# Quant Boosters - Hemant Malhotra - Set 2

• Method 1:
Understand Basic - average speed = total distance/total time

car 1__________________12d _______________ car2

let distance between them = 12d
speed of fly is double so when they both will cover 4d distance fly will cover 8d distance
now speed is triple so when both cars will travel d then fly will travel 3d and meets ann again so
so distance travelled by fly in forward direction is 8d and in backward direction = 3d
so avg speed = (8d + 3d)/(8d/20 + 3d/30)
11/(8/20 + 3/30)
11 * 60/30 = 22 km/hr = 110/18 m/s
time taken by ann and anne to collide = 50/(100/18) = 9 seconds
fly will cover= 9 * 110/18 = 55 m

Method 2 :
Suppose that at a given instant the fly is at Ann and the two cars are 12d apart. Then, while each of the cars travels 4d, the fly travels 8d and meets Anne. Then the fly turns around, and while each of the cars travels d, the fly travels 3d and meets Ann again. So, in this process described, each car travels a total of 5d while the
fly travels 11d.
So the fly will travel 11/5 times the distance traveled by each bumper car = 11/5 * 50/2 = 55 meters

• Q11) A father starts every day from home to pick up his son from school at 3:30. One day, son leaves the school at 2:30 and hence he starts walking towards home at 6km/hr. He meets his father (who starts at his normal time) on the way and they reach 24 min earlier than usual. Find father's speed.

• They saved 24 minutes from both sides it means save 12 minutes on one side
therefore father reached to his son 12 minutes before than he use to be at 3:18
His son walked for 48 minutes. (from 2:30 to 3:18)
so distance travelled by son in 48 minutes is equal to distance travelled by father in 12 minutes
hence ratio of their speed is 1 : 4 and thus 24 km/h.

• Q12) How many distinct real roots the following equation has
x^4 + 8x^2 + 16 = 4x^2 - 12x + 9
a) 0
b) 1
c) 2
d) 3
e) 4

• x^4 + 8x^2 + 16 = 4x^2 - 12x + 9
(x^2 + 4)^2 = (2x - 3)^2
x^2 + 4 = ± (2x - 3)
x^2 - 2x + 7 = 0 or x^2 + 2x + 1 = 0
First equation has no real roots and second has equal roots so only 1 different root

• Q13) if sum of the 6th powers of the roots of the equation x^2 - x - a = 0 is 19721 and the sum of the 4th powers of the roots is 881 find the sum of the 7th power of the roots?
a) 60740
b) 64720
c) 62826
d) 61741

• a^4 + b^4 = 881
Since sum of roots is 1
a = -4 and b = 5
Just check is a^6 + b^6 = 19721
Since 4096 + 15625 = 19721, -4 and 5 are the roots
a^7 + b^7 = -16384 + 78125 = 61741

• Q14) If r, s, t, and u denote the roots of the polynomial f(x) = x^4 + 3x^3 + 3x + 2, find the value of
1/r^2 + 1/s^2 + 1/t^2 + 1/u^2

• Method 1- Transformation of Equation put x = 1/sqrt(x) and solve

Method 2 - Put x = 1/y
so 1/y^4 + 3/y^3 + 3/y + 2 = 0
2y^4 + 3y^3 + 3y + 1 = 0
sum of factors = -3/2
and product of factors taken two at a time = coeff of y^2 =0
so sum of squares of factors + (a + b + c + d)^2 - 2ab - 2bc - 2cd - 2da
= (-3/2)^2 - 2 * (0)
= 9/4

• Q15) Pascal High School organized three different trips. Fifty percent of the students went on the first trip, 80% went on the second trip, and 90% went on the third trip. A total of 160 students went on all three trips, and all of the other students went on exactly two trips. How many students are at Pascal High School?
a) 1400
b) 600
c) 1200
d) 800
e) 1600

• II + 160 = x
II = x - 160
2II + 480 = 2.2x
2 (x - 160) + 480 = 2.2x
so 2x + 160 = 2.2x
so 0.2x = 160
so x = 800

• Q16) It is given that a straight line L intersects the curve y = 3x^3 - 15x^2 + 7x - 8 at three distinct points (x1, y1), (x2, y2), (x3, y3). Find the value of x1 + x2 + x3
a) 3
b) 4
c) 5
d) 6
e) 7

• Suppose straight line is y = mx + c
Then 3x^3 - 15x^2 + 7x - 8 = mx + c
so 3x^3 - 15x^2 + (7 - m)x - (8 + c) = 0
x1, x2, x3 are roots of the above equation and sum of roots of the above equation is 15/3 = 5

• Q17) How many permutations (x1, x2, x3, x4) of the set of integers {1, 2, 3, 4} have the property that the sum x1x2 + x2x3 + x3x4 + x4x1 is divisible by 3?
a) 8
b) 12
c) 14
d) 16
e) 24

• ( x1 + x3 )( x2 + x4 )
( x1, x3 ) = (1, 2) -----> 2 * 2 = 4
( x1, x3 ) = (2, 4)------> 2 * 2 = 4
similarly for (x2, x4)
total = 2 * 8 = 16

• Q18) Find a + b + c if √(6 + 2√2 + 2√3 + 2√6) = √a + √b + √c

• Method 1 :
6 + ( 2 root2 * 1 )+ ( 2 root3 * 1) + ( 2 * root2 * root3 )
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac
( root3 + root2 + root1 )^2 = 6 + ( 2 root2 * 1 )+ ( 2 root3 * 1) + ( 2 * root2 * root3 )
so root3 + root2 + 1

Method 2 :
Here we have three square roots inside the square root, we can write
√(6 + 2√2 + 2√3 + 2√6) = roota + rootb + rootc, now square and compare to get the values of a, b and c.

• Q19) Connie has a number of gold bars, all of different weights. She gives the 24 lightest bars, which weigh 45% of the total weight, to Brennan. She gives the 13 heaviest bars, which weigh 26% of the total weight, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive?
a) 14
b) 15
c) 16
d) 17
e) 18

• The average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya (heavy)
Let the total weight of all the rods be a.
The weight of the bars given to Brennan = 45% of a = 0.45a
The weight of the bars given to Maya = 26 % of a = 0.26a
The weight of the bars given to Claire = rest = 29% of a = 0.29a
The average weight of the bars given to Brennan = Weight / number of bars = 0.45a/24
The average weight of the bars given to Maya = Weight / number of bars = 0.26a/13
Similarly, if the number of bars given to Blair = x, then the average weight of the bars given to Blair = 0.29a/x
As, the average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya (heavy),
So, 0.45a/24 < 0.29a/x < 0.26a/13
=> 15.4666 > x >14.5
as x is an integer, it's 15

• Q20) Ramu chooses five distinct numbers. In how many different ways can he assign these numbers to the variables p, q, r, s, and t so that p < s, q < s, r < t, and s < t?
a) 4
b) 5
c) 6
d) 8
e) 15

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