Quant Boosters  Hemant Malhotra  Set 7

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 = 16


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 value

Q24) A 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=1/2

Q25) Find the largest integer N , if N + 496 = a^2 , N + 224 = b^2.

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=4265

Q26) 2x + 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=81

Q27) In 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 above

Method 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 triplet

Q28) gcd(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 = 189

Q29) For 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 9

Q30) How many integral values of (x, y) will satisfy the following inequality :
√(x^2  6x + 25) + √(y^2  8y + 25) ≤ 7
a) 0
b) 1
c) 2
d) 3
e) more than 3

x^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