Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

4 = 2^2.

Let n = 2^a x 5^b.

4n = 2^(a+2) x 5^b.

Total factors = (a+2+1) (b+1)

15 = (a+3) (b+1) .... (i)

5n = 2^a x 5^(b+1)

Total factors = (a+1) (b+1+1)

12 = (a+1) (b+2) ..... (ii)

From equation (i) & (ii)

a = 2 , b =2.

n = 2^2 x 5^2

3n = 2^2 x 3 x 5^2

Total factors = 3 x 2 x 3 = 18. ]]>

(N+2+100) / (N+2)

1 + 100 / N+2.

100 = 2^2 x 5^2.

Total factors = (2+1) (2+1) = 9.

But here N+2 > 2.( N must be positive)

We need to remove 1 & 2 from total factors.

Required values = 9 – 2 = 7.

Check :-

Factors of 100 are 1,5,25,2,10,50,4,20,100.

After removing 1,2 it can be checked by putting N+2 = 5,25,10,50,4,20,100.

N=3,23,8,48,2,18,98.

(N+102) / (N+2) = 21,5,11,3,26,6,2. (Total 7 values) ]]>

(N+8+20) / (N+8 )

1 + (20 / N+8 )

60 = 2^2 x 5.

Total factors = 3 x 2 = 6.

But in question it is given that for total values, so factors can be negative too.

Total = 2 x 6 = 12 .

Check :-

Factors of 20: – (1,2,4,5,10,20).

Negative factors :- (-1,-2,-4,-5,-10,-20).

N + 8 = -1,-2,-4,-5,-10,-20,1,2,4,5,10,20.

N = -9,-10,-12,-13,-18,-28,-7,-6,-4,-3,2,12.

(N+28 ) / (N+8 ) = -19,-9,-4,-3,-1,0,21,11,6,5,3,2.

Total 12 integer values. ]]>

(a) Find the sum of divisors.

(b) Find the sum of even divisors.

(c) Find the sum of odd divisors. ]]>

(a) Sum of divisors

(2^0 + 2) (3^0 + 3)(5^0 + 5 + 5^2)

= (1+2)(1+3)(1+5+25)

=3 x 4 x 31

= 372.

Check :-

1+150+2+75+3+50+5+30+6+25+10+15 = 372.

(b) sum of even divisors

2 (1+3)(1+5+25)

= 2 x 4 x 31

= 248.

Check :-

150+2+50+30+6+10 =248.

(c) sum of odd divisors

(1+3)(1+5+25)

= 4 x 31

= 124.

Check :-

1+75+3+5+25+15=124. ]]>

1296 = 2^4 x 3^4

Perfect square = [ (4/2) + 1] [(4/2) + 1] = 3 x 3 = 9.

To find perfect square :-

(1,2,4,8,16,32)(1,3,9,27,81)

(1,4,16)(1,9,81)

(1,9,81,4,36,324,16,144,1296)

Total 9. ]]>

If N = p^a x q^b x r^c

E(N) = N (1 – 1/p)(1 – 1/q) (1 – 1/r)

1200 = 2^4 x 3 x 5^2.

Numbers not divisible by 2,3,5

= 1200 (1 – 1/2)(1 – 1/3) (1 – 1/5)

= 1200 (1/2)(2/3)(4/5)

= 320. ]]>

7x-3-6x-15 / 2x+5 = < 0

x-18 / 2x+5 < = 0.

x = 18, x = -2/5.

on line -2/5 will lie on negative side while 18 will lie on positive side.

-inf to -5/2 ~ positive value

-5/2 to 18 ~ negative

18 to inf ~ positive.

So here on -5/2 value can be positive which will not satisfy to the question.

While on 18 it will give perfect 0.

Hence x € ( -5/2, 18] ]]>

hence (x-1) is a factor of numerator.

= > (x^3 - x^2 - 9x^2 + 9x + 20x - 20)/(x^2 + 3x + 4x + 12) < 0.

= > [x^2(x-1) - 9x(x-1) + 20(x-1)] / [x(x+3) + 4(x+3)] < 0.

= > (x-1) (x^2-9x+20) / (x+3)(x+4) < 0.

= > (x-1)(x-4)(x-5)/(x+3)(x+4) < 0.

= > x = 1,4,5,-3,-4.

on line -3 & -4 will lie on left side while 1,4,5 will lie on right side.

-inf to -4 ~ negative

-4 to -3 ~ positive

-3 to 1 ~ negative

1 to 4 ~ positive

4 to 5 ~ negative

5 to inf ~ positive.

x € (-inf,-4) U (-3,1) U (4,5). ]]>

x = 3/2.

x = 5.

x = -1.

-7 & -1 will lie left side of line while 3/2 & 5 will lie right side of line.

-inf to -7 ~ positive

-7 to -1 ~ positive

-1 to 3/2 ~ positive

3/2 to 5 ~ negative

5 to inf ~ positive

x € (-inf,-7)U(-7,-1)U(-1,3/2)U(5,inf). ]]>

total terms = (100-8 )/2 + 1 = 47.

value of x for which total product should be negative.

so for 8-10,if we put x = 9 then one value will be positive,rest 46 values will be negative. Product of negative nos(even no of negative terms) give a positive value.

For x = 11, 2 positive & 45 negative

hence product negative.

so negative value of x = {(100-8 )/2}/2 = 23.

Hence 23 values of x for which product < 0. ]]>

x = 3,7 ... 99

for x € (-1,-99) product will come negative.

total terms = (-1+97)/4 + 1

= 25.

but in between two terms there will be few values for which product will become negative.

Let's take any two term -1 & -5.

in between -1 & -5, 3 values x=-2,-3,-4 product will become negative. It means only 3 values in between any two terms.

Hence total values = 3 x 25 = 75. ]]>

x = -5,-13.....-61.

here terms of x which will give product in negative.

total term = (-5+61)/8 + 1

= 7 + 1

= 8.

total values in between -5 & -13 which will give negative values are :-

-6,-7,-8,-9,-10,-11,-12

total 7 values.

Total = 8 x 7 = 56. ]]>