find the sum of the two middlemost terms of the AP: -4/3, -1 , -2/3, .... , 4 1/3.

about 3 years ago 0 Answer 420 views

A sum of Rs 2700 is to be given in the form of 63 prizes. If the prize is of either Rs 100 or Rs 25, find the number of prizes of each type. Can anyone help me please that how to do it?

about 3 years ago 1 Answer 1736 views

there are 20 terms in a an as .sum of the first and last term is  88.

a)if the 10 th term is 42,what is the 11 th term?

about 2 years ago 2 Answer 332 views

As we made triangles using dots, assume that squares are made using dots. Find the number of dots needed for each square.Write it as a sequence.

about 1 year ago 1 Answer 256 views

In an arithmetic sequence the sumof 1stnine terms 279and the sum of the 1st twenty terms is 1280.then (a)question. What is the fifth term of the sequence. (b)questions.what is the sixtenth term of the sequence. (c)question.write the sequence

about 1 year ago 1 Answer 249 views

Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

about 1 year ago 0 Answer 120 views

Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

about 1 year ago 3 Answer 149 views

Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

about 1 year ago 0 Answer 175 views

Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

about 1 year ago 1 Answer 200 views

SUM OF ALL ANGLES OF A TRIANGLE GIVES 180. PROVE

about 1 year ago 1 Answer 142 views

prove that in any parallelogram the sum the squares of all sides is equal to the sum of the squares of the daiganol

about 5 months ago 6 Answer 89 views

Common difference of an arithmetic sequence is 8 and its one term is 45 can the sum of any 15 terms of this sequence be 2018?

about 5 months ago 0 Answer 197 views

Common difference of an arithmetic sequence is 8 and its one term is 45 can the sum of any 15 terms of this sequence be 2018?

about 5 months ago 1 Answer 132 views

Write the sum of 0.overline{3} and 0.overline{4}.

0.overline{3}+0.overline{4}=left ( 0.333.... right )+left ( 0.444.... right )

                   = 0.777...

               x  = 0.777

           10x  =  7.777

    10 x – x  = ( 7.777.... ) – ( 0.777....)

           9 x  =  7

               x = frac{7}{9}

A student amit of class IX is unable to write in his examination, due to fracture in his arm. Akhil a student of a class VI writes for him. The sum of their ages is  25 years.

(i) Write a linear equation for the above situation and represent it graphically. 

(ii) Find the age of of Akhil from the graph, when age of Amit is 14 years.

Let  Age of Amit = x years

Age of Akhil =  Y years 

(i) According to the question the linear equation  for above situation is 

Rightarrow x+ y = 25

y = 25-x

X 0 10 15
Y 25 15 10

 

(ii) From the graph when Amit's age = 14  years, then Akhil's age = 11 years. 

 

Show that 5sqrt{6} is an irrational number.

 

Let  5sqrt{6}  be a rational number, which can be put in the form frac{a}{b} , where   b neq 0 , a and b   are co-prime

5sqrt{6}=frac{a}{b}

sqrt{6}=frac{a}{5b}

Rightarrow sqrt{6}= rational 

But, we know that sqrt{6} is an irrational number.

Thus, our assumption is wrong. 

Hence, 5sqrt{6} is an irrational number.

 

Prove that sqrt{2} is an irrational number  

Let sqrt{2} be a rational number

therefore sqrt{2}=frac{p}{q}

Where p and q are co-prime integers and  q neq 0

Rightarrow 2 =frac{p^{2}}{q^{2}}

Rightarrow p^{2} = 2q^{2}

Rightarrow P^{2} is divisible by 2.

therefore  P is divisible by 2. 

Let p = 2r for some positive integer r

Rightarrow p^{2} = 4r^{2}

therefore 2q^{2} = 4r^{2}

Rightarrow q^{2} = 2r^{2}

Rightarrow q^{2} is divisible by 2.

therefore q is divisible by 2.

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes. 

Hence, our assumption is false.

So, sqrt{2} is irrational

 

 

 

Prove that sqrt{3}  is an irrational number.

Let sqrt{3}  be a rational number

sqrt{3}= frac{a}{b}

(a and b are integers and co-primes )

On squaring both the sides, 

3 = frac{a^{2}}{b^{2}}

Rightarrow 3b^{2}= a^{2}

Rightarrow a^{2} is divisible by 3

therefore a  is divisible by 3

We can write  a = 3c for some integer

Rightarrow a^{2}=9c^{2}

Rightarrow 3b^{2}=9c^{2}

Rightarrow b^{2}=3c^{2}

b^{2} is divisible by 3

b is divisible by 3.

From equations (i) ad (ii),  we get 3 as a factor of "a"  and  "b" 

Which is contradicting the fact that a and b are co-primes. Hence our assumption that  sqrt{3} is a rational number, is false. So  sqrt{3} is an irrational number. 

 


 

 

if p is a prime number, then prove that sqrt{p} is irrational.

Let p be a prime number and if possible, let sqrt{p} be rational.

Let  sqrt{p}= frac{m}{n}  where m and n are  integers having no

common factor other than 1 and nneq 0

Then,  sqrt{p}= frac{m}{n} 

Squaring on both sides, we get 

frac{(sqrt{p})^{2}}{1}= left ( frac{m}{n} right )^{2}

Rightarrow frac{p}{1} = frac{m^{2}}{n^{2}}

Rightarrow pn^{2} = m^{2}

therefore P divides  m2 and p  divides m . [ because p divides pn2 ]

[because P is prime and p divides m2 Rightarrow p divides m]

Let              m = pq  for some integer q

 on putting  m = pq  [in eq.(i) we get ]

                pn2 = p2q2

                 n2 = pq2

 p divides n2

and p divides n.

[ because p is prime and p divides nRightarrow p divides n ]

Thus p is a common factor of m and n but this contraficts  the fact that m and n have no common factor other than 1.

The contradiction arises by assuming that sqrt{p} is a rational .

Hence, sqrt{p} is irrational.

 

 

 

 

Prove that  sqrt{3}  is an irrational number. Hence show that 7+2sqrt{3}  is also an irrational number 

If possible let sqrt{3}  be a rational number.

(i)   therefore frac{a}{b} = sqrt{3}, where a  and b are integers and co -primes

squaring both sides, we have 

       frac{ a^{2}}{b^{2}} = 3

Rightarrow a^{2} = 3b^{2}

Rightarrow a^{2} is divisible by 3

therefore a is divisible by 3    .............................(1)

We can write a = 3c for some c (integer)

(3c)2 = 3b2

Rightarrow 9c^{2} = 3b^{2}

Rightarrow b^{2} = 3c^{2}

Rightarrow b^{2} is divisible by  3.

 Rightarrow b is divisible by 3   ...........................(2)

From eq (i) and (ii) we have,

3 is a factor a and b which is contradicting the fact that 'a' and 'b' are co-prime.

Thus our assumption that  sqrt{3}  is rational numbers is wrong.

Hence, sqrt{3} is an irrational number.

(ii) Let us assume to the contrary that 7+2sqrt{3}  is a rational number.

7+2sqrt{3}=frac{p}{q},: q: neq 0: and : p: ,: qepsilon : Integer

7+2sqrt{3}=frac{p}{q}

2sqrt{3}=frac{p}{q}-7

2sqrt{3}=frac{p-7q}{q}

sqrt{3}=frac{p-7q}{2q}

p- 7q and  2q both are integers hence sqrt{3}  is a rational number. 

But this contradicts the fact that is sqrt{3} is  irrational number. Hence is 7+2sqrt{3} is an irrational number .

 

If the zeroes of the polynomial x^{2}+px+q   are double in value to the zeroes of  2x^{2} -5x -3 , find the value

of  p and q. 

Let, f (x) = 2x^{2}-5x -3

Let the zeroes of polynomial be alpha  and beta then,

Sum of zeroes   =alpha+ beta = frac{5}{2}

product: : of: : zeros=alpha beta = -frac{3}{2}

According to the question, zeroes of  x^{2}+px+q are 2 alpha and 2 beta

Sum of zeroes  = frac{coeff. of x}{coeff.of x ^{2}} = frac{-p}{1}

                  -p=2alpha +2beta = 2 (alpha +beta )

                 -p=2 times frac{5}{2} = 5 Rightarrow p=-5

Product of zeroes  =frac{constant: : term}{coeff.: of : x ^{2}} = frac{q}{1} 

                                

                      Rightarrow q = 2alpha +2beta = 4alpha beta

                      Rightarrow q = 4 left ( -frac{3}{2} right ) = -6

                          p = -5  and q = -6

 

 

If  alpha   and  beta are zeroes of the polynomial  p(x)=3x^{2}-4x-7 then form a quadratic polynomial whose zeroes are  frac{1}{alpha }  and  frac{1}{ beta }

Given, p (x) = 3x2 -4x-7 and alpha and beta  are its zeroes. 

    sum: : of: zeroes=alpha +beta = frac{coefficient: : of: : x}{coefficient: : of: : x^{2}} 

                                          = -left ( -frac{4}{3} right )= frac{4}{3}

  product: : of: : zeroes= alpha beta = frac{Constant: : term}{ Coefficient of x^{2}}

                                         = left ( frac{-7}{3} right ) = -frac{7}{3}

For the new polynomial 

sum: : of: : zeroes: : =frac{1}{alpha }+frac{1}{beta }=frac{alpha +beta }{alpha beta }=frac{frac{4}{3}}{-frac{7}{3}}=frac{-4}{7}

product: : of: : zeros= frac{1}{alpha }times frac{1}{ beta }=frac{1}{alpha beta }=frac{1}{frac{-7}{3}}=frac{-3}{7}

Required quadratic polynomial = x2 - ( Sum of zeroes ) x + Product of zeroes 

                                                                   = x^{2}- left ( frac{-4}{7} right ) x +left ( frac{-3}{7} right )

                                                                  = frac{1}{7} (7x^{2}+4x-3)

                                                                 =( 7x^{2}+4x -3) frac{1}{7}.

 

 

If alpha  and  beta  are the zeros of the polynomial  x^{2}+4x+3,  find the polynomial whose zeroes  are   1+frac{beta }{ alpha }   and  1+frac{ alpha }{ beta }

 Since alpha and beta are the zeroes of the cubic polynomial  x^{2}+4x+3

then  alpha +beta = -4

and  alpha beta = 3

sum: : of: : zeroes: : =1+frac{beta }{alpha }+1+frac{alpha }{beta }

                          =frac{alpha beta +beta ^{2}+alpha beta +alpha ^{2}}{alpha beta }

                         =frac{alpha^{2}+ beta ^{2}+2alpha beta }{alpha beta }

                         =frac{(alpha+ beta )^{2} }{alpha beta }=frac{(-4)^{2}}{3}=frac{16}{3}

product: : of: : zeroes= left ( 1+frac{beta }{alpha } right ) left ( 1+frac{alpha }{beta } right )

                             = 1+frac{beta }{alpha }+frac{alpha }{beta }+frac{alpha beta }{alpha beta }

                             =frac{alpha ^{2}+beta ^{2}+2alpha beta }{alpha beta } = frac{(alpha +beta )^{2}}{alpha beta }

                             = frac{(-4)^{2}}{3}=frac{16}{3}

But required polynomial = x2 - (Sum of the zeroes)x + product of the zeroes 

                                     = x^{2}-left ( frac{16}{3} right )x+frac{16}{3}

                                    = left (x^{2} -frac{16}{3} x+frac{16}{3} right )

                                   = (3x^{2}-16x+16)frac{1}{3}

The minimum age of children to be eligible to participate in a painting competition is  8 years it is observed that the age of youngest boy was 8 years and the ages of rest of participant are having a common difference of  4 months. If the sum of ages of eldest participant in the painting competition.  

a = 8, d = 1/3 years, Sn = 168

 S_{n}= frac{n}{2} [2a+(n-1)d]

168 = frac{n}{2}left [ 2 (8)+(n-1)frac{1}{3} right ]

              n^{2}+47n-1008 = 0

Rightarrow n^{2}+63n-16n -1008 = 0

         Rightarrow (n-16) (n+63)=0

                 Rightarrow n = 16 or n = -63

                                     n = 16

Age of oldest participant  = a + 15 d  = 13 years

The digits of a positive number of three digits are in AP and their sum is  15. The number obtained by reserving the digits is  594 less than the original number find the number.  

Let the three digits be  a -d, a , a+d

Sum  = a-d+a+a+d = 3a=15 given 

therefore the  three digits are  5 -d, 5,5+d.

Original number = 100 (5-d) +10 x 5 + 1(5+d)

                          = 555+99d

Revered number = 100 (5+d)+10 x 5+1 (5-d)

                          = 555+99 d

According to question , 

(555-99d) - (555+99d)  = 594

                         -198 d = 594

                          Rightarrow  d = frac{594}{-198} = -3

The three digits are 

5-(-3),5,5+(-3)

8,5 and 2

therefore Original number is  8 x 100 + 5x 10 +2 x 1 = 852

Find the mean of the following distribution :

class interval            0-6     6-12     12-18     18-24     24-30

frequency                 5         4            1           6          4

 

   xi                                 fi                                 xifi

 

   3                                 5                                  15

   9                                 4                                  36

 15                                 1                                  15

 21                                 6                                126

 27                                 4                                108

Total              sum fi: : = 20          sum xi fi: : = 300

 

Mean = frac{sum xi fi }{sum fi}: : : : = frac{300}{20} = 15

The mean of the following frequency distribution is  25. Find the value of p.

Class interval                  0-10              10-20               20-30              30-40            40-50

Frequency                        5                   6                    10                     6                  p
 

 

      Class - Interval                       Midxi                    fi                    fixi

            0-10                                   5                       4                    20

           10-20                                15                       6                     90

           20-30                                25                     10                   250

           30-40                                35                       6                   210

           40-50                                45                       p                   45p

                                                                        26+p               570 + 45p

 

x = frac{sum fixi}{sum fi}

25=frac{570+45p}{26+p}

650 +25 p = 570 +45 p

650 -570= 45p -25p 

 p = 4

 

 

In the given figure, AB is a diameter of the circle with centre O. if AC and BD are perpendicular on  a line PQ and BD meets the circle at E, then prove that AC = ED.

 

Proof: angle AEB = 90^{0}

                      = angle AED        (semi-circle)

angle EAC +angle ACD +angle CDE +angle AED = 360^{0}     (sum of angles of a quad.)

                  angle EAC +90^{0}+90^{0}+90^{0}= 360^{0}

                                                                angle EAC = 360^{0}-270^{0}                                           

                                             angle EAC = 90^{0}

                                                    each angle   =90^{0}

          therefore EACD is a rectangle     AC = ED.

Find the mean of the following distribution :

Height (In cm)              Less than 75            Less than 100            Less than 125           less than 150       Less than 175         

No. of students                    5                              11                           14                             18                       21                          

 

Height (in cm )             Less than  200         Less than 225          Less than 250             Less than 275        Less than  300 

No. of students                   28                           33                             37                               45                       50


 

By short-cut Method  (any method )  For making correct table  A = 187.5

sum fiui = -12 ; N = 50, h = 25

Mean  = A + frac{sum fiui }{N} times h

Rightarrow  Mean  = 187.5 + = 187.5 + frac{-12}{50} times 25 = 187.5 -6 = 181. 5

Alternative method  :

By short cut method  (any method)  A = 187.5

Class- interval Height (in cm )             Frequency                   xi                ui = frac{xi-a}{h}                fiui

              50-75                                       5                         62.5                           -5                      -25

            75-100                                       6                         87.5                           -4                      -24

          100-125                                       3                        112.5                          -3                        -9

          125-150                                       4                        137.5                          -2                        -8

         150-175                                        3                        162.5                          -1                        -3

         175-200                                        7                        187.5                           0                         0

        200-225                                         5                        212.5                           1                         5

        225-250                                        4                         237.5                           2                         8

        250-275                                        8                         262.5                           3                       24

       275-300                                         5                         287.5                           4                       20


                                              sum fi = 50                                                                  sum fiyi = -12


 

sum fiui = -12; N = 50, h = 25

Mean  A + frac{fiyi}{N} times h

Mean  = 187.5 + frac{-12}{50}times 25= 187.5 -6 = 181 .5

If the mean of the following is  14.7, find the value of p and q.

Class                  0-6      6-12      12-18      18-24      24-30      30-36      36-42      Total

Frequency           10         p           4             7            q            4             1          40

 

 

     xi                                        fi                                               xifi

     3                                        10                                              30

     9                                        p                                               9p

   15                                       4                                                60

   21                                       7                                               147

   27                                       q                                               27q

  33                                        4                                              132

 39                                         1                                               39

Total               sum fi = 26 + p+q                    sum xifi = 400+9p +27q

 

sum fi = 40

26 + p +q = 40 

Rightarrow p+q = 14

therefore mean, x = frac{sum cifi}{sum fi}

Rightarrow 14.7=frac{408+9p+27q}{40}

Rightarrow 588 = 408 +9p +27q

Rightarrow 180=9p+27q

Rightarrow p +3q = 20

Substracting eq (i)  and (ii),

Rightarrow 2q = 6

Putting this value  q in eq (i), 

P  = 14-q = 14 -3 = 11

p = 11, q = 3.

 

Find the mean and mode of the followin frequency distribution :

classes                  0-10           10-20          20-30           30-40           40-50           50-60           60-70

Frequency                3                8                10                15                7                 4                3

     Class interval                             xi                           fi                                  xifi


         0-10                                      5                            3                                  15

        10-20                                   15                            8                                 120

        20-30                                   25                          10                                 250

        30-40                                   35                          15                                 525

       40-50                                    45                            7                                 315

       50-60                                    55                           4                                  220

      60-70                                     65                           3                                  195
 
                                                                 sum fi = 50                   sum fi xi = 1640

 

Mean  = frac{sum fi xi}{sum fi} =frac{ 1640}{50}=32.8

Modal class  = 30-40

l = 30, f_{1} = 15,f_{2}=7, f_{0}= 10, h = 10

Mode = l +frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}times h

= 30 +frac{15-10}{30-10-7} times 10

= 30 +frac{5}{13} times 10

=30+ frac{50}{13}

=30+3.85

=33.85

ABC is a right-angled triangle in which angle A = 90^{0} and AB = AC. Find angle B, , and, , angle C 

 

We have,    angle A = 90^{0}

                 AB = AC

                 angle B = angle C

( because, , Angles, , opp., , to, , equal , , sides, , of , , a , , triangle , , are , , equal )

Also,, , angle A +angle B+angle C = 180^{0} , , , , , (Angle, , sum, , property)

                 90^{0} +2angle B = 180^{0}

                          2angle B = 180^{0}- 90^{0}

                                  = 90^{0}         (because , , angle C= angle B, )

                            angle B = frac{90^{0}}{2} = 45^{0}

                        therefore , , angle C=angle B = 45^{0}

In the figure AD = BD. Prove that BD

                                  AD = BD

   angle ABD =angle DAB = 59^{0}                         (Angles opposite to equal sides are equal)

Delta ABD

59^{0}+ 59^{0}+angle ADB =180^{0}

                   angle ADB = 180^{0}-118^{0} = 62^{0}

                   angle ACD = 62^{0}-32^{0} =30^{0}     (Exterior angleis equal to the sum of interior opposite angles)

     In   Delta ABD    AB >BD                          ( Side opposite to greatest angle is the longest )

Also in  Delta ABC  AB < AC

                         BD < AC.

Construct a triangle whose angles are in the ratio 1:3:5 and length of sides included by first and last angles is 6 cm.

Let angles x, 3x, and 5x

    therefore x+3x+5x = 180^{0}           (angles of sum property)

                     9x = 180^{0}

                       x = 20^{0}

    therefore First angle  =  20^{0}

   Second angle  =  600

    Third angle    = 1000

Steps of construction:

(i)  Draw a line segment BC  = 6 cm 

(ii)  At point A draw angle XAB = 20^{0}  and at point B, draw angle YAB = 100^{0}

(iii) XA and YB intersect at point C. 

       Thus, ABC is the required triangle.

 

Find the mean overline{x} of first ten prime numbers and hence show that 

sum_{i=1}^{10} (x_{1}-overline{x})=0

First ten prime numbers are   2,3,5,7,11,13,17,19,23,29.

Mean = frac{2+3+5+7+11+13+17+19+23+29}{10}

          =frac{129}{10}= 12.9

Now,   sum_{i=1}^{10} (x_{i}-overline{x})= (2-12.9)+(3-12.9)+(5-12.9)+(7-12.9)+(11-12.9)                                                                                                                            + (13+12.9)+(17-12.9)+(19-12.9)+(23-12.9)+(29-12.9)

                            = -10.9-9.9-7.9-5.9-1.9+0.1+4.1+6.1+10.1+16.1

                           = 36.5+36.5 =0

        sum_{i=1}^{10} (x_{i}-overline{x})=0

From the following frequency distribution, find the median class : 

Cost of living index          1400 - 1550      1550-1700      1700-1850     1850 - 2000

Number of weeks                      8                      15                     21                   8

 

C.I          1400 - 1550     1550-1700     1700 - 1850       1850 - 2000

f                      8                      15                   21                       8

c.f                   8                      23                   44                        52

frac{sum f}{2} = 26 Rightarrow    Median class = 1700- 1850 

In the following frequency distribution, find the median class.

Height (in cm)      140 - 145      145 -150      150 - 155     155 - 160         160 - 165       165 - 170

Frequency                   5                   15                  25                 30                    15                 10

Height                   Frequency            c.f

140-145                       5                   5

145-150                      15                 20

150-155                      25                 45

155- 160                     30                 75

160-165                      15                 90

165-170                      10                 100

                                     sum f = 

 N = 100

frac{N}{2} = frac{100}{2} = 50

Hence median class is 155 - 160

Find the sum of the lower limit of the median class and the upper limit of the modal class:

Classes       10 -2 0        20 - 30       30 - 4 0      40-50        50- 60         60-70

Frequency      1                   3                 5               9                7                3

Classes                        10 -2 0        20 - 30       30 - 4 0      40-50        50- 60         60-70

Frequency                        1                   3                 5               9                7                3

Cumlative Frequency       1                   4                 9                18             25             28

Median Class : 40- 50    Lower limit  = 40

Modal Class :  40-50       Upper limit  = 50

                                    Their limit   = 40 + 50 = 90

(i) Do Euclid fifth postulates imply the existence of parallel lines? Explain 

(ii) Which mathematical concept is used in this problem 

(iii) What is its value 

(i) If a  staright line l falls on two straight lines m and n such that the sum of the interior angles on one side of l is equals to two right angles then by Euclid's fifth postulates the lines will not meet on this side of l. Next we know that the sum of the interior angles on the other side of line l will also be two right angles. Therefore, they will not meet on the other side also. so the lines m and n never meet  and are therefore, parallel.

(ii) Introduction to euclids Geometry 

(iii) Universal truth.

In  a parallelogram PQRS of the given figure the bisectors of

Given : A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.

To prove :   angle POQ = 90^{0}

Now, since PQRS is a parallelogram. Therefore PS II QR

Now, PS II QR and transversal PQ intersects them. 

therefore angle P +angle Q = 180^{0}

(because sum of consecutive interior angles is 1800)

   therefore frac{1}{2} angle P+frac{1}{2} angle Q=90^{0}

angle 1 + angle 2 = 90^{0}  (becauseOP is  bisector of

therefore angle 1=frac{1}{2}angle P and angle 2=frac{1}{2}angle Q

Now, in  Delta POQ

angle 1 + angle POQ + angle 2 = 180^{0}rightarrow 90^{0}+angle POQ = 180^{0}

rightarrow angle POQ = 90^{0}

Construct a right triangle with base  4.5 cm and the perimeter as 11.7 cm 

Given: In right Delta ABC , base BC = 4.5 cm and perimeter  = 11.7 cm 

i.e.,  AB + BC = 11.7 cm or AB  + AC = 7.2 cm Delta ABC = 90^{0}

Required : TO construct the Delta ABC with angle ABC = 90^{0}  base BC = 4.5 cm 

 and sum of other two sides as  7.2 cm 

Steps of construction:

(i)  Draw  BC  = 4.5 cm 

(ii) Draw BY such that  < CBY  = 900

(iii)  From BY cut  off BD = 7.2 cm 

(iv) Join DC.

(v) Draw the perpendicular bisector of DC intersecting BD at A.

(vi) Join AC then, Delta ABC is the required triangle.

 Vikas has to cut out of a diameter 11/4 cm from a rectangular aluminum strip of imenstion 8 3/4  and 1 1/4. How many full circles cikas cut? Also find out the wastage of aluminium strip?

                                                       OR

Deepak, Rahul and Sonu received a total of Rs. 2,016/-  as a monthly allowance from their mother such that Sonu gets 1/2  of what Deepak gets and Rahul gets  1 2/3 times Sonu's share.

How much money do the three brothers get individually?

 

 

Area of rectangular strip =  

= 8 frac{3}{4} times 1 frac{1}{4 } 

= frac{35}{4} times frac{5}{4} = frac{175}{4}

Diameter = = 1 frac{1}{4} = frac{5}{4} rightarrow radius  frac{5}{8}

Area of the circle cut  = pi r^{2}

= frac{22}{7} times frac{5}{8} times frac{5}{8} = frac{550}{448}

No. of circle cuts  = = frac{17}{4} times frac{440}{550} = frac{112}{22}

=7 times frac{112}{22} = frac{7times 56}{11}

=7 times frac{392}{11} = 35frac{7}{11}

Hence total no: of ...... circle is 35

To strip & crap = = frac{175}{4} - 35 times frac{550}{440} = frac{275}{224}

= 43.75 - 42.70

= 1.05 cm (approx)         

(b) Let as assume deepaks share is x Rs.

Sonu's share  = frac{x}{2}  

Rahul's  = 1frac{2}{3} times frac{x}{2} = frac{5x}{6}

Therefore  x + frac{x}{2} +frac{5x}{6} = 2016

frac{6x+3x+5x}{6} = 2016 Rightarrow frac{14x}{6} = 2016

Rightarrow x = frac{2016 times 6}{14} = 264

Deepak share  Rightarrow , , : : : : x = 264

Sanus share  frac{x}{2} = frac{264}{2} = 432

Rahul's share frac{5x}{6} = frac{5times 264}{6} =720

 

The algebraic form for the sum of first n terms of an arithmetic sequence is 2n2 + 8n. How many consecutive terms of the sequence, starting from the first, are to be added to get  330?

 

Given, the sum of the first n terms of an arithmetic sequence = 2n2 + 8n

ie., 2n2 + 8n = 330  Rightarrow 2n^{2} + 8n - 330 = 0  dividing by 2

                                Rightarrow n^{2} + 4n -160= 0

                                Rightarrow (n + 15) (n , , , 11) = 0

ie., n + 15 = 0 or n - 11 = 0

     n = -15  or n = 11 ; -15 rejected

     n = 11

ie., The sum of the first n number be 11.

There are 20 terms in an arithmetic sequence. Sum of the first and last terms is 88.

a. What is the sum of the 2nd and 19th terms? 

b. If the 10th term is 42, What is the 11th term? 

c. What is the common difference of the sequence?

d. What is the first term?

Given n =20

Sum of the first and last term  = 88

a. The sum of the 2nd and 19th terms. 

ie., x2  + x19 = x1 + d + x20 - d

                     = x1 + x20

                     = 88.

b. Given 10th term 42.

x10 + x11 = (x1 + 9d) + (x20-9d)

ie., 42 + x11 = x1 + x20

therefore x11 = x1 + x20 - 42

         = 88 - 42 = 46.

Hence 11th term = 46

c. Commom difference (d) = x11 - x10 = 46 - 42 = 4

d. The first term

x1 = x1 +9d = 42 (given 10th term 42)

     = 42 - 9d

     = 12 - 9  times 4 = 6

Hence the first term  = 6

The table below shows the members in '' Stree-sakthi Kudambasree'' sorted according to their ages.

Age group                  Number of members

20-30                                      4

30-40                                      8

40-50                                      10

50-60                                       7

60-70                                      4

70-80                                      2

Total                                       35       

a. if the members are arranged in increasing order of ages, the age of the member at what position is taken as the median?

b. What is the assumed to be an age of the member at the 13th position?

c. Find the median of the ages.

Class Frequency (f) Cumulative frequency (cf)
20-30 4 4
30-40 8 12
40-50 10 22(N/2)
50-60 7 29
60-70 4 33
70-80 2 35
TOtal 35  

a. frac{35+1}{2}= 18

b. 13th position = 40 + 1/2 = 40.5

c. Median 

l_{1} + frac{frac{N}{2}-C}{f} times h,, , , , , , , , , , l_{1} = Lower limite of the median class, 

N= Total number of frequencies,    C = Above the cf of the median class, 

f = Frequency of the median class, h = class interval

N/2 = 35/2 = 17.5. Median class = 40 - 50    (N/2 = 17.5 included class)

 l_{1} = 40. , , , , c = 12, f = 10

Hence, median = l_{1} + frac{frac{N}{2}-c}{f} times h = 40+ frac{17.5-12}{10} times 10

                        = 40 + 5.5 = 45.5.

Consider the numbers between 100 and 300 which leave remainder 2 on division by 3.

a. Which is the first number in this sequence?

b. Which is the last numbers in this sequence?

c. How many such numbers are there in this sequence?

d. Find the sum of all numbers in the sequence?

Numbers b/w 100 and 300 which leave remainder 2 on division by 3

First term = 99 + 2 = 101.    d= 3

So, the sequence be 101, 104, 107, 110..........................299.

Last term = 300 - 1 = 299

a. The first term = 101

b. Last term  = 299.

c. n = frac{X_{n} - X_{1}}{d} + 1 = frac{299 - 101}{3} + 1 = 67

d): : sum: =frac{n}{2}left ( X_{1} +X_{n}right )=frac{67}{2}left ( 101+299 right )

                 =frac{67}{2}times 400=13400

A rope of length 40 meters is cut into two pieces and two squares are made on the floor with them. The sum of the areas enclosed is 58 square meter.

a. If the length of one piece is taken as x, what is the length of the other piece?

b. What are the lengths of the sides of the squares?

c. Write the given fact about the area as an algebraic equation.

d. What is the length of each piece?

Length of the rope = 40 cm 

Let the length of the one piece = x

a. Length of the other piece = 40 - x

b. Length of the side of the square  =  frac{x}{4} and frac{40-x}{4}

c. The algebraic form, Given, sum of the area = 58 cm2

ie., a2 + a2 =58 cm2.

Rightarrow left ( frac{x}{4} right )^{2} + left ( frac{40-x}{4} right )^{2} = 58

Rightarrow frac{x^{2}}{16} + frac{1600-80x +x^{2}}{16} = 58

Rightarrow 2x^{2}-80x + 1600 = 58 times 16

Rightarrow 2x^{2}-80x + 1600 = 928  dividing throught out by 2,

ie., x2 - 40x + 800 = 464.

Rightarrow x^{2}-40x + 800 - 464 = 0

Rightarrow x^{2}-40x + 336 = 0

 

d)  Length of each  piec, 

x ^{2} - 40x + 336 = 0

Rightarrow (x-28 ) (x-12) = 0

Rightarrow x - 28 = 0 , , , , , , or, , , , , , , , x - 12 = 0

x = 28  or x = 12 

The length of each pieces = 28 m or 12 m 

What is the sum of the angles of a 62 side polygon

 

 

Sum of angle of 62 sided polygon  =  (n-2) 180 = (62 - 2) x 1800 

                                                  = 60 x 180

                                                  = 108000

The terms of an arithmetic sequence with common difference 4 are natural numbers. 

a. If x is a term in this sequence, what is the next term?

b. If the sum of reciprocals of two consecutive terms of this sequence is 4/15, find those terms.

OR

a. Lengths of sides of a right-angled triangle are in arithmetic sequence with common differenced. if the length of the smallest side of the triangle is x-d write the length of its other two sides.

b. Show that any right-angled triangle with sides in an arithmetic sequence is similar to the right-angled triangle with sides 3,4 and 5.

Common difference = 4

a. next term after x = x  +4

b. Let the first term be x, then next  term = x + 4

frac{1}{x} + frac{1}{x+4} = frac{4}{15}

frac{(x+4)+x}{x (x+4)} = frac{4}{15}

frac{2x+ 4}{x^{2} +4x} = frac{4}{15}

15 (2x+4) = 4 (x2 + 4x) 

30x + 70 = 4x^{2} + 16x

4x^{2} + 16x-30x -70 = 0

4x^{2} - 14x-70 = 0

x = frac{-bpm sqrt{b^{2}-4ac}}{2a}

frac{-(-4)pm sqrt{(14)^{2}-4times 4 (-70)}}{2times 4}

=frac{14pm sqrt{196+1120}}{8}

=frac{14pm sqrt{1316}}{8} = frac{14pm 36.27}{8}

x = 6.28 or x = -2.78

x = 6.28

x + 4 = 6.28 + 4 = 10.28

 

OR

 

a.

Common difference = d

Length of smallest side = x-d

Then other two sides = x - d + d = x

and  x + d

b. According to pythagoras theorem

AC = sqrt{3^{2} + 4^{2}}

= sqrt{25} = 5

Similarity consider 

PR = sqrt{x^{2}+(x-d) ^{2}}

PR = sqrt{x^{2}+x^{2} -2xd+d^{2}}

= sqrt{2x^{2}-2xd+d^{2}}

PR = x + d

(x+d)^{2} = 2x^{2} -2xd + d^{2} 

x^{2} +2x d + d^{2} = 2x^{2}-2xd + d^{2}

4xd = 2x^{2} -x^{2}

x^{2} = 4x d

x = 4 d 

When d = 1, ABC will be formed 

Hence these two triangles are similar

Consider the arithmetic sequence 9,15,21............

a. Write the algebraic form of this sequence.

b. Find the algebraic form of this sequence.

c. Find the sum of terms from twenty-fifth to fiftieth of this sequence.

d. Can the sum of some terms of this sequence be 2015? why?

a. Common difference = 6

    Let the first term be x  then the sequence = x, x+6, x + 12, x + 18.....

b.

25th term = f + (n-1)d

                =  9 + (25-1)6

                = 9 +24 x 6

                = 153.

c. 50th term = f + (n-1)d 

                    = 9 + (50-1) 6

                    = 9 + 49 x 6

                    = 303

Sum of n terms of arithmetic sequence 

                   = frac{n}{2}left ( x_{25}+x_{50} right )

                   = frac{25}{5}left ( 153 +303 right )

                   = 2280

The table on the right shows 30 workers sorted according to their daily wages.

a. What do we take as the mean dailly wages of the four workers earning wages between 300 rupees and 400 rupees?

b. According to this. What is the total daily wages of the workers in this class?

c. Making such assumptiuons. Calculate the total wages of workers in other classes also.

d. Calculate the mean daily wages for the entire group ?

 

Wages         Workers

300-400        4

400-500       8

500-600       10

600-700       6

700-800       2

Wages Workers  Mid-value Total wages
300 - 400 4 350 1400
400 - 500 8 450 3600
500 - 600 10 550 5500
600 - 700 6 650 3900
700 - 800 2 750 1500
Total 30   15900

a) Mean daily wages of 4 workers earning wages between 300 and 400 rupees = 350

b) Total wages of workers in this class = 1400

c) Total wages of workers in other classes

class           Total wages

400 - 500     3600

500 - 600     5500

600 - 700     3900

700 - 800     1500 

d) Mean daily wages =   frac{Sum, , , of , , , wages }{Number , , , , of, , , workers} = frac{15900}{30} = 530

Examine the pattern:

frac{4}{9}+frac{5}{9} = 1 ; frac{9}{4} + frac{9}{5} = frac{9}{4} times frac{9}{5}

frac{3}{8}+frac{5}{8} = 1 ; frac{8}{3} + frac{8}{5} = frac{8}{3} times frac{8}{5}

frac{3}{7}+frac{4}{7} = 1 ; frac{7}{3} + frac{7}{4} = frac{7}{3} times frac{7}{4}

(i) Write two fractions whose sum is 1. Check whether the sum and the product of the reciprocals of these fractions are      equal.

(ii) Prove that the sum of the and product of the reciprocals of two such fractions are equal.

 

                                                                     OR

 

(i) If  frac{x}{y}=frac{7}{5}, then find frac{x+y}{x-y}

(ii) The sum of the square of a number and 2 divided by the difference of 2 from the square of that number gives 99/97       Find the number?

(i)  frac{3}{5}, frac{2}{5} 

Sum of the reciprocals

               frac{5}{3}+ frac{5}{2}=frac{25}{6}

Product of the reciprocals

              =frac{5}{3} times frac{5}{2}=frac{25}{6}

They are equal

 

(ii) Let the number be frac{y}{x} and frac{x-y}{x}

     Sum of the reciprocals 

                = frac{x}{y}+frac{x}{x-y} = frac{x(x-y) + xy}{y(x-y)}

               =frac{x^{2} -x y + xy}{xy-y^{2}}= frac{x^{2}}{xy - y^{2}}

Product of the reciprocals 

              = frac{x}{y } times frac{x}{x-y}=frac{x^{2}}{xy - y^{2}}

They are equal

 

                                     OR

 

(i)  frac{x}{y } = frac{7}{5}= k (let)

x = 7k, y = 5k

frac{(x+y)}{(x-y)} =frac{(7k+5k)}{(7k-5k) } = frac{12k}{2k}=6

 

(ii)  frac{(x^{2} +2)}{(x^{2}-2)}= frac{99}{97}

99(x^{2}-2) = 97 (x^{2}+2)

99 x^{2}-198 = 97x^{2} +194

2x^{2} = 392, x^{2}=frac{292}{2}=196

x= sqrt{196}=14,Number = 14

(a) What are the numbers x which satisfy the equation left | x-2 right | + left | x-6 right | = 4 ?

(b) What are the numbers x satisfying the equation left | x-2 right | + left | x-6 right | = 5

(c) Are there numbers x satisfying the equation left | x-2 right | + left | x-6 right | = 3 Write the reason?

(a) left | x-2 right | + left | x-6 right | = 4

The distance between x and 2+ the distance between x and 6 should be 4.

Distance between 2 and 6 left | 6-2 right | = left | 4 right | = 4

So x can be anywhere between 2 and 6 including 2 and 6.

So x can be any number 2 or greater than 2 and 6 or less than 6.

 

(b) left |x -2 right | + left | x-6 right | = 5

The distance between the x and 2+ the distance between x and 6 should be 5.

Distance between 2 and 6= left |6 -2 right |= left | 4 right | = 4

For this distance to be 5, x should move 1/2 unit to the left of 2 and 1/2 unit to the right of 6.

Then the distance = 4 + frac{1}{2} + frac{1}{2} = 5

When x is moved frac{1}{2} unit to the left of 2, it is 1 , , frac{1}{2}.

When x is moved  frac{1}{2}  unit to the right of 6, it is  6 , , frac{1}{2} .

So values of x are 1 , , frac{1}{2} and 6 , , frac{1}{2}.

 

(C)

The sum of the distance between x and 2 and x and 6 is equal to or greater than the distance between 2 and 6. The distance between 2 and 6 is 4. So this distance cannot be less than 4.

therefore There are no numbers for x satisfying

left | x-2 right | + left | x-6 right | = 3

Sides of a right triangle ABC are AB = 10 cm, BC = 8 cm and AC = 6 cm. Semicircles are drawn with sides of the triangle as diameters. 

i. Find the areas of semicircles.

ii. Prove that sum of areas of two small semicircles is equal to the area of the largest semicircle.

i) Area of the semicircle with AC as diameter= frac{1}{2} pi times 3times 3 = frac{9}{2}pi sq.cm 

Area of the semicircle with BC as the diameter = frac{1}{2} pi times 4times 4 = 8pi , , sq. cm

Area of the semicircle with AB as the diameter = frac{1}{2} pi times5times 5 = frac{25}{2} pi , , , sq. cm

 

ii) Sum of the areas of the two small semicircles = frac{9}{2} pi + 8pi

                                                                      = frac{9pi +16pi }{2} = frac{25}{2} , , pi , , sq, , cm

therefore Area of the large semicircle is equal to the sum of the areas of the two small semicircles.

5^{1}times 5^{3}times 5^{5}times ...................5^{2n-1} = (25)^{72}find n.

5^{1}times 5^{3}times 5^{5}times ............5^{2n-1}=25^{72}

5^{1+3+5+.......+2n-1}= 5^{2n-1} = (5^{2})^{72}

5^{n2} = 5^{144} (Sum of the first n odd numbers = n2)

therefore n^{2} = 144, nsqrt{144} = 12

 a) What is the sum of the first 20 natural numbers. 

b) Find the sum of the first 20 terms of 4, 8, 12....

c) If 3 is added to each of term in the above sequence write down the algebraic expression of the new sequence.

c)  Find the sum of the first 20 terms of the new sequence

a) Sum of the first n natural numbers = frac{n(n+1)}{2}

 Sum of the first 20 natural numbers  =  frac{20times 21}{2} = 210

b) 4 + 8 + 12 + ................(20 terms)

    = 4 (1 + 2 + 3 + .....+20) = 4 x 210 = 840

c) Algebraic expression of the arithmetic sequence 4, 8, 12...........= 4n

    Algebraic expression of the arithmetic sequence got by adding 3 to each term of the above sequence = 4n + 3

d) 7 + 11 + 15  + ..........(20terms) + 3 x 20

     = 840 + 60 = 900

 23rd term of an arithmetic sequence is 32.35th term is 104. Then,

a) What is the common difference? 

b) Which is the middle term of the first 35 terms of this sequence?

c) Find the sum of first 35 terms of this sequence 

a) Difference between the terms = 104 - 32 = 72

Difference between the term positions = 35 - 23 = 12

72  = 12 x Common difference 

Common difference = 72div 12 -= 6

b) Middle term of the first 35 terms = frac{35+ 1}{2} = frac{36}{2} = 18^{th} term

18th term = 23rd term  - 5 common difference 

                                               = 32 - 5 x 6 = 32 - 30 = 2

c)  Sum of the first 35 terms  = 35 x middle term 

                                                 = 35 x 2 = 70

 1

2   3    4

5   6    7     8   9

a) How many numbers are there in the 30th row of this number pyramid?

b) which is the last number in the 30th row?

c) Which is the first number in the 30th row?

d) What is the sum of all terms in the first  30 rows? 

a) Sequence of the number of terms in each row = 1,3,5,7.........

Xn = 2n -1

Number of numbers in the 30th row = 2 x 30 - 1 = 60 - 1 = 59

 

b) Last number in the 1st row = 1

Last number in the 2nd row = 4 = 22

Last number in the 3rd row = 9 = 32

......................................................

......................................................

Last number in the 30th row  = 900 = 302

 

c) First number in the 30th row  = last number in the 30th row  - 58 common difference 

= 900 - 58 x 1 = 842

 

d) Sum of all terms in the first 30 rows

= 1 + 2 + 3 + 4 +........ + 900 = frac{900times 901}{2} = 405450

n, 3n,5n .........  is an arithmetic sequence. 

a) What is the common difference? 

b) Prove that the sum of first n terms of this sequence is n3.

c) Then find the sum of 15 terms of the sequence 15, 45,75.......

Arithmetic sequence = n, 3n, 5n.......

Common difference = 5n - 3n = 2n

b) Sum of the first n terms =  n + 3n + 5n + .. (n terms)

                       =  n (1 + 3 + 5 + .....+ 2n - 1) = n x n2 = n3

c) 15 + 45 + 75 + ...........(15 terms)

     = 15 (1 + 3+5+......(15 terms)

    = 15 x 152 = 3375

Sum of the first five terms of an arithmetic sequence is 45. What is the third term? The common difference of the sequence is 4, write the first two terms. write another arithmetic sequence having the sum of the first five-term is 45.

Sum of the first terms = 45

5 x 3rd term = 45, 3rd term = 45 div 5 = 9

If the common difference is 4, the second term = 9 - 4 = 5

The first  term = 5 - 4 = 1, First two terms = 1, 5

If the sum of 5 terms is 45, the third term is 9. another arithmetic sequence  = 7,8,9,10,11

a) Find the least and highest three-digit number which leave a remainder 1 on division by 9.

b) How many three-digit numbers are there, which leave a remainder one on division by 9.

c) Find the sum of all such numbers.

a) The least three digit number which leave a remainder 1 on division by 9 = 99 + 1 = 100

The highest three digit number like this  = 999 - 8 = 991

b) Let there  be n numbers between 100 and 991 with common difference  9

100 + (n-1) 9 = 991, 100 + 9n - 9 = 991

9n = 991 + 9 - 100 = 900, n = 900 div9 = 100

c) Sum  = n/2 (first term + Last term)

= 100/2 (100+991) = 50 x 1091 = 54550

In the circle shown, the chords AQ and BP passes through C. 

a. The central angle of arc AXB is 100^{circ} Calculate  angle Q  The central angle of arc PYQ is 60^{circ} Find all angles of the triangle BQC.

b. In the picture, prove that angle APC is half the sum of the central angle of arc AXC and arc BYD.

a) angle Q = frac{100}{2} = 50^{0}

angle B = frac{60}{2} = 30 ^{circ}

angle BCQ = 180 - (50+30) = 180 - 8 0 = 100^{circ}

b) Draw BC.

In triangle CBP, angle B = half of the central angle of arc AXC.

angle C =  Half of the central angle of arc BYD. in a triangle the measure of an outer angle is equal to the sum of the measures of the inner angles at the other two vertices.

So, angle APC = angle B + angle C

frac{1}{2} the central angle of arc AXC + frac{1}{2} the central angle of arc BYD

= frac{1}{2} (Central angle of arc AXC + central  angle of arc BYD)

 

In the figure, PR is the angle bisector of angle APQ. Prove that ABleft | CD.

                angle APR=angle RPQ=38^0

                                 (Angle bisector)

therefore             angle APQ=38^0 +38^0 =76^0

              angle FQD=angle PQC=104^0

                                   (Vertically opposite angles

angle APQ+angle PQC=76^0 +104^0 =180^0

Since sum of interior angles is 180^0.

thereforeABleft | CD

Prove that sqrt{2} is an irrational number. Hence show that frac{3}{sqrt{2}} is alos an irrational number.

Let sqrt{2} be a rational number 

sqrt{2} = frac{a}{b}

(a, b are co-prime integers and bneq 0)

a = sqrt{2}b

a^{2} = 2b^{2}

Squaring, 

Rightarrow 2 divides a2

Rightarrow 2 divides a

So we can write a = 2c for some integer c,, substitute for a, 2b^{2} = 4c^{2} , b^{2} = 2c^{2}

This means 2 divides b2, so 2 divides b.

a and b have '2' as a common factor

But this contradicts that a, b have no commom factor other than 1.

Our assumption is wrong.

Hence, sqrt{2} is irrational.

Let frac{3}{ sqrt{2}} be rational 

frac{3}{ sqrt{2}} = frac{a}{b} where a and b are integers, b neq 0

Rightarrow 3b = sqrt{2a}

Rightarrow sqrt{2a} = frac{3b}{a}

frac{3b}{a} is rational but sqrt{2} is not rational

Our assumption is wrong

frac{3}{ sqrt{2}} is rational.

Show that there is no positive integer n, for which sqrt{n-1}+sqrt{n+1} is rational.

Let us assume that there is a positive integer n for  whichsqrt{ n-1} + sqrt{ n + 1} is rational and equal to  frac{P}{Q},where  P and Q are positive integers (Qneq 0)

sqrt{ n-1} + sqrt{ n + 1} = frac{P}{Q}................(i)

Rightarrow frac{P}{Q} = frac{1}{sqrt{n-1} + sqrt{n+1}}

= frac{sqrt{n-1} -sqrt{n+1}}{(sqrt{n-1} + sqrt{n+1}) (sqrt{n-1} - sqrt{n+1})}

= frac{sqrt{n-1} -sqrt{n+1}}{(n-1) - (n+1) } = frac{sqrt{n-1} - sqrt{n+1}}{-2}

Rightarrow sqrt{n +1} - sqrt{n-1} = frac{2Q}{P}................(ii)

Apply (i) + (ii) we get

2 sqrt{n+1} = frac{P}{Q} + frac{2Q}{P} = frac{P^{2} + 2Q^{2}}{PQ}

Rightarrow sqrt{n+1} = frac{P^{2}+2Q^{2}}{2PQ}.............(iii)

Apply (i) and (ii) we get 

sqrt{n-1} = frac{P^{2} + 2Q^{2} }{2PQ}.................(iv)

From (iii) and (iv), we can say sqrt{n +1} and sqrt{n -1} both are rational because P and Q both are  rational. But it is possible only when (n+1) and (n-1)  both  are perfect squares. But they differ by 2 and two perfect square never differ by 2. So both (n+1) and (n-1) cannnot be perfect squares, hence there is no positive integer n for which sqrt{n -1} + sqrt{n+1} is rational.