MATHEMATICS PREPARATORY TEST..!!!!

Dear Readers,
How was your test????
If you are having any doubt in any question, then please feel free to discuss over here..!
                                                                                                                  Do share your views with us..!!!

Important Unsolved Questions!!....

Q1. If the eighth term of an A.P. is 31 and fifteenth term is 16 more than the eleventh term , find the A.P.


Q2. Which term of the A.P. 5 , 15 , 25 , ....... will be 130 more than its thirty first term ?


Q3. If the tenth term of an A.P. is 52 and seventeenth term is 20 more than the thirteenth term , find the A.P.


Q4. The sum of fifth and ninth terms of an A.P. is 72 and the sum of seventh and twelfth terms is 97 . Find the A.P.


Q5. Determine the general term of an A.P. whose seventh term is -1 and sixteenth term is 17.


Q6. If the sum of m terms of an A.P. is same as the sum of its n terms , show that the sum of its (m+n) terms is zero .


Q7. Find the sum of all odd integers between 2 and 100 divisible by 3.

Try out these Questions and feel free to ask the queries...................!!!!!

A FEW IMPORTANT QUESTIONS (A.P.)

Ques1: Find the sum of first two 40 positive integers divisible by 6.

Ques2:In an A.P., find n and S_n if a = 5, d = 3, a_n = 50

Ques3: Is -150 a term of 11, 8 , 5 ,2 .. ? Why?

Ques4: If the 8^th term of an A.P. Is 31 and the 15^th term is 16 more than the 11^th term, find the A.P.

                                     Please feel free to share your problems ...!!! 

Arithemetic Progressions ( A.P. ) !!......

Introduction :-

An Arithemetic Prgression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

e.g:- 1 , 4 , 7 , 10 , 13..........

In this example , first term is 1 and all other terms are obtained by adding a fixed number to the preceding term.

First term is signified by :- a

Common Difference between each term is signified by :- d

Here , a = 1 i.e. first term and d = 3 i.e. common difference.

The fixed number i.e. common difference (d) in A.P. Can be positive , negative or zero.

General form an A.P. :-

An Arithemetic Prgression where a is the first term and d is the common difference is referred to as General form an A.P.

Finite A.P. :-

Some A.P.s have only finte number of terms . Such an A.P. is called as Finite A.P. .

Such Finite A.P.s have the last term (l) .

Infinite A.P. :-

The A.P.s having infinite number of terms is referred to as Infinte A.P.

Such A.P.s doesn't have the last term (l).

nth Term of an A.P. :-

Let a₁ , a₂ , a₃ , …. be an A.P. whose first term a₁ is a and the common difference is d .

Then ,

the second term (a₂) = a + d = a + (2 – 1) d

the third term (a₃) = a₂ + d = (a + d ) + d = a + 2d = a + (3 – 1 ) d

the fourth term (a₄) = a₃ + d = (a + 2d ) + d = a + 3d = a + (4 – 1 ) d

Looking at the pattern above , we can say that the nth term a_n = a + (n – 1 ) d .

So , the nth term a_n of the A.P. with first term a and common difference d is given by:-

a_n = a + ( n – 1 ) d .

an is also called the general term of an A.P. . If there are m terms in the A.P. , then a_m represents the last term which is sometimes donated by l .

Sum of first n terms of an A.P. :-

The sum of first n terms of an A.P. with first term ' a ' and common difference ' d ' is given by :-

S_n = n / 2 [ 2a + (n – 1 ) d ]

It can also be written as :-

S_n = n / 2 [ a + a ( n – 1 ) d ]

i.e. S_n = n / 2 (a + a_n )

So , if there are n terms in the A.P. , then an = l , the last term.

Therefore , the sum can also be given by :-

S_n = n / 2 ( a + l )

TRY THESE QUESTIONS!!........

Q1. Find the roots of the following equation 4x² + 4bx – (a² – b²) = 0 by the method of completing the square.

Q2. Solve the quadratic equation :- x² + 3x – (a² + a – 2) = 0 by factorization method.

Q3. By using the method of completing the square , show that the equation 4x² + 3x + 5 = 0 has no real roots.

Q4. Find the values of k for which the given equation i.e. 2x² – 10x + k = 0 has real and equal roots.

Q5. The sum of two numbers is 15. If the sum of their reciprocals is 3 / 10 , find the numbers.

Q6. The product of Rohan's age five years ago with his age 9 years later is 15. Find Rohan's present age.

Q7. One year ago , a man was 8 times as old as his son. Now his age age is equal to the square of his son's age. Find their present ages.

Try to sort out these questions and share your solutions..

Ask the queries in case it occurs in any question!!!!

A FEW IMPORTANT QUESTIONS (Quadratic Equations)

Ques1: For what values of  'k' does the quadratic equation (k-5)x² + 2(k-5)x + 2 = 0 have equal roots?

Ques2: A plane left 30minutes later than the schedule time and in order to reach its destination 1500km away in time, it has to increase its speed by 250km/hr from its usual speed. Find its usual speed.

  

 FEEL FREE TO SHARE YOUR PROBLEMS..!!!

DO COMMENT BELOW IF YOU ARE HAVING ANY DOUBT RELATED THESE QUESTIONS AND CHAPTER..!!

  

CHAPTER :- QUADRATIC EQUATIONS!!.....

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0 , where a , b , c are real numbers , a ≠ 0 .

In fact , any equation of the form p(x) = 0 , where p(x) is a polynomial of degree 2 , is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees , then we get the standard form of the equation .

Therefore the standard / general form of quadratic equation is ax² + bx + c = 0 , a 0 and a , b , c are real numbers. In this equation , variable is 'x' and the degree of the equation is '2'.

If x =  satisfies the equation ax² + bx + c = 0 , then is known as the root of the quadratic equation.

The roots and zeroes of the quadratic equation ax² + bx + c = 0 , a ≠ 0 are same.

If the given quadratic equation ax² + bx + c = 0 , a ≠ 0 can be factorised into two linear factors, then its roots can be found by equating each factor equal to zero.

We can solve quadratic equation by completing the square.

The roots of the quadratic equation ax² + bx + c = 0 , a ≠ 0 can be found by using the following formula, if its discriminate (D = b² – 4c) is greater than or equal to zero.

x = - b  b2 – 4ac / 2a

A quadratic equation ax² + bx + c = 0 , a ≠ 0 has:-

1. two distinct real roots , if D > 0.

2. two equal roots (i.e., repeated roots), if D = 0.

3. no real roots. if D < 0.

Questions related to this chapter will be posted soon!!..

TRY OUT THESE.....!! (CIRCLES)

Q1. Two tangents are drawn to a circle from an exterior point A, touching the at B and C. From another point K a third tangent is drawn to a circle at Q. If AB = 20cm, then find the perimeter of triangle APR.

Q2. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Q3. Find the length of a tangent drawn from a point whose distance from the centre of a circle is 25cm. Also the radius of a circle is 7cm.

Q4. From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

Q5. In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Q6. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Try to solve these questions and share your solutions with us....

You are free to ask the problems that occurs while solving......!!

SOME IMPORTANT QUESTIONS (Chapter:Circles)

Q1. ABC is an isosceles triangle in which AB = AC, circumscribed about a circle. Prove that the base is bisected by the point of contact. i.e. if AB = AC, prove that BE = EC.                 CBSE 2008

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AD = AF - ( Tangents from A)

BD = BE - ( Tangents from B)

CE = CF - ( Tangents from C)

Now,

AB = AC

AB- AD = AC- AD - (Subtracting AD from both sides)

AB- AD = AC – AF - (Since, AD=AF)

BD = CF

BE = CF

BE = CE , Hence proved.

Q2. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

                                                                                                                               CBSE 2008,2009

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AP = AS - (Tangents from A) - (i)

BP = BQ - ( Tangents from B) - (ii)

CR = CQ - ( Tangents from C) - (iii)

DR = CQ - ( Tangents from D) - (iv)

Adding (i) , (ii) , (iii) , (iv) , we get :-

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

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

AB + CD = AD + BC

AB + CD = BC + DA , Hence proved.

Q3. Prove that a parallelogram circumscribing a circle ia a rhombus.                   CBSE 2002, 2008

Solution:-

Let ABCD be a parallelogram such that its sides touch a circle with centre O.

Since the tangents to a circle from an exterior point are equal in length , therefore:-

AP = AS - (Tangents from A) ….......(i)

BP = BQ - (Tangents from B) …........(ii).

CR = CQ - (Tangents from C) ….........(iii)

DR = DS - (Tangents from D) …...........(iv).

Adding (i) , (ii) , (iii) , (iv) , we get:-

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

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

AB + CD = AD + BC

2AB = 2BC ….....................(Since, ABCD is a parallelogram, therefore AB = CD and BC = AD)

AB = BC

AB = BC = CD = AD

Therefore, ABCD is a rhombus, Hence proved.

Q4. A circle is inscribed in a triangle ABC having sides 8cm , 10cm and 12cm. Find AD , BE and CF.                                                                                                                                 CBSE 2001

Solution:-

Since, the tangents drawn from an external point to a circle are equal , therefore:-

AD = AF = say , x

BD = BE = say , y
CE = CF = say , z

Now,

AB = 12cm , BC = 8cm and CA = 10cm

Therefore,

x + y = 12 , y + z = 8 and z + x = 10

(x+y)+(y+z)+(z+x) = 12 + 8 + 10

2(x+y+z) = 30

x+y+z = 15

Now,

x+y = 12 and x+y+z = 15

12 + z = 15

z = 15 – 12 = 3

Similarly,

y+z = 8 and x+y+z = 15

x+8 = 15

x = 15 – 8 = 7

Similarly,

z+x = 10 and x+y+z = 15

y+10 = 15

y = 15 – 10 = 5

Therefore, x = 7cm , y = 5cm ad z = 3cm i.e. AD = 7cm , BE = 5cm and CF = 3cm.

Q5. Prove that the tangents at the extremities of any chord make equal angles with the chord.

                                                                                                                             CBSE 2000,2001,2002

Solution:-

Let AB be a chord of a circle with centre O , and let AP and BP be the tangents at A and B.

Suppose the tangents meet at P. Join OP. Suppose OP meets AB at C .

We need to prove that PAC = PBC.

In triangles PCA and PCB , we have:-

PA = PB ….......(Since, tangents from an external points are equal)

  APC =  BPC ….........(Since, PA and PB are equally inclined to OP)

PC = PC ….......(Common)

therefore, by SAS criterion of congruence , we have:-

triangle PAC = triangle PBC

PAC = PBC , Hence proved.

                                                       More important questions will be posted soon..!!

                                                    Feel free to discuss your problems related any question with us..!!

.

YOU ASK WE REPLY!

This post is dedicated to solutions of questions asked to us by our valuable members. We request our members that they can ask their questions on respective posts and find their answers here. In case you want to ask more on solution provided by us, please comment on the original post where you asked the question. Thank You!

Question 1:  Area of a sector of a circle is 1/6 to the area of circle. Find the degree measure of its minor arc.
Asked by: mansi on 10 January 2012 07:54
Solution: 

Since, Area of sector = 1/6 of Area of circle

Therefore, r²/ 360 = 1/6 * r²

/ 360 = 1/6

6/ 360 = 1/ 

1/ 60 = 1/ 

 = 60

So, the degree measure of its minor arc = 60  

Question 2: If the diameter of a semi circle protractor is 14 cm. Find its perimeter.
Asked by: mansi on 10 January 2012 07:54
Solution: 
 Diameter of semicircle protractor = 14cm

radius = 14/ 2 = 7cm

Perimeter of semicircle protractor = r + 2r

Area = 22/ 7 * 7 + 2(7)

Area= 22 + 14

Area = 36cm

Therefore, perimeter of semicircle protractor = 36cm

Question3:If the pth term of an A.P. us q, and the qth term is p, show that rth term is p + q - r

Asked by Raman Singh on Feb 1, 2012 05:22 AM

Solution:

a+(p-1)d=q

a+(q-1)d=p

by solving these equations we will get, (p-q)d=q-p

                                                                d=q-p/p-q

                                                                d= -1

now, a=q+p-1

        a+(r-1)d=q+p-1+(r-1)-1

                     =q+p-r

Hence proved!

CHAPTER : CIRCLES

CIRCLE:A circle is a collection of all points in a plane which are at a constant distance from a fixed point.
 The fixed point is called the centre and the constant distance is known as the radius.

O is the centre and OA is the radius of the circle

SECANT: A line which intersects a circle in two distinct  points is called a secant of a circle.

Line PQ is a secant of a circle

TANGENT:  A tangent to a circle is a line that  intersects the circle in exactly one point.

 line PQ intersects a circle at a single point O, so it is tangent of a
              circle

THEORM 1: A tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given:-  A circle (O,r) and a tangent AB at point P.

To Prove:- OP┴ AB

Construction:- Take any point Q, other than P, on the tangent
                        AB. Join OQ. Suppose OQ meets the circle at R.

Proof:-  Clearly,  OP = OR
                           OQ = OR + RQ
                           OQ > OR
                           OQ > OP
                           OP < OQ

Thus, OP is the shorter distance than any other segment joining O to any point of AB.

Hence, OP ┴ AB 

Length of tangent: The length of the segment of the tangent  from the external point and the point of contact with the circle is called the  length of the tangent.

THEORM 2: The lengths of tangents drawn from an external point to a circle are equal.

Given:-  PQ and PR are two tangents from point P
to a circle (O,r)

To Prove:- AP = AQ

Construction:- Join OP, OQ and OA.

Proof:- By  joining OP, OQ and OR,
            angle OQP and ORP are right angles.

            Now, in triangle OQP and ORP,
            OQ = OR            (radii of same circle)
            OP = OP             (Common)
            OQP= ORP         (RHS)
            Therefore, PQ = PR  ( by CPCT)                 

SOME IMPORTANT POINTS TO REMEMBER (Areas related to circles)

• If two circles touch internally, then the distance between their centres is equal to the difference of their radii.
• If two circles touch externally, then the distance between their centres is equal to the sum of their radii.
• Distance moved by a rotationg wheel in one revolution is equal to the circumference of the wheel.
• The number of revolutions completed by a rotating wheel in one minute = Distance moved in one minute/ Circumference.
• A minor sector has an angle , subtended at the centre of the circle, whereas a major sector has no angle.
• The sum of the arcs of the major and minor sectors of a circle is equal to the circumference of the circle.
• The sum of the areas of major and minor sectors of a circle is equal to the area of the circle.
• The boundary of a sector consists of an arc of the circle and the two radii.

TRY OUT THESE!...

Q1. A race track is in the form of a ring whose inner circumference is 352m, and the outer circumference is 396m. Find the width of the track.

Q2. Two circles touch externally. The sum of their areas is 130 sq.cm. and the distance between their centres is 14cm. Find the radii of the circles.

Q3. A copper wire, when bent in the form of a square, encloses an area of 4484cm². If the same wire is bent in the form of a circle, find the area enclosed by it.

Q4. An arc of a circle is of length 5cm and the sector it bounds has an area of 20cm². Find the radius of the circle.

Q5. A paper is in the form of rectangle with sides 20cm and 14cm. A semicircular portion is cut off with with one one side of rectangle as diameter. Find the area of remaining part.

Q6. The minute hand of a clock is 10cm long. Find the area of the face of the clock described by the minute hand between 9 A.M. And 9:35 A.M.

Try these Questions and share your solutions with us!

You are free to ask in case any problem occurs........

Some Important Questions..!!!

Q1. The area of an equilateral triangle is 49 3 cm^2.. Taking each vertex of a centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded portion.                                                                                                                        (CBSE 2009)

                                                                                                                         

Area of ABC = 49√3 cm²

√3 / 4 a2 = 49√3

a² = 49 * 4 = 14cm

radius of each circle = 7cm

Required area = Area of ABC – 3 * ( area of sector )

{ 49√3 – 3( 60/ 360 * 22/ 7 * 7²)}cm²

(49√3 – 77)cm²

(49 * 1.73 – 77)cm²

7.77cm²

Q2:AB and CD are two diameters of a circle intersecting each other at centre O and OD is the diameter of the smaller circle. If OA= 7cm, find the area of the shaded region.       (CBSE 2010)

Area of shaded region = (Area of circle with OD as diameter) + 

Area of semicircle with AB as diameter – Area of ABC

∏ * (7/ 2)2 + 1/ 2 * ∏ * (7)² – 1/ 2 * AB * OC

{ ∏/ 4 * 49 + ∏/ 2 * 49 – 1/ 2 * 14 * 7}cm²

( 3∏ / 4 * 49 – 49)cm²

(3/ 4 * 22/ 7 * 49 – 49)cm²

231- 98 / 2 cm²

66.5cm²

Q3:Find the area of the shaded region if PQ= 24cm, PR= 7cm and O is the centre of the circle.

                                                                                                                                           (CBSE 2009)

As, RPQ is angle in a semicircle

therefore, it is a right angle.

Using Pythagoras Theorm:-

RQ² = RP² + PQ²

RQ² = 7² + 24²

RQ² = 625

RQ= 25cm

Radius of circle = 1/ 2 RQ = 25/ 2cm

Area of shaded region = Area of semicircle – Area of RPQ

= 1/ 2 ∏r² – 1/ 2 * PR * PQ

= {1/ 2 * 22/ 7 * (25/ 2)² – 1/ 2 * 7 * 24}cm²

= {6875/ 28 – 84}cm²

= 4523/ 28cm² 

Q4: In the adjoining figure, ABC is a right angled triangle at A. Find the area of the shaded region if AB= 6cm, BC= 10cm and O is the centre of circle.                                                 (CBSE 2009)

Using Pythagoras Theorm in ABC:-

BC² = AB² + AC²

AC² = BC² – AB²

AC² = 100 – 36 = 64

AC= 8cm

Therefore, Area of triangle ABC = 1/ 2 * AB * AC

= 1/ 2 * 6 * 8 = 24cm²

Area of ABC = area of OBC + area of OCA + area of OAB

24 = 1/ 2 (BC* r) + 1/ 2 (CA * r) + 1/ 2 (AB * r)

24 = 1/ 2 r ( BC + CA + AB)

24 = 1/ 2 r (10 + 8 +6)

24 = 12r

r = 2cm

Area of shaded region = Area of ABC – Area of circle

= 24 - ∏r²

= (24 – 22/ 7 * 4)cm²

= 80/ 7cm²

                                          More important questions will be posted soon..!!!

Areas related to Circles..!

Circumference of a circle Or Perimeter of a circle:

The distance around the circle or the length of a circle is called its circumference or perimeter.

Circumference (perimeter) of a circle = пd or 2пr, where r is the radius of the circle and п =

22/7 or 3.14

        FORMULAS

                                                                                                                                                                                                   
SECTOR:The part of a circle enclosed by two radii of a circle and their intercepted arc. A pie-shaped part of a circle.

Minor Sector: A sector of a circle is called a minor sector if the minor arc of the circle is a part of its boundary.

Major Sector:A sector of a circle is called a major sector if the major arc of the circle is a part of its boundary.

SEGMENT: Region enclosed by an arc and a chord is called the segment of the circle.

 

Minor Segment: If the boundary of a segment is a minor arc of a circle, then the corresponding segment is called a minor segment.

Major Segment: A segment corresponding to a major arc of a circle is known as the major segment.

You can ask your problems related this chapter by commenting below..!!! 

                                                              Important questions related this chapter will be posted soon!

             

WELCOME...!!!

Dear Students! We welcome you to our new blog i.e. QUADDo Maths. As you all know our board exams are approaching very soon. We are having our mathematics exam on 2nd March,2012. So, friends its STUDY TIME..!! The main motive of our blog is to solve the problems, clear the doubts which we are facing during the preparation of our exam. We will try to post the most expected questions in board exam, will solve sample papers, test papers etc..
We will start posting from tomorrow.
Hope you will enjoy being part of our blog..!!!