1:07:42 - 27^(12x+3) = (25+2)^12x+3 (25+2)^12x+3/5 for 25 remainder is zer0 for 2^(12x+3)/5 [2^(12x)*2^3]/5 for 8/5 remainder is 3 for 2^12x , divide 12x/4 if remainder is zero use 4 as new power otherwise remainder as the new power since 12x will give remainder as zero 2^4 = 16 16/5 will give remainder as 1 so 2^(12)x remainder is 1 and 8 is 3 since they were in multiplication multiply them for answer which is 3
sir your videos are helpful but are arranged in very unstructured way ;like few topics you covered in 22, 23 and 24 series can u please arrange all , would be very hepful for aspirants.
@@nisheshchoudhary4792 (100c+10b+a)−(100a+10b+c)=495 Solve this and C-A=5 then That means A can be 5 possibilities 01234 as 0-9 is the possibility And for be there are 10 possibilities 0-9 SO -5*10=50
Sir please make a playlist of 2024 NMAT Videos
1:07:42 - 27^(12x+3) = (25+2)^12x+3
(25+2)^12x+3/5 for 25 remainder is zer0
for 2^(12x+3)/5
[2^(12x)*2^3]/5
for 8/5 remainder is 3
for 2^12x , divide 12x/4 if remainder is zero use 4 as new power otherwise remainder as the new power
since 12x will give remainder as zero
2^4 = 16
16/5 will give remainder as 1
so 2^(12)x remainder is 1 and 8 is 3 since they were in multiplication multiply them for answer which is 3
14:40 IN THIS QUESTION WHY DIDNT WE CONSIDER 10+A+22=15+B WALA CASE wha se b toh 9,8 (1) value aari h
sir your videos are helpful but are arranged in very unstructured way ;like few topics you covered in 22, 23 and 24 series can u please arrange all , would be very hepful for aspirants.
Sir for 13:41 For A we can take 9,8,7,6,5 and for B we can take 4,3,2,1,0 so total 5 options right?
Sir scored 251 in nmat today.
Thank you for your lectures
Congrats
could you please share what are different types of number system questions are there in nmat
Yo can we connect right now I have my exam tomorrow?
sir at 10:02 when you open -(a-b)^2, shouldn't it be +2ab?
very helpful. thank you so much sir
01:07:35 but 12X will also be a multiple of 3 so can it nor be option D
same dat
40, 7, 1390,
thanks for the lecture sir, could you provide more content for the same
sir questions without variabel ke bhi solve ho sakte ha kya ? variabel use se difficult lagta ha question
sir mixture and allegation video pls
33:50 can it not be 217 since yhe last digit has to be 7
Even if you just add the last digit it’s 1, so 211 (9+8+7+7)
Sir will you launch any course for snap n nmat?????
We already have a course for it. You can check here: www.mbakaro.in/learn/NMAT-SNAP-2024-Course
HW 2. B
3. C
1)40
2)7
3)1290
1st ka answer 40. How?
40,7,1390
1. 40
2. 7
3. 1390
1st ka answer 40. How?
Thanks sir. Much valuable session
Sir 40
6
1390?
1st ka answer 40. How?
@@nisheshchoudhary4792 (100c+10b+a)−(100a+10b+c)=495
Solve this and C-A=5 then That means A can be 5 possibilities 01234 as 0-9 is the possibility
And for be there are 10 possibilities 0-9
SO -5*10=50
Bakwaas lecture no concept only questions
If you want concepts you can enrol in our course: www.mbakaro.in/learn/NMAT-SNAP-2024-Course
pls be respectful
40,7,1390