# Quant Boosters - Hemant Malhotra - Set 7

• a^2-496=b^2-224
so (a-b)(a+b)=272
now N will be max when a and b are as max possible
a=69 and b=67
so N=69^2-496=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=2m-1 and z=2k-1
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 * (H-M)
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=540

Method 2
S = 4 * (H-M)
so 8 = 4 * (17-15)
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/(3x-2) is also an integer

• 3x-2=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 = (x-3)^2 + 16
so min of sqrt (x^2-6x+25) = 4
y^2 - 8y + 25 = (y-4)^2 + 9
so min of sqrt (y^2-8y+25) will be 3
so min sum will be 7
for x=3 and y=4

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