The Most important Gate Aptitude Crash Course I Number System | By Saurabh Thakur

[@aliusama899]

Last year candidate Now in IITK.....just missing ur lecture..... Thank you sir ... From zero knowledge of apti to my dream institute.....😻😻😻

[@pratikkhale5670]

branch ??

[@aliusama899]

@@pratikkhale5670 I m in M.Des IITK

[@basudevmahato4869]

1) A Logic:- A+2=C, D+3=G , then increase the no in the next term. J+5=O P+6=V Hence OV is the correct answer. 2)B R+3=U O+3=R ....... similarly for SWAN it will be VZDQ. 3)D We add 8 to first letter to get second letter then we add 6 to the 2nd letter to get 3rd letter and then we add 4 to 3rd letter to get 4th letter. Every option follows this logic except option D.

[@sunnybherwani5912]

Brilliant you are genius einstein

[@sunnybherwani5912]

Basudev mahato is God of aptitude

[@basudevmahato4869]

@@sunnybherwani5912 genuine compliment or mocking me?

[@hirenthakkar8531]

=======Student Dharma's Answers======== Question-1 [Gate-2014, IIT-KGP] Find the next term of the sequence: 13M, 17Q, 19S, ____? Answer:- Here, the numbers are prime numbers. So, in the sequence the next prime number will be 23. and 23rd alphabet is W. Thus, the missing number in the sequence will be 23W(i.e., Option-C). --------------------------------------------------------------------------------------- Question-2 [GATE - 2012, IIT Delhi] Given the sequence of terms AD CG FK JP, the next term is ? Answer:- Here, the pattern is as follow: A + 2 = C & D + 3 = G C + 3 = F & G + 4 = K F + 4 = J & K + 5 = P Therefore now, next term will be calculated as follow: J + 5 = O & P + 6 = V Thus, the next term will be OV(i.e., Option A). --------------------------------------------------------------------------------------- Question-3[GATE-2015, IIT-KANPUR] If ROAD is written as URDG, then SWAN should be written as: Answer:- Here, ROAD is written as URDG. There is a +3 pattern in this. Means, R + 3 = U O + 3 = R A + 3 = D D + 3 = G Similarly, SWAN should be written as: S + 3 = V W + 3 = Z A + 3 = D N + 3 = Q Therefore, SWAN should be written as VZDQ(i.e., Option B). ------------------------------------------------------------------------------------------- Question-4[GATE-2014 , IIT-KGP] Find the odd one from the following group (W,E,K,O), (I,Q,W,A), (F,N,T,X), (N,V,B,D) Answer:- Here, we can find the odd one out by observing the difference in position as follow: For W,E,K,O: W + 8 = E E + 6 = K K + 4 = O For I,Q,W,A: I + 8 = Q Q + 6 = W W + 4 = A For F,N,T,X: F + 8 = N N + 6 = T T + 4 = X For N,V,B,D: N + 8 = V V + 6 = B B + 2 = D So, we had observed that in first three groups the difference between first two position is 8 and 2nd and 3rd position is differ by 6 similarly last 2 position is differ by 4. But in last group that is N,V,B,D the difference in last two position is two. Thus, the group N,V,B,D is odd one among all other groups.(i.e., Option-D). --------------------------------------------------------------------------------------------- Please do reply if anything is wrong. 😀

[@acko9262]

Mujhe hamesa maja aata hee aub aptitude koo padh nee mee, thanking you sir👍🙏

[@amankumaramar271]

1:03:37 sir iska ans F,N,T,X bhi ho sakta hai kyuki baki sare me minimium ek vowel hai or isme ek bhi vowel nahi hai number ke according iska ans NVBD hoga kyuki first letter - last letter = 8 and middle ke liye 6 ka difference hai but N - D = 10, issliye N,V,B,D hoga

[@gojosensei1698]

1.A) OV 2. B)VZDQ 3. Nvbd

[@shilpaambalkar2151]

A-c=3, c-f=4, f-j=5 j-o=6 D-G=4, G-k=5, K-p=6, p-v=7 Next term 6-7=ov

[@dnyaneshwarikudi1963]

SD 1. 23W = as 23 is prime no. 2. N,V, B,D = difference between all sets are 7 , 5 , 3 and in N, V, B, D it is not applied that's why ans is D 3. VZDQ = in this each word is replaced with 3rd word after it. So that ans is B 4. OV = in this the difference between two words are increasing like 2,3,4,5,6 that's why ans is A

[@hemanthkumar240]

1.ANS --> SWAN VZDQ 2. ANS ---> AD CG FK JP OV 3. ANS ---> NVBD ( 1st letter-last letter=8 and 3rd -2nd letter = 6)

[@rajatsaxena629]

Student Dharma solution Q1. series 13M, 17Q, 19S, now need to find next term, so, if we observe closely these are consecutive prime numbers and also the alphabets are written according to their order, so, next term will be 23 and since 23rd alphabet is W, so 23W Ans. Q2. Series is AD, CG, FK, JP., Ist letter has gap of +1,+2,+3 ...and for IInd letter, +2,+3,+4... therefore answer will be J+4=O and P+5=, so, OV Ans Q3. Given ROAD=URDG, i.e, each letter are increasing by +2, so for SWAN, add +2 alphabets , we get VZDQ Ans Q4. In given words, alphabets have gap like +7, +5, +3, , but in NVBD, they have +7,+5,+1 so this is odd

[@gururajr3041]

tqq sir

[@harshitmathuriya702]

Harshit OV (A) VZDQ (B) (B)

[@sonalimeena2847]

1 answer OB 2 answer VZDQ 3 answer NVB

[@saurabh2405]

23W

[@angellakshmi438]

Good morning sir

[@ritikarai8233]

1)OV 2)VZDQ 3)NVBD

[@sktausifrahman1496]

I live to express not impress

[@shilpaambalkar2151]

W-E-K-0=I-Q-w-A=F-N-T-X=8-6-4 But in N-V-B-D=8-6-2 Due to this reason N-V-B-D is odd one

[@none2967]

Well explained

[@sktausifrahman1496]

I am the biggest project of my life

[@sachinbamniya8209]

SD ans c 23W

[@brijeshyadav7211]

Will it be suitable for civil also

[@sameerpathan7617]

Last 2nd is (D) & last is (B)

[@sachinbamniya8209]

SD 3 ans B VZDQ

[@khushboobajaj3961]

5 DEC 2022 Second day

[@mukulpal5049]

1) 23W 2)OV 3)VZDQ 4)NVBD

[@NaveenKumar-gg1ii]

Sir koi video sequence and series pr uploade kijye

[@Er.Kamalpratapsingh]

Answer A B A

[@idiosyncratic5468]

1-A=OV 2- B=VZDQ 3-D =N, V, B, D

[@Professorkhatri]

64

[@saurabh2405]

97

[@sachinbamniya8209]

SD 2 ans A OV

[@sanam7445]

1) OV 2 )VZDQ 3)NVBD

[@nayanbijwe7600]

1) OV 2) VZDQ 3)WEKO

[@sumedhshingade3173]

1.OV 2.VXDQ 3.NVBD