Quant Boosters by Hemant Malhotra  Set 19

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
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 * F1
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 = 47Find the smallest positive integer such that (n13)/(5n+6) is a nonzero reducible fraction.
hcf(n13,5n+6)
hcf((n13,5n+65n+65))
hcf(n13,71))
so reducible form for min n=
n  13 = 71
so n = 84Find 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 =40Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B, how many almond cookies are there ?
A gets twice So he has even number of cookies
even = even + even + even
even = odd + odd + even
three even is not here so he will choose two odd and one even
so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25
so 74 or 76 or 78 or 78 or 82
now B has 37 or 38 or 39 or 41
37 = 18+19 (19 is already used so not possible)
38 is not possible
39 = 18+21 (this is possible combination)
41 = 23 + 18
so 23 almond cookies are thereIf 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 valuesFind the number of integral solutions to x + y + z = 15
a. 902
b. 728
c. 734
d. 904x + y + z = 15
so positive solutions = 151c31 = 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 =902If x² + xy +y² = 41, y² + yz + z² = 73 and z² + xz + x² = 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)
add all three
41x  41y + 73y  73z + 9z  9x = 0
32x + 32y = 64z
so (x+y)/z = 2Find all primes p, q, so that p^2 − 2q^2 = 1.
p^22q^2=1
(p^21) = 2 * q^2
(p1)(p+1) = 2 * q^2
when p1=1
then p+1 = 2q^2
so 2q^2=3 not possible
when p1=2
then p+1=2q^2
4=2q^2 so q^2=2 no possible
when p+1=2 and p1=q^2
p=1 and 0=q^2 not possible
now when p1=2q
and p+1=q
so 2p=3q
so p=3q/2
so q=2 and p=3 will satisfyLet F(x) =x^5+x^4+x^3+x^2+x+1 . Find remainder when F(x^12) is divided by F(x)
x^na^n=(xa)(x^(n1)+x^(n2)+ .... a^(n1)
f(x^12)=1+(x^12)+(x^12)^2+(x^12)^3+(x^12)^4+(x^12)^5
f(x^12)=1+(((x^12 1)+1+(x^12)^2 1 +1 +((x^12)^3 1 +1 +(x^12)^4 1 +1 +(x^12)^5 1+1
f(x^12)=6+((x^121)+(x^12)^21+(x^12)^3 1+(x^12)^4 1 +(x^12)^51
now x^121=(x^61)(x^6+1)
x^61=(x1)(1+x+x^2+x^3+x^4+x^5)
so every term will contain x^61 and will be divisible by 1+x+x^2+x^3+x^4+x^5
so OA=6The 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/1r)
so aar=7ar
so 1r=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