# Quant Boosters - Sagar Gupta - Set 1

• x = 0 => c = 4
x = -1 => a - b + 4 = 4 => a = b
x = -2 => 4a - 2b + 4 = 6 => a = b = 1
a,b,c = 1,1,4

• Q10) Find the number of negative integral solution in (x +1)/x + |x + 1| = (x + 1)^2/x
a) 0
b) 1
c) 2
d) 3

• Case 1 : x + 1 > 0 => x > -1
Would not give any negative integral solution.
Case 2 : x + 1 < 0 => x < -1
Expression becomes :
-(x + 1) = (x + 1)^2 / x
=> -x = x + 1 => x = -1/2 => No negative integral
Case 3 : x = -1 satisfies. So 1 solution only.

• Q11) If one root of the equation (I-m)x^2 + Ix +1 = 0 is double of the other and is real, find the greatest value of m.

• Lets root be x and 2x
Sum of roots => I/(m-I) = 3x ------ (1)
Prod of roots => 1/(I-m) =2x^2 ------ (2)
(2) /(1) => x = -3/2I
Substitute in (1) => 2I^2-9I+9m=0
D >= 0 for m to be max, D=0
81=72m => m =9/8

• Q12) The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is
(1) b >= a^4
(2) a >= 4b^4
(3) a >= b^4
(4) b >= 4a^4
(5) none of these

• x^2+y^2+2xy = a^2
Now x^2+y^2 >=2xy
=> 2(x^2 + y^2) >=a^2
Similarily working on 2(x^2+y^2) >=a^2
we get a^4/8 < = (x^4 + y^4)
=> a^4/8 None of these

• Q13) If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?

• Case 1 : 30p (2.3.5)
30 * 7, 30 * 11, 30 * 13, 30 * 17, 30 * 19, 30 * 23 => 6 elements
Also,30 * 7 * 2, 30 * 7 * 3 => 2 elements
Case 2 : 42p (2.3.7)
42 * 11, 42 * 13 => 2 elements
Case 3 : 66p (2.3.11)
66 * 13 => 1 element
Total : 11 elements !

• Q14) 17!=355687ab8096000. Find the value of ab

• Divisibility of 11
33 + a - (24 + b) = 0 or 11
9 + a - b = 0 or 11
9 + a - b = 11
a - b = 2 ......(1)
Divisibility of 9
57 + a + b = 6 or 15 .....(2)
a + b = 15 => a = 17/2 Not possible
a + b = 6 => a = 8/2 , b=4/2 => a,b=(4,2)

• Q15) Consider the increasing sequence 1, 3, 4, 9, 10, 12, 13… and so on. The sequence consists of all those positive integers which are powers of 3 or sum of distinct powers of 3. Find the 100th term of the sequence.

• Each number is of the form 3^0, 3^1, 3^1 + 3^0, 3^2, 3^2+3^0, 3^2 + 3^2 + 3^2+3^1+3^0
convert in base 3 :
1,10,11,100,101 and so on
100 in base 2 is 1100100
100th term will be 3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981

• Q16) There exist three positive integers P, Q and R such that P is not greater than Q, Q is not greater than R and the sum of P, Q and R is not more than 10. How many distinct sets of the values of P, Q and R are possible?

• q = p + x, r = q + y, where p is positive integer and x, y are non-negative integers
=> y + 2x + 3p < = 10
=> y + 2x + 3p' < = 7 (p' = p + 1)
So we have (4 + 3 + 1) + (4 + 2 + 1) + (3 + 2) + (3 + 1) + (2 + 1) + (2) + (1) + (1)
= 31 solutions

• Q17) India and Brazil play a football match in which India defeats Brazil 5-2. In how many different ways could the goals have been scored if Brazil never had a lead over India during the match ?

• First goal always India will score : 1-0
Now :
I / B B I I I I
Only 1 way brazil can take a lead :
( 6! / 2! 4! ) -1
14

• Q18) For any positive integer n, P(n) is the product of digits of n, then find the value of P(1) + P(2) + ...... + P(999).
Note:- P(1) = 1, P(6) = 6, P(23) = 6, P(900) = 9 and so on

• 0-9 : 1 + ( 1+2+3+.....9)
10-19 : 1 + ( 1+2+3+.....9)
20-29 : 2 + 2 ( 1+2+3......9)
.
.
90-99 : 9 + 9 ( 1+2+3...... 9 )
total =46 + 46 *45 = 46^2

100-109 : 1 + ( 1+2+3...9)
110 - 119 : 1 + ( 1+2+3+.....9)
120 - 129 : 2 + 2( 1+2+3......9)
.
.
190-199 : 9+ 9 ( 1+2 + 3 +.........9)
total = 46 + 46 *45 = 46^2

200-209 : 2 + 2 [ 1 + 2 +.......9 ]
210 -219 : 2 + 2 [ 1 + 2 +........9 ]
220 - 229 : 4 + 4 [ 1 + 2 +......... 9 ]
.
.
290 -299 : 18 + 18 [ 1 +2 +3 ...... 9
total = 92 + 92 [ 45 ] = 92 [ 46 ] = 2*46^2

If you see the pattern :
Series will go like :
46^2 + ( 46^2 + 2 * 46^2 +.........9 * 46^2 )
46^2 + 46^2 [ 45 ] = 46^3
Now subtract 1 from it as we considered 0 in the beginning
Answer = 46^3 - 1 = 97735

• Q19) The numbers a1, a2,...,a108 are written on a circle such that the sum of any 20 consecutive numbers equals 1000. If a1 = 1, a19 = 19, and a50 = 50, find a100

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