Quant Boosters by Hemant Malhotra - Set 6
Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975
A circus troupe was performing in a village. One evening all the 120 members of the village went to the circus and Rs.120 were collected in all. The charges were Rs.5 per male, Rs.2 per female & 10 paisa, per child. How many ladies were there?
a=17 and b=13
if N=(5 * sqrt5 + 11)^7 find N * f where f is fractional part of N .
Note – X is conjugate to R
- in R = (a + b*rtc)^k
if k is odd then f = X
fractional part will be f=(a-b * rtc)^n
- and if k is even then f+X=1
f = 1 - (a-b * rtc)^k
so f = (5 sqrt5 - 11)^7
so N * F = (5sqrt5 + 11) * (5sqrt5-11)))^7 = 4^7
(5 * rt5 + 11)^2n+1 find fractional part of this number.
let f is fractional part of this f = R - [R]
let a number X= (5 * rt5 - 11)^2n+1
X will lie in 0 < X < 1
so R-X=(5 * rt5+11)^2n+1-(5 * rt5-11)^2n+1
so R-X=2 * (2n+1c1 * (5 * rt5)^2n * 11+2n+1c3 * (5rt5)^2n-2 * (11)^3......
so R-X=even number
so f-X=an integer (odd or even both possible)
bcz 0< f < 1
0 < X < 1
so f = X = (5rt5-11)^2n+1
so fractional part is (5rt5-11)^2n+1
Find the fractional part of (5+2 * rt6)^n
R=(5+2 * rt6)^n=[R]+f
so[R}+f+X=((5+2 * rt6)^n+(5-2rt6)
so R+f+G=2 * (nc0 * 5^n+nc2 * 5^n-2 * (2rt6)^2....
so f+X=integer so f+X=1
so f=1-x so 1-(5-2rt6)^n
Let F(x) = max (5 − x, x + 2). The smallest possible value of F(x) is
D] None of these.
so at x=1.5 it will give smallest value
A function f is even if f(t) = f(–t), and it is odd if f(t) = –f(–t). Let f(x) = g(x) + h(x) and f(–x) = g(x) – h(–x). Which of the following is definitely correct?
a) f is an even function and g is an odd function.
b) Both f and g are even functions.
c ) f is an even function and h is an odd function.
d ) g is an even function and h is an odd function.
G(x) is an even function et f(x)=g(x)+h(x) (1)
so g(x)=g(-x) so even
For how many positive integer values of ‘x’ is ||||x – 1| – 2| – 3| – 4| < 5
||||x – 1| – 2| – 3| – 4| < 5
so |x-1| < 5 + 4 + 3 + 2
so |x-1| < 14
so -14 < x-1 < 14
so -13 < x < 15
so max 14 and min=-12
Let f be a function such that
f(a,b) = f(a-b , b) if a=>b
= a if a < b.
Now, if f(n,5)= 3 and f(n,6)= 5, then n
f(a,b)=f(a-b,b) if a>=b
=a if a < b
If f(n,5)=3 and we need 3
so n will be in form of 5k+3 for 3
and n should be also in form of 6k+5
so first number will be 23 and So on so OA=CBD
f(x, y) is a polynomial function such that f(2, 3) = 30, f(3, 4) = 84 and f(4, 5) = 180. Find f(5, 6).
(3+2) * 3 * 2 = 30
(3+4) * 4 * 3 = 84
(4+5) * 4 * 5 = 180
f(5,6) = (5 + 6) * 5 * 6 = 330
To measure the weight between 1 to 300 how many minimum weighs are required so that all weight can be measured?
for any weight we have 3 possibilities:-
- Place it on the weight pan
- Place it on other pan, along with something whose weight we need to find
- We don't use it
So, it is similar to base 3, hence by taking weight of form 3^n, we can ensure that we are using least number of weights and when we just have two possibilities. so 3^n >=300 so n=6
- in R = (a + b*rtc)^k