Topic - Quant Mixed Bag

Solved ? - Yes

Source - CAT Preparation With IIMpossible Forum. ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - CAT Preparation With IIMpossible Forum. ]]>

=> C^2 - A^2 = B^2 - C^2

=>(C - A) * (C + A) = (B-C) * (B+C)

Now Let (B-C) = X/Y * (C-A) , --------(1)

Then , (B+C) = Y/X * (C+A) --------(2)

GCD (X,Y) = 1

this way we get to 169 ]]>

a) 1/10

b) 1/12

c) 1/14

d) 1/16 (Credits - Test Funda) ]]>

All zeros occur at 1/n (n is a natural number less than 20)

Let say minimum of E occurs at x = 1/t

This is analogous to considering that there are 1 + 2 + 3 ... + 20 = 210 points on the number line. We need to select a point such that the sum of the distances of this point from all the 210 point will be a minimum. Thus, we need to take the median of these 210 points. i.e, any point between 105th point and 106th point. If we move to the left on the 105th point or to the right of the 106th point, the sum of the distances will increase.

Now the 105th point will be 1 + 2 + 3 + ... + 14 = 105

Thus the 105th point is 1/14 and the 106th point is 1/15. The expression, E will be minimum if we take x as any of these two values. i.e, at x = 1/14 or x = 1/15 (or any value in between)

Here, answer is 1/14

]]>1 - 1/x > = 0, x > 0

So (x - 1)/x > = 0 , (x^2 - 1)/x > 0

x > = 1 is Domain.

now squaring twice n simplifying

Replace x - 1/x = t

t^2 + 2 = x^2 + 1/x^2 so

t^2 + 2 - 2t - 1 = t^2 - 2t + 1 = (t-1)^2 = 0

So t = 1, x-1/x = 1

x^2 - 1 = x, x^2 - x - 1 = 0

x = ( 1 + root(5))/2 ]]>

To get co-efficient of a^204 we have to find the constant term of expression having power a^6 or remove factors of a^6" , which are

- (x^1-1) * (x^5-5)" product of constant term = 5
- (x^2-2) * (x^4-4)" product of constant term = 8
- (x^1-1)" (x^2-2)" (x^3-3)" product of constant term = -6
- (x^6 - 6) * 1 .. product of constant terms = -6

so your required coefficient = 5 + 8 - 6 - 6 = 1

]]>now a should be at-least 1

(x+1) + b + c + d + e = 12

=> x + b + c + d + e = 11 so 15c4 cases

now remove the cases when x > = 9

so, (x + 9) + b + c + d + e = 11

=> x + b + c + d + e = 2 i.e 6c4 cases

and when b, c , d , e > = 10 , we have to remove those cases as well .. isn't it ?

so , x + (b+10) + c + d + e = 11

=> x + b + c + d + e = 1 i.e 5c4 cases

Required numbers = 15c4 - 4 * 5c4 - 6c4

= 1365 - 20 - 15

= 1365 - 35

= 1330

N=a^p * b^q

N^2 = a^(2p) * b^(2q)

So (2p + 1)(2q + 1)= 5 * 401

p = 200, q = 2

N = 2^200 * 3^2

[LogN] + 1 = [200log2 + 2log3] + 1 = [61.16] + 1 = 62 digits ]]>

=> a + b + c = 3

5c2

Total no . of integral solutions = 10

Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3

= x^12 in (1 - x^6)^3 * (1-x)^-3

= 14c2 - 3c1 * 8c2 + 3c2

= 10

If you are good in binomial then do it directly. Else you can also think this in these lines. Is this not same as 3x + 2y + z = 14 and is this not same as finding the positive distinct integral solutions of the equation a + b + c = 14? so (13c2 - 3 * 6)/6 = 10

Method 2 :

If you want to solve it by binomial then check below

coefficient of x^14 in (x^3 + x^6 + x^9 + ...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)

= coefficient of x^14 - 6 (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)

= coefficient of x^8 in (1 + x^2 + x^3 + x^4 + ... + x^8 +...) * (1 + x^2 + x^3 + x^4 + ... + x^8 + ....)

= 1 + 1 +1 + 1 + 1 + 2 + 1 + 2 = 10

So A + 2B can take values 5 , 15, 25 ,......95

The number of solutions corresponding to this will be 2 + 7 + 12 ... + 47 = 245

Also, A + 2B can be 10 , 20 , 30 ... 90

The number of solutions corresponding to this will be 4 + 9 + 14 ... + 44 = 216

so total = 245 + 216 = 461. ]]>

These 5 primes to be distributed into 3 places.

Total : 3^5 ways , but this is ordered ( arranged) and we need unordered.

We know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.

Except for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways

So remove this from total and we are left with only distinct, is divide by 3! ,

So (3^5 - 3)/3! , and just add that one case of 1, 1, 2310

So (3^5-3)/3! + 1 = 41 ]]>

now for "n" to be odd , (n-48) has to be odd

so n - 48 can be either 1 or 3 or 3^2 (because we need ODD)

(m -12) * (n - 48) = 2^6 * 3^2 * 1

OR (m -12) * (n - 48) = 2^6 * 3 * 3

or (m -12) * (n - 48) = 2^6 * 3^2

when n - 48 = 1, n = 49 (acceptable)

when n - 48 = 3, n = 51 (acceptable)

when n - 48 = 3^2, n = 57 (acceptable)

only these 3 values are possible for n < 60 and odd ]]>

= coff of x^20 in (1 - x^9)^4 * (1 - x)^-4

= 23c3 - 4 * 14c3 + 4c2 * 5c3

= 1771 - 1456 + 60

= 375

For more detailed explanation of the concept, refer the video.

]]>a) 6

b) 7

c) 8

d) 9 ]]>