Dear Sir,

Pl refer to the attached files for a solution to the problem.

Thanks and regards

Yours Faithfully

Rajamani Ganesan

Let the one way distance be 180 km(LCM of 10&36).

Time taken for the up journey=  180/10=  18 hrs.

 Time taken for the down journey= 180/36=5 hrs.

Total distance(bothways) = 360 Km

Total time taken=  18+5 Hrs= 23 Hrs.

Average Speed= Total Distasnce/Total time= 360/23= 15..65 km/hr 

Let Cost Price=100

Mark up= 40%

List Price=100+40= 140

Discount @10% of Rs 140= 140x10/100= 14

Therefore Selling Price = List Price-Discoun t+ 140-14=126

Profit=   Selling .price-cost Price= 126- 100= 26%

Profit=26%

A function f can be visualized as  a rule which produces new elements out of some given elements..The function is said to MAP the given elements of one set A to another set B as per the rule set out in the function.The function can be mainly of  two types:

i) ONE –ONE( injective): 

 A function f:     X-------Y is defined as ONE—ONE (Injective) function , if the Y images of distinct elements of X under f are distinct. This means that every element of  X has one and only one image in Y.

 In case two different elements of X produce same  image in Y , that is x1≠ x2 but f(x1)=f(x2)  then the function is called   as MANY-ONE.

Trigonometric and polynomial functions like   y=Sin x ,y= x2 have same  values  (images) for different  of  elements of X  and these are  examples of MANY-ONE functions.

ii) ONTO (Surjective) Function:

    A function f: X---  Y is said to be ONTO (Surjective) if every element of  Y is the image of some element

    of X under f,  that is for every element of Y, there exists an element in X such that f(x)=y and there is  

    no  image of  Y for which a corresponding element does not exist in X.

    Here again we can have a  MANY-ONE case where two(or more) different elements of X can produce a

    Single image in Y.

Note: A  ONE-ONE  function  can have images in Y for which there are no corresponding elements in X. However, a  ONTO function cannot have any images in Y for which no corresponding elements  exist in X. A function which is ONE-ONE and also ONTO is called a ONE-ONE ONTO (Bijective) function.

I hope the above will help you to comprehend better  the relevant chapters on functions in XI and XII std

NCERT Maths text books. 

Volume is proportional to the cube of radius of a sphere.

Therfore Vol.of Jupiter/Volume of earth = (Radius of Jupiter /Radius of earth)^3= (70000/6371)^3= 1326.39 say 1326

So by volume approximately 1326 earths can fit into Jupiter.

Now mass is proprtional to the density.The density is of earth is 5.514 gms/Cu,Cm and the density of Jupiter is 1326 kg/Cu.mtr which is same as 1.326 gms/Cu,=Cm.

Massof Jupiter /Mass of earth= (Vol of Jupiter/Vol.of Earth)X (Density of Jupiter/Density of earth=1326.39X1.326/5.514

=  318.97 say 319. Jupiter is 319 times as heavy as Earth.

Toatal No. of cards in the pack=                                                                                     52

Total No.of Black cards(clubs+Spade)  including 2 black Jacks                                   =    26

Red Jacks in the pack                                                                                            =       2

Total Black cards plus 2 Red  Jacks                                                                       =     2 8 (26+2)                

Total No. of outcomes favourable to get either  a black card or Jack                        =       28             

Total No. of possible outcomes                                                                             =       52

Probabilty for the card to be either black or Jack                                                     =      28/52=  7/13     Answer

Art integration in mathematics is an approach to use arts in teaching mathematics. It uses colours,shapes,graphics and other visuals/audiovisuals so that children learn the concepts of mathematics as a fun and entertainment activity.

Art is already inherent in maths as Geometry ,Graphs etc.and this integration makes maths teaching both interesting and interactive bringing out the creativity in the student.

The answer file attached

The present ratio of no. of men to women in the committee=   5:6

After 20% increase in the number of men and 10% increase in number of women

the new ratio of men to women= (5*120/100) : (6*110/100)=(600/100) : (660/100)=(60/10): (66/10)=60: 66=10;11

Hence the new Ratio of men to women is 10:11

Answer:

Will you please put these books in my cupboard  and those ones on my table?

Explanation: 1.The first blank is filled with the preposition "in"  which connects the two nouns "these books" and 'my cupboard".It is a preposition of place.

                   2. The second blank is filled with "those" which is a demonstrative determiner as it differentiates between     books that are close and books that are not close by. 

This is problem from Boolean Algebra

A*~B*C+A*B*C+A*C= A*~B*C+A*C(B+1)  (by commutative and distributive laws of Boolean Algebra)

=A*~B*C+ A*C*1   =  ( because B+1= 1 as per laws of Boolean Algebra)

=A*~B*C +A*C

=A*C( ~B+1)=  ( by commutative and distributive laws)

=A*C *1           ( because ~B+1=1 as per laws of Boolean algebra)

=A*C               

Answer: A*C

The three equations x-y=0, x+y=0 and x=2 represent 3 straight lines which intersect with each other at 3 distinct points.The points of intersections are as below:

1. Intersection of x-y=0 & x+y=0. x=0 and y=0 are the only values that satisfy these equations.So the point of intersection is  0,0 which is the origin.                                                                                                                         

2. Intersection of x-y=0 and x=2.    x=2 and y= 2 are the only values that satisfy these equations. So the point of intersection is  2,2.    3. Intersection of x+y=0 and x=2.   x=2 and y=(-)2 are the only values that satisfy these equations. Hence 2, (-) 2 is the point of intersection.Hence the three sets of solutions are  x=0,y=0. x=2, y=2. and x=2, y= (-)2

Please refer to the attached files for the answer.

Answer:

Let the number=x.   Multiplying the number with 25% of itself = x*x/4=x^2/4

Given the above product is 200% more than the number.

That is x^2/4= x+200%*x= 3x

Therefore x^2/4= 3x or x^2= 12x or x^2-12x=0

That is x*(x-12)=0 which gives x=0 or x=12.Rejecting the trivial answer x=0 

we get x=12 which nis the correct answer. 

Answer:

Let the number of votes polled by A=X

Let the number of votes polled by B=   Y

Total Votes  polled                             =   X+Y

No. of votes polled A in excess of B= X-Y= 600  ------------------ EQUATION 1   

If B got  40% more votes then  votes polled by B= 1.4*Y= 1.4Y

But in that case A also would have polled =1.4Y  because it would have been a tie in that event

So total votes polled by both = 1.4Y+1.4 Y = 2.8Y

But this should equal to  X+Y

Therefore   X+Y= 2.8Y  or X=1.8Y

But X=Y+600 from EQUATION 1

Therefore 1.8Y=Y+600

Or  0.8 Y=600

Therefore Y= 600/0.8= 750

X= Y+600=  750+600=1350   from EQUATION 1

tHEREFOREV total votes polled= X+Y=1350+750=2100

But percentage of total  registered voters who voted =60%

Therefore Total voters registered = 2100/0.6= 3500  

Answer:

Let the number of people polled by A=X

Let the number of votes polled by B=   Y

Total Votes  polled                             =   X+Y

No. of votes polled A in excess of B= X-Y= 600  ------------------ EQUATION 1   

If B got more 40% more votes then  votes polled by B= 1.4*Y= 1.4Y

But in that case A also would have polled =1.4Y  

So total votes polled by both = 1.4Y+1.4 Y = 2.8Y

But this should equal to  X+Y

Therefore   X+Y= 2.8Y  or X=1.8Y

But X=Y+600 from EQUATION 1

Therefore 1.8Y=Y+600

Or  0.8 Y=600

Therefore Y= 600/0.8= 750

X= Y+600=  750+600=1350

Total votes polled= X+Y=1350+750=2100

Percentage of total  registered voters who voted =60%

Therefore Total voters registered = 2100/0.6= 3500  

Answer:

Given f(x)= - e^-(a+x).

 First derivative of f(x)= df(x)/dx= -e^-(a+x)(-1) = e^-( a+x) 

( -1) being  d/dx of -(a+x) ;

Second derivative  =  d{e^-(a+x)}/dx= e^-(a+x)(-1)= -e^-(a+x)

Third derivative= d{-e^-(a+x)}/dx= -e^(a+x)(-1)=  e^-(a+x) 

:Answer:

The dimensions of the plasticine  cuboid= 5cms, 2cms, 3cms

Volume of above cuboid  =  5*2*3=             30 Cu.cms

For a cube all the three dimensions are equal.    

Therefore the dimensions of a cube made from the

 above cuboids will be an integral multiple of   5,2,3 .

The minimum dimensions of the cube will be LCM of 5,3,2 which is 30

Therefore the size of the smallest possible cube=    30cms

Volume of such cube=30*30*30=27000Cu.cms

No. of cuboids of 30 Cu.cm volume needed

to form the above cube   =  Volume of the cube/ Volume of one cuboid

                =27000/30=900 

                         Answer

Let the original cost of one book         =  Rs x

Then no.of books purchased                = 60/x

The new cost of one book                    =Rs (x-1)

No.of books purchased at this price   = 60/(x-1)

It is given that the number of books that could be purchased at the

new price is 5 more than at the original price.

Therefore 60/(x-1) - 60/x           =  5

That is   {60x - 60(x-1)}/{x(x-1)}   = 5

or 60/{x(x-1)}   =       5

that is 12/{x(x-1)}= 1 or x(x-1)= 12

The 2 factors of 12 in which one factor is 1 more than the other

are  4,3   or 4*3=12

Hence original cost of the book=  x         = Rs 4

The number of books purchased= 60/4=15.

Given 23 +X  =345

Subtracting 23 from bothn sides we get

23-23 + X =  345-23

Therefore X= 322

Please see the attached file for the answer.

Answer:

Ratio of Shares of A,B,C  = 3:2:5

Sum of the ratios= 3+2+5 = 10

That is out of a total amount of   Rs 10 , C will get  = Rs 5

Therfore for Rs 1260  C will get  =  (5/10)*1260=  Rs 630

Both are very much related.

Interesting: something  which arouses your curiosity or holds your attention .

Interested:  Having or showing a liking for  something  

The  following example will bring out the difference between the two

Ex: I am interested in reading books..I find crime novels interesting.

ANSWER:

Let the average age of the  2 persons   =  x years

Then the   total age of the 2 persons=  2x

Given the total age of the other 8 persons  =  488

Therefore total age of all 10 persons=  488+2x

Average  age of all 10 persons=  (488+2x )/10 =  60 (given)

That is 488+2x=  600

That is 2x  =   600-488=112 or x=112/2=56

The average age of the 2 persons =  56 years   

ANSWER:

It is presumed that there is sufficient side clearance between the two cars so that the second car can safely overtake the first car.

The distance between the two cars at start = Lead of  first car over the second + length of the first car

                                                                             =  1.5  +2 = 3.5  metres( As the length of the second car was not given  the same is not considered here.)

Speed of first car                                          =   55Km/hr

Initial speed of second  car                        =  52km/hr

The acceleration of second car                  = 0.5km/hr per sec

The second car accelerates to a speed of 60km per hr from an initial speed of 52km/hr in 16 seconds and thereafter maintains a constant speed of 60km/hr. through out.

The speed of the 2nd car at any instant of time’  t’ seconds during acceleration   = (52+ 0.5t) km/hr

Therefore the average speed of the car during any instant ‘t’=  (52+52+0.5t)/2=  (52+0.25t) km/hr

Relative speed of 2 cars at any time’ t’ during acceleration= (52+0.25t-55)= (0.25t-3 )km/hr

Let the time taken for the second car to overtake first car =  t seconds

Time ‘t’ taken for the second car to overtake the first car= Initial distance during start/Relative speed

That is   t   ={ 3.5*10^(-)3}/(0.25t-3)*(1/3600) ( converting speeds  into km/s and distance into kms.)

that is ,t=       3.5*3.6/(0.25t-3) . By  cross multiplying we get ( 0.25t-3)*t= 3.5*3.6 =12.6

that is 0.25t^2-3t=12.6

that is 0.25 t^2  -3t-12.6=0 ; or t^2-12t=50.4=0  this is a quadratic equation  in ‘t’ and solving for ‘t’ we get

t=  {12+√345.6}/2 and  {12-√345.6}/2. Rejecting the negative value we get

t=15.295 seconds.

There fore the distance travelled  by the second car in overtaking  the first car=

Average speed  * time taken= {52+0.25*15.295}*15.295/3600=    0.23717 km=237.17 metres.

Please see the attachment for reply. 

Answer:

The question is answered in 2 scenarios.One with simple interest and the other with compound interest.

Simple Interest: Let the sum of Rs 500/-  be  lent on simple interest

Then principal =Rs 500.00

Time  = 10 years

Total interest earned in 10 years=Rs 500.00

Therefore simple interest earned on Rs 500 in one year=500/10= Rs 50.00

Hence rate of interest =(50/500)x 100= 10%

Compound  interest:    Let the amount be lent at a compound interest of r% per annum,

with interest added to the principal at the end of every year. If P is the principal ,n is the period

 in years and A is the total amount payable(Principal+interest) then

A=P(1+r/100)^n

Here P= Rs 500.00,n= 10 years and A= Principal+Interest= 500+500=Rs 1000.00

Therefore  1000= 500(1+r/100)^10

That is   (1+r/100)^10= 1000/500=2

Hence 1+r/100=  2^1/10  =2^0.1  =1.07177

Hence r=  Rate of Interest= (1.07177-1)*100= 7.177 or 7.18%

Pl see the attachment for the answer

Answer:

As balls are drawn with replacement ,the probability of balls of a particular colour  coming up in each draw will remain same.The probability of drawing one  white ball from 16 balls is 4/16=1/4 and the probability of a drawing  one ball of any other colour  is 12/16= 3/4. These  will remain same during every draw.

The probability of drawing at least 2 white balls will be the sum of the probabilities of drawing

exactly 2 ,3 and 4 white balls.

1.   Drawing of exactly  2 white balls:  The combined probability of drawing 2 white balls ( and 

      2 non  white balls)  = (1/4)^2*(3/4)^2. But as the balls are replaced after they are drawn,  we can

      have   6 different combination of pairs of white balls. Suppose  the white balls are numbered

       w1,w2,w3&w4, then we can have 6 pairs:  w1w2,w1w3,w1w4,w2w3,w2w4,w3w4.Selection of

      2 items out of 4 is denoted as nCr=4C2=  (4*3)/(1*2)=6 ways.

    Therefore the probability of drawing  2 white balls from the bag = 6*(1/4)^2*(3/4)^2= 54/256.

2.   Drawing of exactly 3 white balls: The combined probability of drawing 3 white balls(and

      1 nonwhite ball)   = (1/4)^3*(3/4).Here we can form four different combinations of a triplet of white

       balls .We can have  4 triplets: w1w2w3 ,w2w3w4,w3w4w1, w4w1w2 .Selection of 3 items out of 4 is

       nCr=  4C3= =(4*3*2)/(1*2*3) =4 ways.

       Therefore the probability of drawing  3 white balls from the bag= 4*(1/4)^3*(3/4)=      12/256

3.  Drawing of all 4 white balls: The combined probability of drawing 4 white balls(and 0 non white ball )

      = (1/4)4*(3/4)^0.Here  we can draw all the 4 white balls in only 1 way(w1w2w3w4)

        Therefore the probability of drawing  4 white balls from the bag= 1*(1/4)^4*(3/4)^0= 1/256

     Therefore the probability of getting at least 2 white balls=   54/256+ 12/256 + 1/256=   67/256.             

Answer:

As a corollary of   Pythagoras theorem, the area of a square drawn on the hypotenuse  of an isosceles right angled triangle is twice the area of the squares drawn on any one of its sides.  In a square, the diagonal and the two sides of the square form an isosceles triangle. Hence the area of the square is half of the area of the square drawn on the diagonal.

Given length of the diagonal= 2.5 mtrs.

Therefore area of the square drawn on the diagonal= 2.5*2.5 = 6.25 Sq.Mtrs.

Hence area of the square whose diagonal is 2.5 mtrs=     (1/2)*6.25=   3.125 Sq.Mtrs

pl see the attached file for answer

Answer: "C"  Cobalt. All others are non magnetic materials

Answer:

DC Power or instantaneous AC power is defined as Voltage*Current= V*I

But as per Ohms law current  I=  V/R .where R is the resistance of the circuit.

Hence Power  P=V*I= V*V/R= V^2/R  or interms of current  P=I^2*R

Ans:  LHS=   5^m* 5^(-)2=  5^(m-2) (as per laws of indices)

         RHS = 5^3

       Therfore  5^(m-2)=  5^3

The two expressions are equal and their bases are also equal.Hence the indices must also be equal.

Hence  m-2= 3 or m=5

Answer is m=5 

Please see the attached file for the answer.

Answer:

Let the capacity of the tank  = 60 litres.

Time taken by first tap to fill the tank= 10 hrs.

Quantity filled by first  tap in 1 hour=  60/10 = 6 litres

Time taken by the second tap to fill the tank= 15 hours

Quantity filled by second tap in 1 hour=  60/15= 4 litres

Quantity filled in one hour when both taps are open   =6+4= 10 litres

Therfore time taken for the tank to be filled when both taps are open= 60/10= 6 Hours. 

Answer :

Let food  consumed by 1 animal in 1 day=            1 unit

Food required for 30 animals for 1 day=                30 units

Food required for  30 animals for 4 days=       30*4 = 120 units.

Now the number of animals in the form=         20

Food required for 20 animals for 1 day=      20 units

No.of days 20 units of  food will last for  20 animals=     1

So 120 units of food  will last for 20 animals       =  120/20=  6 days 

See the attached file for the answer

  Answer:  The two adjectives are as below:

1.  Indian  . It describes the noun Artisan

2. Exquisite: It describes the quality of the abstract noun Workmanship

Answer: Pl refer to the attached file.

Answer:

The question wants to find: Given 2 non empty sets A&B,  if A∩B=0 then whether B is a subset of A.

But this is not true   according to set theory. In fact, the opposite is true. That is if A∩B =0

then B is not a subset of A. The following will make this clear:

Given that  A∩B =0 means that both have no common elements between them  and so

no element present in B is present in A. But B can be a subset of A, if and only if  

all the elements of B lie in A. Thus it is clear that B is not a subset of A.  

Please refer to the attached file for the answer.

Decimal expansion of any rational number of the form p/q will be either terminating or  non –terminating and repeating. The expansion will be terminating if the prime factorization of q is of the form 2^n X 5^m.where n and m are non negative integers. The expansion will be non terminating and repaeting(recurring) if prime factorization of q is not of  the form of

 2^n X 5^m. The following  examples will make the above clear;

1. 32/25 =  1.28.The denominator 25 is of the form 2^0 X5^2..So the decimal expansion  terminates at 1.28

2. 47/20=   2.35   .Here also the denominator is of the form 2^2 X5^1. So the decimal expansion   terminates at 2.35.

Now consider the following:

1. 4/3= 1.333 . The denominator is not of the form 2^n X 5^m. . Hence decimal expansion is non- terminating and  

     repeating. The dot placed above the digit “3” indicates that the expansion  repeats from here  onwards.

2.  29/7=  4.142857142857 .Here also the denominator is not of the form 2^n X 5^m. . Hence decimal    expansion  is

      non terminating and repeating. The dot placed above the digit “7”indicates that the expansion repeats from here

       onwards.

  And finally any number that does not fall into either of the above is not a rational number.

                                 .

Answer:

The equation of motion of the particle is given  as x= 5 Sin (2π/3t). Sine is a periodic function which repeats it self after 2π radians. In one second the particle covers 2π/3 radians.

So to cover  2π radians the time taken T= Time period of 1 oscillation = 3 seconds.

In T seconds the particle completes one oscillation or 4 amplitudes. The distance travelled by the particle  from its mean position to one extreme equals one amplitude.

Therefore time taken to traverse one amplitude is   T/4= 3/4 seconds  

Please refer the files attached

The square root of 45000 is 212.13.Pl refer to the attached files for a step by step desription of the

method. 

Answer:

The unit of resistivity in MKS  units is Ohms Mtr..It is the resistance of a cube of the material of side 1 metre.. 

Answer:  a,b,c,d will be in Harmonic Progression provided their reciprocals 1/a,1/b,1/c and 1/d are in Arithmetic

Progression.     That is( 1/a-1/b) = (1/b-1/c)  =(1/c-1/d).

Answer:

1. Separately Excited Generator:

Let E be the induced EMF, Ia be the armature current, Ra  be  the armature resistance and Ig be  the load current of the separately excited generator. As the excitation current of the  generator does not pass through the motor circuit, the current through the armature will be same as load current Ig and so voltage  drop will be  Ig * Ra.  Hence terminal voltage V of the generator   is given by the equation:

          V= E-   Ig * Ra

2. Shunt excited Generator:  Let Ish  be the shunt field current. This current and load current Ig pass through armature      circuit. The armature current will therefore be Ish+ Ig.. So the voltage drop is (Ish+ Ig) * Ra.. Hence terminal voltage of the generator   is given by the equation:   

V= E- (Ish+ Ig) * Ra

3. Series Excited Generator: Let Rse be the resistance of  the series winding .The same load current Ig  flows through both the armature and the series field winding. Therefore the total voltage drop will be Ig * (Ra+.Rse)..Hence terminal voltage of the generator  is given by the equation:  

V=  E- Ig * (Ra+. Rse)

4.Compound Excited Generator:  The current in the armature  circuit/ the series field winding will be sum of   the load current Ig and shunt field current Ish. Therefore   the voltage drop will be   (Ish+ Ig) * (Ra+ Rse).  Hence terminal voltage of the generator  is given by the equation:

V=  E- ( Ish+ Ig) * (Ra+ Rse)

Answer:

The phrase “to kill a mocking bird is a sin ” is used in the novel as a metaphor to mean that innocent and defenseless people should not be harmed because of prejudices or racial discrimination. The two characters in the novel that are likened to mocking birds  are Tom Robinson ,a black man and Arthur(Boo) Radley, a white man. Tom Robinson was a victim of black discrimination whom the jury  convicted despite being innocent.Arthur Radley, a white man,  was a victim of prejudice because of circumstances.

Arthur(Boo) Radley led a secret and private life,never spoke to any body and mostly kept himself confined to his room. The Finch children believed the town gossip that he killed the pets of his neighbors during night and were very much afraid of him. They thought that he was a devil and made up strange and horrific stories about him. But this man was not bad. He had a consideration for the Finch Children.He left gifts in the tree for them. In fact, he saved Jem and Scout from  Bob Ewell and even killed him to protect the children.

At the end Scout realizes that Boo Radley is not a bad man and in fact an angel who had saved her life.Any prejudice she entertained of him was gone and she understood how tough it would be for anybody  to be in his shoes and carry on with life. Past incidents like secret gifts for them in the tree hole ,blanket thrown  by an unknown man over her to protect her from the neighborhood fire ran through her mind. When Boo Radley asked  her to escort her home then she realizes that the man of whom she was so afraid of was in fact  afraid of people and darkness. She is filled with empathy for the man .She realizes that exposing  a reserved man like Boo Radley to the unknown world outside  will be like killing the  mocking bird which saved her from Bob Ewell. One mocking bird was already killed. She would ensure that the other mocking bird is saved at any cost.. The climax  thus brings out the main theme of the story succinctly.

ANSWER:

Given the equation of the line AB is   X+Y =   5.Let it intersect the Y axis at D.

At the point of intersection with y axis, the value of the x co ordinate of D is 0.

Putting x= 0 in the equation X+ y= 5  we get 0+Y=5 or Y= 5.

Therefore  the coordinates of D are  (0,5)

Also given that the equation of line AC is 7X- Y = 3.

Let it intersect the Y axis at E. At the point of intersection with y axis the value of x co ordinate of E is 0.

Putting x=0 in the equation  7X-Y=3 , we get 7*0-Y=3  or Y=(- )3. Therefore the co ordinates of E are (0, -3).

As both the points D and E lie on the Y axis ,the length of the y intercept cut by the side BC =  DE=  5-(-3)= 5+3=  8 .

Answer :  option (b) 

ANSWER:

Given:  Radius of the cylindrical portion of tent= r =2.8 mtrs

           Ht of the cylindrical portion of the tent= h1 = 3.5 mtrs

           Radius of the conical top of the tent=      r = 2.8 mtrs

           Ht of the conical top of the tent=             h2=  2.1 mtrs  

          Therefore slant ht of  the conical top of the tent=  l = (r^2+h2^2)^0.5= { (2.8^2)+(2.1)^2}= 3.5 mtrs                 

The cloth required for the tent is equal to the lateral surface area of the cylinder &top cone.

Cloth required for 1 tent= 2πrh1 + πrl=   πr(2h1+l)= π*2.8*( 2*3.5+3.5)= (22/7)*2.8*10.5= 92.4 Sq.mtr

Cost of cloth per tent @Rs 120 per Sq.mtr= 92.4*120= Rs 11088

Total cost of 1500 tents=  11088*1500= Rs 1,66,32,000

No. of schools  bearing the cost=  50

Therefore average cost per school = 16632000/50=Rs 3,32,640 

Answer: Pl see the attached file

Answer: Such angles can be drawn with a protractor and scale.as below

1.Draw  a line segment AX of any convenient length.2.Set the centre point of the protractor exactly above the point A

3.Now set  the edge of the protractor in such a way that the 0 degree graduation of the protractor is exactly aligned with the line AB.  4.Now measure exactly 130 degrees on the protractor scale in an anticlock wise direction from AB and mark this point as C.5. Now from A draw line AX joining  the points A,C .6.Angle XAB= 130 degrees.  

Answer; Option B= Specific Gravity.   Specific gravity is the ratio of density of a substance to the density of water,. Because it is a ratio of 2 identical units of meaurement it is dimensionless.  

Answer:

Let the sum  given on 10% rate of  compound interest for 3 years is Rs 100.

Interest for Rs 100 for one year at the above interest rate is Rs 10

Therefore total amount accrued  at the end of first year=Rs100+Rs10= Rs 110

Now this becomes the principal for the second year. Interest for  one year for Rs 110= Rs 11

Therefore total amount accrued at the end of second year=Rs110+Rs11=  Rs 121

Now this becomes the principal for the third year. . Interest for one year for Rs 121=Rs12.10

Therefore total amount accrued at the end of third year = Rs 121+Rs 12.10=  Rs 133.10

For  accrued amount of Rs 133.10, the principal sum =    Rs 100

For accrued amount of Rs 6655/- ,the principal sum=      (6655x 100)/ 133.10)= Rs 5000 

The required sum= Rs 5000

Answer

Let V,W,R be the voltage , power rating and  resistance respectively of each lamp.

Then power rating of each lamp W= V^2/ R  =200 watts

Now when the 2 identical lamps are connected in series the voltage gets equally divided between the 2 lamps

and so voltage across each lamp=   V/2.

Therefore power consumed  by each lamp=  {(V/2)^2}/R= V^2/(4R)

Therefore power consumed by both  lamps= 2*(V^2)/(4R)= V^2/(2R) =  (V^2 /R)/2=W/2= 200/2=100 watts

Please refer the attached file.

Please refer to the attached files for the answer

Answer: Please refer to the attached files

Answer

The angle described by a particle about the axis of rotation in a given time is called angular displacement.It  is denoted as θ. The angular displacement is measured in radians.

The rate of change of angular displacement is called the angular velocity and is denoted by ω. If θ is the angular displacement in time t ,then angular velocity= rate of change of angular displacement=  ω = dθ/dt. Angular velocity is measured in radians/sec. It is a vector quantity.

Answer:

Pl refer to the attached file

Answer:  Option (d)

Pl refer to attached file for detailed answer

Answer:  5,12&13 form a pythagorean triplets.Given  PR=13 cms amd PQ= 12 cms it follows that side 

              QR of the triangle =5cms. As PR =13 cms is the hypotenuse , the other two are the sides

             Therfore  the area  of the triangle PQR = 1/2*PQ*QR  =1/2*12*5= 30 Sq.cm

3^x  - 3^(x-2) =  72 .That is 3^(x-2){ 3^2-1}= 72

Therfore 3^(x-2)*(9-1) =72 or  3^(x-2)=  72/8= 9

Therfore 3^(x-2)= 9= 3^2 wich gives  x-2= 2  or  x=4

Ans:  x-4 

 Answer:

The number of ways of choosing  8 students out of 17 students to stand in a circle of 8 is same as the number of ways of choosing 9 students out of 17 students to stand in another circle of 9 students.

That is 17C 8= 17 C9

  The  number of ways  of choosing 8 students out of 17 students is 17 C8= 17!/( 8!*9!)= (17*16*15*14*13*12*11*10)/(1*2*3*4*5*6*7*8)=24310

Once you choose 8 students the other 9 students will be automatically selected.

So total number of ways of choosing  17 students in 2 circles of 8 and 9 students is 23,410.   

Answer:   The minimumm value is V3. Pl refer to the attached file for a detailed solution. 

Answer:

The equations of the cross roads where the elephant is standing are

  x- y+2=0    ………  A

  y- 1=o.     …………B   that is y= 1

  putting y= 1   in equation A we get x-1+2=0 or x=  (-)1

Therefore the co ordinates of the junction where elephant is standing is   (-)1, 1.

Given the equation of the road where the elephant wants to go is x - y-3=0 or y = x-3.

The shortest distance of a point to a given line is the perpendicular distance of the point to the given line. The slope of a line perpendicular to a given line  will be equal to the negative reciprocal of the given line. That is if the slope of a line is “m “, then the slope of the line perpendicular to this line will be “(-1/m)”.

Equation  of the line representing the road where the elephant wants to go is  y=x-3.Its slope = 1.

Therefore the slope of the  line perpendicular to the above line is (-)1.Therfore the general equation of the perpendicular is   y=  - x +k, where ‘k” is a constant        …….C .

Now the junction (where the elephant is standing) is a point on the perpendicular whose co ordinates we have found as (-)1 , 1.Therfore this point should satisfy the equation of the perpendicular. Putting these values in equation “C” ,we get  1= -{ (-)1} +k = 1+k which gives the value of k=0. 

Therefore the equation of the perpendicular is y= - x+0   or  x +y=0.

Equation of the path to be followed by the elephant is x+ y=0

Answer: Option (i)  

Answer::

We shall consider both simple interest and compound interest for calculation of interest rates.

i) Simple interest: The formula for simple interest is I= P*T*R/ 100 ,where  P is the principal,

   T =Time period in years and R= rate of interest per year per 100.The principal will double

   if the interest, I, accrued equals the principal P.

   Here Principal= P, Time ,T= 5 years and rate of interest per year per 100 = R

   Therefore  P= P*T*R/100=  P*5*R/100;  that is P= PR/20; which gives R= 20%

ii) Compound Interest: Here we consider that interest is compounded every year The formula for total amount (inclusive of interest) ,A=  P{1+(R/100)}^T. When the principal doubles in 5 years ,the amount

A becomes 2P. Therefore   2P= P{1+(R/100)}^5 ;  which gives {1+(R/100)}^5  = 2

That is { 1+(R/100)}  =  2^(1/5)= 1.1487 or  R= 1.1487 -1 = 0.1487 or 14.87%

Answer: i) Simple  interest R  = 20%  ii) Compound interest R= 14.87%.     

Answer: metrers

153 rails stretch to=  680 metres.Therefore length of one rail= 680/153 metres.

Hence length of 135 rails placed end to end= (680/153)*135 =600 metres.

Answer: 600 metres 

Answer:;

A counter is an instance where a  rule/theorem/law, which is generally valid otherwise in most of the cases, is not valid in the particular case. Counter of any rule is an example of an instance, where the rule is not applicable or an exception. to the rule.To illustrate ,let us consider the rule for finding Leap Year. 

We know that there are 366 days in a leap year, that is the month of February has 29 days in a leap year. The general rule is, if the year is divisible by 4,then it is a leap year. That is if the numerical value of any year is a multiple of 4,then the year is a leap year.Thus,1980,1992,2012 ,2020 etc. are leap years. But this rule is not valid ,if the year is a century year, that is the last 2 digits of the year are 0,0.Thus the century years 1800,1900,2100,2200,2300  etc. are not leap years ,though they are multiples of 4. These are counter examples for the general rule of leap year.

Again this rule is not valid if the  century years are multiples of 400 ,because century years which are multiples of 400 are leap years .Thus century years 1600,2000,2400 etc. are century leap years having 366 days. These  are again counter examples for the general rule of  leap years for century years.   

Answer:

Divisibility test:

 i)    for  2:   All  even numbers are  divisible by 2.

ii)    for  3:   If the sum of the digits of the given number is divisible by 3,then the number is divisible by 3.

iii)   for 4:    If the last two  digits of the given number are  divisible by 4,then the number is divisible by 4.

iv)   for 9:    If the sum of the digits of the given number is divisible by 9,then the number is divisible by 9.

 v)   for 10:  All numbers ending with 0 are divisible by 10.

Note:   If a number is divisible by 9 then it is also divisible by its sub multiple 3.

As per the above divisibility rule the set of numbers

a) 128, 342,405,510,616,840,918,1440,1523  can be grouped on the basis of divisibility by 2,3,10as below:

128,616: Both are  even numbers and hence they are divisible by 2

 405: This  is an odd number and so not divisible by 2. But the sum of its digits add up to 9 which is

          divisible by 3. Hence 405 is divisible by 3.

 342,918 :Both are even and so divisible by 2.Further the sum of their digits  (9,18)  are divisible by 3 and

                  hence  they are divisible also by 3.Hence these 2 numbers are divisible by both 2 &3..

510,840,1440:  All the three numbers are even and so divisible by 2.Sum of the digits of all the three numbers are divisible by 3 and so these three numbers are also divisible by 3.Further ,the last digit of all the three numbers is 0 and so these three numbers are also divisible by 10 .Thus these three numbers are divisible by 2,3&10.

1523 : This number  cannot be grouped into any of the above as it is not divisible byany of the numbers  2,3,10.

The results are summarised below:

Divisible by 2 only:      128,616

Divisible by 3 only:       405

Divisible by 2&3:          342,918

Divisible by 2,3,10:       510,840,1440

Not divisible by any of the above: 1523:               

b)  128, 342, 405,510,918 can be grouped on the basis of divisibility by 3,4,9 as below:

510:  The sum of the digits of this number is 6 which is divisible by 3 and so this number is divisible by 3 .

128:   The last two digits of this number is 28 , which is divisible by 4.So this number is divisible by 4.

342,405,918: The sum of the digits of this number are 9,9 and 18  and all are divisible by 9.Hence these three numbers are divisible by 9.As 3 is a sub multiple of 9,these numbers are also divisible by 3.

Hence these three numbers are divisible by both 3&9.  

The results are summarised below:

Divisible by 3 only;       510

Divisible by 4 only:       128

Divisible by both 3&9:   342,405,918,

Answer: Pl refer to the attached file

Answer: Pl refer to the attached file

Answer:

A force of 8 KN is acting horizontally and a force of Q KN is acting at an angle of 30 degrees to the vertical .A third  force of 5 KN is acting at an angle of x degrees to the force Q ,that is at an angle of (x+30) degrees to the vertical.

The horizontal component of the resultant of these 3 forces should  equal to 0, so that there will be no force acting in a horizontal direction to the pole. The vertical component of the resultant force should equal 12KN acting in a vertically downward direction.

Resolving the forces horizontally, we get   Q sin 30+ 5 sin(x+30)= 8   or 

5 Sin(x+30) = 8- Qsin30= 8 - Q*1/2  =      8- 0.5Q                  ------------------------------------A

Resolving the forces vertically, we get   Q cos 30+ 5 cos(x+30) =  12 or

5 cos (x+30)  = 12-Qcos30= 12- (Q*√3/2) =   12  - 0.866Q         ………………………………………..B

Squaring and adding the equations “A &B” we get

25 { sin^2(x+30)+cos^2(x+30)}= (8 -0.5Q)^2+ (12-0.866Q)^2

25=   (8-0.5Q)^2 +( 12-0.866Q)^2=  64+ 0.25Q^2 -8Q +144+ 0.75Q^2 -20.78Q

That is Q^2 – 28.78Q+ 208 = 25.   Or  Q^2- 28.78Q+183= 0

 This is a quadratic equation in Q ,whose two solutions are

Q=  [28.78+√{(28.78)^2 -  4*183}]/2 ,  [28.78-√{(28.78)^2 -  4*183}]/2

    = { 28.78+√96.29}/2,     {28.78 - √96.29}/2

     =   (28.78+9.81)/2,   (28.78- 9.81)/2= 

     =   19.295 and  9.485  that is

           19.30,     9.49 (say)

Putting these values  of Q in equation “A” ,we get 

5 sin(x+30)=  8- 19.30*0.5 =   8 – 9.65  =  (-)1.65 and

5 sin(x+30)=  8- (9.49)*0.5  =   8-  4.745 =    3.255= 3.25 say

Sin(X+30) cannot be negative as this will make x an obtuse angle and hence the value of Q= 19.30  is not acceptable.  Hence Q=  9.49     which gives  

Hence 5 sin(x+30) = ..3.25 or sin(x+30)= 0.65  or ( X+30) =   40.54 degrees or x= 10.54 degrees

Therfore :   Q=  9.49 KN , x= 10.54 degrees. 

The block moves a ditance of 10 metres in 2 seconds under the action of force.As there is no friction,we can assume that the block is moving with uniform  acceleration. So the distsnce trasvelled by the block with uniform acceleration is given by the formula s=ut+(1/2)*a*t^2, whwre s= distance ravelled by the block= 10  metres , u= ininitial velocity= 0         ( because block was under rest before application of force) ,  a= uniform acceleration of the body, t= time taken to attain the final velocity= 2 seconds  and m=mass of the block= 2Kgs. Applying these values in the equation,we get                  10 = (1/2)*a*2*2= 2a which gives a=5 metres/sq.sec.Therfore Force f= m*a= 2*5= 10  Newtons

Ans: 10 Newtons

Answer:

Initially the quantity of pure milk  = 60 litres

Water    quantity                        =   0 litres

                                                    --------

Total liquid                                   60 litres

Afterwards pure milk of 6 litres is removed from the vessel and 6 litres of water is added.

Now pure milk in the vessel               =   54 litres

Water in the vessel                          =      6 litres

                                                            ----------

total quantity of liquid                       =     60 litres

The ratio of pure milk to water        =   54:6=   9:1

Now again 6 litres of milk is removed.

This 6 litres of milk contains pure milk =   (9/10)*6=  5.4 Litres and 0.6 litres of water

Therfore pure milk in the vessel after the above will be 54.0 - 5.4 =           48.6 litres

Water content                                                           =  6-0.6=                5.4 litres

Now water added(after removal of milk as above = 6 litre

Total water content                                 =     5.4+6    = 11.4 litres

Therefore   final ratio of Pure milk to water=   48.6:11.4  =  81:19  

Total milk in the vessel  =  48.6+11.4      =  60 litres

Answer:

  No. of atoms of Chlorine present in 710 gms of chlorine gas  =   X

  35.5 gms (approx.) of chlorine gas =  mass of 1 mole

  Therefore  710 gms of chlorine=   710/35.5= 20 moles

  1 mole of sodium reacts with 1 mole of chlorine to form 1 mole of Sodium Chloride..                                                        That is 20 moles of sodium will       react with 20 moles of chlorine to form 20 moles of Sodium Chloride.

    mass of 1 mole of Sodium             =    23gms

   Therefore  mass of 20 moles of sodium= 460gms

   Total mass of sodium chloride   =710+460=   1170 gms

   Hence 1170 gms of Sodium Chloride contains  710 gms of chlorine which in turn contains  X   number of  chloride        atoms

.

Answer:

Given  f(x)   =  e^(ax+b).  Differentiating with respect to x we get

dy/dx=  f'(x)  =   a*e^(ax+b). Again differentiating with nrespect to x  we get

d/dx(dy/dx)=   f"(x) =  a*a*e^(ax+b)  =   a^2*e^(ax+b). Continuing upto n times we get

nth derivative of f(x)   =  a^n* e^(ax+b)

Answer: 0.375 Amp  Option 1

Pl refer to the attachment for detailed explanation 

Answer: The answers are shown in bold script

FRACTION                               DECIMAL                                  PERCENTAGE

3/5                                          0.60                                                60%

32/100   = 8/25                        0.32                                                 32%

2/100     =  1/50                       0.02                                                  2%

47/100                                    0.47                                                 47%

55/100   =   11/20                    0.55                                                 55%

8/10       =    4/5                      0.80                                                 80%

9/10                                        0.9                                                   90%

17/20                                      0.85                                                  85%

350/1000   = 7/20                    0.350                                                 35%

Answer:

Two things are  to be noted while solving the problem.

1. As  Virat has to keep the strike through out the over, he can score only even runs like 2,4,6 or no runs with each ball.

2. As he has to play all the 6 balls of the over ,  some runs have  to be scored from the last ball. He can score the 12 runs from 6 balls in various sequences and the no. of ways of selecting the balls depends accordingly.

1. Only 2 balls in sequence of 6+6 =12:

  One ball is already pre selected as last ball. The remaining 1 ball can be selected out of 5 balls in 5 ways.( That is 12 runs will be made from ball 1&6  or 2&6 or 3&6 or 4&6 or 5&6.).So no.of ways for this option =  5

2. Only 3 balls in sequence of 4+4+4= 12.:

   Again as one ball is reserved as last ball, we have to select remaining 2 balls   out of 5 balls . As per laws  of  permutations, the selection of " r" objects out   of "n" objects can be done in nPr ways= n ! / (n-r)!. If out of the " n" objects , " m" objects are identical then the number of ways of selecting "r" objects is  nPr/ m! = n!/ { ( n-r)! * m!}.             As Virat has to score  identical runs of 4 each from these 2 balls they both constitute 2 identical objects.                         So number of  ways of selecting 2 identical  balls out 5 balls is  5P2/ 2!=  5!/{ ( 5-2)! * 2!} =  5!/( 3!*2!). = 10 ways.

3. Only 3 balls in combination of 6+4+2 = 12

    Here the last can be hit either for 6, 4, 2 runs. So it can be selected in 3 ways. After that we can select 2 balls out of  5 balls in 5P2 ways. As both the balls will be hit for different amount of runs they are not identical. Therefore the number of ways of selection for the 2 balls is 5P2= 5! /3!= 5*4= 20'ways. As these 20 ways are possible for each of the 3 ways of    selecting the last ball , the total number of selection for this option is 3*20 = 60 ways.

4. Only 4 balls in combination of 6+2+2+2  =12:

    Here the last ball can be hit either for  2 runs or 6 runs.

    Case1: The last ball is hit for 6 runs. Then we have to select 3 balls out of 5 balls in which all the 3 balls will be hit for 2 runs each and hence all 3 are identical. Therefore number of ways of selecting the 3 balls is 5P3 / 3! = 5*4*3/ 6= 10

    Case 2: The last ball is hit for 2 runs. Then we have to select 3 balls out of 5 balls in which 2 balls will be hit for 2 runs each and hence they 2 are identical. Therefore the number of ways of doing this is     5P3/2!  = 5*4*3/2  = 30.         So total no. of ways for this option = 10+30  =40

5. Only 4 number of balls in combination of 4+4+2+2='12.                   

 Here last ball can be selected in 2 ways.It  can be hit for 4 or 2 runs.     

 Case1: Last ball will be hit for 2 runs.From the remaining 5 balls we can select  3 balls out of which 2 balls would be hit for 4 runs each and hence identical..Therefore the number of ways of doing this is 5 P 3/2!=5*4**3/2. = 30

Case 2:  Last ball will be hit for 4 runs.

From the remaining 5 balls we can select 3 balls out of which 2 balls would be hit for 2 runs each and hence identical. This situation is same case 1 and so the number of ways of doing this is 5P3/2!= 30. So total number of ways of selection for this option would be 30+ 30  = 60 ways.

6. Only 5 balls in combination of 4+2+2+2+2  = 12.  

 Here last ball can be hit either for 4 or 2 runs.

Case 1: Last ball will be hit for 4 runs.

From the remaining 5 balls we can select 4 balls in 5 P4 ways and as all the 4 balls will be hit for 2 runs each and so there will be 4 identical balls. Therefore number of ways of selecting 4 balls = 5P4/ 4! = 5!/ ( 1!*4!)  = 5 ways.

Case2:  Last ball will be hit for 2 runs.

From the remaining 5 balls we can select 4 balls in 5P4 ways.But in this 3 balls will be hit for 2 runs each and so there will be 3 identical balls. Therefore number of ways of selecting 4 balls  = 5P4/ 3! = 5!/ ( 1!*3!)  = 20 ways. So total number of ways of selection for this option is  5+ 20  = 25 ways

7. All the 6 balls in combination of 2+2+2+2+2+2 =12.

  In this option all the 6 balls will be selected at a time and so the number of ways for this option =1

So total number of ways of hitting exactly 12 runs= 5+10+60+40+60+ 25+ 1 = 201

Answer:

NAND  and NOR gates are called universal gates because  each of them can be used  by itself to produce  the three basic operators  NOT,OR and  AND. gates. The three basic operators  can be obtained by suitable  combinations of either of NAND or NOR gates alone and there is no need to use any other  gates for the same.

I shall illustrate this with one example of obtaining a NOT gate  by using a single NAND gate.The output of a NAND gate will be at  "1" level when either of its two inputs are at "0" level and the output will be at "0" level if both the inputs are at "1" level.

 The truth table of a NAND gate is shown below

   Input A                Input B              Output

      0                          0                      1

      0                          1                      1

      1                          0                      1

       1                          1                      0

The NOT gate will give an output of "1" when its input is"0" and an output of "0" if its input is "1" 

The truth table of a NOT gate is shown below:

Input                      Output

   0                            1

   1                            0

It can be seen that shorting the inputs A and B of the NAND gate will make  both the inputs  equal to either a “0 level or “1” level and so the NAND gate output will be “1” and “0” respectively making it effectively a NOT gate. 

Similarly we can obtain the other two gates from the NAND gate and also the NOR gate can be used to obtain all the three basic gates.

Answer

The denominator of the given fraction is 6 which has factors 2&3. We know that any rational number whose denominator is not in the form of" 2^n*5^m "will have a recurring decimal expansion.As the denominator of the given fracrion 5/6 contains a factor "3" which is different from "2" and "5" ,the decimal expansion  is recurring and non terminating.

Answer:

The expression for the Centripetal force acting on a particle of mass "m" moving with a velocity "v" along a cicle is F=m*v^2/r.Please refer to the attached file for the derivation.

Answer:

Solids moving on a surface experience  resistance to the movement which is known as Friction..Similarly fluids also experience resistance to the motion which is called Viscosity.

Coefficient of Viscosity is defined as the ratio of Shearing stress to the rate of change of shear strain. If F is the shear force applied over surface of  area "A" units, dx/l is the shear strain caused in time dt seconds then                               Shear  stress  =   F/A . Rate of change of shear strain=  (dx/l) / dt =  dx/(dt*l)= v/l where v is the velocity of motion. Therfore  Coefficient of viscosity=  (F/A)/(v/l)  = (F*l)/(v*A)

The units of viscosity is Poiseiulle. It is also measured in Newtonsec/Sq.mtr. and in pascal seconds.

.

Answer:

Ampere's swimming rule gives the direction of magnetic field that will be set up by a current carrying conductor.

The rule states that if a man swims along the current carrying conductor ( that is as if the current is entering his feet and leaving from his head) such that his face is always towards the magnetic needle,then the North Pole of  the magnetic  needle is always deflected towards his left hand.

Answer:

The distance between the sisters at the time of starting  =   385 Miles

The average relative speed of the sisters  when travellinmg towards each other  =  60+50 = 110 MPH

Time taken to meet =              Distance at the time of starting/  Relative Speed =  385/110 = 3,5 Hrs

Therefore  time taken for the sisters to meet each other  after they set out = 3..5 Hrs  

Answer:

Area remained unpainted after Anil and Umar =  60%

Therfore area painted combinedly by Anil and  Umar  =40%

Of the above 40% ,work completed by Anil=  25%

So work completed by Umar=  40-25%     =   15%.

But work done by Umar in terms of area=     30 Sq.Metres.

Thefore percentage of work done by Umar=  15% =  30 Sq.Metres

Hence work done by Anil in terms of area=    (25/15)*30  = 50 sq.Metres 

Answer:  It is given that (x-2) is a factor of  f(x) but not a factor of  g(x). This means that (x-2)  is not a common  factor between  f(x) and g(x). In that case (x-2) can only be a factor of  f(x)*g(x) or f(x) /g(x). Out of the four options given only option (a) satisfies this condition. Hence the correct  answer is option (a).  

Answer:

A point c in the domain of a function f(x) is said to be critical if ,the value of its first derivative at that point f’(c)=0.

Further (a)  If f’(x) changes sign from positive to negative as x increase through c, then c is said to be                              a point of local maximum.

(b)  If f’(x) changes sign from negative  to positive as x increase through c,then c is said to be a point of local minimum.

(c) If f’(x) does not change sign ,as x increases through c, then c is said to be a point of inflection..

In the present case f(x)= x^3. Therefore   f’(x) =  3 x^2 . By putting f”(x) =3 x^2 = 0 ,we get the

critical point c= 0. It can be seen that f’(x) does not change sign and remains positive as  x

increases through 0{ because f’(x) positive for both negative and positive values of x}.

Hence x= 0 , is a point of inflexion. Moreover x=10 is not a point of inflexion as f'’(10) is not equal to zero ,when x =10.    

Answer: 

Total number of students =   720

Ratio of Boys to Girls=        7:5

Therfore number of boys =     (7/12)*720   =   420

Number of Girls    =              (5/12)*720   =   300

The ratio of Boys and Girls  will be 1:1 if the number of girls equal the number of boys.

At present number of girls less than that of boys=  420- 300  =120.

So 120 girls have to be admitted to make the ratio of boys and girls 1: 1

Hence Correct  option is   120

Answer:

The numbers are in the ratio of  1:1/2: 1/3   or we can say the ratio =  6:3:2.

So the numbers will be in the form of 6x, 3x and 2x respectively.

Average of these numbers,=   (6x+3x+2x)/3   =  (11x)/3  =   121

Therfore  11x=    121*3   or   x=   33.

The difference between first and last number is 4x=  4*33=  132

Answer:

When a train crosses another train the distance travelled by the overtaking train = Its own length+ length of the other train.

However,as the slower train was also moving in the same direction, the faster train had travelled  the above distance 

relative to the slower train.(as if the slower train had not moved.. The relative speed will be difference of the speeds of the 

two trains.

The length of the faster train=         120 metres

Lenght of the slower train=                80 metres

The faster train  after over taking  the slower train,had left it behind a  ditance of    = 100 metres

So total distance travelled by the faster train relative to the slower train =    120+80+100=  300 metres

Total time taken    =  20 secs

Therfore relative speed of the trains =  300/20   =  15 metres/sec       --------------A 

But the speed of the faster train  =       40 metres/sec

Let the speed of the slower train  = x metres/sec

Therfore relative speed of both the trains=  (40 -x) metres sec------------------------B

Equating    A    and  B we get     40-x    = 15 or      x=   25 metres/sec

Hence speed of the secon train    =     25 metres/sec

Answer:    Option   E

A function has to be continuous first for it to be differentiable. It can be easily seen that  the given function ,f(x) is discontinuous whenever the denominator {1-|x|}  is zero.

{1-|x|} will be zero for x= (-)1 and 1. Therefore f(x) will be continuous and differentiable

at all values of x except for x=  (-) 1 and 1.

Hence the exhaustive set of values of x for which the given function is differentiable is

(-∞,-1)U(-1,1)U(1,∞).

Answer  Option C  

Answer Pl refer to attached file

Answer:

We know that the diagonals of a parallelogram bisect each other. Given BD= 7cms, AC=4.8cms and AB= 5 cms. Let the diagonals AC and BD bisect each other at O. Because of bisection we have AO= AC/2 =  4.8/2 = 2.4 cms. Now proceed as follows

1.Draw line segment BD= 7cms.

2.Bisect BD at O.

3.With  O as centre and radius AO= 2.4 cms draw an arc.

4.With B as centre and radius AB = 5cms draw another arc intersecting the above arc at A.

5. Join  AB. Join AD

6.Produce AO to C such that AO=OC=2.4 cms.

7.Join BC and  CD..

ABCD is the required Parallelogram   

Please check the answer attached.

Answer:

Pl refer to the file enclosed

 The nth term Tn=  2n-1.

The nth term (or the gneral term rth term) should be so written that it should yield the value of any term of the sequence correctly..Here we know that the first term is 1,second term is 3 and 3rd term is 5 and so on.

So to find the First term of the sequence you put n=1 in the above expression 2n-1.This yields the value 2*1-1 =1

Similarly for the second term you put n=2 which will yield  a value of 2*2-1=3 and the third term will be 2*3-1 =5 and so on. So we find that  the expression Tn = 2n-1 correctly represents the given sequence.

On the other hand if you represent Tn= 2n+1 ,it will yield first term as 2*1+1 =3, second term as 2*2+1=5 and third term 

as 2*3+1=7 and so on which does not yield correct values for the term of the series and also misses the first term "1" altogether.

So the correct answer is Tn =2n-1

The n th term of the sequence is 6 n^2+2n

Sum of n terms of the equation Sn= Sigma(6n^2+2n) = 6 Sigma(n^2) + 2 Sigma(n) 

Sigma (n^2) is the sum of  Squares of n natural numbers  = n(n+1) (2n+1) /6

Sigma (n) is the sum of n natural. numbers 

= n(n+1) /2.Therefore Sn =

{6*n(n+1) (2n+1) /6 } + 2*n(n+1) /2  = n(n+1) (2n+1) +n((n+1) = n(n+1) (2n+1+1) = n(n+1) ( 2n+2) =

2n(n+1) (n+1)  = 2n(n+1) ^2

Ans : Option C

Answer:

Let the distance between Musau and the town = 240kms

Time taken to reach the town at 60kmph =  240/60 =        4hrs.

Time taken to reach the town at 80 kmph =  240/80 =        3hrs

Time difference =     4- 3 =  1 hr.

Distance travelled for time difference of 1 hr  =                240 kms

Time difference  in the present case = 20 minutes=  1/ 3 hrs( 10 minutes early to 10 minutes late)

Therefore distance travelled for time difference of 1/ 3 hrs =    240*1/ 3=  80 kms

Therefore distance between Musau and the town =  80 kms

Answer:

The first circle is divided into 12 regions and has 8 of them shaded.So this represents fraction: 8/12

The second circle also is divided into 12 regions and has 5 of them shaded.

So this represents a fraction: 5/12

Combinedly they represent a fraction (8/12) +(5/12) =  13/12 = 1 1/12.

So the answer is 1 1/12 =option 4

Answer

Given  y= 3x(X-2)^2.  This is in the form of product of two functions U and V where

U =3x and V= (x-2)^3.Therefore derivative 

dy/dx = U.dV/dx + V.du/dx .That is

dy/dx = 3x{3(x-2)^2} + 3.(x-2)^3 = 9x(x-2)^2+3(x-2)^3

= 3.(x-2)^2(3x+x-2) = 3.(x-2)^2(4x-2)= 6(2x-1)(x-2)^2

Answer:  A vector having zero magnitude with only a particular direcvtion is known as Zero  Vector or Null Vecor.

Correct Option = B

Answer:

It is given that the bodies are kept on a smooth horizontal table.So the frictional forces can be neglected. Also it is assumed  that the  vertical forces are balanced by the reaction offered by the table .     

Thus the motion of the bodies will be as a result of the net horizontal forces.

Body A: The body was moving towards left with a constant speed before application forces.It can be seen thatt the net force acting on the body is 10+5-10 = 5N towards the left.This net force accelerates the body towards left and so the body moves with an increasing speed towards leftt.  

Body B: The body was moving towards right with a constant speed before application of forces.It can be seen that the 

 net force acting on the body is 2+8-10 = 0.As the net force is zero,  the body continues to move with the same initial  speed towards the right.

Thus while body A will continue to move towards left with increasing speed, Body B will continue to move towards right 

 with its initial speed.

Hence Correct Option:  D

Answer: It is given that the measures of the angles of the quadrilaterat are in the ratio of 2:4:5:7.We know that the sum of the 4 angles of the quadrilateral is 360 degrees.Of the 18 parts into which the 360 degress are divided, the angle D will be  the greatest because it mesures 7 of the 18 parts.

Therfore measure of angle D = (7/18)X360  =140 degees.

Answer:  216 is obtained by mutiplying the number 6 3 times repeatedly.

216 =  6x6x6

Answer: We know power dissipated in a resitor is V^2/R,where V is the applied voltage and R is the total resistance in the circuit. Here applied voltage =  V= 120 Volts DC.

Total Resistance = 5.6k+5.6k = 11.2K Ohms=  11200 Ohms

Total power consumption in the circuit =  V^2/R =   120*120/11200= 14400/11200 =9/7= 1.29 Watts

Answer: Please refer to the attached file

Pl refer to the attached file

Answer:

Let x be the maximum marks. Then 10% of the marks =    0.1x = Pass marks -  40.

Therefore Pass marks =   0.1x+40  ………..A

Now 30% of the marks = 0.3x = Pass marks +20  or

Pass marks =  0.3x - 20 …………………….B

Equating the expressions at A and B we get 

0.3x - 20  = 0.1x + 40 or 0.2x = 60.

Therefore   x  = 60/ (0.2x) =  300.

Hence Maximum marks =  300

Answer:

No.of boys =  4

No.of girls =  3.

Total no.of persons =7

As two girls are together,let us treat them as one,then the numbers of persons to be considered 

will be 6.

6 persons can be arranged in a row in 6 ! ways =  720

Now out of 3 girls ,2 girls can be selected in 3 ways .

That is if you denote the girls as G1,G2,G3 you can have 

3 combinations for keeping 2 girls together as G1G2,G2G3 &G3G1

Now these two girls can be arranged between them in 2 ways.

Hence the total number of ways 4 boys and 3 girls can be arranged9with 2 girls always together=

720x3x2 = 4320 ways   

Answer:

4^(x+y) =  64 = 4^3.Comparing the indices of 4 on both sides we get 

 the equation  x+y =    3      ------------------------A

2^(2x-y)  =  128   =2^7 .Comparing the indices of 2 on both sides, we get

the equation    2x-y  =  7     .........................B

Adding  the both the equations "A" and "B" we get   3x = 10 ,that is x = 10/3

and from equation  "A"   y =    3 - x =  3 - (10/3) = (-) 1/3.  

Hence  x= 10/3  and y= (-)1/3

Answer:

Let AB be the line    5x +3y  =  9,     BC be  the line   x = 3y ,  CD be the line 2x=y

and AD be the line x=4y+2 = 0

AB  is 5x+3y= 9 or 3y = 9 - 5x  or  y = (9 - 5x)/3   =  3- (5/3)x.

Slope of AB =  coefficient of x =  (-)5/3

Similarly   BC is x = 3y or  y= x/3. Therefore slope of BC =   coefficient of x =  1/3

 CD is y= 2x. So  slope of CD =  coefficient of x = 2 and

AD is  x+4y+2 =0 or 4y = 2 - x or  y = (2/4) - x/4. So  slope of AD = coefficient of x =  -1/4 

If the above form the four sides of a quadrilateral ABCD,then the intersection of lines

AB& BC will form  angle B, intersection of BC& CD will form  angle C,intersection of

CD&AD will form angle D and finally intersection of sides AD&AB will form angle A

of the quadrilateral  ABCD.

If m1 and m2 are slopes of sides AB&BC then angle B between them is given by formula

Tan B = (m1-m2)/ (1+m1m2.) 

Hence Tan B   = (m1-m2)/(1+m1m2) = ( - 5/3 -  1/3 )/ {1 +(-5/3)(1/3)} =  -2/(4/9) = (-9/2) 

Similarly If m1 and m2 are slopes of sides BC&CD, then 

Tan C = (m1-m2)/ (1+m1m2.)  = (1/3  - 2)/ {1+ 1/3* 2) = (- 5/3 )/ (5/3 )  =  -1

Tan D  = angle between CD& AD  = {2-(-1/4 )}/ {1+2(-1/4 )}  =  (9/4)/ (1/2 )=  9/2

Finally If m1 and m2 are the slopes of AD&AB then   

Tan A = {-1/4 - (-5/3)}/{1+(-1/4)*(-5/3 )}  = ( 5/3- 1/4 )/(1+5/12) = (17/12)/(17/12) = 1 

If ABCD is a cyclic quadrilateral,then the sum of the each pair opposite angles should be 180 degrees .That is   Angle A+ Angle  C  =180 and  Angle C+Angle D = 180

That is tan(A+C) =Tan 180 =  0   and Tan(B+D) = Tan 180 =  0

We know by Trogonometry  Tan (A+ C) =    ( Tan A+ tanC)/(1- Tan A*Tan C)=

{ 1+(-1)}/{1-1*(-1)}  = 0/2 =  0. Hence Angle A+Angle C =  180 Degrees

Similarly Tan( B+D) = (Tan B+ Tan D)/(1-TanB*Tan D)=

{(-9/2)+(9/2)}/1-(-9/2)*(9/2)=  0/ (85/4)= 0.Hence  Angle B+Angle D =180 Degrees

Therefore  ABCD is a cyclic quadrilateral 

Answer:

The elements that appear in a co domain depend entirely on the way we define “Domain” and "Codomain"  of a function. So,it is perfectly alright for a co domain  to have elements which have no  preimages in the domain.That is to say that,if you so desire you can define a domain amd codomain of a function in such a waythat there will be no element in codomain without its corresponding preimage in the domain.

For instance take the function  Y=f(x) = 2x.Suppose you define the domain and co domain as “N”( the set of natural numbers), then you can see that “y” will comprise only of even numbers 2,4,6,8 etc.None of the odd numbers in the codomain  will have any preimage in the domain ,which a set of natural numbers.This happened because we defined both Domain and codomain of the function as "N".This could have been avoided ,if we had defined the domain of the function as "N" but defined the codomain as the set of      non zero even numbers.

By defining co domain specifically  in this manner not to include any odd number in  the co domain,we ensure that there are no elements in the codomain without corresponding pre images in the domain.  In fact,”range”( a subset of co domain )  of a  function   includes only those elements which have pre images in the domain.

Sometimes ,it so happens that either the function is very complex or we do not know enough about the function that the exact range of the function is not known .As a result we have to resort to a codomain which may include some elements for which preimages do not exist in the domain.  

Answer:

The  ith row and jth element of the 3*4 matrix is given by the expression (-3i+4j).

Now (aii)^i   =  (a11)^1 when i =1 and (a22)^2 when i= 2 and (a33)^3 when i =3

Therefore Sigma (aii)^i   where i = 1 to 3 is    {  (a11)^1 +(a22)^2 +(a33)^3}   —----A

Now a11 = ( -3*1+4*1)  =  1         a22  = (-3*2+4*2)  =2   and  a33 = (-3*3+4*3) =3

Putting the above values in A,we get

Sigma(aii)^i where i =1 to 3 is {1^1+2^2+3^3)  = 1+ 4+ 27  =32 = 2^5 = Option C

Answer:

There are only two types of combinations  of chocolates available for distribution that would   

satisfy the given conditions. As there are 8  chocolates of different brands  each chocolate can be treated  as different.    .

a.  5,2,1 combination:  First of let us call the three children  as C1,C2,C3. They can be seated        

  in 6 different orders  as below:

 i)  C1,C2,C3    ii)  C1,C3,C2      iii)  C2,C1,C3    

 iv) C2,C3,C1   v) C3,C1,C2 &   vi) C3,C2,C1

  When the children  are seated in the order C!,C2,C3 ,the first child C1  will receive 5 chocolates,

  second child  C2 will receive 2 chocolates and the last child C3  will receive 1 chocolate.

  With the arrangement C1,C2,C3, the first child C1  can select 5 chocolates out of 

  8  chocolates in 8 C 5 ways =  8C3=  8*7*6/( 1*2*3) ways =  56 ways. 

  The second child C2 can select 2 chocolates out of remaining 3  chocolates in 3 C 2 ways

  that is 3*2/(1*2) =   3 ways. Finally the last child can select the remaining 1 chocolate in 1 way. 

As all these selections are independent of each other,the total number of ways the 

three children sitting in the order C!,C2,C3 can select the 8 chocolates is = 56*3*1= 168 ways

But we have 6 different orders  in which the 3 children can be seated. With every order 

the children can select the chocolate in 168 ways.Therefore total number of ways for

the 3 children selecting 8 chocolates in 5,2,1 combination  is =  168*6 =  1008 ways.   --------A 

b.   4,3,1 combination : Here again the children can be seated in 6 different orders as above.       When children are seated in C1,C2,C3 order , C1 can select 4 chocolates out of 8   Chocolates in 8C4 ways = 8*7*6*5/(1.2*3*4)   =   70 ways. C2 can select 3 chocolate  out of remaining 4 chocolates in 4 C3 = 4C1 ways = 4 and finally C3 can select the remaining   one chocolate in 1 way.                

As all these selections are independent of each other,the total number of ways the 

children sitting in the order C!,C2,C3 can select the 8 chocolates is = 70*4*1= 280 ways.

Again,as we have 6 different  orders  in this case also,therefore total number of ways for the three children selecting 8 chocolates in 4,3,1 combination is 280*6  =  1680 ways ----------------------B

Further, when you select 5,2,1 combination for distribution of chocolates,you cannot select the 

4,3,1 combination and if you select 4,3,1 combination,you cannot select 5,2,1 combination.

Thus both the selections are mutually exclusive. Therefore total number of ways of distributing

8 chocolates in either of the combinations =  A+B=  1008+1680  = 2688

Therefore total number of ways of distributing 8 chocolates among 3 children = N= 2688

Sum of  the digits of N = 2+6+8+8 =24   

Answer:

Please refer to the attached file

Answer:

There are only two combinations in which 22 employees can be arranged in 3 shifts such that at no time there will  more than 8 employees in one shift. They are i) 8+8+6  and ii) 8+7+7.

That is in the first combination, a group of 8 employees will be arranged ,say in I shift,another group of      8 employees  in II shift and the remaininmg employees in the third shift

In the second combination a group of 8 employees will be kept,say  in I shift,another group of 7 employees will be kept in II shift and the remaining 7 employees in the III shift.Of course,you can put any group in I shift or II shift or III shift as per your convenience. . 

i) 8+8+6: Let us call the first group of 8 employees as A and the second group of 8 employees B and the third group of 7 employees as C. Now let these 3 groups of employees  be arranged in First ,second and third  shift in the order of A,B,C respectively.It can be ensured that each employee gets 3rd shift once in 3 weeks by rotating their shifts in a cyclic manner.That is in the First week A,B,C come in I,II,III shiftsrespectively and in the second week the order will be B,C,A and in the third week the order will be  C,A,B.In fourth week the order A,B,C gets repeated and for the fifth  week, the order will be again B,C,A and so on. As already stated earlier, you can put any group in any shift as per your convenience.However after putting them in shifts in a particular order ,you have to  follow the cyclic order so as to ensure that each group gets one I shift,one II shift and one III shift once in three weeks. The following table shows the shift pattern for the above combination for 12 weeks.It can be easily seen that each employee gets one I shift,one II shift and one III shift once in 3 weeks...

Week  I Shift     II shift    III Shift    I Shift    II Shift   III Shift            

1           A          B             C             D       E           F   

2           B          C            A              E      F            D

3            C         A            B              F      D            E

4           A          B             C             D       E           F   

5           B          C            A              E      F            D

6            C         A            B              F      D            E

7           A          B             C             D       E           F   

8           B          C            A              E      F            D

9            C         A            B              F      D            E

  10         A          B             C             D       E           F   

11           B          C            A              E      F            D

12           C         A            B              F      D            E

Now there are 6 ways you can select the order of the groups.They are  ABC, ACB, BAC,BCA,CAB,CBA That  is you can choose ABC to  be in I,II and III shift respectively and then rotate them in cyclic order asabove.Or you can choose ACB to be in I,II and III shifts and rotate them in shifts in cyclic manner.Thus you have 6 patterns to begin with.  

ii) 8+7+7: Let us call them D,E,F.You can arrange these three  groups in I,II and III shift respectively and rotate their shifts in a cyclic order of DEF,EFD,FDE for the first  3 weeks and repeat them for the next 3 weeks and the next 3 weeks and so on.Here again you have 6 ways to select the order of the groups.They are DEF,DFE,EFD,EDF,FE and FDE. Thus you have in all 6+6 =12 patterns in which you can prepare the shift schedule for the 22 employees.

Answer:  The given expression is 4n^3+ 6n^2+4n+1 =  4n^3+2n^2+4n^2+4n+1

= 2n^2(2n+1) +(2n+1)^2   = (2n+1){2n^2+(2n+1)} =  (2n+1)(2n^2+2n+1).

That is the given expression can be expressed as a product of 2 factors and both of these  factors are

natural numbers for all natural(number) values of "n". As the expression is a product of two natural numbers  it is a composite number for all natural values of n.

Answer:

The pressure at the base of a liquid column is given by the formula P =hdg. Or h=P/(dg).  If, h ,the height of the liquid column is in metres , d =density of the liquid in  Kgs/metre^3 and g= acceleration due to gravity in metres/sec then the      pressure P is measured in Pascals

It is given that the pressure desired at the base of the water tank tower =  60 Psi.By using the conversion formula                1 Psi = 6895 Pascals ,d= density of water =1000Kgs/Metre^3 and g= acceleration due to gravity =9.8 metres/sec^2 ,we 

have h =  P/(dg) 60*6895/(1000*9.8)  = 60*6895/9800=  42.21 metres.Hence the water tank should be at a minimum height of 42.21 metres to provide a pressure of 60Psi at the base of the water tower

Answer:

Let the speed of the first train =  x metres/sec. When a train crosses a pole it travels a distance equal to its own length.The time taken by the train to cross a pole =  5 secs.                                            Therefore length of the first train =  5*x =  5x metres.

When the train crosses a platform it  travels the entire length of the platform and also its own length . The time taken by the train to cross the platform = 15 secs .                                                                                                                        

Therefore the distance traveled by the train =  15*x = 15x.Hence length of the platform=                    

Total distance traveled  - length of the train = 15x -  5x =  10x metres.                                    

Now when two trains cross each other the distance traveled by any one train equals the sum of the lengths of both the trains. Length of the second train =  520 metres

Therefore  total distance traveled by the first train = ( x+520) metres.

Speed of the second train = 44kms/hr =44*1000/3600 =110/9 metres/sec

When two  trains  cross each other,they will cross each other with their relative speed.As both the trains are traveling  in opposite directions,their  relative speed will be sum of the   speed of  individual trains. Therefore relative speed of both the trains = (x+110/9) metres/hr

Time taken to cross each other = Total distance traveled/ Relative speed of the trains= 12

= (5x+520)/{x+(110/9)} =  9(5x+520)/(9x+110) = 12. By cross multiplying we get

9(5x+520) =12(9x+110).Simplifying and solving for x we get x = (1120/21) metres.

Now length of platform = 10x metres. Length of first train = 5x metres. 

Therefore difference between length of the  platform and length of first train = 10x - 5x =5x

=  5*1120/21 = 5600/21 = 800/3 =266.66 metres

Answer:

306  =  3*3*34

657 = 3*3*73

Therefore LCM   =  3*3*34*73 = 22338

Answer:

For the quadratic equation  ax^2+ bx+c =0, the sum of the roots=  -b/a; that is sum of the roots= 

  negative  coefficient of x/ co efficient of x^2 .

Here the quadratic equation is 2x^2 -6kx+6x-7 =0. that is 2x^2 -6(k-1)x - 7 = 0

Thetfore  sum of the roots =   6(k-1)/2 = 3(k-1). As one root is the negative of the other the sum of the roots = 0

Hence3(k-1) = 0 or k=1

Therfore k=1

Answer: Please refer to the attached file

Answer: Please refer to the file attached

Answer:   Pl refer to the attached file