Quant Boosters by Hemant Malhotra  Set 18

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Let f be a oneto one function from the set of natural numbers to itself such that f(mn) = f(m) x f(n) for all natural numbers m and n. What is the least possible value of f(111)
one to one so single image for every element
now f(1 * 1)=f(1) * f(1)
so f(1)(f(1)1)=0
so f(1)=1 bcz f(1)=0 not possible we are taking about natural number
now f(111)=f(3 * 37) = f(3) * f(37)
we want f(3) and f(37) as min as possible and different from each other
so put f(3)=2 and f(37)=3
so 2 * 3=6If a three digit number ‘abc’ has 3 factors, how many factors does the 6digit number ‘abcabc’ have?
square of prime
so abc = k^2
now abcabc = abc * 1000 + abc
abc * 1001
so k^2 * 7 * 11 * 13
now when k=any prime other that 11,13 till 31 then 24 factors
when k is prime 11 or 13 let k=11 then 11^3 * 7 * 13 = 4 * 2 * 2 = 16Find number of real values of x satisfying 3[x]+2{x}=x where [x] greatest integral & {x}= fractional part
3I + 2F = I + F
so F = 2I
so 0 < = F < 1
so 0 < = 2I < 1
so 1/2 < I < = 0
so I = 0 is possibility
so F = 0
so x = 0 is only valueA fare dice is thrown 5 times. What is the probability that a prime number appear even number of times
2,3,5 three prime numbers
so out of 6 we have 3 prime number
so prob of prime=1/2 and prob of non prime=1/2
so 5c0 * (1/2)^0 * (1/2)^55c2 * (1/2)^2 * (1/2)^3 + 5c4 * (1/2)^4 * (1/2)
132 + 10 * 1/4 * 1/8 + 5 * 1/16 * 1/2
16/32=1/2Find the largest integer N , if N + 496 = a² , N + 224 = b².
a^2496=b^2224
so (ab)(a+b)=272
now N will be max when a and b are as max possible
a=69 and b=67
so N=69^2496=42652x + y + z = 20 Find number of positive integral solution.
y and z both even then y=2k and z=2m
then x+k+m=10 so 9c2=36 cases
when y and z both odd then y=2m1 and z=2k1
then 2x+2m+2k=22
so x+m+k=11 so 10c2=45
so 36+45=81In a right angled triangle with integral sides, smallest side is four times of difference of other two sides and sum of all sides is 120 then find area of this triangle
a. 60
b. 120
c. 180
d. 240
e. None of the aboveMethod 1
S = 4 * (HM)
Smallest side is multiple of 4 and S+M+H=120 so triplet will be
S+M+(S+4M)/4=120
so 5S+8M=480
so M=(60)((5S/8))
so S=24 and M=45 so H=51 so area=540Method 2
S = 4 * (HM)
so 8 = 4 * (1715)
so multiple of this will give our ans
so 24, 45, 51 will be tripletgcd(1,55) + gcd(2,55) + ... + gcd(55,55) = ?
E(55)=40
now factors of 5 and 11
till 50 , multiple of 5 are 10 so 5 * 10 = 50
and multiple of 11 will be 11,22,33,44 so 11 * 4 = 44
now 55 also
so 40 + 50 + 44 + 55
= 90 + 99 = 189For how many integral x , (3x+8)^2/(3x2) is also an integer
3x2=y
so 3x+8=y+10
so (y+10)^2/y
y+100/y +20
100/y
now factors of 100=2^2 * 5^2
so 9How many integral values of (x, y) will satisfy the following inequality :
√(x²  6x + 25) + √(y²  8y + 25) ≤ 7
a) 0
b) 1
c) 2
d) 3
e) more than 3x^2  6x + 25 = (x3)^2 + 16
so min of sqrt (x^26x+25) = 4
y^2  8y + 25 = (y4)^2 + 9
so min of sqrt (y^28y+25) will be 3
so min sum will be 7
for x=3 and y=4