Quant Boosters  Swetabh Kumar  Set 9

Q1. A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minutehand and hourhand meet in the next 24 hours ?
(a) 22
(b) 26
(c) 24
(d) 2510 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 timesQ2. 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. 280b=a+5
(xa)(xa5)=1 let (xa)=t
t (t5) = 1
t^25t =1
divide by t
t1/t= 5
cube both sides.
t^31/t^3 3*5 = 5^3
t^31/t^3 = 125+15=140Q3. (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) 108/125S= 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= 108/125Q4. 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 13Q5. Semi perimeter of a right angled triangle is 154 cm and smallest median is 72 cm. Area of Triangle= ??
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 {164^2144^2}=1/2 * 308 * 20=3080
so area = 1/2 * xy = 1/2 * 3080 = 1540.Q6. A polynomial p(x) leaves remainder 75 and 15 respectively, when divided by (x1) and (x+2). Then the remainder when f(x) is divided by (x1)(x+2) is
a)5(4x+11)
b) 5(4x11)
c) 5(3x+11)
d) 5(3x11)p(x)= (x1)(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)Q7) 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 these1 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=20Q8) 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) 175NB = (N25)(B+3)
3N25B = 75
N=50, B=3 satisfies, so NB=50 * 3=150.Q9) 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: y5 = m(x3)
x intercept: 5/m + 3
y intercept: 3m+5
area = 1/2*(35/m)(53m)
=1/2 (3025/m9m)
Diff.
=1/2 (25/m^29) = 0
m= +5/3 since its first quadrant so negative slope, so 5/3Q10) 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=1x in other two,
m^2 x4x+13 and mx2x1=0
from 2nd, x= 1/(m2)
putting in 1st,
m^2/(m2) 4/(m2) +13=0
m^2+13m30=0
(m+15)(m2) = 0
m = 2, m = 15
The first two equations are inconsistent for m = 2.
So one possible value.

for question 10)
for lines to be concurrent,
X + Y = 1......eq1
mX + 2Y = 3.....eq2
solve eq1 nd eq2
X = 1 / ( m  2) and Y = (m  3) / ( m  2)
put this in m^X + 4Y + 9 = 0
m^2 / ( (m  2) + 4 (m  3) / (m  2) + 9 = 0
m^2 + 13m  30 = 0
so, m = (13 + 17 ) / 2
m = 13  17 / 2 = 15
or, m = 13 + 17 / 2 = 2

@deep OA seems to be one value.
m = 2 is not possible.

@deep for m = 2, 1st equation becomes x + y = 9/4 which will contradict 2nd equation, x + y = 1

Yeah Guys you are correct I was lazy in cross checking values at the end .
Thanks Zabeer Sir, harris sir for the correction !

@deep Neat explanations Deepak :slight_smile: