Quant Marathon by Gaurav Sharma  Set 5

Solve the inequality x^3 – 8x^2 + 21x – 18 > 0
 (3,infinity)
 ( infinity, 2) U (3, infinity)
 (2,3) U (3,infinity)
 ( infinity, 2)
x^3 – 8x^2 + 21x – 18 > 0
x^3 – 8x^2 + 21x – 18 = (x – 2) (x – 3) (x – 3)
(x – 2) (x – 3) (x – 3) > 0
Since (x – 3) is repeated, the sign on the number line won’t change when we will go from 3 to 2
So, our answer will be (2,3) U ( 3, infinity)
A sequence is defined as 2 = an – a(n1) = a(n+1) – an. Sn is the sum up to n terms of the sequence, a4 = 7. How many values of x, y exist such that Sx – Sy = 33
 2
 3
 4
 More than 4
Putting n = 4, we get 2 = a4 – a3 = > a3 = 7 – 2 = 5
Similarly, a1 = 1, a2 = 3, a3 = 5, a4 = 7 etc
S1 = 1, S2 = ( 1 + 3 ) = 2^2, S3 = ( 1 + 3 + 5) = 3^2
So Sx = x^2 and Sy = y^2
x^2 – y^2 = 33
Number of ways in which 33 can be written as difference between two perfect squares
= number of factors of 33/2 = 4/2 = 2
In an art gallery, paintings are put on two walls facing each other and they are numbered 1, 2, 3 etc. The painting opposite to painting 6 is 19. If there are an even number of paintings, how many paintings are there in total?
 22
 24
 26
 Can’t be determined
Let say there are n paintings ( n even )
Opposite painting 2 is (n1)th painting
So sum of painting = n + 1
6 + 19 = n + 1 = > n = 24
But this is the answer if it is given that the order or placing the painting number is reverse in one row compared to other. Because this information is not given and we assume that order is increasing in both the rows then the total number of paintings comes out to be 26. Hence option d is correct
Consider (AB )^2 = CDA, where A, B, C and D represent distinct digits. Then D equals
 1
 6
 3
 9
Unit’s digit of squares of numbers can only be 1,4,5,6 and 9
A can be 1 or 4 or 5 or 6
Now AB will be a three digit number only when A is 1
Hence B can be 1 or 9
But if B = 1 = > AB = 11 = > (AB )^2 = 121, where C and A are not distinct.
A = 1 and B = 9
19^2 = 361, D = 6
a = 1^2, b = 2^3, c = 3^4 ... z = 26^27. How many zeroes will the product of all the alphabets end in
 100
 104
 80
 106
Number of zeroes in this product will be equal to the number of 5s
= 5^6 x 10^11 x 15^16 x 20^21 x 25^26
= 5^(6 + 11 + 16 + 21 + 52)
= 5^106
Number of zeroes will be 106
The graph of y = f(x) is symmetrical about both x = 2 and x = 4 then f (4 + x) =
 f (x)
 f (2 – x)
 f (x – 2)
 f (x – 3)
y = f(x) is symmetrical about both x = 2 and x = 4
Means, f (2 + x) = f (2 –x) and
f ( 4 + x ) = f ( 4 – x) = f [ 2 + ( 2 – x) ]
= f [2 – (2 – x)]
= f(x)
So, f (4 + x) = f(x)
If the coefficients a, b, c of quadratic equation ax^2 + bx + c = 0 are chosen at random with replacement from the set S = { 1, 2, 3, 4, 5, 6}. Find out the probability that the roots of the quadratic equation are real and distinct.
For roots of ax^2 + bx + c = 0 to be real and distinct, b^2 – 4ac > 0
Total ways = 6 x 6 x 6 = 216
Favorable cases = 38
Required probability = 38/216 = 19/108
If the equations ax^2 + bx + c = 0 and x^2 + 2x + 3 = 0, have a common root then find a : b : c
 1 : 2 : 3
 2 : 1 : 3
 3 : 2 : 4
 1 : 4 : 5
Since roots of x^2 + 2x + 3 = 0 are imaginary, both roots will be common.
Hence both equations are identical.
a : b : c = 1 : 2 : 3
 631
 730
 731
 733
S = (3/2)^2 + (5/2)^2 + (7/2)^2 + … + (25/2)^2
= (3^2 + 5^2 + 7^2 + … + 25^2)/4 = [ ( 1^2 + 3^2 + … + 25^2 ) – 1) / 4
= [ n (2n – 1) ( 2n + 1) ] / 3 = [ ( 13 x 25 x 27 ) / 3 – 1] / 4 = 2924/4 = 731
Find the nth term of the series 2 + 5 + 12 + 31 + 86 + …
Given series : 2, 5, 12, 31, 86 …
First consecutive difference : 3, 7, 19, 55
Second consecutive difference are in GP with common ration 3 : 4, 12, 36 …
Tn = a + bn + c x 3^(n – 1)
T1 = a + b + c = 2
T2 = a + 2b + 3c = 5
T3 = a + 3b + 9c = 12
We get, a = 0, b = 1 and c = 1
Tn = n + 3^(n – 1)