Quant Boosters  Vikas Saini  Set 5

Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source 

Q1) If 4n has 15 factors, 5n has 12 factors. Find number of factors in 3n

Important part here is multiplication of 4 & 5.
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.

Q2) For how many positive values of N, (N + 102) / (N +2) must be an integer

For such questions we need to work on factors only.
(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)

Q3) For how many total values of N, (N + 28) / (N + 8) must be an integer

We need to work on factors.
(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.

Q4) A composite number is given 150.
(a) Find the sum of divisors.
(b) Find the sum of even divisors.
(c) Find the sum of odd divisors.

150 = 2 x 3 x 5^2.
(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.

Q5) How many divisors of 1296 are perfect square

1296 = 36^2 = (2^2 x 3^2)^2
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.

Q6) How many of the first 1200 natural numbers are not divisible by 2, 3 or 5 ?

Solve it by Euler’s theorem
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.

Q7) If (7x3)/(2x+5) ≤ 3 then x lies between

(7x3)/(2x+5)  3 = < 0
7x36x15 / 2x+5 = < 0
x18 / 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]

Q8) If (x^3  10x^2 + 29x  20)/(x^2 + 7x + 12) < 0 then x lies between

here for x = 1, numerator can give value 0.
hence (x1) is a factor of numerator.
= > (x^3  x^2  9x^2 + 9x + 20x  20)/(x^2 + 3x + 4x + 12) < 0.
= > [x^2(x1)  9x(x1) + 20(x1)] / [x(x+3) + 4(x+3)] < 0.
= > (x1) (x^29x+20) / (x+3)(x+4) < 0.
= > (x1)(x4)(x5)/(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).

Q9) If [(x+7)^2 (2x3)^3] / [(x5) (x+1)] < 0 then x lies between

x = 7.
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).

Q10) How many positive integer values can x take that satisfy the inequality (x8)(x10)(x12) ... (x100) < 0