Quant Boosters  Gaurav Sharma  Set 2

Number of Questions  30
Solved ?  Yes
Topic  Quant Mixed Bag
Source  Genius Tutorial Preparation Forum

Q1) In a convex octagon, two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the octagon os
a) 8/37
b) 2/31
c) 7/19
d) 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

Q2) 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
a) 72
b) 56
c) 48
d) 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: 0Total case = 24 + 12 + 8 + 4 = 48

Q3) 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?
a) 25%
b) 30%
c) 28%
d) 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
v  x  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.

Q4) If 9^(x^2 – 2) – 3^(x^2 – 2) = 6, then the sum of all possible values of x is
a) 0
b) Root(3)
c) 3
d) 2root(3)

9^(x^2 – 2) – 3^(x^2 – 2) = 6
Let 3^(x^2 – 2) = y
y^2 – y – 6 = 0
(y3) (y+2) = 0
y = 3 and y = 2
But y cannot be a negative number, so y = 3
x^2 – 2 = 1 = > x^2 = 3 = > x = root(3) or – root(3)
Sum = root(3) + ( root(3)) = 0

Q5) log (5x + 6) > log (x^2 + 10) then (log base is 0.6)
a) 6/5 < x < 1 or x > 4
b) 1 < x < 4
c) 0 < x < 0.6
d) 0 < x < 4

log(5x + 6) > log (x^2 + 10) (base is 0.6)
Also base 0.6 < 1
5x + 6 < x^2 + 10
x^2 – 5x + 4 > 0
(x – 1)(x – 4) > 0
x < 1 or x > 4
Also, log of negative numbers is not possible
So 5x + 6 > 0 = > x > 6/5
And x^2 + 10 > 0 for all values
Hence range will be 6/5 < x < 1 or x > 4

Q6) In an isosceles triangle ABC with AB = AC = 20 cm and BC = 24 cm. Circle with center 0 touches BC at P, CA at Q and AB at R. Find the area of ARPQ

AC = 20, PC = 12
So AP = 16 (Pythagoras theorem)
Now, PC = QC = 12
AQ = 8ARQ ~ ABC
AQ/AC = RQ/BC
8/20 = RQ/24 = > RQ = 48/5
ARPQ is a Kite (AR = AQ, RP = QP)
Area = 1/2 x d1 x d2
= 1/2 x AP x RQ
= 1/2 x 16 x 48/5
= 76.8 sq cm

Q7) x, y and z are integers that are sides of an obtuse – angled triangle. If xy = 6, find z
a) 2
b) 4
c) 3
d) Both (a) & (b)

If xy = 6
Then possible cases for x, y are: 1 x 6 = 6 and 2 x 3 = 6Case 1:
When the sides of triangle are 1, 6 and z
Here (6 – 1) < z < (6 + 1) = > 5 < z < 7
Hence z = 6 but (1, 6, 6) does not form an obtuse triangle.Case 2: Sides are 2, 3 and z
Here (3 – 2) < z < (3 + 2) = > 1 < z < 5
z = 2, 3, 4Triangle 1: (2, 3, 2)
3^2 > 2^2 + 2^2, hence it is obtuse = > z = 2 is possibleTriangle 2: (2, 3, 3)
3^2 # 2^2 + 3^2, hence it is not an obtuse triangle
z = 3 is not possibleTriangle 3 (2, 3, 4)
4^2 > 2^2 + 3^2 hence it is obtuse.
z = 4 is possible.Option d is correct

Q8) The product of digits of a five digit number is 600. How many such numbers are possible?

600 = 2^3 x 3 x 5
Leaving aside 5^2
Arrangements of 2^3 x 3 = 24 are
( 1 x 8 x 3 ), ( 1 x 4 x 6), (2 x 4 x 3), (2 x 2 x 6)Note: We will not consider [1 x 2 x 12] as 12 is a 2 digit number
Now possible cases for
(1, 8, 3, 5, 5) are 5!/2! = 60
(1, 4, 6, 5, 5) are 5!/2! = 60
(2, 4, 3, 5, 5) are 5!/2! = 60
(2, 2, 6, 5, 5) are 5!/2!2! = 30Total = 60 + 60 + 60 + 30 = 210

Q9) If all the four digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 and 9 without repetition are arranged in ascending order, what will be the rank of number 5293 ?
a) 897
b) 898
c) 908
d) None of these

For four digit numbers from (1, 2, 3, 4, 5, 6, 7, 9)
Starting with 1 – 7C3 x 3! = 35 x 6 = 210
Similarly for 4digit numbers starting with 2, 3 and 4 we have 210 cases each.
Total = 210 x 44 digit numbers starting with 5 1   = 6C2 x 2! = 30
4 digit numbers starting with 5 2 1  = 5C1 = 5
Similary for 4 digit numbers starting with 5 2 3  , 5 2 4  , 5 2 6  , 5 2 7  : we will have 5 cases each
Total 25 cases
Now the next number will be 5291 – 1 case
Required number is 5293 – again 1 case
Total cases = 840 + 30 + 25 + 2 = 897
Hence rank of 5293 is 897th

Q10) Bells A and B ring 8 times and 38 times in a minute respectively. If they start ringing simultaneously after how much time (in seconds) will B ring exactly 10 times more than A ?
a) 8
b) 10
c) 15
d) 20