[Aarav]

Some of you have got the answers wrong. I'm posting the official solution from last year.

#4
Note that 2/3 + 2^2/3 = 2 or 2^a is of the form 3k-1, and 2^(a+1) is of the form 3k+1 when a is odd.

[2/3] + [2^2/3] = [(3-1)/3] + [(3+1)/3] = 1 = (2 + 2^2)/3 - 1
[2^3/3] + [2^4/3] = [(9-1)/3] + [(15+1)/3] = 7 = (2^3 + 2^4)/3 - 1
and so on...

Thus, [2/3] + [2^2/3] + [2^3/3] + [2^4/3] +...+ [2^100/3] =

(2/3 + 2^2/3 -1) + (2^3/3 + 2^4/3 -1) +...+ (2^99 + 2^100/3 -1) =
2/3 + 2^2/3 + 2^3/3 + 2^4/3 +...+ 2^100/3 - (50)

1/3(2 + 2^2 + 2^3 + ...+ 2^100) - 50 = 2/3( 2^100 - 1) - 50.

Hence, choice (c) is the right option

#5

Its same as asking we have 7 books in a shelf and we have to choose place

for 4 more books so that no two of these 4 will be together.

Answer 8C4 = 70.

Hence, choice (a) is the right option

#6

We can always have a system of equations with unique, infinite and zero solutions.

Consider a system in 2 variables.

ax + by = c, dx + ey = f.
Let (m,n) be its solution then am + bn = c, dm + en = f
Let (p,q) be another solution then ap + bq = c, dp + eq = f
=> a(m-p) = b(q-n) and d(m-p) = e(q-n); either m=p and q=n or a/b = d/e in which case we have infinite solutions, or 0 solutions depending on the value of variables c and f.

Hence, choice (d) is the correct answer

[krsh.vik]

Originally Posted by warrior

we have 51 even and 50 odd terms

51/3+ 100/3 = 151/3

Shouldn't we subtract 101 instead of 151/3
we are adding 1/3 or 2/3 to the fractions which will result in 1 being added for each term ?

[vineet.nitd]

Next 3 problems ...

7>Let m = 1^2/1 + 2^2/3 + 3^2/5 + ... + 500^2/999, and n = 1^2/3 + 2^2/5 + 3^2/7 + ... + 500^2/1001, then the integer closest to m - n is

(a) 250 (b) 500 (c) 1000 (d) 1500


8>If the equation (x^2 - 2ax -4(a^2 + 1))(x^2 - 4x -2a(a^2 + 1)) = 0 has exactly 3 different roots, then how many values can "a" take?

(a) 4 (b) 3 (c) 2 (d) 1


9> In a wedding 101 people meet. The 1st person shakes hand with only 1 other person, 2nd with 2 others,..., 100th with 100 others. With how many persons does the 101th person shake hands?

(a) 100 (b) 1 (c) 51 (d) none of these

[skylark]

Originally Posted by vineet.nitd

Next 3 problems ...

7>Let m = 1^2/1 + 2^2/3 + 3^2/5 + ... + 500^2/999, and n = 1^2/3 + 2^2/5 + 3^2/7 + ... + 500^2/1001, then the integer closest to m - n is

(a) 250 (b) 500 (c) 1000 (d) 1500

(a) 250

series will be something like:

1^2/1 +
2^2/3 -1^2/3 +
3^2/5 - 2^2/5 ....
+500^2/999 -499^2/999
+ 500^2/1001

ok, not all that clear...but typing from office

[krsh.vik]

wrong post

[Destiny's_Child]

My answers to 7 to 9

7) here m-n = 1 + 1/3 [2^2 - 1^2]+ 1/5 [3^ - 2^2]..... (499 terms) - 500^2/1001
Each of the 499 terms reduces to 1 and taking 1001 approx as 1000
So m-n = 500 - 250 = 250

Here we have three conditions

Eq 1 has one root
Eq 2 has one root
Both the equations have 2 distinct roots but one of them is common to both the equations.

Condition 1 ) For x^2 - 2ax -4(a^2 + 1) = 0
20a^2 + 16 = 0
=> a^2 = -ve
So not possible

Condition 2 ) For x^2 - 4x -2a(a^2 + 1) = 0
16 + 8a(a^2 + 1) = 0
=> a^3 + a + 2 = 0
=> (a+1)(a^2 - a + 2) = 0
a = -1
For a^2 - a + 2, D is negative, so complex solution.

Condition 3) x^2 - 2ax -4(a^2 + 1) = x^2 - 4x -2a(a^2 + 1)
We get x = a^2 + 1
Putting this value of x in any of the equations, the eqn reduces to a^2 - 2a - 3 = 0
we get a = -1, 3
But when a = -1 total number of solutions becomes 2
So a = 3 is the only solution

So answer (d)

9)This is a very interesting question and the logic behind is simply superb....

Here the 100th person shakes hand with all except itself.
In this way one handshake of 1st person is completed.
the 99th person shakes hand with 2 to 101except itself.
In this way two handshakes of 2nd person is completed.
Continuing like this
the 99th person shakes hand with 2 to 101except itself.
the 99th person shakes hand with 2 to 101except itself.
Now all the handshakes are covered

So 101th person shakes hand from 51st to 100th person i.e total 50 persons
So option (d) (Not sure though)

[Aarav]

Official solutions

-----------------------------------------------------------
Quant Answer # 7
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Minus the sequence from m to n at diagonal and you will get , 1^2/1 + (2^1 - 1^1)/3 + (3^2 - 2^2)/5 + ... + (500^2 - 499^2)/999 - 500^2/1001
= 500 - 500^2/1001 which is close to 250.

Hence, choice (a) is the right answer.


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Quant Answer # 8
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We have 2 quadratic equations (x^2 - 2ax -4(a^2 + 1)) = 0, (x^2 - 4x -2a(a^2 + 1)) = 0
which can be written as (x-a)^2 = 5a^2 + 4, and (x-2)^2 = 2(a^3 + a +2)

Now, we can have 3 different roots in the following scenarios
(a) one of the equations has a double root while the other doesn't
(b) both the equations have a common root

Case 1: when (x-a)^2 = 5a^2 + 4 has double root
Not possible as RHS can never be zero.

Case 2: when (x-2)^2 = 2(a^3 + a +2) has double root
Thus, 2(a^3 + a +2) = 0, (a+1)(a^2 - a + 2) = 0, thus a = -1

putting a = -1 quadratic equations (x^2 - 2ax -4(a^2 + 1)) = 0, (x^2 - 4x -2a(a^2 + 1))
we get the 1st equation roots as 2, -4 and 2nd eq. has double root as 2.

Thus, in all we have 2 different roots. Hence, a = 1 is also ruled out.

Case 3: both the equations have a common root
Let p be the common root.
If p is the root of A(x) and B(x), then it must be the root of A(x) - B(x).

Hence p is the root of (x^2 - 2ax -4(a^2 + 1)) - (x^2 - 4x -2a(a^2 + 1)) = 0
hence, (p^2 - 2ap -4(a^2 + 1)) - (p^2 - 4p -2a(a^2 + 1)) = 0
solving, (2a - 4)p = (2a - 4)(a^2 + 1)
so, a =2 or a^2 + 1 = p.

putting a = 2 in our 2 quadratic equations we get (a-2)^2 = 24 for both, and hence
we have only 2 different roots.

putting p = a^2 + 1, we get (a^2 - 1)^2 = 2(a^3 + a + 2) which can be factored as
(a+1)(a-3)(a^2 + 1) = 0, hence a = -1 or a = 3.

But a = -1 is already ruled out. putting a = 3 we get the equations as x^2 - 6x - 40 = 0, and x^2 - 4x - 60 = 0.
the roots of first eq. are 10, -4 and that of 2nd are 10, -6 and we have 3 different
roots.

For a = 3 only the pair of quadratics have 3 different roots.

Hence, choice (d) is the correct answer.


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Quant Answer # 9
------------------------------------------------------------

Note a very simple fact.

Let us denote degree of a vertex by the number of edges passing through it.

In a triangle the degree of each vertex is 2 and sum of the degrees of vertices in 6.

In any graph, the sum of degrees of vertices is double the number of edges in that graph.

Let us consider each person as a vertex.
100th person(vertex) has 100 edges passing through it
=> an edge from 100th vertex meets 99th also.

99th person(vertex) has 98 new edges passing through it
but 99th hasn't shaken hands with 1st person as 1st has already with 100th.
=> 99th has shaked hands with 98th person

98th person(vertex) has 96 new edges passing through it
so on ...

Thus, total number of edges is 100+98+96+ ... + 2 + 0 = 50*51

Let 101th person(vertex) has degree n i.e. the number of edges passing through it or the number of persons he has shaken hands with.

Then, 2*50*51 = 1+2+3+ ... + 100 + n
=> 100*51 = 101*50 + n => n = 50

Hence, choice (d) is the correct answer

[rahul_85]

Originally Posted by Destiny's_Child

My answers to 7 to 9

7) here m-n = 1 + 1/3 [2^2 - 1^2]+ 1/5 [3^ - 2^2]..... (499 terms) - 500^2/1001
Each of the 499 terms reduces to 1 and taking 1001 approx as 1000
So m-n = 500 - 250 = 250

Here we have three conditions

Eq 1 has one root
Eq 2 has one root
Both the equations have 2 distinct roots but one of them is common to both the equations.

Condition 1 ) For x^2 - 2ax -4(a^2 + 1) = 0
20a^2 + 16 = 0
=> a^2 = -ve
So not possible

Condition 2 ) For x^2 - 4x -2a(a^2 + 1) = 0
16 + 8a(a^2 + 1) = 0
=> a^3 + a + 2 = 0
=> (a+1)(a^2 - a + 2) = 0
a = -1
For a^2 - a + 2, D is negative, so complex solution.

Condition 3) x^2 - 2ax -4(a^2 + 1) = x^2 - 4x -2a(a^2 + 1)
We get x = a^2 + 1
Putting this value of x in any of the equations, the eqn reduces to a^2 - 2a - 3 = 0
we get a = -1, 3
But when a = -1 total number of solutions becomes 2
So a = 3 is the only solution

So answer (d)

9)This is a very interesting question and the logic behind is simply superb....

Here the 100th person shakes hand with all except itself.
In this way one handshake of 1st person is completed.
the 99th person shakes hand with 2 to 101except itself.
In this way two handshakes of 2nd person is completed.
Continuing like this
the 99th person shakes hand with 2 to 101except itself.
the 99th person shakes hand with 2 to 101except itself.
Now all the handshakes are covered

So 101th person shakes hand from 51st to 100th person i.e total 50 persons
So option (d) (Not sure though)

HEY destiny's child you have done the last question brilliantly thank you very much
n its much easier to understand than that given as an official answer

keep up the good work!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[Destiny's_Child]

Originally Posted by rahul_85

HEY destiny's child you have done the last question brilliantly thank you very much
n its much easier to understand than that given as an official answer

keep up the good work!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Hey Rahul!!
I dont deserve the credit for it.....
It all goes to the brilliant people you will find at pg..
I am just trying to learn from them the tricks and shortcuts.....

As Aarav said, anybody can post the questions on the thread, so i am putting q10- 12

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Quant Question # 10
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Two statements are made about the set A = {1, 2, 3, ..., 100}.
(1) More 4 element subsets of A have sum greater than 201 than have sum less than 201.
(2) More 4 element subsets of A have sum greater than 203 than have sum less than 203.
Then which among the following is right?
(a) Only (1) is true (b) Only (2) is true
(c) Both (1) && (2) are false (d) Both (1) & (2) are true

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Quant Question # 11
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For all real x, f(x) satisfies 2f(x) + f(1-x) = 2*x^2 + 1. Then which among the following is
true for all x?
(a) f(x) <= 1 (b) f(x) >= -1 (c) f(x) >= 1 (d) f(x) <= -1

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Quant Question # 12
------------------------------------------------------------
Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c)
exist such that the given 18 digit number is divisible by 4?
(a) 62 (b) 74 (c) 86 (d) none of these

All those who were not active during QQAd D 06, i would suggest them to take part now....
You wil learn some mind blowing concepts....

[krsh.vik]

Originally Posted by Destiny's_Child

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Quant Question # 11
------------------------------------------------------------
For all real x, f(x) satisfies 2f(x) + f(1-x) = 2*x^2 + 1. Then which among the following is
true for all x?
(a) f(x) <= 1 (b) f(x) >= -1 (c) f(x) >= 1 (d) f(x) <= -1

B
put x = 0 and x =1 in the eqn
calculate f(0) and f(1)