## Maharashtra Board Practice Set 37 Class 6 Maths Solutions Chapter 16 Quadrilaterals

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 37 Answers Solutions.

Question 1.
Observe the figures below and find out their names:

Solution:
i. Pentagon (5 sides)
ii. Hexagon (6 sides)
iii. Heptagon (7 sides)
iv. Octagon (8 sides)

#### Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 37 Intext Questions and Activities

Question 1.
Observe the figures given below and say which of them are quadrilaterals. (Textbook pg. no. 81)

Solution:

Question 2.
Draw a quadrilateral. Draw one diagonal of this quadrilateral and divided it into two triangles. Measures all the angles in the figure. Is the sum of the measures of the four angles of the quadrilateral equal to the sum of the measures of the six angles of the two triangles? Verity that this is so with other quadrilaterals. (Textbook pg. no. 84)
Solution:

m∠PQR = 104°
m∠QRP = 26°
m∠RPQ = 50°
m∠PRS = 34°
m∠RSP = 106°
m∠SPR = 40°
∴ Sum of the measures of the angles of quadrilateral = m∠PQR + m∠QRP + m∠RPQ + m∠PRS + m∠RSP + m∠SPR
= 104° + 26° + 50° + 34° + 106° + 40°
= 360°
Also, we observe that
Sum of the measures of the angles of quadrilateral = Sum of the measures of angles of the two triangles (PQR and PRS)
= (104°+ 26°+ 50°)+ (34° + 106° + 40°)
= 180° + 180°
= 360°
[Note: Students should drew different quadrilaterals and verify the property.]

Question 3.
For the pentagon shown in the figure below, answer the following: (Textbook pg. no. 84)

1. Write the names of the five vertices of the pentagon.
2. Name the sides of the pentagon.
3. Name the angles of the pentagon.
4. See if you can sometimes find players on a field forming a pentagon.

Solution:

1. The vertices of the pentagon are points A, B, C, D and E.
2. The sides of the pentagon are segments AB, BC, CD, DE and EA.
3. The angles of the pentagon are ∠ABC, ∠BCD, ∠CDE, ∠DEA and ∠EAB.
4. The players shown in the above figure form a pentagon. The players are standing on the vertices of

Question 4.
Cut out a paper in the shape of a quadrilateral. Make folds in it that join the vertices of opposite angles. What can these folds be called? (Textbook pg. no. 83)

Solution:
The folds are called diagonals of the quadrilateral.

Question 5.
Take two triangular pieces of paper such that . one side of one triangle is equal to one side of the other. Let us suppose that in ∆ABC and ∆PQR, sides AC and PQ are the equal sides. Join the triangles so that their equal sides lie B side by side. What figure do we get? (Textbook pg. no. 83)

Solution:
If we place the triangles together such that the equal sides overlap, the two triangles form a quadrilateral.

## Maharashtra Board Practice Set 3 Class 6 Maths Solutions Chapter 2 Angles

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 3 Answers Solutions.

## 6th Standard Maths Practice Set 3 Answers Chapter 2 Angles

Question 1.
Use the proper geometrical instruments to construct the following angles. Use the compass and the ruler to bisect them:

1. 50°
2. 115°
3. 80°
4. 90°

Solution:

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 3 Intext Questions and Activities

Question 1.
Construct an angle bisector to obtain an angle of 30°. (Textbook pg. no. 11)
Solution: .
In order to get a bisected angle of a given measure, the student has to draw the angle having twice the measurement of required bisected angle.

For getting measurement of 30° (for the bisected angle), one has to make an angle of 60° (i.e. 30° × 2).

Step 1:
Draw ∠ABC of 60°.

Step 2:
Cut arcs on the rays BA and BC. Name these points as D and E respectively.

Step 3:
Place the compass point on point D and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point E and cut the previous arc. Name the point of intersection as O

Step 4:
Draw ray BO.
Ray BO is the angle bisector of ∠ABC.
i.e. m∠ABO = m∠CBO = 30°

Question 2.
Construct an angle bisector to draw an angle of 45°. (Textbook pg. no. 11)
Solution:
For getting measurement of 45° (for the bisected angle), one has to make an angle of 90° (i.e. 45° × 2).
Step 1:
Draw ∠PQR of 90°.

Step 2:
Cut arcs on the rays QP and QR.
Name these points as M and N respectively.

Step 3:
Place the compass point on point M and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point N and cut the

Step 4:
Draw ray QO.
Ray QO is the angle bisector of ∠PQR.
i.e. m∠PQO = m∠RQO = 45°

Question 3.
Ask three or more children to stand in a straight line. Take two long ropes. Let the child in the middle hold one end of each rope. With the help of the ropes, make the children on either side stand along a straight line. Tell them to move so as to form an acute angle, a right angle, an obtuse angle, a straight angle, a reflex angle and a full or complete angle in turn. Keeping the rope stretched will help to ensure that the children form straight lines. (Textbook pg. no. 6)
Solution:

Question 4.
Look at the pictures below and identify the different types of angles. (Textbook pg. no. 8)

Solution:
i. Complete angle
ii. Reflex and Acute angle
iii. Acute and Obtuse angle

## Maharashtra Board Practice Set 15 Class 6 Maths Solutions Chapter 5 Decimal Fractions

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 15 Answers Solutions.

## 6th Standard Maths Practice Set 15 Answers Chapter 5 Decimal Fractions

Question 1.
Write the proper number in the empty boxes.

Solution:

Question 2.
Convert the common fractions into decimal fractions:
i. $$\frac { 3 }{ 4 }$$
ii. $$\frac { 4 }{ 5 }$$
iii. $$\frac { 9 }{ 8 }$$
iv. $$\frac { 17 }{ 20 }$$
v. $$\frac { 36 }{ 40 }$$
vi. $$\frac { 7 }{ 25 }$$
vii. $$\frac { 19 }{ 200 }$$
Solution:
i. $$\frac { 3 }{ 4 }$$

ii. $$\frac { 4 }{ 5 }$$

iii. $$\frac { 9 }{ 8 }$$

iv. $$\frac { 17 }{ 20 }$$

v. $$\frac { 36 }{ 40 }$$

vi. $$\frac { 7 }{ 25 }$$

vii. $$\frac { 19 }{ 200 }$$

Question 3.
Convert the decimal fractions into common fractions:
i. 27.5
ii. 0.007
iii. 90.8
iv. 39.15
v. 3.12
vi. 70.400
Solution:
i. 27.5
= $$\frac { 275 }{ 10 }$$

ii. 0.007
= $$\frac { 7 }{ 1000 }$$

iii. 90.8
= $$\frac { 908 }{ 10 }$$

iv. 39.15
= $$\frac { 3915 }{ 100 }$$

v. 3.12
= $$\frac { 312 }{ 100 }$$

vi. 70.400
= 70.4
= $$\frac { 704 }{ 10 }$$

## Maharashtra Board Practice Set 27 Class 6 Maths Solutions Chapter 10 Equations

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 27 Answers Solutions.

## 6th Standard Maths Practice Set 27 Answers Chapter 10 Equations

Question 1.
Rewrite the following using a letter:
i. The sum of a certain number and 3.
ii. The difference is obtained by subtracting 11 from another number.
iii. The product of 15 and another number.
iv. Four times a number is 24.
Solution:
i. Let the number be x.
∴ x + 3 represents the sum of a certain number x and 3.

ii. Let the number be x.
∴ x – 11 represents the number obtained by subtracting 11 from another number x.

iii. Let the number be x.
∴ 15x represents the product of 15 and another number x.

iv. Let the number be x.
∴ 4x = 24 represents four the product of a number x four times.

Question 2.
Find out which operation must be done on both sides of these equations in order to solve them:

1. x + 9 = 11
2. x – 4 = 9
3. 8x = 24
4. $$\frac { x }{ 6 }$$ = 3

Solution:

1. Subtract 9 from both sides.
2. Add 4 to both sides.
3. Divide both sides by 8.
4. Multiply both sides by 6.

Question 3.
Given below are some equations and the values of the variables. Are these values the solutions to those equations?

 No. Equation Value of the Variable Solution (Yes/No) i. y – 3 = 11 y = 3 No ii. 17 = n + 7 n = 10 iii. 30 = 5x x = 6 iv. $$\frac { m }{ 2 }$$ = 14 m = 7

Solution:

 No. Equation Value of the Variable Solution (Yes/No) i. y – 3 = 11 y = 3 No ii. 17 = n + 7 n = 10 Yes iii. 30 = 5x x = 6 Yes iv. $$\frac { m }{ 2 }$$ = 14 m = 7 No

i. y – 3 = 11
∴ y – 3 + 3 = 11 + 3
…. (Adding 3 to both sides)
∴ y + 0 = 14
∴ y = 14

ii. 17 = n + 7
∴ 17 – 7 = n + 7 – 7
…. (Subtracting 7 from both sides)
∴ 17 + (-7) = n + 7 – 7
∴ 10 = n
∴  n = 10

iii. 30 = 5x
∴ $$\frac{30}{5}=\frac{5x}{5}$$
…. (Dividing both sides by 5)
∴  6 = 1x
∴ 6 = x
∴  x = 6

iv. $$\frac { m }{ 2 }$$ = 14
∴ $$\frac { m }{ 2 }$$ × 2 = 14 × 2
…. (Multiplying both sides by 2)
$$\frac { m\times2 }{ 2\times1 }$$ = 28
∴ m = 28

Question 4.
Solve the following equations:
i. y – 5 = 1
ii. 8 = t + 5
iii. 4x = 52
iv. 19 = m – 4
v. $$\frac { p }{ 4 }=9$$
vi. x + 10 = 5
vi. m – 5 = -12
vii. p + 4 = -1
Solution:
i. y – 5 = 1
∴y – 5 + 5 = 1 + 5
…. (Adding 5 to both sides)
∴y + 0 = 6
∴y = 6

ii. 8 = t + 5
∴8 – 5 = t + 5 – 5
……(Subtracting 5 from both sides)
∴8 + (-5) = t + 0
∴ 3 = t
∴t = 3

iii. 4x = 52
∴$$\frac{4x}{4}=\frac{52}{4}$$
…. (Dividing both sides by 4)
∴ 1x = 13
∴ x = 13

iv. 19 = m -4
∴ 19 + 4 = m – 4 + 4
…. (Adding 4 to both sides)
∴ 23 = m + 0
∴ m = 23

v. $$\frac { p }{ 4 }$$ = 9
∴ $$\frac { p }{ 4 }$$ × 4 = 9 × 4 …. (Multiplying both sides by 4)
∴ $$\frac { p\times4 }{ 4\times1 }=36$$
∴ 1p = 36
∴ p = 36

vi. x + 10 = 5
∴ x + 10 – 10 = 5 – 10
…. (Subtracting 10 from both sides)
∴ x + 0 = 5 + (-10)
∴ x = -5

vii. m – 5 = -12
∴m – 5 + 5 = – 12 + 5
…. (Adding 5 to both sides)
∴m + 0 = -7
∴m = -7

viii. p + 4 = – 1
∴p + 4 – 4 = -1 – 4
…. (Subtracting 4 from both sides)
∴p + 0 = (-1) + (-4)
∴P = -5

Question 5.
Write the given information as an equation and find its solution:
i. Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?

ii. Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?

iii. Archana bought some kilograms of wheat. She requires 12 kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?
Solution:
i. Let the number of sheep before selling be x.
∴ x – 34 = 176
∴ x – 34 + 34 = 176 + 34 ….(Adding 34 to both sides)
∴ x + 0 = 210
∴ x = 210
The number of sheep with Haraba before selling is 210.

ii. Let the total number of bottles be x.
∴ x – 7 = 12
∴ x – 7 + 7 = 12 + 7 ….(Adding 7 to both sides)
∴ x + 0 = 19
∴ x = 19
Weight of jam in each bottle = 250g
∴ Total weight of jam = 19 × 250g = 4750 g = $$\frac { 4750 }{ 1000 }$$kg = 4.75 kg
∴ The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

iii. Let the total wheat bought by Archana be x kg.
Wheat used in 1 month = 12 kg
∴ Wheat used in 3 months = 3 × 12 = 36 kg
∴ x – 36 = 14
∴ x – 36 + 36 = 14 + 36 ….(Adding 36 to both sides)
∴ x + 0 = 50
∴ x = 50
∴ The total amount of wheat bought by Archana was 50 kg.

## Maharashtra Board Practice Set 36 Class 6 Maths Solutions Chapter 15 Triangles and their Properties

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 15 Triangles and their Properties Class 6 Practice Set 36 Answers Solutions.

## 6th Standard Maths Practice Set 36 Answers Chapter 15 Triangles and their Properties

Question 1.
Observe the figures below and write the type of the triangle based on its angles:

Solution:
i. right angled
ii. Obtuse angled
iii. acute angled

Question 2.
Observe the figures below and write the type of the triangle based on its sides:

Solution:
i. equilateral
ii. scalene
iii. isosceles

Question 3.
As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.

Solution:
The two roads which Avinash can choose to go to school are

The three roads together form ∆ABC.
Road AC is shorter because the sum of the lengths of any two sides (side AB + side BC) of a triangle is always greater than the third side (side AC).

Question 4.
The lengths of the sides of some triangles are given. Say what types of triangles they are.

1. 3 cm, 4 cm, 5 cm
2. 3.4 cm, 3.4 cm, 5 cm
3. 4.3 cm, 4.3 cm, 4.3 cm
4. 3.7 cm, 3.4 cm, 4 cm

Solution:

1. Since, no two sides have equal lengths, the given triangle is a scalene triangle.
2. Since, two sides have equal length, the given triangle is an isosceles triangle.
3. Since, all the three sides have equal lengths, the given triangle is an equilateral triangle.
4. Since, no two sides have equal lengths, the given triangle is a scalene triangle.

Question 5.
The lengths of the three segments are given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer.
i. 17 cm, 7 cm, 8 cm
ii. 7 cm, 24 cm, 25 cm
iii. 9 cm, 6 cm, 16 cm
iv. 8.4 cm, 16.4 cm, 4.9 cm
v. 15 cm, 20 cm, 25 cm
vi. 12 cm, 12 cm, 16 cm
Solution:
i. The lengths of the three sides are 17 cm, 7 cm, 8 cm.
a. 7 cm + 17 cm = 24 cm, greater than 8 cm
b. 8 cm +17 cm = 25 cm, greater than 7 cm
c. 7 cm + 8 cm =15 cm, not greater than 17 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 17 cm, 7 cm, 8 cm.

ii. The lengths of the three sides are 7 cm, 24 cm, 25 cm.
a. 7 cm + 24 cm = 31 cm, greater than 25 cm
b. 25 cm + 7 cm = 32 cm, greater than 24 cm
c. 24 cm + 25 cm = 49 cm, greater than 7 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 7 cm, 24 cm, 25 cm.

iii. The lengths of the three sides are 9 cm, 6 cm, 16 cm.
a. 9 cm + 16 cm = 25 cm, greater than 6 cm
b. 6 cm + 16 cm = 22 cm, greater than 9 cm
c. 9 cm+ 6 cm =15 cm, not greater than 16 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 9 cm, 6 cm, 16 cm.

iv. The lengths of the three sides are 8.4 cm, 16.4 cm, 4.9 cm.
a. 8.4 cm + 16.4 cm = 24.8 cm, greater than 4.9 cm
b. 4.9 cm + 16.4 cm = 21.3 cm, greater than 8.4 cm
c. 8.4 cm + 4.9 cm = 13.3 cm, not greater than 16.4 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 8.4 cm, 16.4 cm, 4.9 cm.

v. The lengths of the three sides are 15 cm, 20 cm, 25 cm.
a. 15 cm + 20 cm = 35 cm, greater than 25 cm
b. 25 cm + 20 cm = 45 cm, greater than 15 cm
c. 15 cm + 25 cm = 40 cm, greater than 20 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 15 cm, 20 cm, 25 cm.

vi. The lengths of the three sides are 12 cm, 12 cm, 16 cm.
a. 12 cm + 12 cm = 24 cm, greater than 16 cm
b. 12 cm + 16 cm = 28 cm, greater than 12 cm
c. 12 cm + 16 cm = 28 cm, greater than 12 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 12 cm, 12 cm, 16 cm.

#### Maharashtra Board Class 6 Maths Chapter 15 Triangles and their Properties Practice Set 36 Intext Questions and Activities

Question 1.
In the given figure, some points and some line segments joining them have been drawn. Which of these figures is a triangle? Which figure is not a triangle? Why not? (Textbook pg. no. 77)

Solution:
ABC it is a closed figure with three sides. Hence, ABC is a triangle.
PQRS has three sides but it is not a closed figure. Hence, PQRS is not a triangle.

Question 2.
As seen above, ∆ABC has three sides. Line segment AB is one side. Write the names of the other two sides. ∆ABC has three angles. ∠ABC is one among them. Write the names of the other angles. (Textbook pg. no. 77)
Solution:
The names of other two sides are: seg BC and seg AC
The names of other angles are: ∠BCA and ∠CAB

Question 3.
Measure the sides of the following triangles in centimeters, using a divider and ruler. Enter the lengths in the table below. What do you observe? (Textbook pg. no. 77)

 In ∆ABC In ∆PQR In ∆XYZ l (AB) =       cm l (QR) =       cm l (XY) =       cm l (BC) =       cm l (PQ) =       cm l (YZ) =       cm l (AC) =       cm l (PR) =        cm l (XZ) =       cm

Solution:

 In ∆ABC In ∆PQR In ∆XYZ l (AB) = 2.6 cm l (QR) = 2.8 cm l (XY) = 2.8 cm l (BC) = 2.6 cm l (PQ) = 3.8 cm l (YZ) = 2.6 cm l (AC) = 2.6 cm l (PR) = 3.8 cm l (XZ) = 4.3 cm

We observe that,

1. ∆ABC is an equilateral triangle,
2. ∆PQR is an isosceles triangle, and
3. ∆XYZ is a scalene triangle.

Question 4.
Measure all the angles of the triangles given below. Enter them in the following table. (Textbook pg. no. 78)

 In ∆DEF In ∆PQR In ∆LMN Measure of ∠D = m ∠D =___ Measure of ∠P = m ∠P =___ Measure of ∠L =__ Measure of ∠E = m ∠E =___ Measure of ∠Q =___=___ Measure of ∠M =___ Measure of ∠F = ___=___ Measure of ∠R =___=___ Measure of ∠N =___ Observation: All three angles are acute angles. Observation: One angle is right angle and two are acute angles. Observation: One angle is an obtuse angle and two are acute.

Solution:

 In ∆DEF In ∆PQR In ∆LMN Measure of ∠D = m ∠D = 60º Measure of ∠P = m ∠P = 45º Measure of ∠L = 30º Measure of ∠E = m ∠E = 68º Measure of ∠Q = m = 90º Measure of ∠M = 116º Measure of ∠F = m = 52º Measure of ∠R = m ∠R = 45º Measure of ∠N = 34º
1. ADEF is an acute angled triangle,
2. APQR is a right angled triangle,
3. ALMN is an obtuse angled triangle.

Question 5.
Observe the set squares in your compass box. What kind of triangles are they? (Textbook pg. no. 78)

Solution:
The first set square is a scalene triangle and also a right angled triangle.
The second set square is an isosceles triangle and also a right angled triangle.

Question 6.
Properties of a triangle. (Textbook pg. no. 79)
Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?

Solution:
The three angles of the triangle form a straight angle.
∴ m∠A + m∠B + m∠C = 180°
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 7.
Properties of a triangle (Textbook pg. no. 79)
Take a triangular piece of paper and make three different types of marks near the three angles. Take a point approximately at the center of the triangle. From this point, draw three lines that meet the three sides. Cut the paper along those lines. Place the three angles side by side as shown. See how the three angles of a triangle together form a straight angle, or, an angle that measures 180°.

Solution:
The three angles of the triangle form a straight angle.
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 8.
Draw any triangle on a paper. Name its vertices A, B, C. Measure the lengths of its three sides using a divider and scale and enter them in the table. (Textbook pg. no. 79)

 Length of side Sum of the lengths of two sides Length of the third side l (AB) =         cm l (AB) + l (BC) =         cm l (AC) =         cm l (BC) =         cm l (BC) + l (AC) =         cm l (AB) =         cm l (AC) =         cm l (AC) + l (AB) =        cm l (BC) =         cm

Solution:

 Length of side Sum of the lengths of two sides Length of the third side l (AB) = 2.7 cm l (AB) + l (BC) = 6.6 cm l (AC) = 5.6 cm l (BC) = 2.9 cm l (BC) + l (AC) = 9.5 cm l (AB) = 2.7 cm l (AC) = 5.6 cm l (AC) + l (AB) = 8.3 cm l (BC) = 3.9 cm

## Maharashtra Board Practice Set 14 Class 6 Maths Solutions Chapter 5 Decimal Fractions

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 14 Answers Solutions.

## 6th Standard Maths Practice Set 14 Answers Chapter 5 Decimal Fractions

Question 1.
In the table below, write the place value of each of the digits in the number 378.025.

 Place Hundreds Tens Units Tenths Hundredths Thousandths 100 10 1 $$\frac { 1 }{ 10 }$$ $$\frac { 1 }{ 100 }$$ $$\frac { 1 }{ 1000 }$$ Digit 3 7 8 0 2 5 Place value 300 $$\frac { 0 }{ 10 }=0$$ $$\frac { 5 }{ 1000 }$$ = 0.005

Solution:

 Place Hundreds Tens Units Tenths Hundredths Thousandths 100 10 1 $$\frac { 1 }{ 10 }$$ $$\frac { 1 }{ 100 }$$ $$\frac { 1 }{ 1000 }$$ Digit 3 7 8 0 2 5 Place value 300 7 × 10 = 70 8 × 1 = 8 $$\frac { 0 }{ 10 }=0$$ $$\frac { 2 }{ 100 }$$ = 0.02 $$\frac { 5 }{ 1000 }$$ = 0.005

Question 2.
Solve :
i. 905.5 + 27.197
ii. 39 + 700.65
iii. 40 + 27.7 + 2.451
Solution:
i. 905.5 + 27.197

ii. 39 + 700.65

iii. 40 + 27.7 + 2.451

Question 3.
Subtract:
i. 85.96 – 2.345
ii. 632.24 – 97.45
iii. 200.005 – 17.186
Solution:
i. 85.96 – 2.345

ii. 632.24 – 97.45

iii. 200.005 – 17.186

Question 4.
Avinash traveled 42 km 365 m by bus, 12 km 460 in by car and walked 640 m. How many kilometers did he travel altogether? (Write your answer in decimal fractions)
Solution:
Distance traveled in bus = 42 km 365 m
= 42 km + $$\frac { 365 }{ 1000 }$$ km
= 42 km + 0.365 km

= 42.365 km
Distance travelled in car = 12 km 460 m
= 12 km + $$\frac { 460 }{ 1000 }$$ km
= 12 km + 0.460 km

= 12.460 km
Distance walked = 640 m
= $$\frac { 640 }{ 1000 }$$ = 0.640 km
∴ Total distance travelled = Distance travelled in bus + Distance travelled in car + Distance walked
= 42.365 + 12.460 + 0.640

= 55.465 km
∴ Distance travelled altogether by Avinash is 55.465 km.

Question 5.
Ayesha bought 1.80 m of cloth for her salwaar and 2.25 for her kurta. If the cloth costs Rs 120 per metre, how much must she pay the shopkeeper?
Solution:
Total length of cloth bought = 1.80 m + 2.25 m
= 4.05 m

Cost of 1 m of cloth = Rs 120
∴ Cost of 4.05 m of cloth = 4.05 x 120

∴ Amount to be paid to the shopkeeper is Rs 486.

Question 6.
Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750 g to the children in her neighbourhood. How much of it does she have left?
Solution:
Total weight of watermelon = 4.25 kg
Weight of watermelon given to children = 1 kg 750 g
= 1 kg + $$\frac { 750 }{ 1000 }$$ kg
= 1 kg + 0.75 kg

= 1.75 kg
∴ Weight of watermelon left = Total weight of watermelon – Weight of watermelon given to children
= 4.25 kg – 1.75 kg

= 2.5 kg
∴ Weight of watermelon left with Sujata is 2.5 kg.

Question 7.
Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?
Solution:
Speed at which Anita is driving = 85.6 km per hr.
Speed limit = 55 km per hr.
∴ Anita should reduce her speed by 85.6 km per hr – 55 km per hr.

= 30.6 km per hr.
∴ Anita should reduce her speed by 30.6 km per hour to be within the speed limit.

#### Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 14 Intext Questions and Activities

Question 1.
Nandu went to a shop to buy a pen, notebook, eraser and a paint box. The shopkeeper told him the prices. A pen costs four and a half rupees, an eraser one and a half, a notebook six and a half and a paintbox twenty-five rupees and fifty paise. Nandu bought one of each article. Prepare his bill.
If Nandu gave a 100 rupee note, how much money does he get back? (Textbook pg. no. 29)

Nandu will get __ rupees back.
Solution:
100 – 38 = 62.00
Nandu will get Rs 62 rupees back.

Question 2.
Take a pen and notebook with you when you go to the market with your parent. Note the weight of every vegetable your mother buys. Find out the total weight of those vegetables. (Textbook pg. no. 30)
Solution:
(Students should attempt this activity on their own.)

## Maharashtra Board Practice Set 2 Class 6 Maths Solutions Chapter 2 Angles

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 2 Answers Solutions.

## 6th Standard Maths Practice Set 2 Answers Chapter 2 Angles

Question 1.
Match the following:

 Measure of the angle Type of the angle i. 180° a. Zero angle ii. 240° b. Straight angle iii. 360° c. Reflex angle iv. 0° d. Complete angle

Solution:
(i – Straight Angle),
(ii – Reflex Angle),
(iii – Complete Angle),
(iv – Zero Angle).

Question 2.
The measures of some angles are given below. Write the type of each angle:

1. 75°
2. 215°
3. 360°
4. 180°
5. 120°
6. 148°
7. 90°

Solution:

1. Acute angle
2. Zero angle
3. Reflex angle
4. Complete angle
5. Straight angle
6. Obtuse angle
7. Obtuse angle
8. Right angle

Question 3.
Look at the figures below and write the type of each of the angles:

Solution:
a. Acute angle
b. Right angle
c. Reflex angle
d. Straight angle
e. Zero angle
f. Complete angle

Question 4.
Use a protractor to draw an acute angle, a right angle and an obtuse angle:
Solution:

[Note: Students may draw acute and obtuse angles of measure other than the ones given.]

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 2 Intext Questions and Activities

Question 1.
Look at the angles shown in the pictures below. Identify the type of angle and write its name below the picture: (Textbook pg. no. 6)

Solution:

Question 2.
Complete the following table: (Textbook pg. no. 6)

Solution:

 Sr. No. i. ii. iii. Name of the angle ∠PYR or ∠RYP ∠LMN or ∠NML ∠BOS or ∠SOB Vertex of the angle Y M O Arms of the angle YP and YR ML and MN OB and OS

## Maharashtra Board Practice Set 29 Class 6 Maths Solutions Chapter 11 Ratio-Proportion

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 11 Ratio-Proportion Class 6 Practice Set 29 Answers Solutions.

## 6th Standard Maths Practice Set 29 Answers Chapter 11 Ratio-Proportion

Question 1.
If 20 metres of cloth costs Rs 3600, find the cost of 16 m of cloth.
Solution:
Cost of 20 metres of cloth = Rs 3600
∴ Cost of 1 metre of cloth = $$\frac{\text { cost of } 20 \text { metres of cloth }}{20}=\frac{3600}{20}$$
= Rs 180
∴ Cost of 16 metres of cloth = Cost of 1 metre of a cloth × 16
= 180 x 16 = Rs 2880
∴ The cost of 16 metres of cloth is Rs 2880.

Question 2.
Find the cost of 8 kg of rice, if the cost of 10 kg is Rs 325.
Solution:
Cost of 10 kg rice = Rs 325
∴ Cost of 10 kg rice

Cost of 8 kg rice = Cost of 1 kg rice x 8

∴ The cost of 8 kg rice is Rs 260.

Question 3.
If 14 chairs cost Rs 5992, how much will have to be paid for 12 chairs?
Solution:
Cost of 14 chairs = Rs 5992
∴ Cost of 1 chairs

= Rs 428
∴ Cost of 12 chairs = Cost of 1 chair x 12
= 428 x 12 = Rs 5136
∴ The amount to be paid for 12 chairs is Rs 5136.

Question 4.
The weight of 30 boxes is 6 kg. What is the weight of 1080 such boxes?
Solution:
Weight of 30 boxes = 6 kg
∴ Weight of 1 box

∴ Weight of 1080 boxes = Weight of 1 box x 1080

∴ The weight of 1080 boxes is 216 kg.

Question 5.
A car travelling at a uniform speed covers a distance of 165 km in 3 hours. At that same speed,
a. How long will it take to cover a distance of 330 km?
b. How far will it travel in 8 hours?
Solution:
Distance covered in 3 hours = 165 km
Distance covered in 1 hour

= 55 km
a. Time required to covered a distance of 330 km

= 6 hours
∴ The time required to cover a distance of 330 km is 6 hours.

b. Distance traveled in 8 hours = Distance covered in 1 hour x 8
= 55 x 8 = 440 km
∴ The distance traveled in 8 hours is 440 km.

Question 6.
A tractor uses up 12 litres of diesel while ploughing 3 acres of land. How much diesel will be needed to plough 19 acres of land?
Solution:
Diesel required to plough 3 acres of land =12 litres
∴ Diesel required to plough 1 acre of land

= 4 liters
∴ Diesel required to plough 19 acres of land = Diesel required to plough 1 acre of land x 19
= 4 x 19 = 76 litres
∴ Diesel needed to plough 19 acres of land is 76 litres.

Question 7.
At a sugar factory, 5376 kg of sugar can be obtained from 48 tonnes of sugarcane. If Savitatai has grown 50 tonnes of sugarcanes, how much sugar will it yield?
Solution:
Sugar obtained from 48 tonnes of sugarcane = 5376 kg
∴ Sugar obtained from 48 tonnes of sugarcane

∴ Sugar obtained from 50 tonnes of sugarcane = Sugar obtained from 1 tonne of sugarcane x 50
= 112 x 50 = 5600 kg
∴ 50 tonnes of sugarcane will yield 5600 kg of sugar.

Question 8.
In an orchard, there are 128 mango trees in 8 rows. If all the rows have an equal number of trees, how many trees would there be in 13 rows?
Solution:
Number of mango trees in 8 rows =128
Number of mango trees in 1 row

∴ Number of mango trees in 13 rows = Number of mango trees in 1 row x 13
= 16 x 13 = 208
∴ The number of mango trees in 13 rows are 208.

Question 9.
A pond in a field holds 120000 litres of water. It costs Rs 18000 to make such a pond. How many ponds will be required to store 480000 litres of water, and what would be the expense?
Solution:
Capacity of 1 pond = 1,20,000 litres
Total quantity of water = 4,80,000 litres
∴ Number of ponds required

Amount required to make 1 pond = Rs 18,000
∴ Amount required to make 4 ponds = Amount required to make 1 pond x 4
= 18,000 x 4 = Rs 72,000
∴ The number of ponds required to store 4,80,000 litres of water is 4, and the expense incurred in making the ponds is Rs 72,000.

#### Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 29 Intext Questions and Activities

Question 1.
Vijaya wanted to gift pens to seven of her friends on her birthday. When she went to a shop to buy them, the shopkeeper told her the rate for a dozen pens.
i. Can you help Vijaya to find the cost of 7 pens?
ii. If you find the cost of one pen, you can also find the cost of 7, right? (Textbook pg. no. 59)

Solution:
Cost of 12 pens = Rs 84.
∴ Cost of 12 pens

∴ Cost of 7 pens = Cost of one pen x Number of pens = 7 × 7
∴ Cost of 7 pens = Rs 49
∴ The cost of 7 pens (Rs 49) can be found by unitary method.

## Maharashtra Board Practice Set 26 Class 6 Maths Solutions Chapter 10 Equations

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 26 Answers Solutions.

## 6th Standard Maths Practice Set 26 Answers Chapter 10 Equations

Question 1.
Different mathematical operations are given in the two rows below. Find out the number you get in each operation and make equations.

1. 16 ÷ 2,
2. 5 × 2,
3. 9 + 4,
4. 72 ÷ 3,
5. 4 + 5,
1. 8 × 3,
2. 19 – 10,
3. 10 – 2,
4. 37 – 27,
5. 6 + 7

Solution:

1. 16 ÷ 2 = 8
2. 5 × 2 = 10
3. 9 + 4 = 13
4. 72 ÷ 3 = 24
5. 4 + 5 = 9
6. 8 × 3 = 24
7. 19 – 10 = 9
8. 10 – 2 = 8
9. 37 – 27 = 10
10. 6 + 7 = 13

∴ The equations are

1. 16 ÷ 2 = 10 – 2
2. 5 × 2 = 37 – 27
3. 9 + 4 = 6 + 7
4. 72 ÷ 3 = 8 x 3
5. 4 + 5 = 19 – 10

## Maharashtra Board Practice Set 1 Class 6 Maths Solutions Chapter 1 Basic Concepts in Geometry

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 1 Basic Concepts in Geometry Class 6 Practice Set 1 Answers Solutions.

## 6th Standard Maths Practice Set 1 Answers Chapter 1 Basic Concepts in Geometry

Question 1.
Look at the figure alongside and name the following:
i. Collinear points
ii. Rays
iii. Line segments
iv. Lines

Solution:

 i. Collinear Points Points M, 0 and T Points R, O and N ii. Rays ray OP, ray OM, ray OR, ray OS, ray OT and ray ON iii. Line Segments seg MT, seg RN, seg OP, seg OM, seg OR, seg OS, seg OT and seg ON iv. Lines line MT and line RN

Question 2.
Write the different names of the line.

Solution:
The different names of the given line are line l, line AB, line AC, line AD, line BC, line BD and line CD.

Question 3.

Solution:
(i – Line),
(ii – Line Segment),
(iii – Plane),
(iv – Ray)

Question 4.
Observe the given figure. Name the parallel lines, the concurrent lines and the points of concurrence in the figure.

Solution:

 Parallel Lines line b, line m and line q are parallel to each other. line a and line p are parallel to each other. Concurrent Lines and Point of Concurrence line AD, line a, line b and line c are concurrent. Point A is their point of concurrence. line AD, line p and line q are concurrent. Point D is their point of concurrence.

Maharashtra Board Class 6 Maths Chapter 1 Basic Concepts in Geometry Intext Questions and Activities

Question 1.
Complete the rangoli. Then, have a class discussion with the help of the following questions:

1. What kind of surface do you need for making a rangoli?
2. How do you start making a rangoli?
3. What did you do in order to complete the rangoli?
4. Name the different shapes you see in the rangoli.
5. Would it be possible to make a rangoli on a scooter or on an elephant’s back?
6. When making a rangoli on paper, what do you use to make the dots?

Solution:

1. For making a rangoli, I need a flat surface.
2. I can start making a rangoli by drawing equally spaced dots on the flat surface using a chalk.
3. In order to complete the rangoli, I joined the dots by straight lines to make a design.
4. In the rangoli, I find various shapes such as square, rectangle and triangles of two different size.
5. No. It won’t be possible to make a rangoli on a scooter or on an elephant’s back as they do not have a flat surface.
6. When making a rangoli on paper, I made use of scale and pencil to make equally spaced dots.

Question 2.
Write the proper term, ‘intersecting lines’ or ‘parallel lines’ in each of the empty boxes. (Textbook pg. no. 4)

Solution:
i. Intersecting Lines
ii. Parallel Lines
iii. Intersecting Lines

Question 3.
Draw a point on the blackboard. Every student now draws a line that passes through that point. How many such lines can be drawn? (Textbook pg. no. 2)
Solution:
An infinite number of lines can be drawn through one point.

Question 4.
Draw a point on a paper and use your ruler to draw lines that pass through it. How many such lines can you draw? (Textbook pg. no. 2)
Solution:
An infinite number of lines can be drawn through one point.

Question 5.
There are 9 points in the figure. Name them. (Textbook pg. no. 3)
i. If you choose any two points, how many lines can pass through the pair?
ii. Which three or more of these nine points lie on a straight line?
iii. Of these nine points, name any three or more points which do not lie on the same line.

Solution:
i. One and only one line can be drawn through two distinct , points.
ii. Points A, B, C and D lie on the same line. Points F, G and C lie on the same line.
iii. Points E, F, G, H and I do not lie on the same line, points A, B, E, H and I do not lie on the same line.

Question 6.
Observe the picture of the game being played. Identify the collinear players, non-collinear players, parallel lines and the plane. (Textbook pg. no. 4)

Solution:

 i. Collinear Players Players A, B, C, D, E, F, G ii. Non-collinear Players Players I, H, C Players I, A, B etc. iii. Parallel Lines line l, line m, line n, line p, line q, line r and line s iv. Plane The ground on which the boys are playing is the plane

Question 7.
In January, we can see the constellation of Orion in the eastern sky after seven in the evening. Then it moves up slowly in the sky. Can you see the three collinear stars in this constellation? Do you also see a bright star on the same line some distance away? (Textbook pg. no. 4)

Solution:

1. The three stars shown by points C, D and E are collinear.
2. The star shown by point H lies on the same line as the stars C, D and E.

Question 8.
Maths is fun! (Textbook pg. no. 5)
Take a flat piece of thermocol or cardboard, a needle and thread. Tie a big knot or button or bead at one end of the thread. Thread the needle with the other end. Pass the needle up through any convenient point P. Pull the thread up, leaving the knot or the button below. Remove the needle and put it aside. Now hold the free end of the thread and gently pull it straight. Which figure do you see? Now, holding the thread straight, turn it in different directions. See how a countless number of lines can pass through a single point P.

Solution:

1. The pulled thread forms a straight line.
2. An infinite number of lines can be drawn through one point.

Question 9.
Choose the correct option for each of the following questions:
i. ______ is used to name a point.
(A) Capital letter
(B) Small letter
(C) Number
(D) Roman numeral
Solution :
(A) Capital letter

ii. A line segment has two points showing its limits. They are called_____
(A) origin
(B) end points
(D) infinite points
Solution :
(B) end points

iii. An arrow head is drawn at one end of the ray to show that it is _____ on that side.
(A) finite
(B) ending
(C) infinite
(D) broken
Solution :
(C) infinite

iv. Lines which lie in the same plane but do not intersect are said to be ____ to each other.
(A) intersecting
(B) collinear
(C) parallel
(D) non-collinear
Solution :
(C) parallel

Question 10.
Determine the collinear and non-collinear points in the figure alongside:

Solution:
Collinear points:

1. Points A, E, H and C.
2. Points B, E, I and D.

Non-collinear points:
Points B, G, F and I

Question 11.
Look at the figure alongside and answer the questions given below:
i. Name the parallel lines.
ii. Name the concurrent lines and the point of concurrence.
iii. Write the different names of line PV.

Solution:
i. Parallel lines:
a. line l and line n
b. line p, line q, line r and line s

ii. Concurrent Lines: line q, line m, line n
Point of Concurrence: point S

iii. line l, line PT, line PR, line PV, line RT, line RV and line TV.

Question 12.
Name the different line segments and rays in the given figure:

Solution:
Line Segments:
seg UV, seg OY, seg OX, seg OV and seg OU

Rays:
ray OV, ray OX, ray OY and ray UV.