Quant Boosters by Hemant Malhotra  Set 11

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
How many positive integral solutions does the following equation has : a + b + c + d = a × b × c × d
All these types of questions can be solved by taking upper bound of the variables.
Suppose all a, b, c and d are more than 1, then minimum product is 16, which is more than 2+2+2+2, taking bigger numbers will increase the product even further. Thus one of them is 1
then the equation reduces to 1 + a + b+ c = abc.
Again all a, b, c cannot be more than 1 as abc becomes a larger number, thus one of them is also 1,
so 1 + 1 + a+b =ab or ab a b= 2 or (a 1)b1) = 3,
so (a, b) = (2, 4)
so (1,1,2,4)Find number of real roots of x^44x^3+6x^24x+1=0
Pascal triangle (1,4,6,4,1 ) for 4 degree so (x1)^4 so OA=4
In the expansion of (a + b + 2)^6, what will be the coefficient of a^2b^2?
General term of (x+y+z .)^n is n!/(p!) * (q!) * (r!) * (x)^p * (y)^q*(z)^r where p+q+r=n
same for 4,5,6 and so on variables
now 6!/(p!) * (q!) * (r!) * (a)^p * (b)^q * (2)^r
so p=2 and q=2 so r=2
so coefficient
6!/(2! * 2! * 2!) * 2^2A real valued function f is such that f(x + y) = x + f(f(y)) find f(9)
let f(y) is quadratic equation then f(f(y) will be biquadratic which is not possible bcz RHS will not be biquadratic
so F(y) will be linear
so f(y)=ay+b
so f(x+y)=a(x+y)+b
and x+f((ay+b)=x+a*(ay+b)+b=x+a^2y+ab+b
so ax+ay+b=x+a^2y+ab+b
so ax+ay=x+a^2y+ab
so ab=0 so a=0 or b=0 but a=0 not possible so b=0
so ax+ay=x+a^2y
so a=1
so f(x)=ax+b=x
so f(9)=9If x – 1 + x – 5 + x – 9 = 8, what is the average of all the possible values of x?
at x=9 , 8+4+0=12 so x > = 9 will not satisfy
at x=1 ,0+4+8=12 again x < =1 will not satisfy
now check for 1 < = x < = 5 and 5 < = x < = 9
when 1 < = x < = 5
then x1x+5x+9=8
so x+13=8 so x=5 will satisfy
now for 5 < x < 9
x1+x5x+9=8
so 2x+4=8
so x=2 which is not satisfying so only x=5 will satisfyConsider sets containing 3 elements, Sn = { n, 2n + 1, 3n + 2}. For how many values of n, where n is a natural number and n < 1000, both sets Sn and S(n+1) contain exactly one element that is divisible by 7?
S={n,2n+1,3n+2} Sn+1={n+1,2n+3,3n+5}
exact one element that is divisible by 7
when n=7k form
then {7k,14k+1,21k+2,7k+1,14k+3,21k+5} so exact one element is div by 7 so 7k < 1000
so k=143
now n=7k+1
then Sn={7k+1,14k+3,21k+5,7k+2,14k+5,12k+12 so no element is exact div by 7
now check till 7k+6 and find answer.If all the roots of y^5 – 10y^4 + py^3 + qy^2 + ry – 32 = 0 are positive, find the value of (p+q)/r ?
Sum of roots=10
Average=2
Product of roots=32
Geometric mean = 2
so AM = GM
so all roots equal. Roots are 2, 2,2,2,2A is the absolute difference between a three digit natural number and the sum of the digit of the number. If ‘A’ is a multiple of 12 and greater than 504, then how many values can A assume?
let three digit number=100x+10y+z
100x+10y+zxyz=A
so 99x+9y=A
so 911x+y=A
if A is multiple of 12 so A=12k
911x+y=12k
so 12 k > 504
so k > 42
311x+y=4k
so k will be multiple of 3 and greater than 42
so k=45
then 11x+y=60
when k=48 then 11x+y=64
till 11x+y=108Find the remainder if 8533^859^361 divided by 19?
8533 mod 19=2
so 2^(859^361) mod 19
E(19)=18
so 859^361 mod 18=13
so 2^13 mod 19=(16)^3 * 2 mod 19=9 * 32 mod 19=9 * 13 mod 19=3From the numbers 1, 2, 3, .., 9, we can choose six numbers in exactly two ways such that their product is a perfect square, let’s say sum are m^2 and n^2, where m and n are distinct positive integers.
Find (m + n).
a) 108
b) 11
c) 61
d) 56
e) 144The number can never have 5 and 7 so 1 * 2 * 3 * 4 * 6 * 8 * 9
Now we can remove numbers which are x and x*n^2 once each.
Only such pair is 2 & 8
So one number is 36^2 and another one is 72^2
so 108