# Quant Boosters by Hemant Malhotra - Set 14

• Find numerically the greatest term in the expansion of (2+3x)^9 when x=3/2

(n-r+1)/(r) *(a/x) >= 1
(10-r)/r * (3x/2) >= 1
(10-r)/r *(9/4)>= 1
90-9r> = 4r
so 13r< = 90
r < = 6. 9 so r=6
so 7th term

If a quadrilateral with sides 2, 3, 4 and 5 is inscribed in a circle and circumscribed to another circle then

Direct formula : If any quadrilateral with sides a, b, c and d inscribed in a circle and circumscribed to another circle then area of quadrilateral is sqrt(abcd). So here answer is sqrt(120)

Find remainder when 1^99+2^99+3^99+....99^99 is divided by 100

1 to 99 there are 99 terms
1^99+99^99 will be div by 100
2^99+98^99 will be div by 100
now 49^99+51^99 mod 100=0
but there will be one more term
50^99 mod 100
which is also divisible by 100
so OA=0

How many integral solutions (a,b) does the equation a^b=a * b have
(1) 1
(2) 2
(3) 3
(4) 4
(5) More than four

a^b = a * b
let b=1
then a=a
we can put any integral value of a here so More than 4
OA= 5

What is the minimum value of 2x^2+3y^2-4x-12y+18=0 for real x and y

Method1-
partial differentiation
with respect to x
4x-4=0 so x=1
now with respect to y
6y-12=0 so y=2
now put x=1 and y=2
2+12-4-24+18=4
so OA=4

Method2-
try to make perfect square form
2*(x-1)^2+3*(y-2)^2+4
so x=1 and y=2 will give us min value

F(x) and G(x) are two quadratic Functions such that
f(1)-G(1)=1
f(2)-G(2)=2
F(3)-G(3)=5
Find value of F(4)-G(4)
(1) 8
(2) 9
(3)10
(4)12

Method1- f(x)=ax^2+bx+c
g(x)=dx^2+ex+m
now f(x)-g(x)=x^2(a-d)+x(b-e)+c-m
let a-d=M
b-e=N
c-m=T
so f(x)-g(x)= Mx^2+Nx+T
so M+N+T=1
4M+2N+T=2
9M+3N+T=5
now find M,N,T= M=1 N=-2 and T=2
so F(x)-g(x)=x^2-2x+2
so f(4)-g(4)=10

Quick approach = this is double difference AP if u are aware of this... it will take less than 20 seconds to solve. 1, 2, 5 so next will be 10

Centroid of triangle is at (1,-1) while its orthocenter is at (5,3) then circumcentre of triangle could be
(1) (-1,-3)
(2) (8/3 , 0)
(3) (0,8/3)
(4) (7/3,1/3)

Many Methods to tackle this, centroid bisect ortho and circum in 2:1 ratio
distance between (1,-1) and (5,3) is sqrt(16+16)=sqrt32
so distance between centroid and circum should be sqrt32/2
check
sqrt(4+4)=sqrt8
so OA= A

There is a clock that has a special way of telling the time. It does not have any hands or numbers on it, but it has a chimer. If the time is 1 o'clock, it chimes once. If the time is 2 o'clock, it chimes twice, and so forth. The time gap between any two chimes is 4 seconds. How many seconds would it take you to know the time, after the first chime is heard, if it is 11 o'clock?
a) 40 seconds
b) 44 seconds
c) 42 seconds
d) none of these

OA=44 , 40+ extra 4 seconds u have to wait to confirm chimer rung no more

In triangle ABC, AB = 3000
BC = 875, CA = 3125
If I is the in center of the triangle, what is the ratio of AI:ID where AD is angle bisector

so AB/AC=BD/CD
so 3000/3125=BD/CD
let BD=x then CD=875-x
so we can find x from here
now in triangle ABD AI is angle bisector
so BA/BD=AI/ID
3000/x=AI/ID

Alternate -
AI/ID = (AB+AC)/BC, valid for every triangle
so ((6125/875)

The average of three consecutive multiples of 3 is a.
The average of four consecutive multiples of 4 is a + 27.
The average of the smallest and largest of these seven integers is 42.
Determine the value of a.

3(k+1) = a
4p+6 = a+27
4p+3k = 72
3 equations 3 variables.
k = 8 ; p = 12 and a = 27.

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