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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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