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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
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