SSC TIME TABLE 2014 MARCH

DATE	SUBJECTS	TIME
05.03.2014	Hindi	11.00 to 2.00
07.03.2014	English	11.00 to 2.00
11.03.2014	Algebra	11.00 to 1.00
13.03.2014	Geometry	11.00 to 1.00
18.03.2014	Science & technology	11.00 to 2.00
20.03.2014	History & Political Science	11.00 to 1.00
22.03.2014	Geography & Economics	11.00 to 1.00
24.03.2014	Information Communication Technology	11.00 to 1.00
25.03.2014	Marathi	11.00 to 2.00

SSC HSC TIME TABLE 2014 MARCH

SSC TIME TABLE

Day & Date	Time	Subject
Monday, 03rd March, 2014	11.00a.m. to 2.00 p.m.	FIRST LANGUAGE Marathi (01) Hindi (02) Urdu (04) Gujrathi(05) Kannada (06) Tamil (07) Telugu (08) Malayalam (09) Sindhi (10) Bengali(11) Punjabi(12)
Monday, 03rd March, 2014	3.00 p.m. to 6.00 p.m.	SECOND OR THIRDLANGUAGE French(35)
Wednesday, 05th March, 2014	11.00 a.m. to 2.00 p.m	SECOND OR THIRD LANGUAGE Hindi (15)
	11.00 a.m. to 1.00 p.m.	SECOND OR THIRD LANGUAGE (COMPOSITE COURSE) Hindi (Composite) (B)
Friday, 07th March, 2014	11.00 a..m. to 2.00 p.m	FIRST LANGUAGE English (03) THIRD LANGUAGE English(17)
Tuesday, 11 th March, 2014	11.00 a.m. to 1.00 p.m.	MATHEMATICS PAPER I Algebra (71) ARITHMETIC (76) (For Blind, Deaf – Dumb, Spastic, Autism, & Dyscalculia, candidates only)
Tuesday, 11 th March, 2014	3.00 p.m. to 5.00 p.m	General. Mathematics Paper – I (74)
Thursday, 13 th March, 2014	11.00a.m. to 1.00 p.m.	MATHEMATICS – PAPER – II
Thursday, 13 th March, 2014	3.00 p.m. to 5.00 p.m.	General. Mathematics Paper – II
Saturday 15 th March, 2014	11.00 a..m. to 2.00 p.m.	Urdu(18) Sanskrit(27) Pali(28) Ardhamagdhi(29) Persian(30) Arabic(31) Avesta(32) Pahalvi(33) Russian(36)
Saturday 15 th March, 2014	3.00 p.m. to 5.00 p.m	SECOND OR THIRD LANGUAGE (COMPOSITE COURSE) Sanskrit (Comp.) Pali (Comp.) Ardhamagadhi(Comp) Arabic (Comp.) Persian (Comp.) French (Comp.) German (Comp.) Russian (Comp.) Kannada (Comp.) Tamil (Comp.) Telugu (Comp.) Malayalam (Comp.) Sindhi (Comp.) Punjabi (Comp.) Bengali (Comp.) Gujarati (Comp.)
Tuesday 18th March, 2014	11.00 a.m. to 2.00 p.m	SCIENCE AND TECHNOLOGY- (72) Physiology Hygiene & Home Science (77) (For Blind, Deaf – Dumb & Spastics Candidates Only)
Thursday 20th March, 2014	11.00 a.m. to 1.00 p.m.	SOCIAL SCIENCES-PAPER – I History-Political Science (73)
Saturday 22nd March, 2014	11.00 a.m. to 1.00 p.m	SOCIAL SCIENCES-PAPER – II Geography - Economics
Monday 24th March,2014	11.00 a.m to 1.00 p.m.	INFORMATION COMMUNICATION TECHNOLOGY (41)

HSC Commerce, Science and Arts Board Exam Time Table 2015

HSC Timetable 2015
Day and Date	Subject	Time
Saturday 21st February 2015	Marathi (02) Gujarati (03) Kannada (06) Sindhi (07) Malayalam (08) Tamil (09) Telugu (10) Punjabi (11) Bengali (12)	11:00 am to 2:00 pm
Monday 23rd February 2015	English (01)	11:00 am to 2:00 pm
Tuesday 24th February 2015	Hindi (04)	11:00 am to 2:00 pm
Wednesday 25th February 2015	History and Development of India Music (A)	11:00 am to 2:00 pm
Thursday 26th February 2015	Secretarial Practice (C)	11:00 am to 2:00 pm
	Physics (S)	11:00 am to 2:00 pm
Friday 27th February 2015	Sociology (A/S)	11:00 am to 2:00 pm
Saturday 28th February 2015	Mathematics and Statistics (A/S)	11:00 am to 2:00 pm
	Mathematics and Statistics (C)	11:00 am to 2:00 pm
Monday 02th March 2015	Economics (A/S/C)	11:00 am to 2:00 pm
Wednesday 04th March 2015	Organization of Commerce and Management (C)	11:00 am to 2:00 pm
	Chemistry (S)	11:00 am to 2:00 pm
Saturday 07th March 2015	Marathi Literature (A/S/C)	11:00 am to 2:00 pm
Monday 09th March 2015	Book Keeping and Accountancy (A/C)	11:00 am to 2:00 pm
	Biology (S)	11:00 am to 2:00 pm
Wednesday 11th March 2015	Geography (A/S/C)	11:00 am to 2:00 pm
Thursday 24th to 26th March 2015	Online Exam (Information Technology) (A/S/C)	11:00 am to 01:30pm

AS PER NEW SYLLABUS MAHARASHTRA STATE BOARD 

ENGLISH QUESTION PAPER

COMMERCE MATHS PAPER

COMMERCE MATHS PAPER

ECONOMICS PAPER ONE

ECONOMICS PAPER TWO

ECONOMICS PAPER THREE

SECRETARIAL PRACTICE PAPER ONE

SECRETARIAL PRACTICE PAPER TWO

SECRETARIAL PRACTICE PAPER THREE

ORGANISATION OF COMMERCE AND MANAGEMENT PAPER ONE

ORGANISATION OF COMMERCE AND MANAGEMENT PAPER TWO

ORGANISATION OF COMMERCE AND MANAGEMENT PAPER THREE

ACCOUNTS PAPER ONE

ACCOUNTS PAPER TWO

ACCOUNTS PAPER THREE

Try these SSC sample papers for your board exam. 

AS PER NEW SYLLABUS MAHARASHTRA STATE BOARD 

ALGEBRA HOTS FOR PRACTICE

ALGEBRA PAPER A

ALGEBRA PAPER B

ALGEBRA PAPER FIVE

ALGEBRA PAPER FOUR

ALGEBRA PAPER ONE WITH SOLUTION

ALGEBRA PAPER THREE

ALGEBRA PAPER TWO WITH SOLUTION

ENGLISH SAMPLE PAPER OF SSC 

GENERAL MATHS ALGEBRA PAPER ONE

GEOGRAPHY ECONOMICS QUESTION PAPER AS PER NEW SYLLABUS

GEOMETRY PAPER A

GEOMETRY PAPER B

GEOMETRY PAPER C

GEOMETRY QUESTION FOR PRACTICE

HISTORY & POLITICAL SCIENCE PAPER ONE

HISTORY  & POLITICAL SCIENCE PAPER TWO

SCIENCE AND TECHNOLOGY PAPER A

SCIENCE AND TECHNOLOGY PAPER B

SCIENCE AND TECHNOLOGY PAPER C

SCIENCE AND TECHNOLOGY PAPER D WITH SOLUTION

SCIENCE PAPER ONE WITH SOLUTION

Simple Geometry Question paper for Students. [New]

🏛️ Advanced Geometry Paper (Grade X) 📈

Challenge yourself with these problems covering key Grade X topics. Think critically, apply the theorems, and check your work. Best of luck! 🧠

Question 1: Similar Triangles 🔼

The areas of two similar triangles, \(\triangle ABC\) and \(\triangle PQR\), are 81 cm² and 144 cm² respectively. If the longest side of the smaller triangle, \(\triangle ABC\), is 27 cm, find the length of the longest side of the larger triangle, \(\triangle PQR\).

✅ Solution

We use the theorem which states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$ \frac{Area(\triangle ABC)}{Area(\triangle PQR)} = \left(\frac{\text{Corresponding Side of } \triangle ABC}{\text{Corresponding Side of } \triangle PQR}\right)^2 $$

Given information:

• Area(\(\triangle ABC\)) = 81 cm²
• Area(\(\triangle PQR\)) = 144 cm²
• Longest side of \(\triangle ABC\) = 27 cm

Let the longest side of \(\triangle PQR\) be \(x\).

$$ \frac{81}{144} = \left(\frac{27}{x}\right)^2 $$

Taking the square root of both sides:

$$ \sqrt{\frac{81}{144}} = \frac{27}{x} $$ $$ \frac{9}{12} = \frac{27}{x} $$

Now, we solve for \(x\):

$$ 9x = 27 \times 12 $$ $$ x = \frac{27 \times 12}{9} $$ $$ x = 3 \times 12 = 36 $$

Therefore, the length of the longest side of \(\triangle PQR\) is 36 cm. 💯

Question 2: Circles and Tangents 🔘

In the figure below, a circle with center O is inscribed in a quadrilateral ABCD such that it touches the sides AB, BC, CD, and DA at points P, Q, R, and S respectively. If AB = 29 cm, BC = 23 cm, CD = 20 cm, find the length of the side DA.

✅ Solution

The key property here is that the lengths of two tangents drawn from an external point to a circle are equal.

• From point A: \(AP = AS\)
• From point B: \(BP = BQ\)
• From point C: \(CQ = CR\)
• From point D: \(DR = DS\)

A property of a tangential quadrilateral (a quadrilateral with an inscribed circle) is that the sums of opposite sides are equal.

$$ AB + CD = BC + DA $$

We are given:

• AB = 29 cm
• BC = 23 cm
• CD = 20 cm

Substitute these values into the equation:

$$ 29 + 20 = 23 + DA $$ $$ 49 = 23 + DA $$ $$ DA = 49 - 23 $$ $$ DA = 26 \text{ cm} $$

So, the length of the side DA is 26 cm. 🎯

Question 3: Coordinate Geometry 🗺️

Find the ratio in which the y-axis divides the line segment joining the points A(5, -6) and B(-1, -4). Also, find the coordinates of the point of division.

✅ Solution

Let the ratio in which the y-axis divides the line segment be \(k:1\). Let the point of division on the y-axis be P.

Since the point P lies on the y-axis, its x-coordinate must be 0. So, \(P = (0, y)\).

Using the section formula for the x-coordinate:

$$ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} $$

Here, \((x_1, y_1) = (5, -6)\), \((x_2, y_2) = (-1, -4)\), \(m_1 = k\), and \(m_2 = 1\).

$$ 0 = \frac{k(-1) + 1(5)}{k + 1} $$ $$ 0 = \frac{-k + 5}{k + 1} $$ $$ -k + 5 = 0 $$ $$ k = 5 $$

So, the ratio is \(k:1\), which is 5:1.

Now, let's find the y-coordinate of the point of division P using the section formula for y:

$$ y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} $$ $$ y = \frac{5(-4) + 1(-6)}{5 + 1} $$ $$ y = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3} $$

Therefore, the point of division is (0, -13/3). 📍

Question 4: Surface Areas & Volumes ⚗️

A medicinal capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. (Use \(\pi = 22/7\))

✅ Solution

First, let's break down the dimensions of the capsule.

• Diameter = 5 mm, so Radius (r) = 2.5 mm
• Total length = 14 mm

The total length includes two hemispherical ends. The length of the cylindrical part is the total length minus the radii of the two hemispheres.

Height of cylinder (h) = Total length - 2 \(\times\) radius

$$ h = 14 - 2(2.5) = 14 - 5 = 9 \text{ mm} $$

The total surface area of the capsule is the sum of the curved surface area (CSA) of the cylinder and the CSA of the two hemispheres.

$$ \text{Total SA} = (\text{CSA of Cylinder}) + (2 \times \text{CSA of Hemisphere}) $$

Note: Two hemispherical CSAs make up the surface area of one sphere.

$$ \text{Total SA} = (2\pi rh) + (2 \times 2\pi r^2) = 2\pi rh + 4\pi r^2 $$ $$ \text{Total SA} = 2\pi r (h + 2r) $$

Let's substitute the values:

$$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times (9 + 2 \times 2.5) $$ $$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) $$ $$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times 14 $$

Simplify the calculation:

$$ \text{Total SA} = 2 \times 22 \times 2.5 \times \frac{14}{7} $$ $$ \text{Total SA} = 44 \times 2.5 \times 2 $$ $$ \text{Total SA} = 44 \times 5 = 220 \text{ mm}^2 $$

The surface area of the capsule is 220 mm². 💊

Excellent work! 🚀 Consistent practice is the highway to mastering geometry. 🌌

GEOMETRY QUESTION PAPER THREE [New]

📖 Simple Geometry Practice Paper 📐

Test your fundamental geometry skills. Read each question carefully and try to solve it before revealing the solution. Good luck! 🍀

Question 1: Area of a Triangle 📏

A triangle has a base of 14 cm and a corresponding height of 8 cm. What is its area? 🧐

✅ Solution

The formula for the area of a triangle is given by:

$$ Area = \frac{1}{2} \times base \times height $$

We are given:

• Base (b) = 14 cm
• Height (h) = 8 cm

Plugging these values into the formula:

$$ Area = \frac{1}{2} \times 14 \times 8 $$ $$ Area = 7 \times 8 $$ $$ Area = 56 \text{ cm}^2 $$

So, the area of the triangle is 56 square centimeters. 💯

Question 2: Circumference of a Circle 🔴

Calculate the circumference of a circle with a radius of 7 meters. Use the value of Pi (\(\pi\)) as approximately \( \frac{22}{7} \). 🤔

✅ Solution

The formula for the circumference of a circle is:

$$ Circumference = 2 \times \pi \times r $$

We are given:

• Radius (r) = 7 m
• Pi (\(\pi\)) ≈ \( \frac{22}{7} \)

Substitute the values into the formula:

$$ Circumference = 2 \times \frac{22}{7} \times 7 $$

The 7 in the numerator and denominator cancel out:

$$ Circumference = 2 \times 22 $$ $$ Circumference = 44 \text{ meters} $$

Thus, the circumference of the circle is 44 meters. 🌍

Question 3: Pythagorean Theorem 📐

In a right-angled triangle, the length of the two shorter sides (legs) are 6 inches and 8 inches. What is the length of the longest side (the hypotenuse)?

✅ Solution

We use the Pythagorean theorem, which states that for a right-angled triangle with legs 'a' and 'b' and hypotenuse 'c':

$$ a^2 + b^2 = c^2 $$

Here, we have:

• a = 6 inches
• b = 8 inches

Let's find \(c^2\):

$$ 6^2 + 8^2 = c^2 $$ $$ 36 + 64 = c^2 $$ $$ 100 = c^2 $$

Now, we find 'c' by taking the square root of 100:

$$ c = \sqrt{100} $$ $$ c = 10 \text{ inches} $$

The length of the hypotenuse is 10 inches. ✨

Question 4: Volume of a Cylinder 🥫

Find the volume of a cylinder that has a radius of 5 cm and a height of 10 cm. Use the value of Pi (\(\pi\)) as approximately \(3.14\).

✅ Solution

The formula for the volume of a cylinder is:

$$ Volume = \pi \times r^2 \times h $$

We are given:

• Radius (r) = 5 cm
• Height (h) = 10 cm
• Pi (\(\pi\)) ≈ \(3.14\)

Substitute the values into the formula:

$$ Volume = 3.14 \times (5^2) \times 10 $$ $$ Volume = 3.14 \times 25 \times 10 $$ $$ Volume = 78.5 \times 10 $$ $$ Volume = 785 \text{ cm}^3 $$

The volume of the cylinder is 785 cubic centimeters. 🧪

Keep practicing! 🚀 Geometry is the key to understanding the world around us. 🌌

📐 Super Shapes Adventure! 🚀 [New]

📐 Super Shapes Adventure! 🚀

Hello, young mathematician! Let's solve some fun shape puzzles together. Good luck! 🌟

Question 1: Awesome Angles! 🖍️

An angle that is smaller than a right angle (less than 90°) is called what?

🏆 The Answer Is...

An angle smaller than 90° is called an Acute Angle. Think of it as a "cute" little angle!

Question 2: Perimeter Puzzle 📏

A rectangular garden has a length of 8 meters and a width of 5 meters. If you want to put a fence all the way around it, how long will the fence need to be?

🏆 The Answer Is...

To find the length of the fence, we need to calculate the perimeter. The formula for the perimeter of a rectangle is:

Perimeter = \( 2 \times (length + width) \)

So, Perimeter = \( 2 \times (8 + 5) = 2 \times 13 = 26 \) meters.

You would need a fence that is 26 meters long. Great job! 🌿

Question 3: Area Adventure 🖼️

You are painting a square canvas. If one side of the square is 7 inches long, what is the total area you need to paint?

🏆 The Answer Is...

The area of a square is found by multiplying the side by itself.

Area = \( side \times side \)

So, Area = \( 7 \times 7 = 49 \) square inches.

You need to paint an area of 49 square inches. Fantastic! 🎨

Question 4: Shape Sleuth 🧐

I am a shape with 5 sides and 5 corners. What am I?

🏆 The Answer Is...

A shape with 5 sides and 5 corners is called a Pentagon. You solved the mystery! 🎉

You are a Geometry Genius! Keep up the amazing work! ✨

Geometry Practice Paper for Students 9th Grade [New]

📐 Geometry Challenge (Grade IX) 📚

Strengthen your geometry fundamentals with these problems. Take your time, draw diagrams, and think logically before checking the solution. Let's begin! 🚀

Question 1: Lines and Angles 📏

In the figure below, line \(l\) is parallel to line \(m\) (\(l \parallel m\)), and line \(t\) is a transversal. Find the value of \(x\).

✅ Solution

The angles given, \((3x - 10)°\) and \((2x + 15)°\), are consecutive interior angles.

A property of parallel lines states that consecutive interior angles are supplementary, meaning their sum is 180°.

$$ (3x - 10) + (2x + 15) = 180 $$

Combine like terms:

$$ 5x + 5 = 180 $$

Subtract 5 from both sides:

$$ 5x = 175 $$

Divide by 5:

$$ x = \frac{175}{5} $$ $$ x = 35 $$

Therefore, the value of \(x\) is 35. 💯

Question 2: Heron's Formula 🌳

A triangular park has sides of length 40 m, 24 m, and 32 m. Find the area of the park using Heron's formula.

✅ Solution

Heron's formula is used to find the area of a triangle when the lengths of all three sides are known.

First, we find the semi-perimeter (\(s\)) of the triangle.

$$ s = \frac{a + b + c}{2} $$

Given sides: \(a = 40\) m, \(b = 24\) m, \(c = 32\) m.

$$ s = \frac{40 + 24 + 32}{2} = \frac{96}{2} = 48 \text{ m} $$

Now, we apply Heron's formula:

$$ Area = \sqrt{s(s-a)(s-b)(s-c)} $$

Substitute the values:

$$ Area = \sqrt{48(48-40)(48-24)(48-32)} $$ $$ Area = \sqrt{48 \times 8 \times 24 \times 16} $$

To simplify, let's break down the numbers into factors:

$$ Area = \sqrt{(16 \times 3) \times 8 \times (3 \times 8) \times 16} $$ $$ Area = \sqrt{16^2 \times 8^2 \times 3^2} $$

Now, take the square root:

$$ Area = 16 \times 8 \times 3 = 384 $$

The area of the park is 384 square meters. 🌿

Question 3: Quadrilaterals 🔷

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find the measure of all four angles.

✅ Solution

The sum of the interior angles of any quadrilateral is always 360°.

Let the angles be \(3x\), \(5x\), \(9x\), and \(13x\) based on the given ratio.

Set up the equation:

$$ 3x + 5x + 9x + 13x = 360 $$

Sum the terms with \(x\):

$$ 30x = 360 $$

Solve for \(x\):

$$ x = \frac{360}{30} = 12 $$

Now, find each angle by substituting \(x = 12\):

• First angle = \(3x = 3 \times 12 = \textbf{36°}\)
• Second angle = \(5x = 5 \times 12 = \textbf{60°}\)
• Third angle = \(9x = 9 \times 12 = \textbf{108°}\)
• Fourth angle = \(13x = 13 \times 12 = \textbf{156°}\)

Check: \(36 + 60 + 108 + 156 = 360°\). The solution is correct. ✅

Question 4: Circles and Chords ⚪

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the center of the circle.

✅ Solution

Let the circle have center O, the chord be AB, and M be the midpoint of AB. The line segment OM is the perpendicular distance from the center to the chord.

The perpendicular from the center of a circle to a chord bisects the chord. So:

$$ AM = MB = \frac{1}{2} \times AB = \frac{1}{2} \times 16 = 8 \text{ cm} $$

Now, consider the right-angled triangle \(\triangle OMA\). The hypotenuse is the radius OA.

• Radius (OA) = 10 cm
• Half-chord length (AM) = 8 cm
• Distance from center (OM) = ?

Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)):

$$ OM^2 + AM^2 = OA^2 $$ $$ OM^2 + 8^2 = 10^2 $$ $$ OM^2 + 64 = 100 $$ $$ OM^2 = 100 - 64 = 36 $$ $$ OM = \sqrt{36} = 6 \text{ cm} $$

The distance of the chord from the center is 6 cm. 🎯

Great job completing the challenge! Keep practicing to build a strong foundation in geometry. 🌟