# Quant Boosters - Swetabh Kumar - Set 5

• no of integral solution x^2+y^2 =36 / < = 36 / < 36

For < = case,
x=0, y=0,+-1,+-2...+-6 so 13 here
x=+-1, y=0, +-1....+-5 so 2 * 11=22 cases
x=+-2, y=0, +-1...+-5 so 2 * 11=22 cases
x=+-3 y=0,+-1...+-5 so 2 * 11=22 cases
x=+-4 y=0,+-1...+-4 so 2 * 9=18 cases
x=+-5 y=0,+-1,+-2,+-3 so 2 * 7=14 cases.
x=+-6 y=0 so 2 * 1=2 cases
total = 113.

For < case,
remove (0,+-6) and (+-6,0) from above, so 113-4=109.

for = 36 case
only (+-6,0), (0,+-6) so 4 cases

If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

x^2+y^2+z^2 = 25-6=19
(y+z)^2 < = 2*(y^2+z^2) (Cauchy)
(5-x)^2 < = 2(19-x^2) = 38-2x^2
x^2+25-10x < = 38-2x^2
3x^2-10x-13 < = 0
3x^2+3x-13x-13 < = 0
(3x-13)(x+1) < = 0
so max x= 13/3 and min x = -1

If the LCM of 75 and N is 150.Find how many different values can N take?

75=3 * 5^2 150= 2 * 3 * 5^2
so N= 2 * 3^a * 5^b
max (1,a) = 1 and max (2,b)=2
so for (1,a)= (1,0),(1,1)=2
for max (2,b)=2: (2,0),(2,1),(2,2) so 3 so 2*3=6 numbers

Integral solution of rectangle having perimeter 1260

a+b = 630
(1,629)...(315,315) so 315

If N = 2^3 * 3^4 * 5^9, then find the number of trailing zeroes at the end of product of all factors of N which are not divisible by 12.

number formed from 2 * 3^3 * 5^9 = 80 factors.
so product of these = 12^80 * (2 * 3^3 * 5^9)^40 = 2^200 * 3^83 * 5^360.
total product = (2^3 * 3^4 * 5^9)^(100) = 2^300 * 3^400 * 5^900
so not divisible ke liye.
2^100 * 3^x * 5^540 so 10^100

In ∆ABC , I is the incentre of triangle then find the length of AI if AB=4cm, BC= 6cm & CA= 8cm

AP be angle bisector.
AI/IP = (4+8)/6 = 2 --(1)
so AI = 2/3 * AP
and AP^2 = (AB^2+AC^2-BP * PC) = (16+64-2 * 4) = 72 so AP=6 rt 2
so AI= 2/3 * 6 rt 2 = 4 rt 2.

There are 6 balls of different colours and 3 boxes of different sizes. Each box can hold all the 6 balls. The balls are put in the boxes so that no box remains empty. The number of ways in which this can be done is:
a) 534
b) 543
c) 540
d) 528

Method 1: (1,1,4)" 6C1 * 5C1 * 3= 90
(1,2,3)= 6C1 * 5C2 * 3! = 360
(2,2,2)= 6C2 * 4C2 * 1 = 90 adding 90+360+90=540

Method 2: 6 balls 3 boxes.
so by inclusion exclusion. 3^6 - 3C1 * 2^6 + 3C2 * 1^6 = 540

If a set such that S= {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} then how many subset can be made where the product of elements will be even?

toal subsets= 2^n = 2^20
for odd product, all even nos should be left out. so 10 nos left. so 2^10.
so for even product = total-odd = 2^20-2^10

One root of equation x^4−5x^3+ax^2+bx+c is 3+ rt2 and all roots real and a, b and c are rational. Find the max value of a and c and least of b?

a=5/4, b=11/2, c=7/4
(3+rt 2) + (3-rt 2) +a+b = 5
a+b=-1 a=b=-1/2
a=sum of roots two at a time= {7+1/4 -6} = 5/4
b= three at a time= -{1/4*(3+rt 2) + 1/4*(3-rt2) -1/2*(7)-1/2*(7)} = 7-3/2 =11/2
c= product = 7/4

How many ordered pairs of integers (a,b), are there such that their product is a positive integer less than 100?
In the above question, if (a,b) is not different from (b,a), then how many such pairs are possible?

a) when both are +ve:
1 * 1 ... 1 * 99 and reverse = 99 * 2 - 1 = 197
2 * 2...2 * 49 and reverse = 48 * 2 - 1 = 95
3 * 3....3 * 33 = 31 * 2 - 1 = 61
4 * 4...4 * 24 = 21 * 2 - 1 = 41
5 * 5...5 * 19 = 15 * 2 - 1 = 29
6 * 6....6 * 16 = 11 * 2 - 1 = 21
7 * 7...7 * 14 = 8 * 2 - 1 = 15
8 * 8 .... 8 * 12 = 5 * 2 - 1 = 9
9 * 9...9 * 11 = 3 * 2 - 1 = 5
so total = 473
same when both are -ve so 473 * 2 = 946

b) Unordered = 99+48+31+21+15+11+8+5+3= 241 so 241*2= 482

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