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DownloadGiven, AB is the diameter of the larger circle.
P and Q are the Centres of the smaller circles.
M is the point where the circles with centres P and Q touch each other.
So, AB = AM + MB ----------------------------- (i)
Thus, AM is the diameter of the circle with centre P
So, AM = AP + PM
= PM + PM (AP = PM, both being radii of the circle with centre P)
= 2PM ----------------------------- (ii)
And MB is the diameter of the circle with centre Q
So, MB = MQ + QB
= MQ + MQ (MQ = QB, both being radii of the circle with centre Q)
= 2MQ ----------------------------- (iii)
From (i), AB = AM + MB
= 2PM + 2MQ (from ii and iii)
AB = 2(PM + MQ)
or, PM + MQ = AB/2
or, MQ = AB/2 - PM
= (AB - 2PM)/2
Total weight of mixture = 25 kg Total weight of water = 20% of 25 kg = 5 kg. Now, 20% weight = 5 kg Let the extra amount of water be 'x' kg. Thus, (5+x)/(25+x)×100 = 33.33 Or, 500 + 100x = 33.33x + 833.35 Or, (100 - 33.33)x = 833.35 - 500 Or, 66.67x = 333.35 Or x = 5 kg. So, the weight of water required to be added = 5 kg.
Let the total population be 'x'. Now, males:females = 3:2 Sum of ratios = 3+2 = 5 So, males = 3x/5 And females = 2x/5. No. Of male graduates = 20% of 3x/5 = (20/100)(3x/5) = 3x/25 No. Of female graduates = 25% of 2x/5 = (25/100)(2x/5) = x/10 Thus, no. Of graduates = 3x/25 + x/10 = 11x/50. Thus, no. Of non-graduates = x - 11×/50 = 39x/50 Hence, percentage of non-graduates = (39x/50)/(x)×100 = 39×2 = 78% So, the correct answer is Option D.
The No. of marbles kept originally = 15
no. of white marbles = 5
Thus, fraction of white marbles = 5/15 = 1/3
Let the number of white marbles added be 'x'.
Thus, (5+x)/(15+x) = 3/5
or, 5(5 + x) = 3(15 + x)
or, 25 + 5x = 45 + 3x
or, 5x - 3x = 45 - 25
or, 2x = 20
or, x = 10.
So, the correct option is (D).
There are mistakes and missing informations in the question. Where is M? Are you sure that it is AM and not AD? Also, isn't <CAB = 90 degrees?
Surface Area of a Sphere = 4πr^2
Curved Surface Area of Cylinder = 2πrh
Thus, ratio of Surface Area of a Sphere : Curved Surface Area of Cylinder = 4πr2 : 2πrh
= 2r : h
= d : h
So, the correct option is (A).
Option B. OM = 2 cm and OQ = 7 cm
Given, Mode = 2
So, 3x - 10 = 2
or, 3x = 10 + 2
or 3x = 12
or, x = 4
And 4y + 6 = 2
or, 4y = 2 - 6
or, 4y = - 4
or, y = - 1
Thus, x + y = 4 - 1 = 3
So, the correct option is B.
Given, (a + b)/2 = 9
Thus, a = 18 - b
And, b = 18 - a
Also, c = 5a - 4
Thus, c = 5(18 - b) - 4
= 90 - 5b - 4
= 86 - 5b
So, (b + c)/2 = (b + 86 - 5b)/2
= (86 - 4b)/2
= 43 - 2b
= 43 - 2(18 - a)
= 43 - 36 + 2a
= 7 + 2a
So, Option C is correct.
Even numbers between 21 and 43 are:
22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42
Thus, Median = 6th number
= 32
So the correct option is B.
Re-arrenging the series:
1, 2, 4, 5, 7, 8, 10, 10, 12, 14, 15
Since, the series has 11 numbers, so the Median will be the 6th number = 8
And as 10 is the most repeated number so the Mode = 10
Hence, Option A is the correct answer.
Correct option is (B).
Working shown in the attached file.
The question is erroneous. Instead of <QAB = 90, it should be <PQB = <PQA = 90
The solution is provided in the attachment.
Please check the attachment for the solution.
Option C
Option D
Let the monthly salary be R 100.
Money given to mother = R 25
Money remaining = R 75
Money paid for rent = R 11.25
Money kept aside for monthly expenses = R 18.75
Thus, money kept in bank = R [100 – (25 + 11.25 + 18.75)]
= R [100 – 55]
= R 45
Sum of amount given to mother and amount kept in bank = R [25 + 45] = R 70
When assumed sum is R 70, then actual sum is R 42000
So, when assumed sum is R 100, then actual sum is R 60000
Speed of car in the 1st part of the journey = 39(45/60) km/hr = 52 km/hr
Speed of car in the 2nd part of the journey = 25 (35/60) km/hr = 300/7 km/hr = 42.85 km/hr
Average speed = (52 + 42.85)/2 km/hr = (94.85)/2 km/hr = 47.43 km/hr
So, the correct answer is E (None of the above).
Let the duration of the investment be ‘t’ and let the annual profits be divided proportionately by each.
Also, let the total profits be ‘x’.
So, A = R 64000; B = R 52000; C = R 36000
Hence, A:B:C = 64:52:36 = 16:12:9
Hence, the profit shares = 16x : 12x : 9x
If 16x = 35584
And 9x = (9 x 35584)/16 = R 20016
So, C’s share of the annual profits = R 20016
So, the correct option is E.
The solution is worked out in the attached document.
Hence, from the derivation in the attachment,
h = 4a
or h = 4(8) where a = 8
or h = 32 cm
and r = 8root2
Let the price of the bike be R 100.
Thus, increased price = 25% of R100 = R125
Now, decrease in % of price with respect to the new price = (25/125) x 100 = 20%
Let the number be ‘x’.
Therefore the new number = 2x/3
And the actual new number = 3x/2
Difference = (3x/2 – 2x/3) = (9x – 4x)/6 = 5x/6
Thus, percentage error in calculation = (5x/6)/(3x/2) x 100 = (10x/18x) x 100 = 1000/18 = 55.56%
Let the original fraction be x/y
Now, x + 25% of x = 5
Or, x + 25x/100 = 5
Or, 5x/4 = 5
Or, x = 4
And, y – 10% of y = 9
Or, y – 10y/100 = 9
Or, 9y/10 = 9
Or, y = 10
Thus, x/y = 4/10 = 2/5
Let the distance to be covered be x km
Ratio of time taken:
A:B:C = 12:15:10
Sum of ratios = 12 + 15 + 10 = 37
Speed of A = 60 km/hr
Thus, 12x/37 = 60
Or, x/37 = 5
Or, x = 5(37) = 185
Hence, Speed of B = 15(185)/37 = 75 km/hr
Let the charges per hour be R 100
Thus, increase in charges = 25% of 100
So, new charges = R 125
Let the time taken to play a game be ‘T’ hr
Thus, the decreased time = t hr
Now, actual expenditure = R 110
So, 125t = 100T + 10T
Or, 125t = 110T
Or, t = 22T/25
Now, decrease in time = T – t = T – 22T/25 = 3T/25
And % decrease = [(3T/25)/T]100 = 12%
[(4356)/2]/(11/4) = 792
792/6 = 132
So, 132 x 132 = √17424
so, ? = 17424
centre = (5, 7)
x2 + y2 + 4x + 8y – 45 = 0
x2 + 4x + y2 + 8y – 45 = 0
or, x2 + 4x + 4 + y2 + 8y + 16 = 25
or, (x + 2)2 + (y + 4)2 = 52
so, centre = (– 2, – 4)
Let the Time taken be T hr.
So, for T/2, Distance = 80 x T/2
And, for T/2, Distance = 40 x T/2
total distance = 60 = 80T/2 + 40T/2
= 60 = 120 T/2
= 60 = 60T
= 1 hr = T
So, Avg. Speed = 60/1 = 60 km/hr
A
I
I
C ------------------------------------I B
The question is wrong as C cannot be to the West of A. There is some missing data in this case.
The Relative Velocity of the ground with respect to Train B is 90 km/hr.
C/5 = (F-32)/9 where C = Celsius and F = Farenheiht
1. Archemedes Number
2. Chandrasekhar Number
3. Decibel
R = [–(–10) ± √(100 – 1160)]/2x29 = [100 ± √(– 1060)]/58 = [50 ± √(– 265)]/29 = [50 ± 16.28i]/29
A = - 14
T5 = 2
Now, t5 = a + (n-1)d
2 = – 14 + (5 – 1)d
16 = 4d
D = 4
Sn = n/2[2a – (n – 1)d]
40 = n/2[– 28 – (n – 1)4]
80 = n[- 28 – 4n + 4]
80 = n[- 24 – 4n]
20 = n[6 – n]
n2 – 6n + 20 = 0
n = [–(–6) ± √(36 – 80)]/2 = [6 ± √(– 44)]/2 = [6 ± 2√(– 11)]/2 = 3 ± 1(3.32) = 6.32 or 0.32
A = - 14
T5 = 2
Now, t5 = a + (n-1)d
2 = – 14 + (5 – 1)d
16 = 4d
D = 4
Sn = n/2[2a + (n – 1)d]
40 = n/2[– 28 + (n – 1)4]
80 = n[- 28 + 4n - 4]
80 = n[- 32 + 4n]
20 = n[-8 + n]
n2 – 8n - 20 = 0
n2 - 10 n + 2n - 20 = 0
n(n - 10) + 2(n - 10) = 0
(n - 10)(n + 2) = 0
n = 10 or - 2
As n cannot be negetive, na = 10 isthe correct answer.
Let the age of the son be ‘x’ years.
Thus, Mr. Mohan’s age = 4x years
After 10 years,
Son’s age = x + 10 years
Mr. Mohan’s age = 4x + 10
Now, 2(x + 10) = 4x + 10
Or 2x + 20 = 4x + 10
Or 2x = 10
Or x = 5 years
X-ray
Newton's third law of motion states that 'for every action there is an equal and opposite reaction'. When a gun is being shot, there are forces acting within it to make the bullet move forward and out. These forces are in the form of the kinetic energy of the bullet and the potential energy that is tored in it. When the bullet moves out with the help of the combination of force, acceleration and energy, there is also an equal and opposite combination of foce, energy and acceleration working on it, known as the 'recoil'. This recoil leads us to fall back. This is the same principle that provides 'thrust' to rockets before they can lift off to space.
The Archemedes' principle is used in designing ships, submarines, boats and hydrometers.
Archemedes himself used it to find out the purity of gold.
Fishes use the Archemedes priciple to rise up to the surface of waterbodies or to sink doen to their bottom.
Power (D) = 1/Focal Length (F)
or, 0.2 = 1/F
or, F = 1/0.2
= 10/2
= 5 m.
Hence, the options given are wrong.
1 Newton is the measurement of the force exerted when a mass of 1 kg moves with an acceleration of 1 m per second sq.
Hence, 1 N = 1 kg x 1 m/s2
= 1 kgm/s2
The vector product of two vectors a and b is given by a vector whose magnitude is given by |a||b|sinθ (where0∘≤θ≤180∘), which stands for the angle between the two vectors. Note that the direction of the resultant vector is denoted by a unit vector ^n whose direction is perpendicular to both the vectors a and b in a way that a, b and ^n are oriented in right-handed system.
Right-handed orientation happens when vector a is twisted in the direction of vector b, then the direction of the unit vector ^n goes in the direction in which a right-handed screw would spin if moved in a similar manner. Also, these given vectors a and b cannot be called null vectors or non-parallel in nature. The right-hand thumb rule gives a clear picture of the direction of the resultant vector.
Therefore, we can conclude that, a×b = |a||b|sinθ ^n, where a×b stands for the cross product of two vectors. In any given situation, if the vectors are null or both the vectors are parallel to each other, then the cross product cannot be defined.
In this case, we can conclude that a×b = 0
Physical Representation
Let’s assume that a and b are the adjacent sides of the parallelogram OACB and the angle between the vectors a and b is θ. Then, we can say that the area of the parallelogram is denoted by |a×b| = |a||b|sinθ.
Properties of Cross Product of Two Vectors
i) The vector product never has a Commutative Property. It is denoted by,
a×b = – (b×a)
ii) The property given below is true in the case of vector multiplication:
(ka)×b= k(a×b) =a×(kb)
iii) If the vectors mentioned are collinear then
a×b= 0
(Since the angle between both the vectors would be 0, then sin 0 = 0)
iv) As per the above property
We can conclude that the vector multiplication of a vector with itself would be a×a= |a||a|sin0 ^n = 0 Also, when it comes to unit vector notation
^j×^j=^j×^j=^k×^k =0
As per the above discussion
^i×^j=^k=−^j×^i
^j×^k=^i=−^k×^j
^k×^i=^j=−^i×^k
This example can be explained better with the help of the following diagram. When moving in clockwise direction and considering the cross product of any two pair of the unit vectors, we can derive the third one and we get the negativeresultant in anticlockwise direction.
v) a × b in terms of unit vectors can be represented as
a =a1^i+a2^j+a3^k
b =b1^i+b2^j+b3^k
Then →a×→b =(a1^i+a2^j+a3^k)(b1^i+b2^j+b3^k)
When expanded, we would get
|a||b|sinθ ^n = (a2b3–a3b2)^i+(a3b1–a1b3)^j+(a1b2–a2b1)^k
vi) Distributive Law: a×(b+c) = a×b+a×c
Moment of inertia of a rod of length r = 1/3 (mr2)
Given, mass of rod = m, length = l and < of rotation = theta
Let, the radius of the circle of raotation be r.
Thus, r = l cos theta
and I = 1/3 (l cos theta)2
A ray of light bends when it travels from a optically denser medium to an optically lighter one or vice versa. This phenomenon is known as Refraction of Light.
This change of direction of light is because of the change in the speed of light in different media. Also, the angle at which the ray of light moves from one medium to another, influences the angle of refraction of light. The greater the angle, the greater is the refraction.
Let the age of the son be x years.
Thus, Mohan's age is 4x years.
After 10 years,
Son's age is (x + 10) years
ans Mohan's age is (4x + 10) years
Now, 4x + 10 = 2(x + 10)
Or, 4x + 10 = 2x + 20
Or, 2x = 10
Or, x = 5
So, present age of Mohan's son is 5 years.
Your question is incomprehensible. Please rewrite the question for us to answer it properly.
Given a = - 15; d = 9; n = 30
Thus, S30 = 30/2 [2a + (30 – 1)d]
= 15 [2(–15) + 29(9)]
= 15 [–30 + 261]
= 15[231]
= 3465
Let the CP = ₹ 100
Thus, MP = ₹ 140
Discount = 10%
So, SP = ₹ 140 - ₹ 14 = ₹ 126
So, Profit = ₹ 26
When Profit = ₹ 26, CP Profit = ₹ 100
When Profit = ₹ 325, CP Profit = ₹ (325 x 100)/26 = ₹ 1250
Let the CP = ₹ 100
Thus, MP = ₹ 140
Discount = 10%
So, SP = ₹ 140 - ₹ 14 = ₹ 126
So, Profit = ₹ 26
When Profit = ₹ 26, CP = ₹ 100
When Profit = ₹ 325, CP = ₹ (325 x 100)/26 = ₹ 1250
Physical quantities having units but no dimensions are: plane angle, angular displacement and solid angle.
Let mass of the body be ‘m’ and velocity be ‘v’.
So, KE = ½ (mv2)
Now, increase in mass = 10% of m
Thus, new mass = m + m/10 = 11m/10
Increase in velocity = 20% of v
Thus, new velocity = v + v/5 = 6v/5
Hence, new KE = ½[(11m/10)(6v/5)2]
= 198mv2/250
So, increase in KE = 198mv2/250 – mv2/2 = 73mv2/250
Hence, % increase in KE = [(73mv2/250)/(mv2/2)]100 = 58.4%
2. (i) Speed may be described as the distance traversed by a body with rerspect to time.
(ii) Instanteneous velocity may be described as the velocity of a body at a certain point of time.
(iii) Velocity may be defined as the displacement of a body or a the ditance covered by it in a particular direction with respect to time.
(iv) Acceleratio is the rate of change of velocity with rspect to time.
Given focal length = 10 mm = 10/1000 m = 1/100 m = 0.01 m
Thus, P = 1/f = 1/0.01 D = 100D
Pressure = Force/Area = mg/cm2
Density = mass/volume = m/cm3
Thus, m x mg/cm2 x cm3/m = mg cm = mgh = Potential Energy.
Thus, it is the formula for finding Potential Energy.
Since the bodies are of equal dimension and masses, then the centre of mass will be at the centre of the sphere B where the spheres are arranged as A, B and C respectively. Hence, distance of the Centre of Mass from the centre of A is equal to its diametre as its radius is the same as the radius of sphere B.
Given, v = v0; u = 0; height = S and acceleration = g
So, v02 = u2 + 2gS
As u = 0,
v02 = 2gS
S = v02/2g
While, tripling the height, the acceleration due to gravity remains the same.
So, 3S = 3 v02/2g
Or, H = 3S
= (√3 v0) 2/2g
Hence, the velocity in case the height is tripled is √3 v0
Relative density of a substance means the density of an object at a particulart temperature with respect to the density of water at the same temperature.
A pentavalent dopant such as Antimony are known as donor impurities since they donate an extra electron in the crystal structure which is not required for covalent bonding purposes and is readily available to be shifted to the conduction band. This electron does not give rise to a corresponding hole in the valence band because it is already excess, therefore upon doping with such a material, the base material such as Germanium contains more electrons than holes, hence the nomenclature N-type intrinsic semiconductors.
On the other hand when a trivalent dopant such as Boron is added to Germanium additional or extra holes get formed due to the exactly reverse process of what was described in the upper section. Hence this dopant which is also known as acceptor creates a P-type semiconductor.
The portion of the energy band structure of a semiconductor or insulator between the valence andconduction bands. This portion does not contain a continuum of states as do the valence and conduction bands but may contain discrete states introduced by dopant or deep level impurities in the material.
The speed of a film (or film speed), refers to the measure of a film's sensitivity to light. Each film speed is best suited for a different type of photography. The lower the speed, the longer an exposure to light is necessary to produce image density. If the film speed is higher, it requires less exposure but generally has reduced quality in the form of grain and noise. Noise and grain are the abnormalities in brightness and colour in images; they look similar to a layer of "snow" on a television set. They're measured using the ISO system from the International Organization for Standardization, and are the giant numbers typically seen on a box of film. The abbreviation ASA (American Standard Association) is also used in conjunction with film speed. ASA and ISO are interchangeable.
Total time taken = 1 hr and 30 minute = 90 minutes
Total stoppage time = 12 x 2 = 24 minutes
Thus, total travelling time = 90 – 24 = 66 minutes = 1 1/10 hr = 11/10 hr
Speed = 40 km/hr
Thus, distance = Speed x Time = 40 x 11/10 = 44 km
Let the Principal be P and Rate be R
Thus, A = 2P
Hence, SI = A – P = 2P – P = P
T = 10 years
Thus, R = (100 x P)/(P x 10) = 10%
CP = ₹ 12500; Transport = ₹ 300; Installation = ₹ 800
Thus, total CP = ₹ 13600
P% = 15% = (P/CP)100
Hence, P = (15 x 13600)/100 = ₹ 2040
Thus, SP = CP + P = ₹ 13600 + ₹ 2040 = ₹ 15640
Investment of A:B:C = ₹ 12000 : ₹ 15000 : ₹ 18000 = 4 : 5 : 6
Sum of ratios = 4 + 5 + 6 = 15
A’s share = 4x/15
B’s share = 5x/15
C’s share = 6x/15
A’s share of profit = ₹ 3000
Thus, 4x/15 = ₹ 3000
Or, x = ₹ (750 x 15) = ₹ 11250 = Total Profit
Let the CP be ₹ 100.
Therefore, the MP is 40% of CP = ₹ 140.
Discount is 20%of MP = ₹ (20 x 140)/100 = ₹ 28.
Thus, actual SP = ₹ (140 – 28) = ₹ 112.
Thus, Profit = ₹ (112 – 100) = ₹ 12.
Profit % = (P/CP) x 100 = (12/100) x 100 = 12%
Given A = {2, 4, 5} and B = {7, 8, 9}
Thus, A x B = {(2, 7), (2, 8), (2, 9), (4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9)}
Hence, n(A x B) = 9
So, the correct option is B.
Public
Let the distance travelled be ‘x’ km on each side.
Speed (Uphill) = 10 km/hr
Thus, Time (Uphill) = x/10 hr ……………… (i)
Speed (Downhill) = 36 km/hr
Thus, Time (Downhill) = x/36 hr ……………… (ii)
Total time = x(1/10 + 1/36) hr
= x(18 + 5/180) hr
= 23x/180 hr
Average time = 23x/360
Total distance = x km
Hence average speed = (x/23x)360 = 360/23 = 15.65 km/hr
Let the fourth proportional be x
Thus, 100:50 :: 40:x
Or, 100/50 :: 40/x
Or, 2 :: 40/x
Or, x = 40/2 = 20
–2x2 + 7x – 6
= 2x2 – 7x + 6
= 2x2 – 4x – 3x + 6
= 2x(x – 2) – 3(x – 2)
= (x – 2)(2x – 3)
Or, x = 2 and 3/2
The mean age of 25 students is 12 years.
Thus, total age of 25 students = 25 x 12 = 300 years
The mean age of 25 students and 1 teacher = 13 years
Thus, total age of 25 students and 1 teacher = 26 x 13 = 338 years
Thus, the age of the teacher = 338 – 300 = 38 years
Given,
Distance travelled in 30 minutes = 30 km
Thus, v1 = 60 km/hr
And distance travelled in 40 minutes = 30 km
Thus, v2 = 45 km/hr
Thus, average velocity = 52.5 km/hr
Given,
Distance travelled in 30 minutes = 30 km
Or, distance travelled in hour = 30 km
Thus, v1 = = =
And distance travelled in 40 minutes = 30 km
Or, distance travelled in = hours = 30 km
Thus, v2 = = =
Thus, average velocity = = =
As the speed of the stream has not been mentioned, the correct option is (d).
For the given question, all three statements are sufficiaent to answer it.
The dimensions of a/b would be in meters/second.
Time taken to travel first 30 km = 30 min = ½ hours
Thus, Velocity1 = 60 km/hr
Time taken to travel next 30 km = 40 minutes = 2/3 hours
Thus, Velocity2 = 45 km/hr
Thus, average velocity = (60 + 45)/2 = 105/2 = 52.5 km/hr
The other name for Boolean Algebra is Binary Algebra or Logical Algebra. It was developed in 1854.
Ratio of men:women = 5:6. Thus, sum of ratios = 5+6 =11 Let the total number of members = x So, men candidates = 5x/11 And, no. of women = 6x/11 Now, new number of men = 5x/11 + 20/100(5x/11) = 6x/11 And, new number of women = 6x/11 + 10/100(6x/11) = 6x/11 (1 + 1/10) = 6x/10 Hence, new ratio = 6x/11:6x/10 = 11:10
Acceleration can be opposite to velocity if the object experiencing the same is slowing down.
The direction of tension of a string will always be opposite to the direction of the mass on the string. If there is a mass hanging down from a string then the direction of tension will be upward because the string will always try to pull the mass upwards.
A suitable unit for Gravitational Constant is G, where G = 6.673×10^(-11) N m2 kg-2
The first principal focus is defines as a fixed point on the principal axis such that rays starting from this point (in convex lens) or appearing to go towards this point (concave lens), after refraction through the lens, becomes parallel to the Principal Axis. It is represented as F1.
As D = 1/f,
so focal length of the lens will be 1/2f.
Also, as D is negetive, the nature of the lens is concave.
Let the age of Bineesh be 'x' years
Thus, Kiran's age is 'x - 7 ' years.
now, Kiran : Bineesh = 7 : 9
Thus, 9(x - 7) = 7 x
or, 9x - 63 = 7x
or, 9x - 7x = 63
or, 2x = 63
or, x = 31.5 years
Thus, Bineesh's age is 31.5 years
Hence, Kiran's age is 31.5 - 7 = 24.5 years
Given, m : w = 5 : 6
Sum of ratios = 5 + 6 = 11
Thus, no. of men = 5x/11
And no. of women = 6x/11
New Composition = (m + 20m/100) : (w + 10w/100)
= 6m/5 : 11w/10
= 12m : 11w
= 12(5x/11) : 11(6x/11)
= 60x/11 : 66x/11
= 10x : 11x
= 10 : 11
15pq + 15 + 9q + 25p
By rearranging we have,
15pq + 25p + 9q + 15
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5)(5p + 3)
Let the no. of registered voters be x
No. of votes cast = 60% of x = 3x/5
A – B = 600 … … … … … … (i)
And, A + B = 3x/5 … … … … … … (ii)
Now, B + 40% of B = A
Or, 7B/5 = A … … … … … … (iii)
Or, 7B = 5A
From (i) and (iii),
7B/5 – B = 600
Or, 7B – 5B = 3000
Or, 2B = 3000
Or, B = 1500
Thus, A = 1500 + 600 = 2100
Substituting A and B in (ii),
2100 + 1500 = 3x/5
Or, 3600 = 3x/5
Or x = 6000
Thus, no. of registered voters = 6000
x[25% of x] = 200% of x + x
Or, x[x/4] = 3x
Or, x = 12
So, correct option is (i)
At any time t, a projectile's horizontal and vertical displacement are:
x = VtCos θ where V is the initial velocity, θ is the launch angle
y = VtSinθ – ½gt^2
The velocities are the time derivatives of displacement:
Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)
Vy = VSinθ – gt
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g [total flight time = 2t]
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g ….. (1)
The maximum height H is
H = V^2 Sin^2(θ) / 2g ….. (2)
In this case, we know that V = 30m/s and θ = 30°
Therefore, after 1s,
Vx = VCosθ = 30Cos30 = 25.98m/s
Vy = VSinθ – gt = 30Sin30 – 9.81(1) = 5.19m/s
The total velocity after 1s is V(1) = √(25.98^2 + 5.19^2) = 26.49m/s
A. Three classes - 1st class (load > fulcrum > effort); 2nd class (effort > load > fulcrum); 3rd class (load > effort > fulcrum)
N/m or dyn/cm
Let his total income be Rs 100
Thus, Income given to his elder son is 30% of 100 = Rs 30
Income given to younger son = 20% of 100 = Rs 20
Thus, income remaining = Rs 100 - (20 + 30) = Rs 50
Income given to the beggar = 10% of 50 = Rs 5
Income reamining = Rs (50 - 5) = Rs 45
When income left behind is Rs 45, then actual income is Rs 100
When income left behind is Rs 10080, then actual income is Rs (10080 x 100)/45 = Rs 22400
Since both the numbers are prime numbers, then they can either be divided by 1 or by themselves. Thus, the product of the prime numbers is also the LCM ofthe prime numberd.
Hence, the LCM is 493.
Almost all metals form amalgams with Mercury.
The strain experienced by an object is directly proportional to the stress applied. Stress is the force applied on a body to induce change in its shape.
We know that v = (3RT/M)1/2
If T is increased 4 times the,
v = (3R4T/M)1/2
or, v = 2(3RT/M)1/2
v = 2v
Thus, the increase in T to four times its values will lead to increase in RMS velocity to twice its value.
Milligram, centigram, decigram, gram, decagram, hectogram
Given, x = 2t3 – 3t2 + 2t + 2
Thus, differentiating with respect to t, we have,
= 3(2t2) – 2(3t) + 2
= 6t2 – 6t
= 6t(t – 1)
Or, = 12t – 6
Substituting t = 2 sec,
x = 2(2)3 – 3(2)2 + 2(2) + 2 = 16 – 12 + 4 + 2 = 10 m
x' = 6(2)(2 – 1) = 12 m/s
x" = 12(2) – 6 = 18 m/s2
(11.92)^2 + (16.01)^2 = ?^2 + (3.85)^2
Let ? be x
So, (11.92 + 16.02)^2 - 2(11.92x16.01) = x^2 + (3.85)^2
or, (27.94)^2 - (3.85)^2 - 11.92x32.02 = x^2
or, (27.94 + 3.85)(27.94 - 3.85) - 381.68 = x^2
or, 31.79 x 24.09 - 381.68 = x^2
or, 765.82 - 361.68 = x^2
or, 404.14 - x^2
or, x = 20.10
Total amount = Rs 60.
Let the number of books bought initially = x
thus, price per book = Rs (60/x)
Now, if five more books were bought for Rs 60,
price of each book = Rs (60/x + 5)
now, 60/x - 60/(x + 5) = 1
or, 60[1/x - 1/(x + 5)] = 1
or, 60[x + 5 - x] = x(x + 5)
or, 300 = x^2 + 5x
or, x^2 + 5x - 300 = 0
or, x^2 + 20x - 15x - 300 = 0
or, x(x + 20) - 15(x + 20) = 0
or, (x + 20)(x - 15) = 0
thus, x = 15
Given, sides of cuboid = 5cm , 3 cm, 2 cm
Thus, LCM of sides = 5 x 3 x 2 = 30 cm
So, side of cube = 30 x 30 x 30
Hence, no. of cuboids required = (30 x 30 x 30)/(5 x 3 x 2) = 900
(11.29)^2 + (16.01)^2 = x^2 x (3.85)^2
or, (12)^2 + (16)^2 = x^2 x 4^2
or, (12 + 16)^2 - 2 x 12 x 16 = 16x^2
or, 28^2 - 384 = 16x^2
or, 784 - 384 = 16x^2
or, 400 = 16x^2
or, 20^2 = (4x)^2
or, 20 = 4x
or, x = 5
So, the correct option is D.
Given, A:B:C = 2:3:5
So, sum of ratios = 2 + 3 + 5 = 10
total amount = Rs 1260
thus, C's share = 1260 x (5/10) = 1260/2 = Rs 630
23 + x = 345
or, x = 345 - 23 = 322
Let each side of the octagon be s and its area be A
thus, A = 2(1 + root2)a^2
Let the number of books be x
Price of x books = Rs 60
thus, price of 1 book = Rs 60/x ................. (i)
Now, new number of books = x + 5
Price = Rs 60
Hence, price of each book = Rs 60/(x + 5) ....................... (ii)
Now, 60/(x + 5) + 1 = 60/x
or, 60[1/x - 1/(x + 5)] = 1
or, 60[x + 5 - x]/x(x + 5) = 1
or, 300/x(x + 5) = 1
or, x(x + 5) = 300
or, x^2 + 5x - 300 = 0
or, x^2 + 20x - 15x - 300 = 0
or, x(x + 20) - 15(x + 20) = 0
or, (x + 20)(x - 15) = 0
or, x = 15, - 20
Hence, the original number of books were 15.
Given, dimensions of the cuboid = 5 cm x 3 cm x 2 cm
so, volume of the cuboid = 30 cm^3
Now, LCM of 5, 2 and 3 = 30
thus, each side of the cube = 30 cm x 30 cm x 30 cm
hence, number of cuboids required = (30 x 30 x 30)/30 = 900
The second car cannot overtake the first car by moving aside only by 1 m as the first car is 1.5 m wide. In oerder to overtake it, the second car has to move aside at least by 2 m.
Given, average age of 10 persons = 60 years
So, total age of 10 persons = 60 x 10 = 600 years.
Also, given the total age of 8 persons = 488 years.
Thus, the total age of 2 persons = (600 - 488) years = 112 years
Hence, the average age of 2 persons = 112/2 = 56 years.
Te locus of the centres of the circles can be found out by subtracting the second circlefrom the first.
Thus, locus is: 8x - 12y + 5 = 0
Hooke’s law, law of elasticity discovered by the English scientist Robert Hooke in 1660, states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load. Under these conditions the object returns to its original shape and size upon removal of the load.
In equation form, Hooke’s law is given by F = kΔL where ΔL is the amount of deformation (the change in length, for example) produced by the force F, and k is a proportionality constant that depends on the shape and composition of the object and the direction of the force.
The moment of inertia of an object is its resistance to rotational acceleration.
Given, tan A + cot A = 2
Now, tan^2 A - cot^2 A = (tan A - cot A)(tan A + cot A)
= 2(tan A - cot A) ........ (i)
But, tan A - cot A = [(tan A - cot A)^2]^1/2
= [(tan A + cot A)^2 - 4tan A cot A]^1/2
= [4 - 4]
= 0 ......... (ii)
Substituting (ii) in (i), we get,
tan^2 A - cot^2 A = (tan A - cot A)(tan A + cot A)
= 2 x 0
= 0
Hence, tan^2 A - cot^2 A = 0
(a) The amplitude of a sound is responsible for the loudness of the sound. Larger the amplitude, louder the sound. Smaller the amplitude, the softer is the sound.
(b) The frequency of a sound is responsible for the pitch of the sound. A higher frequency sound has a higher pitch while a lower frequency sound has a lower pitch.
(c) The wave form of the sound is responsible for the tone of the sound. It determines whether it is a musical note or a loud jarring noise.
No. of animals Days
30 4
20 x (more)
Lesser the number of animals, leser the daily food conumption and more number of days do the rations last.
Thus, x = (30 x 4)/20 = 6 days
So, if there were 20 animals, the number of days the food would last be 6 days.
Given, tan A + cot A = 2
thus, tan A + 1/tan A = 2
or (tan A)^2 + 1 = 2tan A
or (tan A)^2 - 2tan A + 1 = 0
or (tan A)^2 - tan A - tan A + 1 = 0
or tan A (tan A - 1) - 1(tan A - 1) = 0
or, (tan A - 1)^2 = 0
or tan A - 1 = 0
or tan A = 1
thus, cot A = 1
hence, (tan A)^2 - (cot A)^2 = 1 - 1 = 0
Given, 5^m x 5^-2 = 5^3
thus, 5^m x 1/5^2 = 5^3
or, 5^m = 5^3 x 5^2
or, 5^m = 5^5
thus, m = 5
Given, mass of solid = 40 g Volume of water = 50 cc Volume of cylinder = 70 cc Thus, volume of water displaced = Volume of the object = 70 cc - 50 cc = 20 cc Thus, density of the object = mass/volume = 40/20 g/cc = 2 g/cc Relative density = (mass of 20 cc of object)/(mass of 20 cc or water) = 40/1 = 40
Given, mass of the object = 40 g Volume of water = 50 cc Volume of water + volume of the object = 70 cc Thus, volume of the object = 70 cc - 50 cc = 20 cc Density of the object = (mass of the object)/(volume of the object) = 40/20 g/cc = 2g/cc Also, density of water = 1g/cc Thus, mass of 20 cc of water = volume x density = 20 × 1 = 20 g Thus, Relative Density of the object = (mass of 20 cc of the object)/(mass of 20 cc of water) = 40/20 = 2
As it is an entire chapter, it is difficult to explain the principle here. You may refer to the site: Teach Engineering.
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