# Quant Boosters - Hemant Malhotra - Set 2

• t is definitely the largest no.
now there are 2 cases when r is > s
t r s p q and t r s q p
when r < s then we have t s _ _ _
we have 6 possible cases
hence total 8

• Q21) How many positive integral solutions does the following equations have :-
14x + 21y + 42z = 840
2x + 3y + 6z = 120

• x should be multiple of 3 and y should be multiple of 2
so x = 3k and y = 2m
so 6k + 6m + 6z = 120
so k + m + z = 20
so 19c2 = 19 * 9 = 171

• Q22) If x and y are integers with (y − 1)^(x + y) = 4^3, then the number of possible values for x is
a) 8
b) 3
c) 4
d) 5
e) 6

• (y - 1)^(x + y) = 64
y = 65 and x = -64

(y - 1)^(x + y) = 8^2
y = 9 and x = -7
y = -7 and x = 9

(y - 1)^(x + y) = 4^3
y = 5 and x = -2

(y - 1)^(x + y) = 2^6
y = 3 and x = 3
y = -1 and x = 7

so 1 + 2 + 1 + 2 = 6 values

• Q23) What is the largest power of 2 that divides the number K = 75! – 71!

• K = 71! (75 * 74 * 73 * 72 - 1)
75 * 74 * 73 * 72 - 1 is odd number so no power of 2
so we basically need power of 2 in 71!
[71/2] + [71/4] + [71/8] + [71/16] + [71/32] + [71/64]
35 + 17 + 8 + 4 + 2 + 1 = 67

• Q24) The hypotenuse of a right triangle is 10 cm, and the radius of the inscribed circle is 1.5 cm. Find the perimeter of the triangle.

• x - 1.5 + y - 1.5 = 10
x + y = 13
x + y + 10 = 23

• Q25) A lizard travelled from corner A to corner B of the rectangular room shown below. Determine its shortest distance covered • Now mug this formula too
min(sqrt(a^2 + (b + c)^2), sqrt(b^2 + (a + c)^2), sqrt(c^2 + (a + b)^2))
Here a=4, b=3, c=2 so min will be sqrt((4^2 + (3 + 2))^2) = sqrt41

• Q26) Suppose a function f is defined by
f(1) = 1
f(2) = 2
f(n) = 2f(n - 1) + f(n - 2) for n ≥ 3.
find f(100) - 2f(99) - 2f(97) - f(96)

• f(100) = 2f(99) + f(98)
f(98) = 2f(97) + f(96)

• Q27) In which of the base is the operation 61 × 61 = 4541 valid?

• let base is n
61 * 61 = 4541
(1 + 6n) (1 + 6n) = 1 + 4n + 5n^2 + 4n^3
1 + 12n + 36n^2 = 4n^3 + 5n^2 + 4n + 1
so n = 8

• Q28) What is the remainder when 27^11 + 9^11 + 3^11 + 1 is divided by 3^10 - 1

• 3^10 = x
so 27 * 3^30 + 3^2 * 3^20 + 3 * 3^11 + 1
so 27 * x^3 + 9x^2 + 3x + 1 put x = 1
so 27 + 9 + 3 + 1 = 40

• Q29) What is the last digit of the LCM of 7^5678 - 1 and 7^5678 + 1 ?

• HCF = 2
LCM * HCF = first number * second number
LCM * 2 = ( 7^5678 - 1) * (7^5678 + 1)
unit digit of 7^5678 = 9
so unit digit of 7^5678 - 1 will be 8
unit digit of 7^5678 + 1 will be 0
so unit digit of LCM will be 0

2nd method to solve this
LCM * 2 = (7^5678 - 1) * (7^5678 + 1)
= 49^5678 - 1
(49-1)(1 + 49 + 49^2 + ... 49^5677)/2 = 24 (1 + 49 + 49^2 + ... 49^5677)
49^k mod 10 = 1 when k is even and is -1 when k is odd
now check inside portion is zero

• Q30) Seven Hockey teams A, B, C, D, E, F and G participated in a tournament. Each team played with every other team twice and D won all of its games and F lost all of its games. B and G scored equal points and are ahead of C. C, A and E also scored equal number of points. Each team gets 3 points, 1 point and no point for a win, draw and loss respectively. What is the highest possible number of points that can be scored by team A?

61

60

61

61

61

61

63