10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks

Here,10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks for 2022-23 board part 2 published by Future Study Point are presented for the students of 10 th class who are going to appear in the 2022-23 board exam of CBSE.

All the questions in CBSE class 10 maths most important questions and answers for 2022-23 board Part 2 are designed by the expert of the subject and their solutions are readily explained in an ordered way.

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Solutions of Most Important Class 10 Maths Questions of 3 and 4 marks Part-1

Solutions of the most important maths questions for Class 10 CBSE-Part 3

Solutions of the Most important maths questions for Class 10 CBSE for Part-1 and Part 2

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks

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Q1.Find the sum of 50 terms of an AP whose n^th term is defined as 7n – 6 .

Solution.

The n^thterm is defined as = 7n – 6

First term of the AP is = 7 × 1 – 6 = 1

Second term = 7 × 2 – 6 = 8

Third term = 7 × 3- 6 = 15

Hence the AP is 1, 8, 15,22…….

Where first term,a = 1, common difference, d = 8 -1 = 7

We have to find the sum of n term, let it is S_n}, where n = 50

Applying the formula for sum of the AP

$S_{n} =\frac{n}{2}\left [ 2a + \left ( n-1 \right )d \right ]$

$S_{50} =\frac{50}{2}\left [ 2\times 1 + \left ( 50-1 \right )7 \right ]$

= 25 (2 +49×7) = 25 (2 + 343) = 25 ×345 = 8625

Q2.A bucket is in the form of a frustum of a cone whose height is 42 cm and the radii of its circular ends are 30 cm and 10 cm. Find the amount of milk (in liters) which this bucket can hold. If the milkman sells the milk at the rate of Rs. 40 per liter, what amount he will get from the sale?

Solution.

r₁= 10 cm, r₂= 30 cm, h = 42 cm

The volume of bucket is

$=\frac{1}{3}\pi h\left ( r_{1}^{2}+r_{2}^{2}+r_{1}r_{2} \right )$

$=\frac{1}{3}\times \frac{22}{7}\times 42 \left ( 10^{2}+30^{2}+10\times 30 \right )$

The rate of milk is Rs 40 /liter

$= \frac{22}{21}\times 42 \left ( 100+900+300 \right )=44\times 1300=57200$

Hence the volume of the milk= 57200 cm² = 57200/1000 =57.2 liter

Therefore the amount he will get from the sale of milk = 40 ×57.2 = Rs 2288

Q3.If the coordinates of two points are A(3,4), B (5,-2) and a point P (x,5) is such that PA = PB, then find the area of ΔPAB.

Solution.

We are given that

PA = PB

√[(x -3)² + (5 – 4)²] = √[(x -5)² + (5 + 2)²]

Squaring both sides

(x -3)² + (5 – 4)² = (x -5)² + (5 + 2)²

x ² + 9 – 6x + 1= x ² + 25 -10x+49

10x -6x = 25 + 49 -10 = 64

4x = 64

x = 16

Therefore the co-ordinates of P are (16,5)

Area of Δ = 1/2 [x₁(y₂– y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ]

x₁=3, x₂=5 , x₃=16, y₁=4, y₂=-2 , y₃=5

Area of ΔPAB = 1/2 [3(-2– 5) + 5(5-4) + 16(4+2) ]

= 1/2[ -21 + 5 + 96] = 1/2(80) = 40

Therefore area of ΔPAB = 40 sq.unit

Q4.A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.

Solution.

The diameter of A solid metallic cylinder =12 cm

The radius of the solid metallic cylinder will be(R)= 6 cm

Height of the solid metallic cylinder(H) = 15 cm

Radii of the recasted cones(r) =3 cm

Height of recasted cones(h) = 9 cm

Let the number of toys formed = N

Therefore

The volume of cylinder = N × Volume of cone

N = Volume of cylinder/Volume of the cone = πR²H/(1/3)πr²h = 3R²H/r²h =(3×6²×15)/(3²×9) = 1620/81 = 20

Therefore the number of toys so formed = 20

10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks

Q5. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do the work.

Solution.

In 6 days A and B do = 1 piece of work

In 1 day A and B will do = (1/6) Part of the work

Let B is capable to finish that work in = x days

B will do part of that work in 1 day = 1/x

An alone can do that work in = (x–5) days

An alone do that piece off work in one day = 1/(x -5)

One day’s work of A + One day’s work of B = (A and B)’s one day work

1/(x-5) + 1/x = 1/6

(x +x -5)/x(x -5) = 1/6

(2x – 5)/(x²- 5x) = 1/6

12 x – 30 = x² – 5x

x² – 17x + 30 = 0

x ² – 15x -2x +30 = 0

x(x – 15) – 2 ( x – 15) = 0

( x -15)( x – 2) = 0

x = 15, x = 2

Neglecting 2 because 2 < 6

Therefore B alone can do that work in 15 days.

Q5.A hollow metallic sphere of external and internal diameters 8 cm and 4 cm respectively is melted to form a solid cone of base diameter 8 cm. Find the height of the cone.

Solution. The inner radius of the sphere = 4/2 = 2 cm

The outer radius of the sphere = 8/2 = 4cm

Volume of the sphere = (4/3) π r³

The volume of the metals the sphere is made up of = Outer volume of the sphere – Inner volume of the sphere

=(4/3) π 4³ – (4/3) π 2³

= (4π/3)×64 – (4π/3) ×8

= (4π/3) (64 – 8) = (4π/3) ×56

Therefore, the volume of sphere = Volume of the cone

Diameter of the cone is = 8 cm

The radius of the cone is = 8/2 = 4 cm

Volume of the cone = (1/3) πr²h

Since sphere is recasted into the cone ,therefore

Volume of the cone = Volume of the sphere

(1/3) π4²h = (4π/3) ×56

h = (4×56)/16 = 14

Therefore the height of the cone = 14 cm

Q6. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution.

Given → Parallelogram ABCD circumscribing a circle with center O which is touching it at P,Q,R and S points.

To Prove → The given parallelogram ABCD is a rhombus

Proof → AB = DC……….(i)(Opposite side of the parallelogram)

AD = BC……….(ii)(Opposite side of the parallelogram)

AP = AS ………………(iii) [Tangents drown on a circle from a same external point are equal]

BP = BQ (—do—)………(iv)

DR = DS (—do—)………(v)

CR = CQ (—do—)……….(vi)

HINT: Write the tangents corresponding to opposite sides same side as written above.

Adding (iii) to (vi) all equation we get

AP + BP + DR + CR = AS + BQ + DS + CQ

Arranging these line segments in order to get the side of paralellogram

(AP + BP) + (DR + CR) = (AS + DS )+ (BQ + CQ)

AB + DC = AD + BC

From equation number (i) and (ii)

DC + DC = AD + AD

2DC = 2AD

DC = AD

It is clear that AB = BC = DC= AD

All sides of the given parallelogram are equal therefore it is a rhombus.

Q7. Find the area of the triangle formed by joining the mid-points of sides of triangle whose vertices (2,1), (4,3) and (2,5).

Solution. Let the Δ formed PQR after joining the midpoints of the sides of ABC

We are given a Δ ABC such that the coordinates of each vertex are A(2,1),B(4,3) and C(2,5)

The vertices of the ΔPQR are calculated as follow

P(x₁ ,y₁) = (2+2)/2, (5+1)/2 = (2, 3)

P(x₂ ,y₂) = (2+4)/2, (5+3)/2 = (3, 4)

P(x₃ ,y₃) = (2+4)/2, (1+3)/2 = (3, 2)

Area of Δ = 1/2 [x₁(y₂– y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ]

x₁=2, x₂=3 , x₃=3, y₁=3, y₂= 4 , y₃=2

Area of Δ = 1/2 [x₁(y₂– y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ]

Area of ΔPAB = 1/2 [2(4– 2) + 3(2-3) + 3(3-4) ]

= 1/2[ 4 -3 -3] = 1/2(-2) = -1

Neglecting its negative sign because area can not be negative

Therefore area of ΔPQR = 1 square unit

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NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals