# Quant Boosters - Hemant Malhotra - Set 8

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Elite's Grid Prep Forum

• Q1) Find number of integral solutions of 2/x + 3/y = 1/10

• Concept:
a/x + b/y=1/k
where a, b and k are positive integers and we want number of value of (x,y) satisfying this equation
Approach - first find number of factors of a * b * k^2 let number of factors=F
a) total number of positive integral solutions=F
b) total integral solutions = 2 * F-1
c)total number of negative solution= zero ( bcz if both x and y will be negative than lhs will be negative but rhs is positive so not possible )

solution:

First find total number of factors of 2 * 3 * 10^2
2^3 * 3 * 5^2 so (4 * 2 * 3) = 24 total number of factors
so positive integral solutions = 24
total number of integral solution = 2 * 24 - 1 = 47

• Q2) Find the smallest positive integer such that (n-13)/(5n+6) is a non-zero reducible fraction.

• hcf(n-13,5n+6)
hcf((n-13,5n+6-5n+65))
hcf(n-13,71))
so reducible form for min n=
n - 13 = 71
so n = 84

• Q3) Find the minimum possible value of a + b + c if abc + bc + c = 2014, where a, b, c are positive integers.

• c * (ab+b+1)=2014
c * (ab+b+1) = 2 * 19 * 53
so c=19
ab+b+1=106
ab+b=105
b * (a+1)=7*15
so b=7 and a=14
so sum =40

• Q4) If x = 1! + 2! +3! +4! + ... + n!, how many value of n, x is a perfect square?

• 3,5,6 any perfect square will not leave this remainder when div by 7
so 1!+2!+3!+4!=33 mod 7=5
or 1!+2!+3!+4!+5!=153 mod 7=6
or 1!+2!+3!+4!+5!+6! mod 7=153+620 mod7=6+4=10 mod 7=3
so no value of n>=4 will give perfect square
so check for n=1,2,3
when n=1 then 1!=1 so perfect square
when n=2 then 1!+2!=3 not a perfect square
when n=3 then 1!+2!+3!=9=perfect square
so n=1 and 3 only two values

• Q5) Find the number of integral solutions to |x| + |y| + |z| = 15
a. 902
b. 728
c. 734
d. 904

• x + y + z = 15
so positive solutions = 15-1c3-1 = 14c2 = 7 * 13 = 91
now
when all negative =91
when 2 positive one negative 3c2 * 91
same for 2 negative one positive
91 + 91 + 3 * 91 + 3 * 91
8 * 91= 728
now case when one is zero
|y|+|z|=15
then
14 cases
again 14 * 4=56
so 3c2 * 56 = 3 * 56= 168
now when 2 are zero
|z|=15
so 2 solutions
so 3c2 * 2=6 solutions
so total 728+174 =902

• Q6) If x^2 + xy + y^2 = 41, y^2 + yz + z^2 = 73 and z^2 + xz + x^2 = 9, find (x+y)/z.

• (x^3 - y^3) = 41 * (x - y)
y^3 - z^3 = 73 * (y - z)
z^3 - x^3 = 9 * (z - x)
41x - 41y + 73y - 73z + 9z - 9x = 0
32x + 32y = 64z
so (x+y)/z = 2

• Q7) Find all primes p, q, so that p^2 − 2q^2 = 1.

• p^2-2q^2=1
(p^2-1) = 2 * q^2
(p-1)(p+1) = 2 * q^2
when p-1=1
then p+1 = 2q^2
so 2q^2=3 not possible
when p-1=2
then p+1=2q^2
4=2q^2 so q^2=2 no possible
when p+1=2 and p-1=q^2
p=1 and 0=q^2 not possible
now when p-1=2q
and p+1=q
so 2p=3q
so p=3q/2
so q=2 and p=3 will satisfy

• Q8) The sum of first 2 terms of an infinite GP is 18. Also, each term in the series is seven times the sum of the terms that follow. Find the first term and common ratio of the GP

• a = 7 * ( ar + ar^2 + ar^3 ... )
a=7(ar/1-r)
so a-ar=7ar
so 1-r=7r so 8r=1 so r=1/8
now a+ar=18
a(1+1/8)=18
so a * 9/8=18
so a=16

• Q9) Thirty two men can complete a work in 16 days and 48 women can complete the same work in 12 days. Sixteen men and 36 women started working together and worked for 8 days. If the remaining work has to be completed in 2 days, how many additional men would be required?

• let total work=512 unit
32 men per day work=512/16=32 unit
so 1 men per day=1 unit
in same way per day work of women=8/9 units
now 16 M+ 36 W = 8 day work=16 * 1 * 8 + 36 * 8 * 8/9 = 384 units
so remaining work=128 units
let x more men needed
so (16+x) * 2 + (36 * 8/9 * 2) = 128
so x = 16

• Q10) The roots of x^3 - ax^2 + bx - c = 0 are p, q and r while the roots of x^3 + dx^2 + ex - 90 = 0 are p+3, q+3 and r+3. what is the value of 9a + 3b + c ?

61

65

64

67

61

61

64

64