# 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 (7x-3)/(2x+5) ≤ 3 then x lies between

• (7x-3)/(2x+5) - 3 = < 0
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]

• 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 (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).

• Q9) If [(x+7)^2 (2x-3)^3] / [(x-5) (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 (x-8)(x-10)(x-12) ... (x-100) < 0

• x = 8,10,12 ... 100.
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.

• Q11) Find the number of integer values of x for which (x+1)(x-3)(x+5)(x-7) ... (x+97)(x-99) < 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 (x-1)(x+5)(x-9) ... (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)(x-2)(x+3) ... (x-98) < 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 2|x-3| + 3|x+2| + 5|x-2| = 40.

• Here values of x = 2,3,-2.
Firstly we take extreme value of x = 3.
At x = 3
2|x-3|+3|x+2|+5|x-2| = 20.
At x = 4,
it will increase by 10,because 2+3+5 = 10.
2|4-3|+3|4+2|+5|4-2| = 30.
At x = 5,
it will give value 40.
2|5-3|+3|5+2|+5|5-2| = 40.
So max value of x = 5.
For min value of x = -2.
at x = -2,
2|x-3|+3|x+2|+5|x-2| = 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| + |x-3| + |x-5| < 15.

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

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