# Question Bank - Algebra - Hemant Malhotra

• 0^2=0
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
6^2=36
so 40 is even number
so even + even + even
or odd + odd + even
so only one possible 36 + 4 + 0 = 40
so 2 triplets possible
(1,2,3) and (1,3,2)

• Q60) How many integer solutions exist for the equation 5x - y - 120 = 0 such that values that x assumes has opposite signs compared to the corresponding values of y

• 5x-y=120
So x=24+(y/5)
When y positive then x will be positive so we can no take y as positive
So we are looking for negative values of y such that x is positive
x=24+(y/5)
When y=-5 then x is positive
Till y=-115 , x will be positive
So -5 to -115 with common difference of 5

• Q61) If f(x) is an even function defined on the interval (-5,5) , find total number of real values of x satisfying
f(x) = f(x+1/(x+2)

• even function means f(-x)=f(x)
means if f(a)=f(b) then a=b and a=-b
f(x)=f(x+1/(x+2)
so x=(x+1/(x+2) so 2 values of x by this equation
and x=-(x+1/(x+2) and two values of x by this equations
so total 4 real values of x

• Q62) If (a + b)/(1 - ab), b, (b + c)/(1 - bc) are in AP then a, 1/b, c are in
a) AP
b) GP
c) HP
d) None of the above

• Method-1 If numbers are in AP ., always start from HP in these kind of questions ...and take any random values of a,1/b and c which will make a,1/b and c in HP
let a=1 , b=2 and c=1/3
so (a+b)/(1-ab)=3/1-2=-3
b=2
(b+c)/(1-bc)=2+1/3/(1-2/3)=7
so our assumption is correct so a,1/b, c are in HP

Method 2-
2b=(a+b)/(1-ab)+(b+c)/(1-bc)
by solving
2b=1/a+1/c
so a,1/b and c are in HP

• Q63) In a soccer tournament, if all teams were divided into groups of 7 each, 2 teams got a bye to the next round. If the group size was 9 and 11, 3 and 4 teams got byes respectively. If atleast 500 teams participated in the tournament, what is the minimum possible number of teams?

• 7x + 2 = 9y + 3 = 11z + 4
7k = 2y + 1

k = 2y+1/7 = y = 3 , 10 ...
y = 7p - 4

63p -33 = 11z + 4
63p - 37/11 = z
8p - 37/11 = z
p = 6 , 17 , 28 , 39 , 50

p = 17 gives minimum so y = 115
number = 9 * 115 + 3 = 1038

• Q64) The expression f(x) is cubic in x, in which the coefficient of x^3 is 1. If f(1)=5, f(2)=8. f(3)=11. find f(4)

• Q65) Two equations have a common root which is positive. The other roots of the equation satisfy x^2 - 9x + 18 = 0. The product of the sums of roots of two equations is 40. Find the common root.

• Q66) The number of pairs of integers (a, b) such that (a + 2b)^2 + (2a + 5b - 1/2)^2 < = 2 is

• Q67) If the real numbers x and y satisfy x^3 - 3x^2 + 5x - 17 = 0 and y^3 - 3y^2 + 5y + 11 = 0, then the numerical value of x + y is

• x^3-3x^2+5x-17=0
y^3-3y^2+5y+11=0,
x^3+y^3-3(x^2+y^2)+5(x+y)-6=0
(x+y)((x+y)^2-3xy)-3((x+y))^2-6xy+5(x+y)-6=0
(x+y)^3-3xy(x+y)-3((x+y)^2-6xy+5(x+y)-6=0
((x+y-2))((x+y)^2-(x+y))+((3-3xy))=0
so x+y=2

• Q68) If a, b, c, d, e and f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of (ab + bc + cd + de + ef) is

• Q69) If x and y are positive numbers and 1/x + 8/y = 1, find the minimum value of (x + y) + √(x^2 + y^2)

• x + y + sqrt(x^2+y^2)=a
a-(x+y))=sqrt(x^2+y^2)
a^2+(x+y)^2-2a(x+y)=x^2+y^2
a^2+2xy-2a(x+y)=0
1/x+8/y=1
y+8x=xy so so y=8x/(x-1)
put
a^2+2x((8x/(x-1))-2a((x+8x/(x-1))=0
a^2 +16x^2/(x-1) -2a((x^2+7x/(x-1)=0
so a^2(x-1)+16x^2-2ax^2-14ax=0
so x^2(16-2a)+x(a^2-14a)-a^2=0
x real so D>=0
(a^2-14a)^2 -4 (16-2a)(-a^2)>=0
so a^2((a-14)^2 +4a^2(16-2a)>=0
so a^2((a^2+196-28a+64-8a)>=0
a^2((a^2-36a+260)>=0
so a>=26 min value

• Q70) Aishwarya used a calculator to compute (a + b)/c, where a, b and c are positive integers. She pressed a, +, b, /, c and = in that order, and got the answer 11. When she pressed b, +, a, /, c and = in that order, she was surprised to get a different answer 14. Then she realized that the calculator performed the division before the addition. So she pressed (, a, +, b, ), /, c and = in that order. She ﬁnally got the correct answer, which is

• Q71) a^2(b^2 + 1) + b^2(a^2 + 16) = 448
Find the number of ordered pairs of integers (a , b)

149

136

199

121

102

61

58

42