Quant Boosters - Vikas Saini - Set 5
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Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source -
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Q1) If 4n has 15 factors, 5n has 12 factors. Find number of factors in 3n
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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.
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Q2) For how many positive values of N, (N + 102) / (N +2) must be an integer
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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)
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Q3) For how many total values of N, (N + 28) / (N + 8) must be an integer
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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.
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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.
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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.
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Q5) How many divisors of 1296 are perfect square
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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.
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Q6) How many of the first 1200 natural numbers are not divisible by 2, 3 or 5 ?
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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.
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Q7) If (7x-3)/(2x+5) ≤ 3 then x lies between
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(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]
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Q8) If (x^3 - 10x^2 + 29x - 20)/(x^2 + 7x + 12) < 0 then x lies between
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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).
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Q9) If [(x+7)^2 (2x-3)^3] / [(x-5) (x+1)] < 0 then x lies between
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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).
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Q10) How many positive integer values can x take that satisfy the inequality (x-8)(x-10)(x-12) ... (x-100) < 0
