SIMCAT-1 Discussion
And my (approx.) expected scores are
English – 42
DI – 28.
Quant - 37
My scores are Section 1 – English - 48
Section 2 – DI – 32
Section 3 – Quant - 27
Total 107 I think it should be enough to clear the cutoffs
section 1..
eng......37.5
section 2
di/lr.......46.5
section 3
maths......49
n/a
The definition of digital root was given (if you don’t know what is it - If we add up the digits of a number until there is only one number left we have found what is called the digital root. In other words, the sum of the digits of a number is called its digital root and order of the DR is the number of time the process of adding is done to get the 1 digit number)
Q66. How many three digit numbers have a digital root equal to 6?
Q67. How many three-digit numbers have a digital root equal to 9 with the order 1?
Nice score in DI rajesh
Some how I just mixed up things
me 2...
my accuracy in maths was not up 2 the level
n/a
The digital root of any number n can be calculated as
DR(n) = n mod (9) when there is a reminder
= 9 when the reminder is zero
So question 66 = How many three digit numbers have reminder 6 when divided by 9
and its 105 , 114. . . . . .Total , 900/9 = 100
n/a
Math was horrible, rest was ok
SEC1 - 52
SEC2 - 23
SEC3 - 16 
The solution given in explanatory is okey but I calculated it with simple logic
If the three digit number starts with 1 then possible numbers are 108, 117. . . . .180 . . . i.e 1 number in each tenth number (108 in 100-109 | 117 in 110 – 119) and total numbers = 9
Similarly for 3-digit numbers starting with 2 , total numbers =8
. . . . . . and so on for 3 digit number starting with 8, there is only one possible number i.e. 801
Total numbers = 9 + 8 + 7. . . . + 1 = 45
This won’t take more than 30 seconds