# Quant Boosters - Swetabh Kumar - Set 3

• 10 minutes every hour, so in 24 hours, clock gains 240 minutes.
minute and hour hands meet every 65 5/11 minutes = 720/11 mins
so number of extra meetings: 240*11/720 = 3.xx so 3 more meetings than usual.
so 22+3=25 times

• Q11) If (x−a)(x−b)=1 and (a−b)+5=0, then (x−a)^3−1/(x−a)^3 is
a. 100
b. 140
c. 200
d. 280

• b=a+5
(x-a)(x-a-5)=1 let (x-a)=t
t (t-5) = 1
t^2-5t =1
divide by t
t-1/t= 5
cube both sides.
t^3-1/t^3 -3*5 = 5^3
t^3-1/t^3 = 125+15=140

• Q12) (1 x 2)/6 + (3 x 4)/6^2 + (5 x 6)/6^3 + (7 x 8 )/6^4 + ( 9 x 10)/6^5 + …
a) 24/25
b) 125/116
c) 125/108
d) 116/125
e) 108/125

• S = 2/6 + 12/6^2 + 30/216 + 56/6^4 + 90/6^5....
S/6 = 2/6^2 + 12/216 + 30/6^4 + 56/6^5...
5S/6 = 2/6 + 10/6^2 + 18/6^3 + 26/6^4...
5S/36 = 2/6^2 + 10/6^3+ 18/6^4....
25S/36 = 2/6 + 8 (1/6^2 + 1/6^3....)
25S/36 = 2/6 + 8 ( 1/30) = 18/30 = 3/5
S = 108/125

• Q13) If y = (40/x) + x and x is less than 0, then what is the greatest possible integral value of y?

• Differentiate, -40/x^2 + 1 = 0
x^2 = 40 x < 0 so x = -rt (40)
so y = 40/(-rt40) - rt 40 = -2 rt 40= -rt (160) = bw -12 and -13 so max is -13

• Q14) Semi perimeter of a right angled triangle is 154 cm and smallest median is 72 cm. Area of Triangle is ?

• smallest median is to hypotenuse, which is equidistant from all vertices. so hypotenuse= 2 * 72=144
Let the perpendicular legs be x,y
so x^2+y^2=144^2
and x+y+144=308 so x+y=164
so xy = 1/2{(x+y)^2-(x^2+y^2)} = 1/2 {164^2-144^2}=1/2 * 308 * 20=3080
so area = 1/2 * xy = 1/2 * 3080 = 1540.

• Q15) A polynomial p(x) leaves remainder 75 and 15 respectively, when divided by (x-1) and (x+2). Then the remainder when f(x) is divided by (x-1)(x+2) is
a) 5(4x+11)
b) 5(4x-11)
c) 5(3x+11)
d) 5(3x-11)

• p(x)= (x-1)(x+2)Q(x) + ax+b
Put x = 1
a + b = 75
put x = -2
-2a + b = 15
From both, a = 20, b = 55
so ax + b = 20x + 55 = 5(4x + 11)

• Q16) Gopal went to a fruit market with certain amount of money. With this money he can buy either 50 oranges or 40 mangoes. He retains 10% of the money for taxi fare. If he buys 20 mangoes, then the number of oranges he can buy is
(a) 25
(b) 18
(c) 20
(d) None of these

• 1 orange= 4 Rs, 1 mango=5 Rs, say
so He has Rs 200. and then 180 to spend.
20 mangoes means he spent 100.
80 left, so oranges= 80/4=20

• Q17) In Sivakasi, each boy’s quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?
(a) 200
(b) 150
(c) 125
(d) 175

• NB = (N-25)(B+3)
3N-25B = 75
N=50, B=3 satisfies, so NB=50 * 3=150.

• Q18) Find the slope of the line which passes through the point (3,5) and cuts off the least area from the first quadrant.

• Any such line: y-5 = m(x-3)
x intercept: -5/m + 3
y intercept: -3m+5
area = 1/2*(3-5/m)(5-3m)
=1/2 (30-25/m-9m)
Diff.
=1/2 (25/m^2-9) = 0
m= +-5/3 since its first quadrant so negative slope, so -5/3

• Q19) Straight lines m^2 x + 4y + 9 = 0, x + y = 1 and mx + 2y = 3 are concurrent for exactly how many values of 'm'

• Putting y=1-x in other two,
m^2 x-4x+13 and mx-2x-1=0
from 2nd, x= 1/(m-2)
putting in 1st,
m^2/(m-2) -4/(m-2) +13=0
m^2+13m-30=0
(m+15)(m-2) = 0
m = 2, m = -15
The first two equations are inconsistent for m = 2.
So one possible value.

• Q20) a, b, c are integers such that – 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc?

42

61

61

62

61

61

58

62