Class 10: Maths Chapter 7 solutions. Complete Class 10 Maths Chapter 7 Notes.

## Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 7- Mensuration

Maharashtra Board 10th Maths Chapter 7, Class 10 Maths Chapter 7 solutions

Question 1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Given: For the cone,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:

Volume of cone = 13 πr2h

∴ The volume of the cone is 11.79 cm3.

Question 2. Find the volume of a sphere of diameter 6 cm. [π = 3.14]
Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:

Radius (r) = d2 = 62 = 3 cm
Volume of sphere = 43 πr2
= 43 × 3.14 × (3)3
= 4 × 3.14 × 3 × 3
= 113.04 cm3
∴ The volume of the sphere is 113.04 cm3.

Question 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. [π = 3.14]
Given: For the cylinder,
height (h) = 40 cm
To find: Total surface area of the cylinder.
Solution:

Total surface area of cylinder = 2πr (r + h)
= 2 × 3.14 × 5 (5 + 40)
= 2 × 3.14 × 5 × 45
= 1413 cm2
The total surface area of the cylinder is 1413 cm2.

Question 4. Find the surface area of a sphere of radius 7 cm.
Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:

Surface area of sphere = Aπr2
= 4 × 227 × (7)2
= 88 × 7
= 616 cm2
∴ The surface area of the sphere is 616 cm2.

Question 5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Given: For the cuboid,
length (l) = 44 cm, breadth (b) = 21 cm,
height (h) = 12 cm
For the cone, height (H) = 24 cm
To find: Radius of base of the cone (r).
Solution:

Volume of cuboid = l × b × h
= 44 × 21 × 12 cm3
Volume of cone = 13 πr2H
= 13 × 227 × r2 × 24 cm3
Since the cuboid is melted to form a cone,
∴ volume of cuboid = volume of cone

∴ r2 = 21 × 21
∴ r = 21 cm …[Taking square root of both sides]
∴ The radius of the base of the cone is 21 cm.

Question 6. Observe the measures of pots in the given figures. How many jugs of water can the cylindrical pot hold?

Given: For the conical water jug,
radius (r) = 3.5 cm, height (h) = 10 cm
For the cylindrical water pot,
radius (R) = 7 cm, height (H) = 10 cm
To find: Number of jugs of water the cylindrical pot can hold.
Solution:
Volume of conical jug = 13 πr2h

∴ The cylindrical pot can hold 12 jugs of water.

Question 7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm2. The cone is placed up on the cylinder. Volume of the solid figure so formed is 500 cm3. Find the total height of the figure

Given: For the cylindrical part,
height (h) = 3 cm,
area of the base (πr2)= 100 cm2
Volume of the entire figure = 500 cm3
To find: Total height of the figure.
Solution:
A cylinder and a cone have equal bases.
Area of base = 100 cm2
∴ πr2 =100 …(i)
Let the height of the conical part be H.
Volume of the entire figure
= Volume of the entire + Volume of cone

∴ Height of conical part (H) =6 cm
Total height of the figure = h + H
= 3 + 6
= 9 cm
∴ The total height of the figure is 9 cm.

Question 8. In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Given: For the conical Part,
height (h) = 4 cm, radius (r) = 3 cm
For the cylindrical part,
height (H) = 40 cm, radius (r) = 3 cm
For the hemispherical part,
To find: Total area of the toy.
Solution:
Slant height of cone (l) = h2+r2−−−−−−√
= 42+32−−−−−−√
= 16+9−−−−−√
= 25−−√ = 5 cm
Curved surface area of cone = πrl
= π × 3 × 5
= 15π cm2
Curved surface area of cylinder = 2πrH
= 2 × π × 3 × 40
= 240π cm2
Curved surface area of hemisphere = 2πr2
= 2 × π × 32
= 18π cm2
Total area of the toy
= Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
= 15π + 240π + 18π
= 2737 π cm2
∴ The total area of the toy is 273π cm2.

Question 9. In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Given: For the cylindrical tablets,
thickness = height(h) = 5 mm
For the cylindrical wrapper,
diameter (D) = 14 mm, height (H) = 10 cm
To find: Number of tablets that can be wrapped.
Solution:
Radius of wrapper (R) = Diameter2
= 142 = 7 mm
Height of wrapper (H) = 10 cm
= 10 × 10 mm
= 100 mm
Volume of a cylindrical wrapper = πR2H
= π(7)2 × 100
= 4900π mm3
Volume of a cylindrical tablet = πr2h
= π(7)2 × 5
= 245 π mm3
No. of tablets that can be wrapped

∴ 20 tables can be wrapped in the wrapper

Question 10. The given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.
(π = 3.14)

Given: For the conical part,
height (h) = 4 cm, radius (r) = 3 cm
For the hemispherical part,
To find: Volume and surface area of the toy.
Solution:

Now, volume of the toy
= Volume of cone + volume of hemisphere
= 12π + 18π
= 30π
= 30 × 3.14
= 94.20 cm3
Also, surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
= 15π + 18π
= 33π
= 33 × 3.14
= 103.62 cm2
∴ The volume and surface area of the toy are 94.20 cm3 and 103.62 cm2 respectively.

Question 11.
Find the surface area and the volume of a beach ball shown in the figure.

Given: For the spherical ball,
diameter (d) = 42 cm
To find: Surface area and volume of the beach ball.
Solution:
Radius (r) = d2 = 422 = 21 cm
Surface area of sphere= 4πr2
= 4 × 3.14 × (21)2
= 4 × 3.14 × 21 × 21
= 5538.96 cm2
Volume of sphere = 43 πr3
= 43 × 3.14 × (21)3
= 4 × 3.14 × 7 × 21 × 21
= 38772.72 cm3
∴ The surface area and the volume of the beach ball are 5538.96 cm2 and 38772.72 cm3 respectively.

Question 12.
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

Given: For the metal sphere,
diameter (d) = 2 cm
For the cylindrical glass, diameter (D) =14 cm,
height of water in the glass (H) = 30 cm
To find: Volume of water in the glass.
Solution:
Let the radii of the sphere and glass be r and R respectively.

Volume of water with sphere in it = πR2H
= π × (7)2 × 30
= 1470π cm3
Volume of water in the glass
= Volume of water with sphere in it – Volume of sphere

∴ The volume of the water in the glass is 1468.67 π cm3 (i.e. 4615.80 cm3).

Question 1.
The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm3) (Textbook pg. no.141)

Given: For the cuboidal can,
length (l) = 20 cm,
height (h) = 30 cm
To find: Oil that can be contained in the can.
Solution:
Volume of cuboid = l × b × h
= 20 × 20 × 30
= 12000 cm3
= 120001000 litres
= 12 litres
∴ The oil can will contain 12 litres of oil.

Question 2.
The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap? (Textbook pg. no. 141)

Given: For the conical cap,
slant height (l) = 21 cm
To find: Cloth required to make the cap.
Solution:
Cloth required to make the cap
= Curved surface area of the conical cap
= πrl = 227 × 10 × 21
=22 × 10 × 3
= 660 cm2
∴ 660 cm2 of cloth will be required to make the cap.

Question 3.
As shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is ‘r’,

i. what is the ratio of the radii of the sphere and the cylinder ?
ii. what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
iii. what is the ratio of the volumes of the cylinder and the sphere? (Textbook pg. no. 141)
Solution:
∴ Radius of sphere = r
Also, height of cylinder = diameter of sphere
∴ h = d
∴ h = 2r …(i)

∴ radius of sphere : radius of cylinder = 1 : 1.

∴ curved surface area of cylinder : surface area of sphere = 1:1.

∴ volume of cylinder : volume of sphere = 3 : 2.

Question 4.
Finding volume of a sphere using cylindrical beaker and water. (Textbook, pg. no. 142)

i. Take a ball and a beaker of the same radius.
ii. Cut a strip of paper of length equal to the diameter of the beaker.
iii. Draw two lines on the strip dividing it into three equal parts.
iv. Stick this strip on the beaker straight up from the bottom.
v. Fill the water in the beaker upto the first mark of the strip from bottom.
vi. Push the ball in the beaker so that it touches the bottom.
Observe how much water level rises.
You will notice that the water level has risen exactly upto the total height of the strip. Try to obtain the formula for volume of sphere using the volume of the cylindrical beaker.
Solution:
Suppose volume of beaker upto height 2r is V.
V = πr2 h
∴ V = πr2(2r) …[∵ h = 2r]
∴ V = 2πr3
But, V = volume of the ball + volume of water in the beaker
∴ 2πr3 = Volume of the ball + 13 × 2πr3
∴ Volume of the ball = 2πr3 – 23 πr3
= 6πr3−2πr33
∴ Volume of the ball = 43 πr3
∴ Volume of a sphere = 43 πr3

#### Practice Set 7.2

Question 1.
The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm3)
Given: Radii (r1) = 14 cm, and (r2) = 7 cm,
height (h) = 30 cm
To find: Amount of water the bucket can hold.
Solution:

Volume of frustum = 13 πh (r1+ r22 + r1 × r2)

∴ The bucket can hold 10.78 litres of water.

Question 2.
The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
i. curved surface area,
ii. total surface area,
iii. volume, (π = 3.14)
Given: Radii (r1) = 14 cm, and (r2) = 6 cm,
height (h) = 6 cm
Solution:

i. Curved surface area of frustum
= πl (r1 + r2)
= 3.14 × 10(14 + 6)
= 3.14 × 10 × 20 = 628 cm2
∴ The curved surface area of the frustum is 628 cm2.

ii. Total surface area of frustum
= πl (r1+ r2) + πr12 + πr22
= 628 + 3.14 × (14)2 + 3.14 × (6)2
= 628 + 3.14 × 196 + 3.14 × 36
= 628 + 3.14(196 + 36)
= 628 + 3.14 × 232
= 628 + 728.48
= 1356.48 cm2
∴ The total surface area of the frustum is 1356.48 cm2.

iii. Volume of frustum
= 13 πth(r12 +r22 + r1 × r2)
= 13 × 3.14 × 6(142 + 62 + 14 × 6)
= 3.14 × 2(196 + 36 + 84)
= 3.14 × 2 × 316
= 1984.48 cm3
∴ The volume of the frustum is 1984.48 cm3.

Question 3.
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity. (π = 227)
Solution:

Circumference1 = 27πr1 = 132 cm

Curved surface area of frustum = π (r1 + r2) l
= π (21 + 14) × 25
=π × 35 × 35
= 227 × 35 × 25
= 2750 cm2

#### Practice Set 7.3

Question 1.
Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)
Given : Radius (r) = 10 cm,
Measure of the arc (θ) = 54°
To find : Area of the sector.
Solution:

Area of sector = θ360×πr2
= 54360 × 3.14 × (10)2
= 320 × 3.14 × 100
= 3 × 3.14 × 5
= 15 × 3.14
= 47.1 cm2
∴ The area of the sector is 47.1 cm2.

Question 2.
Measure of an arc of a circle is 80° and its radius is 18 cm. Find the length of the arc. (π = 3.14)
Given: Radius (r) = 18 cm,
Measure of the arc (θ) = 80°
To find: Length of the arc.
Solution:

Length of arc = θ360×2πr × 2πr
= 8360 × 2 × 3.14 × 18
= 29 × 2 × 3.14 × 18
= 2 × 2 × 3.14 × 2 = 25.12 cm
∴ The length of the arc is 25.12 cm.

Question 3.
Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Solution:
Given: Radius (r) = 3.5 cm,
length of arc (l) = 2.2 cm
To find: Area of the sector.
Solution:

Area of sector = l×r2
= 2.2×3.52
= 1.1 × 3.5 = 3.85 cm2
∴ The area of the sector is 3.85 cm2.

Question 4.
Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm2. Find the area of its corresponding major sector, (π = 3.14)
Given: Radius (r) = 10 cm,
area of minor sector =100 cm2
To find: Area of maj or sector.
Solution:

Area of circle = πr2
= 3.14 × (10)2
= 3.14 × 100 = 314 cm2
Now, area of major sector
= area of circle – area of minor sector
= 314 – 100
= 214 cm2
∴ The area of the corresponding major sector is 214 cm2.

Question 5.
Area of a sector of a circle of radius 15 cm is 30 cm2. Find the length of the arc of the sector.
area of sector = 30 cm2
To find: Length of the arc (l).
Solution:

∴ The length of the arc is 4 cm.

Question 6.
In the adjoining figure, radius of the circle is 7 cm and m (arc MBN) = 60°, find

i. Area of the circle.
ii. A(O-MBN).
iii. A(O-MCN).

Given: radius (r) = 7 cm,
m(arc MBN) = θ = 60°
Solution:
i. Area of circle = πr2
= 227 × (7)2
= 22 × 7
= 154 cm2
∴ The area of the circle is 154 cm2

ii. Central angle (θ) = ∠MON = 60°
Area of sector = θ360×πr2
∴ A(O – MBN) = 60360 × 227 × (7)2
16 × 22 × 7
= 25.67 cm2
= 25.7 cm2
∴ A(O-MBN) = 25.7 cm2

iii. Area of major sector = area of circle – area of minor sector
∴ A(O-MCN) = Area of circle – A(O-MBN)
= 154 – 25.7
∴ A(O-MCN) = 128.3 cm2

Question 7.
In the adjoining figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).
Given: Radius (r) = 3.4 cm,
perimeter of sector 12.8 cm
To find: A(P-ABC)

Solution:

Perimeter of sector
= Iength of arc ABC + AP + CP

∴ 12.8 = l + 3.4 + 3.4
∴ 12.8 = l + 6.8
∴ l = 12.8 – 6.8 = 6cm

∴ A(P-ABC) = 10.2 cm2

Question 8.
In the adjoining figure, O is the centre of the sector. ∠ROQ = ∠MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and (π = 227)

Given: ∠ROQ = ∠MON = 60°,
radius (r) = OR = 7 cm, radius (R) = OM = 21 cm
To find: Lengths of arc RXQ and arc MYN.
Solution:
i. Length of arc RXQ = θ360×2πr
= 602 × 2 × 227 × 7
= 16 × 2 × 22
= 7.33 cm
ii. Length of arc MYN = θ360×2πR
= 602 × 2 × 227 × 21
= 16 × 2 × 22 × 3
= 22 cm
∴ The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.

Question 9.
In the adjoining figure, if A(P-ABC) = 154 cm2, radius of the circle is 14 cm, find

i. ∠APC,
ii. l(arc ABC).

Given: A(P – ABC) = 154 cm2,
Solution:
i. Let ∠APC = θ

Question 10.
Radius of a sector of a circle is 7 cm. If measure of arc of the sector is

i. 30°
ii. 210°
iii. three right angles, find the area of the sector in each case.
Given: Radius (r) = 7 cm
To find: Area of the sector.
Solution:
i. Measure of the arc (θ) = 30°

∴ Area of the sector is 12.83 cm2.
ii. Measure of the arc (θ) = 210°

∴ Area of the sector is 89.83 cm2.
iii. Measure of the arc (θ) = 3 right angle
= 3 × 90° = 270°

∴ Area of the sector is 115.50 cm2.

Question 11.
The area of a minor sector of a circle is 3.85 cm2 and the measure of its central angle is 36°. Find the radius of the circle.
Given: Area of minor sector = 3.85 cm2,
central angle (θ) = 36°
To find: Radius of the circle (r).
Solution:

Area of minor sector = θ360×πr2

∴ The radius of the circle ¡s 3.5 cm.

Question 12.
In the given figure, ꠸PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Given: In rectangle PQRS,
PQ = 14 cm, QR = 21 cm
To find: Areas of the parts x, y and z.
Solution:

∠Q = ∠R = θ = 90° …[Angles of a rectangle]

For the sector (Q – PA),
PQ = QA …[Radii of the same circle]
∴ QA = 14 cm
Now, QR = QA + AR … [Q – A – R]
∴ 21 = 14 + AR
∴ AR = 7 cm

Area of rectangle = length × breadth
area of ꠸PQRS = PQ × QR
= 14 × 21
= 294 cm2
Area of part z = area of ꠸PQRS
– area of part x – area of part y
= 294 – 154 – 38.5
= 101.5 cm2
∴ The area of part x is 154 cm2, the area of part y is 38,5 cm2 and the area of part z is 101.5 cm2.

Question 13.
∆ALMN is an equilat triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find,

i. A (∆ LMN).
ii. Area of any one of the sectors.
iii. Total area of all the three sectors.
iv. Area of the shaded region. (3–√ = 1.732 )
Given: In equilateral triangle LMN, LM =14 cm,
radius of sectors (r) = 7 cm
Solution:
i. ∆LMN is an equilateral triangle.

ii. Central angle (θ) = 60° …[Angle of an equilateral triangle]

∴ Area of one sector = 25.67 cm2
iii. Total area of all three sectors
= 3 × Area of one sector
= 3 × 25.67
= 77.01 cm2
∴ Total area of all three sectors = 77.01 cm2
= A(∆LMN) – total area of all three sectors
= 84.87 – 77.01
= 7.86 cm2
∴ Area of shaded region = 7.86 cm

Question 1.
Complete the following table with the help of given figure. (Textbook pg. no. 149)

Solution:

Question 2.
Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. (Textbook pg. no. 150)

Solution:

Thus, if measure of an arc of a circle is θ, then
Area of sector (A) = θ360 × Area of circle
∴ Area of sector (A) = θ360 × πr2

Question 3.
In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. (Textbook pg. no. 151)

Solution:

Thus, if the measure of an arc of a circle is 0, then
Length of arc (l) = θ360 × circumference of circle
∴ Length of arc (l) = θ360 × 2πr

#### Practice Set 7.4

Question 1. In the adjoining figure, A is the centre of the circle. ∠ABC = 45° and AC = 72–√ cm. Find the area of segment BXC, (π = 3.14)

Solution:
In ∆ABC,
AC = AB … [Radii of same circle]
∴ ∠ABC = ∠ACB …[Isosceles triangle theorem]
∴ ∠ABC = ∠ACB = 45°
In ∆ABC,
∠ABC + ∠ACB + ∠BAC = 180° … [Sum of the measures of angles of a triangle is 180° ]
∴ 45° + 45° + ∠BAC = 180°
∴ 90° + ∠BAC = 180°
∴ ∠BAC = 90°
Let ∠BAC = θ = 90°

∴ The area of segment BXC is 27.93 cm2.

Question 2. In the adjoining figure, O is the centre of the circle.
m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region.
(π = 3.14, 3–√ = 1.73)

Given: m(arc PQR) = 60°, radius (r) = OP = 10 cm
To find: Area of shaded region.
Solution:

∠POR = m (arc PQR) …[Measure of central angle]
∴ ∠POR = θ = 60°

∴ The area of the shaded region is 9.08 cm2.

Question 3. In the adjoining figure, if A is the centre of the circle, ∠PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)

Given: Central angle (θ) = ∠PAR = 30°,
radius (r) = AP = 7.5
To find: Area of segment PQR.
Solution:

Let ∠PAR = θ = 30°

∴ The area of segment PQR is 0.65625 sq. units.

Question 4. In the adjoining figure, if O is the centre of the circle, PQ is a chord, ∠POQ = 90°, area of shaded region is 114 cm2, find the radius of the circle, (π = 3.14)

Given: Central angle (θ) = ∠POQ= 90°,
A (segment PRQ) = 114 cm2
Solution:

…[Taking square root of both sides]
∴ r = 20 cm
∴ The radius of the circle is 20 cm.

Question 5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, 3–√ = 1.73)
Given: Radius (r) =15 cm, central angle (θ) = 60°
To find: Areas of major and minor segments.
Solution:

Let chord PQ subtend ∠POQ = 60° at centre.
∴ θ = 60°

= 225 [0.0908]
= 20.43 cm2
∴ area of minor segment = 20.43 cm2
Area of circle = πr2
= 3.14 × 15 × 15
= 3.14 × 225
= 706.5 cm2
Area of major segment
= Area of circle – area of minor segment
= 706.5 – 20.43
= 686.07 cm2
Area of major segment 686.07 cm2
∴ The area of minor segment Is 20.43 cm2 and the area of major segment is 686.07 cm2.

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 7- Mensuration

Download PDF: Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 7- Mensuration PDF

## FAQs

Students can get the Maharashtra Books for primary, secondary, and senior secondary classes from here.  You can view or download the Maharashtra State Board Books from this page or from the official website for free of cost. Students can follow the detailed steps below to visit the official website and download the e-books for all subjects or a specific subject in different mediums.
Step 1: Visit the official website ebalbharati.in
Step 2: On the top of the screen, select “Download PDF textbooks”
Step 3: From the “Classes” section, select your class.
Step 4: From “Medium”, select the medium suitable to you.
Step 5: All Maharashtra board books for your class will now be displayed on the right side.

Who developed the Maharashtra State board books?

As of now, the MSCERT and Balbharti are responsible for the syllabus and textbooks of Classes 1 to 8, while Classes 9 and 10 are under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE).

How many state boards are there in Maharashtra?

The Maharashtra State Board of Secondary & Higher Secondary Education, conducts the HSC and SSC Examinations in the state of Maharashtra through its nine Divisional Boards located at Pune, Mumbai, Aurangabad, Nasik, Kolhapur, Amravati, Latur, Nagpur and Ratnagiri.

## About Maharashtra State Board (MSBSHSE)

The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

The Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC.

The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.