All these questions were asked in CHSL 20, rest questions are of BODMAS or of simple divisiblity.

Total questions asked =
71 in 36 shifts

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Exam held today on 5 September 2021

@thefuckingbest7545 ye kon se exam ka h

y = (99-2x)/5
x = .....,-3 ,2, 7, 12....
x,y = (2,19)...(7,17),
Here, AP is formed with alternate coefficients like this
(7+5k, 17-2k) ..
We want x > y so
7 + 5k > 17 - 2k k ~ 1+ =~ 2
(17,13)
Now y will decrease with gap of 2 and we need it to be y ≥ - 20
Minimum possible = - 19
This will give us number of possibilities..

a =13..... d = -2 an = -19
-19 = 13 + (n-1)-2
n = 17

only possible consecutively odd prime numbers in the number system are only 3, 5, 7.
Just compare with it.
N³ - 1 = 3 or 7
n³ = 4 or 8
possible value of n = +2

Possibilities :
N⁰ = 1,
1ⁿ = 0,
1⁻¹ = 1,
(-1)ᵉᵛᵉⁿ = 1

Case 1 : N⁰ that means base can be anything or doesn't matter.
x² - 13x + 42 = 0 x = 6, 7

Case 2 : 1ⁿ
x² - 7x + 11 = 1 x = 2, 5

Case 3 : 1⁻¹
x² - 13x + 42 = -1 no real values found

Case 4 : (-1)ᵉᵛᵉⁿ = 1
x² - 7x + 11 = -1 x = 3,4
x² - 13x + 42 gives even when x = 3, 4
so both values are valid.

Total possible values are 6.

x² - 13x + 42 = x(x-13) + 42 it will always be even for any +ve integeral value of x.