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Thursday, February 26, 2009

Q) A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0 < t < 1).

If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true?
(a) α cannot remain positive throughout
(b) α cannot exceed 2 at any point in its path
(c) α must be equal to 4 at some point or points in its path
(d) α must change sign during the motion, but no other assertion can be made with the information given


A)
(a) is obviously true, since for motion to initiate and end, acceleration and deceleration must be provided.

Now, for determining the least value of maximum acceleration in this motion, the particle can be considered to be moving in SHM (simple harmonic motion) with equilibrium position x = 1/2 and T (time period) = 2 seconds.

acceleration = -w^2 .x

Max acceleration
in such a motion will be = w^2 . Amplitude = (2pi/2)^2 . (1/2) = pi^2 / 2 = 4.934 m/s^2

And max. acceleration would be higher if any other type of motion occurred (instead of SHM). So the least value of max. acceleration is 4.934 m/s^2 which means that 4m/s^2 acceleration HAS to occur at some point in the path of the particle's motion. (which implies that (c) is true too).

Hence, (a) and (c) are true.

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Thursday, January 29, 2009

Q)  Define a a sequence    as follows: 

, if number of positive divisors of  is odd 
, if number of positive divisors of  is even 

(The positive divisors of  include  as well as .)Let  be the real number whose decimal expansion contains  in the -th place,.Determine,with proof,whether  is rational or irrational.


A)
 


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Saturday, December 13, 2008

Q) Let f(x) = and f(0) = 0 . Then f(1) = ?

A)


=


=


By Applying partial fractions -

=


=


=


=


BY using the statement that f(0) = 0 ........We get C = 0



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Q) Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter 2008 units.
(From RMO - 2008)

A) Let x be the length of the isosceles side.

Perimeter = 2x + k = 2008

Using cosine rule:

k^2 = x^2 + x^2 - 2x^2. cos@ .......... where @ E (90, 180)

k^2 = 2x^2.[2.sin^2 (@/2)] ......... where (@/2) E (45, 90)

k E ( sqrt(2) x, 2x)

Also notice that 2x > k is satisfied automatically here.

Now 2x + k = 2008

Solving,

x E ( 2008/4,
(2008)/(2+ sqrt2) )

x E ( 502, 588.something)

But only integral values are required, so x lies between 503 and 588, inclusive.

So, number of values
= 588 - 503 + 1
= 86 solutions


Thus, there are 86 such triangles.

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Q) Find the number of all 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them.
(From RMO - 2008)

A) The various possibilities of the digits chosen acc. to the conditions mentioned are:

1) Three 2s, One 3, One 1, One 0
2) One 2, Two 3s, Two 1s, One 0
3) One 4, One 2, One 3, One 1, Two 0


Number of permutations for the cases mentioned above:

1) Three 2, One 3, One 1, One 0
(6! / 3!) - (5! / 3!) -----> ie. net permutations - permutations having 0 as first digit since those won't be considered a 6 digit number. Similar for other cases too.
= 5.5! / 3!
= 100 ways

2) One 2, Two 3, Two 1, One 0
(6! / 2!.2!) - (5! / 2!.2!)
= 5.5! / 4
= 150 ways

3) One 4, One 2, One 3, One 1, Two 0
(6!/2!) - 5!
= 240 ways

Total = 100 + 150 + 240
= 490 such numbers


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Q) A body weighs 6 grams when placed on one pan and 24 grams when placed on the other pan of a defective beam balance according to the reading shown on the instrument. If the beam is horizontal when both pans are empty, what is the true weight of the body?

A) Let the left pan have weight m1 and distance l1 from pivot.
Let the right pan have weight m2 and distance l2 from pivot.

Balancing the torques:
m1. l1 = m2.l2 (given that they are balanced when empty)

Also,
(X + m1).l1 = (6+m2). l2 -----------------------------(1)

(24 + m1).l1 = (X+m2).l2 ----------------------------(2)

where X is the unknown weight of the body.

Substituting m1.l1 = m2.l2 in both equations above, we get:
l1/l2 = X/24 = 6/X
so X^2 = 24*6

X = 12 gms


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Q) Find all three-term Harmonic Progressions a, b, c of strictly increasing positive integers in which a = 20, and b divides c .
( from RMO - 2008)


A) Let c = k.b ......... where k is a natural number

From the definition of a harmonic progression:
1/a + 1/c = 2/b

(1/20) + (1/bk) = 2/b
40 - (20/k) = b
Using values of k E [1, 20] we get b = 20, 30, 35, 36, 38, 39. But exclude 20 as we require strictly increasing values (since a = 20 already, b must not be 20).

[ Remember -
use only those values of k which are factors of 20, else you won't get integral values... so k E {1, 2, 4, 5, 10, 20} ]

Then using :
c = 20 / [(40/b) - 1]
we get c = 60, 140, 180, 380, 780 respectively for the aforementioned values of b (excluding 20).


So there are 5 such Harmonic Progressions :

20, 30,60
20, 35, 140
20, 36, 180
20, 38, 380
20, 39, 780

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Q) A circle touches the x axis and also touches another fixed circle with center (0,3) and radius = 2 units. What is the locus of the center of the (first) circle?

A) Consider any such arbitrary circle, with center (h,k).
It touches the x axis, so its radius is k units ( the ordinate).

Now, the distance of this circle's center from the given circle's center = 2 + k units
This distance is the same as the distance between center of the arbitrary circle and the line y = -2 .

Now, since the center of this arbitrary circle is equidistant from a fixed point (the center of given circle ie. (0,3) ) and the line y = -2, it follows the definition of the parabola and hence the locus of (h,k) is a parabola.

So the locus of the center of the circle is a parabola with focus (0,3) and directrix y = -2 .

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Q) Find the number of triangles that can be formed with the vertexes of a 10-sided-polygon (decagon) as its vertexes ,if
1)The triangles can not have more than one side common with the polygon;
2)The triangle can not have any side common with the polygon .

A) Total triangles possible in general = 10C3
Triangles having more than 1 side common with polygon = Triangles with 2 sides common with polygon = 10 (ten triangles with 2 sides as 2 consecutive sides of the polygon)
so
1) 10C3 - 10 triangles

2) Triangles with 1 side common with polygon = 10.(6) = 60
(Fix one side, then 8 vertexes are left, and also ignore 1 consecutive vertex on either side, so we are left with 6. And 10 such triangles are possible due to 10 vertexes)
So, triangles with no sides common with polygon = (10C3 - 10) - 60
= 10C3 - 70
triangles

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Q) How many groups can be made from 5 different green balls, 4 different blue balls and 3 different red balls, if at least 1 green ball and 1 blue ball are to be included?

A)
Choose 1 green ball and 1 blue ball, + extra balls [ 0 to 10 in number]
5C1 . 4C1 . (10C0 + 10C1 + 10C2 ....10C10)
= 5*4* (1+1)^10
= (20). 2^10 groups

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Q) A particle starts from rest x=0 at t=0, moves on a straight-line path and again comes to rest at x=1 at t=1. If the motion is simple harmonic, find the approx. maximum acceleration.

A) Using apt equations of simple harmonic motion:
x = A.cos(wt)
(since the abcissa 1/2 is the midpoint of the given path, we measure amplitude and displacement by considering that as the central point)

At t=1 ---> 1/2 = 1/2* cos (w)
so, w = pi

And since acceleration (a) = -w^2. x
magnitude of a = (pi^2). (1/2) (a is max. at extremities)
= approx 5 units/s^2
--------> max. magnitude of acceleration

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Q) A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually six.

A) Since he has already reported that it's a 6, either he is lying about it or it is actually a six.
That is, either it is truly a six & he's telling the truth, OR it's a non-six & he's lying by saying that it's a six.

Net Probability = (prob. of actual 6 & truth) / [prob. of actual 6 & truth + prob. of non 6 & lies ]
= (3/4)(1/6) / [3/4*1/6 + 1/4*5/6]
= 3/8


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Q) Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by 10%, by what percent does the balance-point shift (and in which direction) ?

A) Since a Metre-Bridge has wire of 100 cm, so if the 2 resistances are equal, the balance point is at 50 cm.

Now if the left resistance is increased by 10%, it becomes 1.1 times its original value.
So balance point L :
1.1 R/ R = L / (100 - L)
110 - 1.1 L = L
110 = 2.1 L
L = 52.4 cm approx.

So the balance point L has shifted right by 2.4 cm
So in terms of %, it has shifted right by (2.4 / 50 x 100) %
= 4.8 % to the right

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Q) Prove that .

A)

log7 10 = 1 + log7 (10/7) = 1 + log7 1.428

log11 13 = 1 + log11 (13/11) = 1 + log11 1.18


Let A = log7 1.428


and B = log11 1.18


So, 7A = 1.428


and 11B = 1.18


Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )


Thus, 1 + log7 1.428 > 1 + log11 1.18

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Sunday, November 2, 2008

Q) Find the coefficient of x^4 in (1 + 2x + 3x^2)^5 .

A) [(1+2x)+3x^2)]^5

'r'th Term = 5Cr (1+2x)^r . (3x^2)^{5-r}

To form net effective 4th powers of x:
from the righthand term, we can only use x^0 x^2 and x^4
so from the lefthand term (1+2x)^r ... we use x^4, x^2 and x^0 respectively.

5C5.(3x^2)^0 . (1+2x)^5 -----------> 5C5. 5C4.(2x)^4
and
5C4.(3x^2).(1+2x)^4 --------> 5C4.(3x^2).[4C2.(2x)^2]
and
5C3.(3x^2)^2 .(1+2x)^3 ---------> 5C3.(3x^2)^2 [3C0. (2x)^0]

So, coefficient of x^4 :

5C5.5C4.2^4 + 5C4.3.4C2. 2^2 + 5C3.3^2. 1
= 5. 16 + 5.3.6.4 + 10.9
= 80+ 360 + 90
= 530

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Q) Find the sum of :
1 + 1/2 + 1/3 + 1/4 .... up to infinite terms.

A)
1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...
= ( 1/1 ) + ( 1/2 ) + ( 1/3 + 1/4 ) + ( 1/5 + 1/6 + 1/7 + 1/8 ) + ...

(where each set of parenthesis stops at a reciprocal of a power of two.) Replacing the terms in each set of parenthesis by the smallest term (in those parenthesis) we get a smaller sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...
> ( 1/1 ) + ( 1/2 ) + ( 1/4 + 1/4 ) + ( 1/8 + 1/8 + 1/8 + 1/8 ) + ...
= 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + ...

which clearly gets arbitrarily large! So it is a diverging series and will not have a definite value since it is not converging. The sum will go on approaching infinity.

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Q) Find the number of real pairs (x,y) satisfying and .

A) y = 2xy

y (1-2x) = 0..... So, either y=0 or x = 1/2

For y = 0,

x = x^3

x^3 - x = 0 = (x - x^1/3) (x^2 + x^{2/3} + x^{4/3} )

x = -1,1, 0 for y = 0 ............ 3 real roots.

AND for x = 1/2

1/2 - 1/8 = y^4

0.375 = y^4

y^4 - 0.375 = 0 = (y^2 + rt(0.375) ) (y^2 - rt(0.375) )

which has 2 real and 2 imaginary roots.


So, total 3 + 2 = 5 real solutions.

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Q) Evaluate the definite integral :



Here, {x} is the fractional part function, defined as {x} = x - [x] ( [.] denotes greatest integer function )


A)


Thus, its value is 40 / ln 3 .

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Q) Find the integral of :


A)


=

=


Now take out the 1 and you get:

e^x + e^x [ f(x) + f'(x) ] ..... where f(x) = -4/(x+4)

And we know that integral of e^x [ f(x) + f'(x) ] = e^x. f(x) + C

So the integral is :

e^x + e^x.[-4 / (x+4) ] + C

OR

e^x . [x/ (x+4)] + C

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Q) Find the number of 5-digit numbers of the form xyzyx in which x is less than y.


A) Number of ways of selecting x and y: 9C2 . 2!

Out of these, 1/2 will be such that x than y, and 1/2 such that x greater than y

So for x less than y: 9C2 selections

And z can be any number from 0 to 9, so 10 ways.

So total number of such numbers is 9C2 . 10

= 360 numbers

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Saturday, November 1, 2008

Q) If the arithmetic mean of two numbers 'a' and 'b' is 'n' times their harmonic mean, find a/b in terms of 'n'.

A) AM = n. HM
(a+b) / 2 = n. [2ab/(a+b)]
a^2 + b^2 + 2ab = 4*n*ab
(a+b)^2 = 4*n*ab
(a/b + 1)^2 = 4*n*a/b
Let a/b = x
(x + 1)^2 = 4nx
x^2 + 1 + 2x - 4nx = 0
x^2 + x(2-4n) + 1 = 0
x = 4n-2 +- sqrt(16n^2 + 4 - 16n -4) / 2
x = 4n-2 +- 4.sqrt(n^2 - n) / 2
x = 2n - 1 +- 2.sqrt (n^2 - n)
OR
a/b = 2n - 1 +- 2.sqrt (n^2 - n)


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Q) A particle is thrown upwards from the ground. It experiences constant air resistance which produces a retardation of 2 m/s^2. Find the ratio of time of ascent to time of descent.

A)
v = u + at
0 = u + -(g + 2)t
t = u / (g + 2)

0 - u^2 = 2as
- u^2 = 2 *-(g + 2)s
s = u^2 / 2(g + 2)

s = 1/2 aT^2
u^2 / 2(g+2) = (1/2) * (g-2)T^2
u / sqrt(g^2 - 4) = T

So t/T = sqrt (g-2) / sqrt (g+2) = sqrt (8/12) = sqrt (2/3)

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Q) Find the integral of:

dx / [(x)(1 + x^n)]

A) 1 / [(x)(1 + x^n)]
= (1 + x^n - x^n) / [(x)(1 + x^n)]
= (1/x) - [ (x^n) / {(x)(1 + x^n)} ]
= (1/x) - [ (x^{n-1}) / (1 + x^n) ]
Integrating now:
ln x - ln (1 + x^n) + C
= ln [x / (1 + x^n)] + C

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Tuesday, September 16, 2008

Q) A car is traveling on a straight road. The maximum velocity that can be attained by the car is 24 m/s. The maximum acceleration and decceleration it can attain is 1 m/s^2 and 4 m/s^2 respectively. Find the minimum time taken by the car to start from rest, cover a distance of 200m and stop.


A) Let the car accelerate through distance S1, and decelerate through distance S2.

S1 + S2 = 200

[v^2 - u^2 / 2a]        +  [ V^2 - U^2 / 2A ]   = 200

v^2 / 2 + (-v^2 / 8) = 200

v^2 = 320

v = sqrt(320) m/s

Also,
v = at

t = sqrt(320)

And,
V - U = AT

T = sqrt (320) / 4


Total time = t + T = sqrt (320) (5/4)

Total time = 10 sqrt(5) seconds

= 22.36 seconds


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Q) If is a 6th root of unity, then prove that :
1)
2)

A)


= 1 × 1 ............. ---> (Modulus of each of the 6th roots of unity is 1)
= 1

[ Part 1 proved]





[ Part 2 proved ]

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Q) Find the equations of the lines parallel to x-axis between which the graph of y=   lies .

A) 
 y=  

This is of the form :


| 2ab / (a^2 + b^2) |   where a is 2x and b is 3.


And we know:

(a - b)^2 > = 0 

a^2 + b^2 > = 2ab


so | 2ab / (a^2 + b^2) |   < = 1


so y < = 1 


and y > = 0    ....(since it is absolute value of something)


So the two boundary lines are y = 1 and y = 0 .


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Q) Six persons A , B , C , D , E and F  are to be seated around a circular table. Find the number of ways in which this can be done if A must have either B or C on his right, and B must have either C or D on his right.

A) Consider three cases:

One case with A, having B to his right, and C to B's right.
Now number of ways the rest can be arranged is 3!.

For the next case, take A, then B to his right, and D to B's right... again the rest can be arranged in 3! ways.

For the last case, you have A, with C to his right. Now three entities are left - E, F and BD (since B must be seated with D to his immediate right 'cos C is already used up) and again we get 3! ways.

So total number of ways is 3. 3! = 18 ways

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Q) A ball of mass 100g is projected vertically upwards from the ground with a velocity of 49 m/s. At the same time, another identical ball is dropped from a height of 98 m to fall freely along the same path as followed by the first ball. After some time, the two balls collide, stick together and finally fall together.Find the time of flight of the masses.

A) Let the 2 balls meet at time T.
Their displacements will be S and 98 - S meters.

S = gT^2/2
98 - S = 49T - gT^2/2
Adding the 2 equations:
98 = 49T
T = 2 seconds.
(and S = 20 m from top height) 

At t = 2 seconds, Velocity of balls have to be found.
v1 = g*2 seconds = 20 m/s downwards
v2 = 49 - g*2 = 29 m/s upwards

Now an inelastic collision occurs and we have to conserve the momentum.
0.1 kg (20 - 29) = 0.2 kg * V
V is 4.5 m/s upwards.
Now from here to final stop, time taken is to be found:
98 - 20 = 78 m = -4.5 t + gt^2/2
78 = -4.5t + 5t^2
t = 4.5 +- root (20.25 + 1560) / 10
t = 4.5 + 39.75 /10 =  4.425 seconds

So the balls collide after 2 seconds, and the lump of the masses takes a further 4.425 seconds to reach the ground.

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Q) Find the remainder when 7^23 is divided by 25 .

A) 7^23 = (7^22) . 7
= [ (7^2)^11] . 7
= [(50 - 1)^11] .7
Now In expansion of (50 - 1)^11 , all terms except the last one will be divisible by 25 since they contain an exponent of 50 in them.
Last term = 11C11 * (-1)
So we are left with
(-1)(7)
= -7
so 25 - 7 = 18 which is the remainder


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Q) Find the sum of the sequence .7 + .77 + .777 + ..... till n terms.

A)
.7 + .77 + .777 ... n terms

= 7 (0.1 + 0.11 + 0.111....)

= (7/9)*(0.9 + 0.99 + 0.999 ...)

= (7/9)*(1 - 0.1 + 1 - 0.01 + 1 - 0.001 ....)

= (7/9)* [n - (0.1*(1 - (0.1)^n) / (1 - 0.1) ) ]

= (7/9)* [n - (0.1)(1 - (0.1)^n) / 0.9]

= (7/9)*[ n - (1 - 0.1^n)/9]

= 7n/9 - 7(1 - (0.1)^n) / 81

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Q) A fly wheel rotates about its central axis . Due to friction at the axis, it experiences angular retardation proportional to its angular velocity. If its angular velocity falls to half its initial value while its makes n revolutions , how many more revolutions will it make before coming to rest?

A) - a = k.w .........(k is a constant of proportionality)
(dw/dQ)(dQ/dt) = k.(dw/dt)
dw = - k.dQ
w = -kQ + c
Initially Q = 0, and w = W0
W0 = C

w = -kQ + W0
After n revolutions,
W0 / 2 = -k(2.n.pi) + W0
W0 / 2 = 2.k.n.pi
k = W0 / 4.n.pi

For final stop, w = 0
w = -kQ + W0
0 = -(W0 / 4.n.pi). Q + W0
Q = 4.n.pi
means 2n revolutions.
And given that n revolutions have already occurred. So number of revolutions that will further occur before the flywheel stops = 2n - n = n revolutions.

Thus, the flywheel will complete 'n' more revolutions before it stops.
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Q) The figure below shows the acceleration-time graph of a particle moving along a straight line. After what time does the particle acquire its initial velocity?




A) Let initial velocity be u

At t =1, velocity = u + at = u + 2

For t >= 1, da/dt = -2 ....(slope)

Using point-slope form of equation of straight line:

(a-2) = (-2)(t-1)

a = 4 - 2t

dv/dt = 4 -2t

V = 4t -t^2 + C

at t = 1, we have V = u+2, so C= u-1

so V = 4t - t^2 + U -1

So for V to equal U:

t^2 - 4t + 1 = 0

t = 2 + root(3) seconds.

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Q) How many 5- digit telephone numbers can be formed that have at least one of their digits repeated.

A)
(A phone number doesn't begin with 0.)

Number of phone numbers with at least one repeated digit
= (Total ways in which a phone number can be made) - (number of phone numbers with no repeated digits)

= 9*10*10*10*10 - 9*9*8*7*6
= 9.(10^4) - 9.(9P4)
= 62784

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Q) There is a regular bus service between 2 towns A and B,with a bus leaving town A and B every T minutes. A cyclist moving with a speed of 20 km h-1 in the direction A to B notices that a bus goes past him every 18 minutes in the direction of his motion,and every 6 minutes in the opposite direction. Find the period T of the bus service and the speed with which the buses ply between A and B. Assume the speed of the bus to be constant.

A)
Let D = distance between A and B
Let velocity of the buses = v
Relative velocity of buses with respect to the cyclist (going from A to B)
v-20 km/hr for buses going from A to B
v+20 km/hr for Buses going from B to A

Now use Time = distance/velocity
Use the 2 equations to get d and v

D/(v-20) = 18/60 ---------------------1
D/(v+20) = 6/60 -----------------------2
=>v=40 km/hr and D=6 km
T = 6/40 hr = 9 mins

Hence the speed of the buses is 40km/hr & the period of bus service is 9 mins . (A new bus is sent after every 9 mins)

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Q) Prove that if abc=1 and a,b,c are positive real numbers.

A) Rewrite the expression as
(a-1 + 1/b + b - b) (b - 1 + 1/c + c - c) (c - 1 + 1/a + a - a)
Max value of the terms will be obtained when 1/b + b, 1/c+ c and 1/a +a are max, since a, b , c are all positive.
And we know that 1/n + n >= 2
(and for max value, n = 1)
So a=b=c=1 for the max value :

(a -b +1)(b- c+1)(c - a+1)
= 1*1*1
= max value --> 1
so the expression is less than or equal to 1.


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Q) Two skaters A and B of weights 40 kg wt and 60 kg wt, respectively, stand facing each other 5 m apart. They then pull a light rope stretched between them. Where do they meet?

A)
Use the fact that center of mass remains stationary in absence of any external forces.
So the length of the rope can be divided in inverse ratio of the masses.

M1 / M2 = L2 / L1
L2 / L1 = 2/3
And the sum of lengths is 5.
Therefore, they meet at 3m from initial position of A.


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Q) If the momentum of a body increases by 0.01%, by how much will its Kinetic Energy increase?

A)
New momentum = m (1.0001 v)
So new velocity = 1.0001 v

So new kinetic energy = 1/2 * m * V^2
= 1/2 * m * (1.0001 v)^2
= 1/2 * m * 1.00020001 v^2
= 1.0002 x original momentum
So the increase is 0.02 %.


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Q) A body falling from a given height H hits an inclined plane in its path at a height h . As a result of this impact, the direction of the velocity of the body becomes horizontal. Find the value of h/H for which the body will take the maximum time to reach the ground.

A)
When it collides with the inclined plane at height h, the body's velocity is sqrt(2g(H-h) ).
And time taken to cover this distance is sqrt(2(H-h) / g).
(By using ---> s = (1/2) g.t^2 )

Now this becomes the horizontal velocity. So now time taken to reach the ground will be :
t = sqrt (2h/g) since downward velocity initially now is 0.

So total time taken = sqrt(2(H-h) / g) + sqrt (2h/g)
Time taken will be max when the two terms are equal (symmetry), so 2(H-h) = 2h
2H = 4h
h/H = 1/2


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Q) There are 3 groups.The 1st group contains 3 girls & and 1 boy. The 2nd group contains 2 girls and 2 boys and the 3rd group contains 1 girl and 3 boys. Another group is formed by picking one student each from the existing groups. What is the probability that the group formed will contain 1 boy & 2 girls?

A)
The 3 groups are A, B and C.
The prob. we have to find is of getting BGG where B is boy and G is girl.

Case 1 - boy is selected from A, and the girls from B and C.
prob. of BGG = 1/4*1/2*1/4 = 1/32

Case 2 - boy is selected from B, and girls from A and C.
prob. of BGG = 1/2*3/4*1/4 = 3/32

Case 3 - boy is selected from C, and girls from A and B.
prob. of BGG = 3/4*3/4*1/2 = 9/32

Total prob. = Sum of prob.= 1/32 + 3/32 + 9/32
= 13/32

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Q) In an increasing Geometric Progression, the sum of the first and the last term is 66, the product of the second and the last but one term is 128, and the sum of all the terms is 126. How many terms are there in the progression ?

A) Let the GP be : a, ar, ar^2 ....a.r^n

a + a.r^n = 66 = a(1 + r^n)
ar.a.r^(n-1) = 128 = a^2.r^n
So 66 = a (1 + 128/a^2)
66a = a^2 + 128
a^2 - 66a + 128 = 0
a = 66 +- rt(66^2 - 4(128)) / 2
a = 66 +- 62 / 2
a = 2, 64
Since, 66 = a(1 + r^n)
66 = 2(1 + r^n) OR 66 = 64 (1 + r^n)
Since r>1 (increasing GP) so we take
66 = 2(1 + r^n)
r^n = 32

And,

a(r^(n+1) - 1) / (r - 1) = 126
2(32r - 1) / (r-1) = 126
63r - 63 = 32r - 1
31r = 62
r = 2
and r^n = 32
2^n = 32
so
n = 5
.

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Q) Find the number of positive integral solutions of x4 _ y4 = 3789108 .

A)
Let A = x^2 and B = y^2
A^2 - B^2 = 3789108
(A + B)(A - B) = 3789108

Let A + B = T
(T)(T-2B) = 3789108
T^2 - 2BT - 3789108 = 0
T = 2B +- root (4B^2 + 4*3789108) / 2
T = B +- root (B^2 + 3789108)
A + B = B +- root (B^2 + 3789108)
A = +- root (B^2 + 3789108)
Now since A = x^2 so A cannot be negative, thus
A = root (B^2 + 3789108)
Now A is x^2, and B is y^2, where x and y are integers.
Squares of integers (ie. x^2 and y^2) end in 0, 1, 4, 5, 9, 6 and squares of these perfect squares (ie. A^2 and B^2) end in 0, 1, 5,6.
Now,
last digit of B^2 + last digit of 3789108 should end in 0,1,5,6
the various sum of last digits possible are 8, 9, 3, 4 none of which is 0, 1, 5 or 6...
So no integral solution exists.

The equation has 0 integral solutions.


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Q) If 12 + 22 + 32 + ..... + 20032 = (2003)(4007)(334) and (1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) = (2003)(334)(x), then find x .


A)


So for 2003 terms the summation is = (2003)(2004)(4007) / 6 = (2003)(4007)(334)



Now,


(1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) =


= [ 2004(2003)(2003 + 1) / 2 ] - [ (2003)(4007)(334) ]


= [ (1002)(2003)(2004) ] - [ (2003)(4007)(334) ]


= 3 [(334)(2003)(2004) ] - [ (2003)(4007)(334)]


= (2003)(334)( 3*2004 - 4007)


= (2003)(334)(2005)



So the answer is 2005.


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Q) Find the sum of 1 + (1+2) + (1+2+4) +....... till n terms .

A)
1 + (1+2) + (1+2+4) + ...... till n terms

= 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ....(1 + 2 + 4 + ...2^(n-1) ) ---->( Concept of sum of infinite Geometric Progression )

= 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ...[1*(2^n - 1) / (2 -1) ]

= 1 + [2^2 - 1 + 2^3 - 1 + 2^4 - 1 .... + 2^n - 1]

= 1 + [ 2^2.(2^(n-1) - 1) / (2-1) - (n-1) ]

= 1 + [ 4. 2^(n-1) - 4 - n + 1]

= 2^(n+1) - 2 - n

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Q) A motorboat going downstream came across a bottle floating in the river at a point A . After 60 minutes of further motion, it turned back and after some time passed the bottle at a distance 6 km from point A. Find the flow speed of the river (in km/hr) assuming the duty of engine to be constant?


A) Let Vf = velocity of flow & V = velocity of boat

Distance covered by boat in 60 mins = (V + Vf) x 1 hour = V + Vf

Also
6 km = Vf * (60 min + T) where T is time taken for the boat to return to the bottle again.
6 = Vf [1 hour + (V + Vf - 6) / (V - Vf) ] ........... V - Vf is net speed for upstream motion
6 = Vf (2V - 6) / V - Vf
6V - 6Vf = 2V.Vf - 6Vf
6V = 2V.Vf
Vf = 3

So the flow speed = 3 km/hr


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Q) If e^^(sin^2x+sin^4x+sin^6x+....\infty)log_e3 satisfies the equation x2 - 28x + 27 = 0, find the value of (cosx+sinx)^^-1. Given that x lies between 0 and pi/2 .

A) The given expression can be simply rewritten as:

= 3^ (sin^2 x + sin^4 x + sin^6 x ....infinity)

= 3^[ (sin^2 x) / (1 - sin^2x)] (Using concept of sum of infinite GP)

= 3^(tan^2 x)


Now,

x^2 - 28x + 27 = 0

x^2 - 27x - x + 27 = 0

(x-1)(x-27) = 0

roots are 1, 27

Given that 3^(tan^2 x) satisfies the equation so it = 1, 27

3 ^(tan^2 x) = 1, 27

tan^2 x = 0, 3

tan x = 0, +- rt(3)

x = 0, +- pi/3

Since for second part of the question x lies between 0 and pi/2,
so we take x = pi/3
So
1/ (cosx + sinx) = 1 / (cos60 + sin60)

= 1/ (1/2 + sqrt(3)/2)

= 2 / ( 1 + sqrt(3) )


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Q) A bird flies with a speed of v=| t-2 | m/s along a straight line , where t is in seconds. Find the distance traveled by the bird during the first 4 seconds of motion.

A)
v = | t-2 | m/s
The critical point is t = 2 (At which the modulus becomes 0)
so from t=0 to t=2 , velocity is (t-2) m/s
and from t=2 to t= 4, velocity is (2 - t) m/s
Total distance covered = mod of ( integral (0 to 2) [ t-2] + integral (2 to 4) [ 2 - t] )
= [t^2/2 - 2t] 0 to 2 + [ 2t - t^2/2] 2 to 4
= | -2 - 2 |
= |-4 |
= 4 m

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Q) If f(x) is sum of all digits of x then find f(101) + f(102) + f(103) .... f(200) .

A)
f(101) + f(102) + f(103) .... f(200)

= 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 (for 200)

= 99 + 20(45) + 2
= 101 + 900
= 1001


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Q) Find the area contained within the graph of the equation: x^2-6x+y^2+12y=0

A)

x^2 - 6x + y^2 + 12y = 0

Form perfect squares:

x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 = 0

(x + 3)^2 + (y + 6)^2 = 45

(x + 3)^2 + (y + 6)^2 = [sqrt(45)]^2

This is the equation of a circle, with radius sqrt(45)

so area = pi.r^2

= 45.pi


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Q) Consider a sequence { } such that .
How many distinct pairs chosen from this sequence have g.c.d.= 6 ?

A) a1 = 2
a2 = 3
a3 = 7
a4 = 43

Now since
a(n+1) = a.n^2 - an + 1
and after a1, the next terms are odd...
so terms will be of the type
odd^2 - odd + 1
odd^2 is always odd... and odd^2 + 1 will become even.
then even - odd = odd always....

So there won't be any distinct pair with GCD = 6... 2 will be the only even number in the sequence.


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