Quant Marathon by Gaurav Sharma  Set 4

Q1) If 9^{(x^2 – 2)} – 3^{(x^2 – 2)} = 6, then the sum of all possible values of x is
 0
 Root(3)
 3
 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
Q2) log0.6(5x + 6) > log0.6(x^2 + 10) then
 6/5 < x < 1 or x > 4
 1 < x < 4
 0 < x < 0.6
 0 < x < 4
log_{0.6}(5x + 6) > log_{0.6}(x^2 + 10)
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
Q3) 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 = ½ x d1 x d2 = ½ x AP x RQ = ½ x 16 x 48/5
= 76.8 sq cm
Q4) x, y and z are integers that are sides of an obtuse – angled triangle. If xy = 6, find z
 2
 4
 3
 Both (a) & (b)
If xy = 6
Then possible cases are: 2 x 3 = 6; 1 x 6 = 6
Case 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, 4
Triangle 1: (2, 3, 2)
3^2 > 2^2 + 2^2, hence it is obtuse = > z = 2 is possible
Triangle 2: (2, 3, 3)
3^2 # 2^2 + 3^2, hence it is not an obtuse triangle
z = 3 is not possible
Triangle 3 (2, 3, 4)
4^2 > 2^2 + 3^2 hence it is obtuse. Z = 4 is possible.
Option d is correct
Q5) 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! = 30
Total = 60 + 60 + 60 + 30 = 210
Q6) 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 ?
 897
 898
 908
 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 4
4 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 897^{th}.
Q7) 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 ?
 8
 10
 15
 20
B will ring 38 – 8 times more than A in a minute
Hence to ring 10 times more it will take 10/30 minutes
10/30 minutes = 10 x 60/30 = 20 seconds
Q8 ) The areas ( in cm^2) of two circles are 576 Pi and 729 Pi. If the distance between their centres is 54 cm what is the number of common tangents?
 4
 3
 2
 1
The areas of two circles are 576 Pi and 729 Pi
The radii of the two circles are 24 cm and 27 cm respectively.
Sum of radii = 24 + 27 = 51
And 51 < 54 (distance between the centres)
When the sum of radii ot two circles is less than the distance between their centres, then they do not intersect. So the two circles do not intersect each other. Hence there are 4 common tangents.
Q9) x has 4 factors and xy has 6 factors. How many factors does y have?
If x has 4 factors  > x is of the form a^3 or ab
xy has 6 factors
If x = a^2 then xy = a^5
Means y = a^2, number of factors will be 3
If x = ab then xy = a^2b or ab^2
Here y = a or b. hence number of factors will be 2
So number of factors of y can be 2 or 3.
Q10) A solid right circular cone with base radius r units and height h units is melted and remolded into a cube of side a units. Which of the following cannot be true?
 h > a > r
 r > a > h
 a > r > h
 h > r > a
Volume of the right circular cone = 1/3 Pi r^2 h
Volume of the cube = a^3
It is given that 1/3 Pi r^2 h = a^3
We know, Pi/3 > 1
1/3 Pi r^2 h > r^2 h
a^3 > r^2 h
Hence the following can be deducted
 Exactly 1 of r and h is less and a ‘OR’
 Both r and h are less than a
So, h > r > a cannot be true.