Future Study Point

# Solutions of the Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-Part 3

This article brings the CBSE Class 10 Mathematics: Most Important questions of 3 Marks and 4 marks Questions to get ready for board exams 2021-22.Each question is explained by a step by step method that offers you an idea to display how to solve maths questions in the most ideal manner in the exam.

Mathematics required skill  that can be developed by practicing it. Examining the solutions to a variety of questions is the only way to be perfect in mathematics. We at Future Study Point, furnish students of class 10  a series of  important 3 and 4 marks questions  with solutions  for helping class 10 students . The students who are planning for  achieving  excellent marks in  maths in CBSE Class 10 Board Exam are encouraged to practice the questions given below.

In this article Future Study Point is delivering the CBSE Class 10 Mathematics most  Important 3 and 4 Marks Questions for the preparation of CBSE board exam 2021-22. Every one of these questions is totally solved in a systematic and ordered way.

In CBSE Class 10 Mathematics Board Exam 2021-22, Section C and D will involve 10 and 8 questions of 3 and 4 marks each consecutively. In the maths question paper pretty much 65 percent of toatal marks  in class 10 maths paper can be accomplished by trying the 3 and 4 marks questions successfully. Along these lines, if one wishes to surpass desires in the appraisal, an incredible everyday practice with respect to 3 and 4 marks questions are fundamental since the prepration of these questions also accomodate 1 and 2 marks questions.

Here, CBSE class 10 maths most important questions and answers for 2021-22 CBSE board exam – III published by Future Study Point are presented for the students of 10 class who are going to appear in 2021-22 board exam of CBSE.

NCERT Solutions Class 10 Science from chapter 1 to 16

All the questions in CBSE class 10 maths most important questions and answers for 2021-22 board-III are designed by the expert of the maths subject and their answers are readily explained in an ordered way.

These  CBSE class 10 maths most important questions and answers for 2021-21 board-III are designed as per the updated syllabus of CBSE, so you need not to have any kinds of doubt in the selections of the questions. Don’t forget to make a comment and to subscribe us.

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-II

Solutions of the Most important maths questions for Class 10 CBSE for SA-1 and SA2

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

## Solutions of the Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-Part 3

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1.Two taps running together can fill a tank in 3 and 1/13 hr. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?

Solution: Let the time is taken by one of the taps to fill the tank = x hr

∴ In 1 hour it will fill the part of tank = 1/x  part

Time taken by another tap to fill the tank = (x +3) hr

∴ In 1 hour it will fill the part of tank = 1/(x+3)  part

Both of tank will fill the tank in one hour = (1/x) + 1/(x+3)

The part of tank filled by one of the tap to fill the tank in 1 hr + The part of tank filled by another tap to fill the tank in 1 hr = The part of tank filled by both the taps to fill the tank in 1 hr

(1/x) + 1/(x+3) = 13/40

(x +x +3)/x(x+3) = 13/40

80x + 120 = 13x² + 39x

13x² – 41x  – 120 = 0

$\textup{x} = \frac{41\pm \sqrt{(-41)^{2}-4\times 13\times-120}}{2\times13}$

$=\frac{41\pm \sqrt{1681+6240}}{2\times 13}$

$=\frac{41\pm \sqrt{7921}}{26}$

Neglecting minus sign because value of time can not be negative

$=\frac{41+89}{26}=\frac{130}{26}=5$

x = 5

Therefore the time taken by one of the tap is 5 hours and by another tap is (5+3) = 8 hours.

Q2. If  α and β are zeroes of the polynomial P(x) = 3x² +2x + 1, find

$\mathbf{{\textup{the polynomial whose zeroes are}\, \frac{1-\alpha }{1+\alpha} \, \textup{and}\, \frac{1-\beta }{1+\beta }}}$

### See the Video-

Solution: The polynomial given to us = 3x² +2x + 1 with its zeroes α and β

The relationship between zeroes and their coefficients is as follows:

$\alpha +\beta = \frac{-b}{a}=\frac{-2}{3}...(i)$

$\alpha \beta = \frac{c}{a}=\frac{1}{3}...(ii)$

The polynomial corresponding to given zeroes in the question can be driven from the following formula:

x² – (sum of the zeroes)x + product of the zeroes

$\textup{x}^{2}-\left ( \frac{1-\alpha }{1+\alpha}+ \frac{1-\beta}{1+\beta }\right )\textup{x}+\left ( \frac{1-\alpha}{1+\alpha} \right )\left ( \frac{1-\beta}{ 1+\beta}\right )$

The sum of zeroes

$\frac{1-\alpha }{1+\alpha}+ \frac{1-\beta}{1+\beta }$

$=\frac{2-2\alpha\beta }{1+(\alpha+\beta )+\alpha \beta }$

From (i) and (ii) we have

$=\frac{2-2\times \frac{1}{3} }{1-(\frac{2}{3} )+\frac{1}{3} }$

$=\frac{2-\frac{2}{3} }{1-\frac{1}{3}}=\frac{\frac{4}{3}}{\frac{2}{3}}=2$

The product of zeroes

$\frac{1-\alpha }{1+\alpha}\, \frac{1-\beta }{1+\beta }$

$=\frac{1-\beta -\alpha +\alpha \beta }{1+\beta+\alpha +\alpha \beta }$

$=\frac{1-(\alpha +\beta)+\alpha \beta }{1+(\beta+\alpha) +\alpha \beta }$

From equation(i) and (ii)

$=\frac{1+\frac{2}{3}+\frac{1}{3}}{1-\frac{2}{3}+\frac{1}{3}}$

$=\frac{\frac{6}{3}}{\frac{2}{3}}=3$

Therefore the sum of the zeroes  and product of the zeroes of the polynomial  = 2 and 3 respectively

So, required polynomial will be = x² – 2x + 3

Q3.If the polynomial  X4 −6×3+ 16×2− 25x + 10   is divided by another polynomial  x²  – 2x + k, the remainders comes out to be x + a,, find k and a.

See the Solution.

CBSE Class lX, X Class, NCERT Solutions and Important questions of Science and Maths

Q4. Prove that the chords of a circle that are equidistant from the center are equal in length.

Solution:

GIVEN. A circle with center O, in which there are two chords such that OP ⊥AB and OQ⊥CD and OP =OQ

TO PROVE. AB = CD

PROOF. Joining A and C to O, we get two Δ’s AOP and QOC, so considering both of these Δ’s

AO = CO (radii of the same circle)

OQ = OP (given)

OQ  ⊥ CD, OP ⊥ AB ( OQ and OP are distances of the chords CD and AB from the center O)

∴ ∠OQC = ∠APO= 90°

Δ AOP ≅ ΔQOC ( RHS rule)

CQ = AP ( By CPCT)

2CQ = 2AP( OQ and OP are perpendiculars on both the chords)

∴ CD = AB, Hence proved

### Solutions of the Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-lll

Q5.If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. Prove it.

Solution.

GIVEN: ΔABC, In which BC∥DE

TO PROVE:

$\fn_cm \frac{AD}{BD}=\frac{AE}{CE}$

CONSTRUCTION: Drawing PE ⊥AB and DQ⊥AC, Joining  C to D and B to E

PROOF.

$\fn_cm Area \: of\: \Delta ADE=\frac{1}{2}\times AD\times PE\: .....\left ( i \right )$

$\fn_cm Area \: of\: \Delta BDE=\frac{1}{2}\times BD\times PE\: .....\left ( ii \right )$

From (i) and (ii)

$\fn_cm \frac{ar\Delta ADE}{ar\Delta BDE}=\frac{AD}{BD}.....\left ( iii \right )$

Area of ΔADE in terms of other dimensions

$\fn_cm Area of \Delta ADE=\frac{1}{2}\times AE\times DQ......\left ( iv \right )$

$\fn_cm Area \: of \: \Delta DCE=\frac{1}{2}\times CE\times DQ......\left ( v \right )$

From (iv) and (v)

$\fn_cm \frac{ar\Delta ADE}{ar\Delta DCE}=\frac{AE}{CE}.....\left ( vi \right )$

arBDE = arDCE (area of Δ’s are equal with a common base and between the same parallel lines)

∴from (vi) we have

$\fn_cm \frac{ar\Delta ADE}{ar\Delta BDE}=\frac{AE}{CE}.....\left ( vii \right )$

Comparing (iii) and (vii) we have

$\fn_cm \frac{AD}{BD}=\frac{AE}{CE}$

HENCE PROVED

Q6.Prove that

$\fn_cm \frac{tan\Theta +sin\Theta }{tan\Theta -sin\Theta }=\frac{sec\Theta +1}{sec\Theta -1}$

Solutions.

LHS

$\fn_cm \frac{tan\Theta +sin\Theta }{tan\Theta -sin\Theta }$

Changing LHS into sin and cos

$\fn_cm =\frac{sin\Theta /cos\Theta +sin\Theta }{sin\Theta /cos\Theta -sin\Theta }$

$\fn_cm =\frac{sin\Theta \left ( 1/cos\Theta +1 \right )}{sin\Theta \left ( 1/cos\Theta -1 \right )}$

$\fn_cm =\frac{sec\Theta +1}{sec\Theta -1}$

=RHS

Q7. Prove that

$\fn_cm \frac{cos^{2 }\Theta }{1-tan\Theta }+\frac{sin^{3}\Theta }{sin\Theta -cos\Theta }=1+sin\Theta cos\Theta$

Solutions.

Taking LHS

$\fn_cm \frac{cos^{2 }\Theta }{1-tan\Theta }+\frac{sin^{3}\Theta }{sin\Theta -cos\Theta }$

$\fn_cm =\frac{cos^{2 }\Theta }{1-sin\Theta/cos\Theta }+\frac{sin^{3}\Theta }{sin\Theta -cos\Theta }$

$\fn_cm =\frac{cos^{2 }\Theta }{(cos\Theta -sin\Theta)/cos\Theta }+\frac{sin^{3}\Theta }{sin\Theta -cos\Theta }$

$\fn_cm =\frac{cos^{3 }\Theta }{(cos\Theta -sin\Theta) }+\frac{sin^{3}\Theta }{sin\Theta -cos\Theta }$

$\fn_cm =\frac{cos^{3 }\Theta }{(cos\Theta -sin\Theta) }-\frac{sin^{3}\Theta }{cos\Theta -sin\Theta }$

$\fn_cm =\frac{cos^{3 }\Theta -sin^{3}\Theta }{(cos\Theta -sin\Theta) }$

Since  x³ – y³ = (x -y)( x² + y² +xy)

$\fn_cm =\frac{\left ( cos\Theta -sin\Theta \right )\left ( cos^{2}\Theta +sin^{2}\Theta +cos\Theta sin\Theta \right ) }{cos\Theta -sin\Theta }$

cos²θ + sin²θ +sinθ.cosθ

= 1 + sinθ.cosθ = RHS, Hence Proved

See the video for most important questions of Trigonometry of Class 10

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

### NCERT Solutions for class 10 maths

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 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

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

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution