Quant Boosters  Sagar Gupta  Set 2

abc = 3/2
a+b+c > 3 [ 1.5 ]^1/3
(a+b)/2 > 1.5[1.5]^1/3  c/2 = 1.5^4/3  c/2
(b+c)/2> 1.5 [1.5]^1/3  a/2 = 1.5^4/3  a/2
(a+c)/2 >1.5[1.5)^1/3 b/2 = 1.5^4/3  b/2
a = b = c = (1.5)^1/3
1.5^4/3  0.5* 1.5^1/3
1.5^1/3 [ 1.5  0.5 ] =1*1.5^1/3 =1.5^1/3
(1.5^1/3) ^3 = 1.5 = 3/2

Q15) The number 'm' and 'n' are reciprocals of each other. Both m and n are positive real numbers. If both m^3 + n^3 = 65/8, determine (m+n).

m = 1/n
mn = 1
m^3 + n^3 +3mn [ m+n ] = (m+n)^3
65/8 + 3 x = x^3
5/2 = x

Q16) The median of five positive integers is 7. If the only mode is greater than the median and the mean is greater than 9, what is the lowest possible value of the mode of these 5 integers?

a < b < c < d < e
c = 7
d and e are same
a + b + c + d + e > 45
5 = a
b = 6
c = 7
18 + 2d > 45
2d > 27
d > 13.5
d = 14
e = 14
Mode =14

Q17) The number of factors of the number N = 4^6 + 6^8 is
a. 18
b. 36
c. 54
d. 72

2^12 + 2^8 * 3^8
2^8 [ 16 + 3^8 ]
2^8 [ 16 + 81 * 81 ]
2^8 [ 16 + 6561 ]
2^8 [6577]
9 * 2 = 18 factors

Q18) When a natural number, N is divided by D, the remainder is 35. When 50N is divided by D, the remainder is 11. Find D
a. 1739
b. 43
c. 47
d. Cannot be determined

n=ad +35
50n=bd +11
50n=50ad+1750
50n=bd+11
50ad+1750 = bd +11
1739= (b50a)d
d can be 1739 or 47
CBD

Q19) In a row at a bus stop, A is 9th from the right and B is 7th from the left. They both interchange their positions. If there are 20 people in the row, what will be the new position of B from the left'?
(1) 11th
(2) 12th
(3) 13th
(4) 10th

1,2,3,4,5,6,B,8,9,10,11,A,13,14,15,16,17,18,19,20
So, 12'th from left

Q20) Find the sum of all positive twodigit integers that are divisible by each of their digits

{11,22...99}
10a+b mod a=0 > b mod a=0
10a+b mod b=0 >10a mod b=0
when a=b , the above 2 conditions will be satisfied > {11,22...99}
also : {12,15,24,36,48 } will satisfy the above conditions
sum =630

Q21) If a + b + c + d + e = 8 & a^2 + b^2 + c^2 + d^2 + e^2 = 16, where a,b,c,d& e are real numbers then
maximum (a,b,c,d,e) = ?
a. 4
b. 2
c. 16/5
d. 6/5
e. None of these

Application of cauchyschwarz inequality
a+b+c+d = 8e
a^2+b^2+c^2+d^2 = 16e^2
now
(a^2+b^2+c^2+d^2)(1+1+1+1) >= (a+b+c+d)^2
or 4*(16e^2) >= (8e)^2
or 644e^2 >= 64+e^216e
or 5e^2 < = 16e
or e(5e16) < = 0
so 0 < = e < = 16/5

Q22) Let f(x) be a function such that f(x).f(y)  f(xy) = 3(x+y+2). Then f(4)=?
(1) can not be determined
(2) 7
(3) 8
(4) either 7 or 8
(5) none of these

f(0)^2  f(0) =6
f(0) =3 or 2
:
f(0) f(2)  f(0) = 3 (0+2+2)
f(0) [ f(2) 1 ] = 12
if f(0) = 3 , f(2) =5
if f(0) =2 , f(2) =5
:
f(2)^2  f(4)= 18
25 18 = f(4) = 7 ( (5)^2 = 5^2 = 25 )
Hence only 7

Q23) Number of non negative solutions for x + y = x^2  xy + y^2

x^3 + y^3 = (x + y)(x^2 xy + y^2)
=> x^3 + y^3 = (x + y)^2
Squares => Sum of cubes
0 => (0,0) ; (x,x) for all x integers
1 => (0,1) ; (1,0)
4 => None
9 => (1,2) ; (2,1)
16 => (2,2)
But (x,x) do not satisfy the original equation.
Total real solutions : Infinite
Total integral solutions : 6
Total non negative solutions : 6
Total positive integral solutions : 3

Q24) There are coins of three denominations 1 rupee, 2 rupees and 5 rupees, in a bag. The number of coins of Re1 is to the number of coins of Re 2 is 1:4 and the number of coins of Rs.2 is to the number of coins of Rs 5 is 7:2. If it is known that the total number of coins in the bag is between 200 and 250, then the total amount of money in the bag must be ?