Quant Boosters by Hemant Malhotra  Set 17

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
A person takes a loan of Rs.200 at 5% simple interest . He returns 100 at the end of one year . In order to clear his dues at the end of 2 years , he would pay :
a) 125.50
b) 110
c) 115.50
d) NOTafter 1 year SI=10 rs
so amount =210 rs
he returned 100 rs so 110 rs remaining
now SI=110 * 5/100=5.5
so amount =110+5.5=115.5For what value of x would y assume the maximum value where y= min (x+3,2x1,2x). Also find the maximum value of y.
just check those are increasing or decreasing
x+3 = increasing
2x1 = increasing
2x decreasing
so compare
x+3=2x so x=1/2
2x1=2x
so x=1
now check for these two values
at x=1 y=min(4,1,1)=1
and x=1/2 , y=min((5/3,2,5/2)=2
so max of min will be 1At what time between 4 to 5 PM Minute hand and hour hand will make angle of 60
90 degree = 15 div
so 60 degree =10 div
At 4 there is a gap of 20
but we need a gap of 10
so either minute hand will be 10 before or after the hour hand
so (2010 )=10 and (20+10 ) = 30
so 4 past ((12 * 10/11 ) or 4 past ((12 * 30/11)In an alloy, zinc and copper are in ratio 1:2. In second alloy the same elements are in ratio 2:3. In what ratio should these two be mixed to form a new alloy in which zinc and copper are in ratio 5:8
zinc in alloy 1 =1/3
zinc in alloy 2 = 2/5
in new alloy = 5/13
1/3  2/5
............ 5/13..............
A ............................BA/B= 2/55/13 / 5/13 1/3
= so 3:10Shoelace Theorem to find area of polygons with given vertices
Find area of triangle (2,4) (3,8) (1,2)
2 .... 4
3 .... 8
1.... 2
2 .... 4 (this is repeated from first point)
then 2 * 8 + 3 * 2 + (1 * 4 ) = 6
and then 4 * 3 + (8) * 1 + (2 * 2) = 8
so take difference of these two
6  8 = 14 take mod 14
so area will be 1/2 * 14 = 7Another example, if we want to find area of (1,1) (2,3) (3,4) (5,6) (1,1)
1……..1
2 …….3
3……..4
5……...6
1…….1
1……...1 (this is repeated from first point)
1 * 3 + 2 * 4 + 3 * 6  5 * 1  1 = 23
now
1 * 2 + 3 * 3 + 4 * 5  6  1 = 24
now take difference = 24  23 = 1
so area = 1/2 * 1 = 1/2Find range of a for which one negative and two positive roots of x^33x+a=0 are possible
Approach x^33x=a
let f(x)=x^33x and g(x)=a
For 2 positive and 1 negative value g(x) will cut positive x axis 2 times and negative x axis one time
this is possible when 2Kangna decides to participate in a beauty contest in her office.There are 123 contestants in all,including her. As all the participants enter a room and sit around a round table,she observes that all the other girls are prettier than her. Getting jealous,she devises a divide and rule policy.Soon a fight breaks out such that the contestants start killing each other,in the order : 1st kills 2nd,hands over the weapon to 3rd, who kills 4th and so on..Find the sum of the number on the seats she should sit if she wants to be among the last 2 alive. Numbering is 1 to 123 clockwise.
For last survivor represent number in form of 2^m +n
where m is as maximum possible here 123 so 2^m +n=123 so 2^6 +59=123
so m=6 and n=59
then last survivor will be 2 * n+1
so last survivor=2 * 59+1=119th number
for 2nd last represent number in form of 3*2^m +n
so 3 * 2^m +n=123
3 * 2^5 +27 =123
then 2nd last survivor will be 2 * n+1=2 * 27+1=55th so sum=119+55=174If you want to read Extension of this Concept Follow this link http://www.scribd.com/doc/97328978/JosephusProblem
If a,b are integers and ab is even then which of the following must always be even?
(A) ab
(B) a^2 + b^2 + 1
(C) a^2 + b + 1
(D) ab  bab=2m
so a & b both odd or a & b both even
a) when both odd then product odd so rejected
b) when both even then even+even+1=odd
c) when both odd then odd+odd+1=odd so rejected
d) when both odd then odd * odd  odd = even or even * even  even = even so D ansLet F(x) =x^5+x^4+x^3+x^2+x+1 . Find remainder when F(x^12) is divided by F(x)
x^na^n=(xa)(x^(n1)+x^(n2) + ... a^(n1)
f(x^12)=1+(x^12)+(x^12)^2+(x^12)^3+(x^12)^4+(x^12)^5
f(x^12)=1+(((x^12 1)+1+(x^12)^2 1 +1 +((x^12)^3 1 +1 +(x^12)^4 1 +1 +(x^12)^5 1+1
f(x^12)=6+((x^121)+(x^12)^21+(x^12)^3 1+(x^12)^4 1 +(x^12)^51
now x^121=(x^61)(x^6+1)
x^61=(x1)(1+x+x^2+x^3+x^4+x^5)
so every term will contain x^61 and will be divisible by 1+x+x^2+x^3+x^4+x^5
so OA=6Two different positive numbers x and y such that each differ from their reciprocals by 1 . Find x + y
x and y will be roots of a1/a=1
so a1/a=1 or a1/a=1
a^2a1=0 or a^2+a1=0
a=(1+√(5))/2 or a=(1+√(5))/2
so 4 values and sum of x and y will be (1+√5)/2 + (1+√5)/2= √5