Quant Boosters  Vikas Saini  Set 5

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

x = 8,10,12 ... 100.
total terms = (1008 )/2 + 1 = 47.
value of x for which total product should be negative.
so for 810,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 = {(1008 )/2}/2 = 23.
Hence 23 values of x for which product < 0.

Q11) Find the number of integer values of x for which (x+1)(x3)(x+5)(x7) ... (x+97)(x99) < 0

Here x = 1,5,9 ... 97.
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.

Q12) Find the number of integer values of x for which (x1)(x+5)(x9) ... (x+61) < 0

x = 1,9....57.
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.

Q13) Find the number of integer values of x for which (x+1)(x2)(x+3) ... (x98) < 0

x = 1,3 ... 97
x = 2,4 .... 98
so total terms which will give negative value = (1+97)/2 + 1
= 49.
The value in between two 1 & 3 which will give negative value is only one 2.
so total values = 49 x 1 = 49.

Q14) Find the sum of all values of x which satisfy the equation 2x3 + 3x+2 + 5x2 = 40.

Here values of x = 2,3,2.
Firstly we take extreme value of x = 3.
At x = 3
2x3+3x+2+5x2 = 20.
At x = 4,
it will increase by 10,because 2+3+5 = 10.
243+34+2+542 = 30.
At x = 5,
it will give value 40.
253+35+2+552 = 40.
So max value of x = 5.
For min value of x = 2.
at x = 2,
2x3+3x+2+5x2 = 30.
at x = 3
It will give 40.
So sum of values of x = 5  3 = 2.

Q15) Find the range of x if x+2 + x3 + x5 < 15.

First we take extreme value of x = 5.
At x = 5,
x+2+x3+x5 = 9.
at x = 6,
x+2+x3+x5 = 12.
at x = 7,
x+2+x3+x5 = 15.
For minimum take x = 2(min)
at x = 2,
x+2+x3+x5 = 12.
at x = 3,
x+2+x3+x5 = 15.
x € (3,7).