Quant Marathon by Gaurav Sharma  Set 3

Q1) Odd integers are grouped as {1} {3,5} {7,9,11} and so on. What will be the sum of all the numbers in the twelfth group?
 1000
 1728
 1536
 1262
Sum of integers in group 1 = 1 = 1^3
Sum of integers in group 2 = 8 = 2^3
Sum of integers in group 3 = 27 = 3^3
Observing the pattern, sum of integers in group 12 = 12^3 = 1728
Q2) An AP, GP and HP have the same first and last terms and the same odd number of terms. Then the middle terms of the three series are in
 AP
 GP
 HP
 None of these
Let the first and term be a & b respectively.
The middle term of AP will be AM i.e, (a + b)/2
The middle term of GP will be GM i.e root(ab)
The middle term of HP will be HM i.e 2ab/(a+b)
Now AM x HM = GM^2
So the middle terms are in GP
Q3) Let x, y and n be positive integers, such that n > 1. How many different solutions are there to the equation x^2 – y^2 = 2^150
 1
 3
 49
 74
2^150 is a perfect square divisible by 4
So number of solutions of x^2 – y^2 = 2^150 are [(number of factors of 2^150/4) – 1]/ 2
= [number of factors of (2^148 – 1)/ 2]
= (149 – 1) /2 = 74
Q4) Let a1, a2 … a11 be an arbitrary arrangement of the integers, 1,2 …. 11. Then the number (a1 – 1)(a2 – 2) … (a11 – 11) is
 Necessarily ≤ 0
 Necessarily 0
 Necessarily even
 None of the above
Necessarily ≤ 0 case : take values as (21)(32)(43) … (109)(110)(1011)
Now, all the terms except the last 2 terms are positive and there are 2 negative terms whose product will be even. Also, the product is greater than 0. Hence the statement is not necessarily negative or equal to 0
Necessarily 0 case: not true. Proved above
Necessarily even case: There are 6 odd and 5 even terms
Let us try to make each term odd so that product is odd.
(even – 1) (odd – 2)(even – 3)(odd – 4) … (even – 9)(odd – 10) after this there are no even terms left.
Hence the last bracket will be (odd – 11) which is even number. So the product will be necessarily be even. Hence Option C.
Q5) In a chess tournament, each of the 5 players plays against every other player. No game results in a draw and the winner of each games gets one point and loser gets zero. Then which of the following sequences cannot represent the scores of the 5 players?
 3,3,2,1,1
 3,2,2,2,1
 2,2,2,2,2
 4,4,1,1,0
Total number of games played by 5 players will be 5C2 = 10
Now in the 4^{th} option it says that 2 of the players have won 4 games each. Let us consider the two players as A and B
Now A has played 4 matches and won all of them = > A won when he played the match with B
Also, B has played 4 matches and won all of them = > B won when he played the match with A
Hence there is a contradiction.
So Option 4 is not valid.
Q6) There are 11 vessels of capacity 2,3,5,7,11,13,17,19,23,29 and 31 liters. All the vessels except one are filled with water, oil or alcohol. The quantity of alcohol is twice the quantity of water and the quantity of oil is thrice the quantity of alcohol. The capacity of the container that is empty is
 13
 23
 7
 9
Let the quantity of vessels containing water be x
Quantity of alcohol = 2x
Quantity of oil = 6x
Total quantity will be 9x
Hence the sum of the numbers must be divisible by 9 and
2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 = 160
Hence vessel kept empty would be 7 kgs as 160 – 7 = 153 is divisible by 9
Q7) One wishes to find positive integers x,y such that (x + y)/xy is also a positive integer, identify the correct alternative
 Not possible
 Possible and the pair (x,y) can be chosen in infinite ways
 Possible and there exist a finite number of ways of choosing the pair (x,y)
 Possible and the pair (x,y) is unique
(x+y)/xy = 1/x + 1/y
Only (1,1) and (2,2) satisfy the equation/
Hence option C is correct.
Q8 ) In a convex octagon, two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the octagon os
 8/37
 2/31
 7/19
 4/9
Total diagonals = 8C2 – 8 = 20
Total points of intersection = 20C2
Every combination of 4 vertices (eg: H, A, C, D) of an octagon corresponds to two diagonals intersecting at the interior point.
Favorable occurrences = 8C4 = 70
Required probability = 70/20C2 = 7/19
Q9) How many numbers can be formed using all the digits 0, 2, 3, 5 and 6 without repetition such that the hundredth digit is more than the ten’s digit
 72
 56
 48
 60
We have 5 digit numbers using 0, 2, 3, 5 and 6 without repetition.
Number of cases when 0 is at hundredth place: 4! = 24
Number of cases when 2 is at hundredth place: 2 x 2 x 1 x 3 x 1 = 12
Number of cases when 3 is at hundredth place: 2 x 2 x 1 x 2 x 1 = 8
Number of cases when 5 is at hundredth place: 2 x 2 x 1 x 1 x 1 = 4
Number of cases when 6 is at hundredth place: 0
Total case = 24 + 12 + 8 + 4 = 48
Q 10) A solution contains milk and water in the ratio 4:3. What part of the solution must be taken out and replaced with water so that the resultant solution contains milk and water in the ratio 2:3?
 25%
 30%
 28%
 23%
Let v be the volume of the solution.
Given that the ratio of milk and water is 4:3
Let x litres of the solution be taken out.
Now the volume v –x will contain milk and water in the ratio 4:3
Now x litres of water is added to this solution and the ratio of milk and water becomes 2:3 i.e 4:6
Hence, to vx litres containing milk and water in the ratio of 4:3, we added x litres of water and the volume v contains milk and water in the ration 4:6
vx 4:3
v 4:6
Clearly, if v is 10 litres, x is 3 litres of water
Hence 30% of the solution must be replaced with water.