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i am in final year B Tech in Mathematics and Computing IIT Delhi . I am very good in Mathematics , Physics and Chemistry .
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58-56 = 2
62-58 = 4 = 2^2
70-62 = 8 = 2^3
86-70 = 16 = 2^4
118-86 = 32 = 2^5
182-118 = 64 = 2^6
Hence 84 is the wrong number shold be replaced by 86.
Always try to write the number corresponding to the alphabets
Z=26 , U=21 , Q=17 , ? , L=12
26-21 = 5
21-17 = 4
17-14 = 3
14-12 = 2
difference decreases by 1
Answer is 14 which corresponds to N
Average acceleration = change in velocity / change in time = (Velocity_final - Velocity_initial)/ T
Now Velocity_final = 0 (since car stops and comes to rest)
Velocity_initial = 40km/hr = 40*1000 metre/ 3600 second = 11.11 metre/second (always check unites)
time = 6 seconds
Average_accelration = (0 - 11.11)/6 = -1.85 metre / second^2 (also take care of negative sign)
Acceleartion is negative since velocity is decreasing
Always write whats given and whats asked
m = 1000kg , u = 100m/s , F = 20kN = 20,000 N , t = 8s
a = F/m = 20,000/1000 = 20 m/s^2
v = u + a*t = 100 + 20*8 = 100+160 = 260m/s
B option will be the answer
(a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+ca)
0 = 20+2*(ab+bc+ac)
ab+bc+ac = -10
let the number of girls in the beggining be x ,
15 girls leave , the number of girls = x-15
2 boys for one girl , the number of boys = (x-15)*2 = 2x-30
45 boys leave , the number of boys = 2x-30 - 45 = 2x - 75
5 girls for each boy , the number of girls = 5*(2x-75) = 10x - 375
So 10x-375 = x-15 => 9x = 360 => x = 360/9 = 40
It's a good question , just start with whats given and keep on writing
Let the popuation be equal after t years
Loss in one year is 1200 , loss after t years = 1200*t
Population X = 68000 - 1200*t
Loss in one year is 800 , loss after t years = 800*t
Population Y = 42000 - 800*t
Now , 68000 - 1200*t = 42000-800t => 1200t-800t = 68000-42000 => 400t = 26000 => t = 26000/400 = 65 years
9*11 = 9^2+11^2-9.11 = 81 + 121 - 99 = 103
say coefficient of friction is u , Normal force is N , weight is mg , and angle with horizontal of incline is A , friction is f .
f_kinetic = uN (only when body is moving)
mg will be broken into two parts one along the incline which is mgsin(A) and perpendicular to incline which is mgcos(A)
mgcos(A) will be balanced by Normal N , Hence N = mgcos(A)
f = uN = umgcos(A) As A increases cos(A) decreases , hence friction decreases.
B option is correct
Force down the incline is mgsin(theta) = 10*9.8*sin(30) = 49N This much force we need to apply to keep the body from sliding down
Accelration vector is g = downwards
Initial velocity is horizontal . Hence intital angle is 90 degrees .
As the gains velocity in the downwards direction velocity vector shifts from horizontal direction to vertical direction and acceleration is still constant in the downwards direction , so angle decreases between them. so the answer is A
2A11
4D13 = (2*2 , (A+3) = D , 11+2)
12G17 = (4*3,(D+3) = G , 13+4)
? = (8*3,G+3, 17+8) = 24J25
let us see the middle numbers 4 , 9, 16 , next is 5*5 = 25 as first is 2*2 = 4 , second is 3*3= 9 , third is 4*4 = 16
F - C = 6-3 = 3
I - F = 9-6 = 3
L-I = 12-9 = 3
first letter is L
Now lets see the last one
X-U = 24-21 = 3
U - R = 21- 18 = 3
R - O = 18-15 = 3
hence the letter is O
Answer is : L15O
Centre of mass should be at the fulcrum
lets say the father sits at a distance x from fulcrum on the other side
(30*2 - 90*x)/(30+90) = 0 => x = 60/90 = 2/3 = 67cm from fulcrum
B option is correct
B option
Distance travelled is the path travelled = 30+40 = 70m
Displacement is the shortest path = 50m northeast (find using pythagoras therorem )
Difference = 70-50 = 20m
7-4 = 3
12-7 = 5
19-12 = 7
28-19 = 9
x - 28 = 11 => x = 28+11 = 39
Cursor
If
D all of the above
LCM of 20, 24, 30 = 120 that means they will ring together after every 120 seconds , ie 2 minutes.
Therefore they will ring next together at 11: 27
Metals react with acid to produce hydrogen gas.
After 1 year , A = 40000 + 10% *40000 - 10000 = 34000
After 2 year , A = 34000+10%*34000 - 10000 = 27400
After 3 year A = 27400+10%*27400 - 10000 = 20140
After 4 year A = 20140+10%*20140 - 10000 = 12154
let the first two digits be x and last two digts be y .
Then x = x + 200%x = 3x and y = y+300% = 4y
given 3x/4y = 12/15 => x/y = 16/15
three digits number greater than 500 containing odd digits will contain , 5 , 7 , 9 , since by statemet II , it must be maximum of all three , the answer is 975 , both statements are needed to answer
A = [53,59,61,67,71,73,79,83,89,97] elements in A = 10
P = 10/50 = 1/5
P = 4C2 * 2C2 / 14C4
5 = 1+1+3 number of permuations = 3!/2! = 3
5 = 1+2+2 number of permuation = 3!/2! = 3
P = (3+3)/6*6*6 = 1/36
P = 10 / (4+6+8+10) = 10/28 = 5/14
correct ways is only one
total number of ways is 3! = 6
P = 1/6
rotten = 1/3 * 12 = 4
P ( all rotten ) = (4/12) * (3/11)*(2/10)*(1/9)
7 = 1+6 = 6+1 = 2+5= 5+2 = 4+3= 3+4
P = 6/36 = 1/6
P(A) = 5/(5+3) = 5/8
P(B) = 6/(6+5) = 6/11
P(A' ^ B') = P(A') * P(B') = (1-P(A))*(1-P(B)) = (1 - 5/8) * ( 1 - 6/11)
6 = 1+2+3
number of ways = 3! = 6
Choose any two person and its a handshake
ie nC2 = 55 => n(n-1)/2 = 55 => n = 11
6 = 1+5 = 2+3
Answer = 6C5 + 6C2 = 6+15 = 21
7 = 1+6 = 6+1 = 2+5= 5+2 = 4+3= 3+4
P = 6/36 = 1/6
3C2*5C1+3C3 =3*5+1 = 16
dimater suspends 90 degree at circle and Area of right angle traingle = 1/2 * hypotenuse * distance of hypotenuese from P = 1/2 * 20 * 10 = 100