac{3}{77}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 4

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. \(\frac{7}{11}-\frac{3}{7}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 5

ii. \(\frac{13}{36}-\frac{2}{40}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 6

iii. \(1 \frac{2}{3}-3 \frac{5}{6}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 7

iv. \(4 \frac{1}{2}-3 \frac{1}{3}\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 22 8

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

Std 7 Maths Digest

Practice Set 17 Class 6 Answers Maths Chapter 5 Decimal Fractions Maharashtra Board

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

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 17 Solutions Maharashtra Board

Std 6 Maths Practice Set 17 Solutions Answers

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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 1

ii. 17.5÷5
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 2

iii. 20.6÷2
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 3

iv. 32.5÷25
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 4

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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 5
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 6
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 17 7
= 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.)

Std 6 Maths Digest

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

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 Maths Chapter 4 Practice Set 20 Solutions Maharashtra Board

Std 7 Maths Practice Set 20 Solutions Answers

Question 1.
Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 20 1
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?Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 20 2
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.

Std 7 Maths Digest

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

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 Maths Chapter 4 Practice Set 19 Solutions Maharashtra Board

Std 7 Maths Practice Set 19 Solutions Answers

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 1

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.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 2

iv.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 3

v.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 4

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)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 19 5

  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.

Std 7 Maths Digest

Practice Set 18 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 18 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 18 Solutions Maharashtra Board

Std 7 Maths Practice Set 18 Solutions Answers

Question 1.
Name the pairs of opposite rays in the figure alongside.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 1
Solution:

  1. Ray PL and ray PM
  2. Ray PN and ray PT

Question 2.
Are the ray PM and PT opposite rays? Give reasons for your answer.
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 2
Solution:
No.
Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 18 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 3

  1. Name the rays in the figure alongside.
  2. Name the origin of the rays
  3. Name the angle in the given figure

Solution:

  1. Ray BA and ray BC
  2. Point B
  3. ∠ABC or ∠CBA

Question 2.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)
Maharashtra Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 18 4

  1. Name the angle in the figure alongside.
  2. Name the rays whose origin is point B

Solution:

  1. ∠ABC or ∠CBA
  2. Ray BA and ray BC

Std 7 Maths Digest

Practice Set 38 Class 6 Answers Maths Chapter 16 Quadrilaterals Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 38 Answers Solutions.

Quadrilaterals Class 6 Maths Chapter 16 Practice Set 38 Solutions Maharashtra Board

Std 6 Maths Practice Set 38 Solutions Answers

Question 1.
Draw ₹XYZW and answer the following:
i. The pairs of opposite angles.
ii. The pairs of opposite sides.
iii. The pairs of adjacent sides.
iv. The pairs of adjacent angles.
v. The diagonals of the quadrilateral.
vi. The name of the quadrilateral in different ways.
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 1
i. a. ∠XYZ and ∠XWZ
b. ∠YXW and ∠YZW

ii. a. side XY and side WZ
b. side XW and side YZ

iii. a. side XY and side XW
b. side WX and side WZ
c. side ZW and side ZY
d. side YZ and side YX

iv. a. ∠XYZ and ∠YZW
b. ∠YZW and ∠ZWX
c. ∠ZWX and ∠WXY
d. ∠WXY and ∠XYZ

v. Seg XZ and seg YW

vi. ₹XYZW
₹YZWX
₹ZWXY
₹WXYZ
₹XWZY
₹WZYX
₹ZYXW
₹YXWZ

Question 2.
In the table below, write the number of sides the polygon has.

Names Quadrilateral Octagon Pentagon Heptagon Hexagon
Number of sides

Solution:

Names Quadrilateral Octagon Pentagon Heptagon Hexagon
Number of sides 4 8 5 7 6

Question 3.
Look for examples of polygons in your surroundings. Draw them.
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 2

Question 4.
We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 3
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 4

Question 5.
Draw any polygon and divide it into triangular parts as shown here. Thus work out the sum of the measures of the angles of the polygon.
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 5
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 6
Hexagon ABCDEF can be divided in 4 triangles namely ∆BAF, ∆BFE, ∆BED and ∆BCD
Sum of the measures of the angles of a triangle = 180°
∴ Sum of measures of the angles of the polygon ABCDEF = Sum of the measures of all the four triangles
= 180° + 180° + 180°+ 180°
= 720°
∴ The sum of the measures of the angles of the given polygon (hexagon) is 720°.

Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 38 Intext Questions and Activities

Question 1.
From your compass boxes, collect set squares of the same shapes and place them side by side in all possible different ways. What figures do you get? Write their names. (Textbook pg. no. 85)
a. Two set squares
b. Three set squares
c. four set squares
Solution:
a. Two set squares
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 7

b. Three set squares
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 8

c. four set squares
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 9

Question 2.
Kaprekar Number. (Textbook pg. no. 86)
i. Take any 4-digit number in which all the digits are not the same.
ii. Obtain a new 4-digit number by arranging the digits in descending order.
iii. Obtain another 4-digit number by arranging the digits of the new number in ascending order.
iv. Subtract the smaller of these two new numbers from the bigger number. The difference obtained will be a 4-digit number. If it is a 3-digit number, put a 0 in the thousands place. Repeat the above steps with the difference obtained as a result of the subtraction.
v. After some repetitions, you will get the number 6174. If you continue to repeat the same steps you will get the number 6174 every time. Let us begin with the number 8531.
8531 → 7173 → 6354 → 3087 → 8352 → 6174 → 6174
This discovery was made by the mathematician, Dattatreya Ramchandra Kaprekar. That is why the number 6174 was named the Kaprekar number.
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 38 10

Std 6 Maths Digest

Practice Set 17 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

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 Maths Chapter 4 Practice Set 17 Solutions Maharashtra Board

Std 7 Maths Practice Set 17 Solutions Answers

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°

Std 7 Maths Digest

Practice Set 16 Class 6 Answers Maths Chapter 5 Decimal Fractions Maharashtra Board

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

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 16 Solutions Maharashtra Board

Std 6 Maths Practice Set 16 Solutions Answers

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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 1
= 3.75

ii. 6.17 × 3.9
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 2
= 24.063

iii. 0.57 × 2
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 3
= 1.14

iv. 5.04 × 0.7
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 4
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 5
Cost of 1 kg of rice = Rs 42
∴ Cost of 94.5 kg of rice = 42 × 94.5
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 6
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 7
= 4.25 m
∴ Cloth required to make 5 curtains = 5 × 4.25
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 8
= 21.25 m
Cloth remaining with Vedika = Total cloth with Vedika – Cloth used
= 23.5 m – 21.25 m
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 16 9
= 2.25 m
∴ The length of cloth remaining with Vedika is 2.25 m.

Std 6 Maths Digest

Practice Set 16 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

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 Maths Chapter 4 Practice Set 16 Solutions Maharashtra Board

Std 7 Maths Practice Set 16 Solutions Answers

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.

Std 7 Maths Digest

Practice Set 15 Class 7 Answers Chapter 4 Angles and Pairs of Angles Maharashtra Board

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 Maths Chapter 4 Practice Set 15 Solutions Maharashtra Board

Std 7 Maths Practice Set 15 Solutions Answers

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 interior point C, point R, point N, point X
Points in the exterior point T, point U, point Q, point V, point Y
Points on the arms of the angles point 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

Std 7 Maths Digest