As many of you might know that I am providing useful study material and practice questions for SBI PO Preliminary exam.
Before solving these practice questions, read Inequality shortcut technique
Directions (Q. 1-5)
Before solving these practice questions, read Inequality shortcut technique
Directions (Q. 1-5)
In the following questions, the symbol §,☆,@,# and $ are used with the following meaning as illustrated below.
‘P ☆Q’ means ‘P is not greater than Q’.
‘P $ Q’ means ‘P is not smaller than Q’.
‘P § Q’ means ‘P is neither smaller than nor greater than Q’.
‘P @ Q’ means ‘P is neither smaller than nor equals to Q’.
‘P # Q’ means ‘P is neither equal to nor greater than Q’.
Now in each of the following questions assuming the given statements to be true, find which of the three conclusions, 1,2,3 given below them is/are definitely true and give your answer accordingly.
#1
Statements N § B, B $ W,W # H, H ☆ MConclusions:
I. M @ W
II. H @ N
III. W § N
IV. W # N
1) Only I is true
2) Only III is true
3) Only IV is true
4) Only either III or IV is true
5) Only either III or IV and I is true
#2
Statements R ☆ D, D $ J, J # M, M @ K
Conclusions:
I. K # J
II. D @ M
III. R # M
IV. D @ K
1) None is true
2) Only I is true
3) Only II is true
4) Only III is true
5) Only IV is true
#3
Statements H @ T, T # F, F § E, E ☆ V
Conclusions:
I. V $ F
II. E @ T
III. H @ V
IV. T # V
1) Only I, II and III are true
2) Only I, II and IV are true
3) Only II, III and IV are true
4) Only I, III and IV are true
5) All I, II, III and IV are true
#4
Statements D @ R, R ☆ K, K @ F, F $ J
Conclusions:
I. J # R
II. J # K
III. R # F
IV. K @ D
1) Only I, II and III are true
2) Only II, III and IV are true
3) Only I, III and IV are true
4) All I, II, III and IV are true
5) None of the above
#5
Statements M $ K, K @ N, N ☆ R, R # W
Conclusions:
I. W @ K
II. M $ R
III. K @ W
IV. M @ N
1) Only I and II are true
2) Only I, II and III are true
3) Only III and IV are true
4) Only II, III and IV are true
5) None of the above
Directions (Q. 6-11)
In the following questions, the symbol @,©,%,☆ and $ are used with the following meaning as illustrated below.
‘P © Q’ means ‘P is not greater than Q’.
‘P $ Q’ means ‘P is not smaller than Q’.
‘P @ Q’ means ‘P is neither smaller than nor greater than Q’.
‘P ☆ Q’ means ‘P is neither greater than nor equals to Q’.
‘P % Q’ means ‘P is neither equal to nor smaller than Q’.
Now in each of the following questions assuming the given statements to be true, find which of the three conclusions, 1,2,3 given below them is/are definitely true and give your answer accordingly.
#6
Statement : D @ M, M $ B, B ☆ R, R % T
Conclusion:
I. B ☆ D
II. B @ D
III. T ☆ M
1) None is true
2) Only I is true
3) Only II is true
4) Only III is true
5) Only either I or II is true
#7
Statement : W © F, F @ D, D ☆ K, K $ J
Conclusions: I. K % W
II. D $ W
III. F ☆ K
1) Only I and II are true
2) Only I and III are true
3) Only II and III are true
4) All I, II and III are true
None of the above
#8
Statements R * K,K © M,M % T,T $ JII. R * M
III. K © J
1) Only I is true
2) Only II is true
3) Only I and II are true
4) All I, II and III are true
None of the above
#9
Statements R @ K, T © K,T $ M,M * WII.M © R
III. T © R
1) Only I is true
2) Only II is true
3) Only III is true
4) All I, II and III are true
5) None of the above
#10
Statements T $ N, N % B,B @ W,K © WII.K $ T
III. T % B
1) Only I and II are true
2) Only I and III are true
3) Only III is true
4) All I, II and III are true
None of the above
#11
Statements Z % V, V * J,J © M,M @ RII.M % V
III. Z % M
1) Only I and II are true
2) Only I and III are true
3) Only II and III are true
4) All I, II and III are true
None of the above
Direction Q (12-16) In these questions relationships between different elements is shown in the statements. These statements are followed by two conclusions.
Give Answer
1) If only conclusion I follows
2) If only conclusion II follows
3) If either conclusion I or conclusion II follows
4) If neither conclusion I or conclusion II follows
5) If both conclusion I and II follow
#12
Statements N ≥ O ≥ P = Q > R
Conclusions
I. N > R
II. R = N
#13
Statements W ≤ X < Y = Z >A;W<B
Conclusion
I. B > Z
II. W < A
#14
Statements : H > I > J > K ; L > M < K
Conclusions
I. I > M
II. L < H
#15
Statements : C < D < E ; D > F ≥ G
Conclusions
I. C ≥ G
II. F > E
#16
Statements : R > S ≥ T ≥ U; V < T
Conclusions:
I. V ≥ U
II. V < R

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Shukriya !
ReplyDeleteKripya quadratic equations ke bhi practice set post karne ka kasht karein esp. the ones involving higher order roots.....
thnxx !
Complaint and suggestion - Dear sir nowadays u never update about ibps RRB's about upcoming vacancy, cutoff and chances in RRB's. plz update about this. if u will update about this many new visitors visit on this website
ReplyDeleteone more question about inequality in mathematics. plz sir discuss this topic because this topic will sure come in upcoming exams
iss mein kuch numbers galat hai jaise ki 1st ka adhaa hai , 4th one is wrong
ReplyDeletethanks alot
ReplyDelete1st Q is WRONG..its not in continuation...Relation between N B W H M cannot be established !
ReplyDeleteplease anyone can explain me the condition 3 of Q8 and condition 1 of Q9
ReplyDeletein qn no. 1 no reln between W and H is mentioned in d qn.. so answer will not be option 5, it shud b option 4
ReplyDeletein qn no 9 : in d solution its mentioned dat iii. is false but actually iii. is true - ofcourse finally its mentioned dat answer is non of diz but going thru each conclusion individually , d 3rd conclusion is actualy true
in qn no 14 : der is no uniq relncip which can be established between H and L according to qn. and hence , it shud be option 1 instead of 5
also., it wud hav been better had u posted qns 12 - 16 first and then from qn 1 - 11 ( as a gradual increase in d level of difficulty)
its very simple actually.. whnevr u find two variables separated by more dan 1 variable , see if u hav atleast two signs of opposite nature ( like >= and <= or > and < or any combination of 4 of diz ) relating the variables , then there cant be any unique relation which can be established between those two variables
ReplyDeletein qn no. 8 : we have R < K <= M > T >= J
and in its 3rd condition : between K and J we have a <= just aftr K and >= just before J - which means no unique relation can be established between K and J.. in short , the proposed relation in the given condition is false
simmilarly, in qn no 9 : we have R = K >= T >=M < W
and in its 1st condition : between W and K we have a < just before W and >= just after K - which again means no unique relation can be established between W and K.. in short , the proposed relation in the given condition is false
1st& 4th edited now
ReplyDelete1st,4th,9th & 14 th edited now
ReplyDeleteno its not possible.. coz acc to d instructions : @ represents neither smaller , nor equal to.. i.e greater than ( or > )
ReplyDeleteso d inequation becomes : D > R <= K > F >= J
but in d solution its given : D < R <= K > F >= J
4th solution is wrong
ReplyDeleteyess... xactly
ReplyDeletePlz post tricks for cube as well as number serie
ReplyDelete4th right plz check once!
ReplyDeleteYup! My mistake u r right @ Prateek.
ReplyDeleteyess.. dats a grt topic u hav took up - cubes - i do struggle wid such qns.. n i c u r a regular visitor of dis site..
ReplyDeletei do hav som tricks for qns from cubes but number series - der r som observations.. but not tricks
thanku very much.. :)
ReplyDeleteplzz do com up wid som tricks for solving machine input and output qns in d fastest possible way :)
Sm1 plz solve this....A man sells an article at 5% above its cost price. If he had bought it at 5% less than what he paid for it & sold it for Rs. 2 less, he would have gained 10 %. Find the CP of the article.
ReplyDeleteits Rs 120 :)
ReplyDeletehow? ans is 400
ReplyDeletehow? d ans is 400
ReplyDeleteoopss sorry.. i made a blunder in my calculation.. yeah its 400 only
ReplyDeleteassume CP = 100
now SP = 105
acc to qn , if CP = 95
and SP = 103
gain = 103 - 95 = 8
acc to qn , Rs 8 = 10% gain
or 100% gain = Rs 800
if 100% gain = 800 , thn CP = Rs 400
i made a blunder in calculating 103 - 95 = 12 instead of 8.. sorry :(
sorry ., i made a blunder in my calculations...
ReplyDeletelet cp = x
sp = 105x/100
new cp = 95x/100
sp = (105x/100 ) - 2 = (105x - 200)/100
profit = profit% of cp = 10% of 95x/100 = sp - cp = (105x - 200)/100 - 95x/100 = (105x - 200 - 95x)/100
solving this, we get x = 400 :) i.e CP = 400
i made an error while solving dis
ohh ya.. got it.. thank u.
ReplyDeleteok...1 more..., Q. find the last 2 digit of: 15x37x63x51x97x17. (shrtcut mtd.).
ReplyDelete1 more...Q. if we know the HCF & LCM of 2 numbers, then can we find their sum??
ok...1 more..., Q. find the last 2 digit of: 15x37x63x51x97x17. (shrtcut mtd.).
ReplyDelete1 more...Q. if we know the HCF & LCM of 2 numbers, then can we find their sum??
Q1: 5*7*3*1*7*7=>cube of 7 * 3*5=>343*15=.>3*15=>45
ReplyDeleteokkk.. so answer to d 1st qn is 35
ReplyDeletesoln :
15 x 37 : if u know d table of no 15 thn u can calculate it as 15 x 7 = 105 , keep 5 and carry 10.. thn 15 x 3 = 45 + 10 = 55 , we want only last 2 digits , so its 55
now 63 x 51 = 63 ( 50 + 1) = _ _ 50 + 63 = 13 ( consider only last 2 digits while calculating coz dats wat we want finaly, no need to multiply 63 x 50 fully nor add 50 and 63 completely to get 113 - just consider d last 2 digits )
and 91 x 17 : agen if u know d table of no 17 , then u can calculate it as 17 x 7 = 119 , keep 9 and carry 11.. 17 x 9 = 153 + 11 = 164 , consider only 4 as we want only d last 2 digits.. so its 49
now d grand multiplication : 55 x 13 x 49 = 55 x 37 ( consider d last two digits obtained aftr multiplying 13 and 49 , considering the table of no 13 )
55 x 37 = 11 x 5 x 37 = 11 x 85 ( considering d last 2 digits obtained aftr multiplying 5 and 37)
11 x 85 = do dis multiplication by adding 8 and 5 and use its sum as d mid digit of d product wid 8 and 5 retaining their places in d product.. so we have 8 13 5.. since only 1 digit can be present between 8 and 5 , carry 1 and add it to 8.. d result of d product wil b 935.. but we r not interested wid any digit oder dan d last 2 digits.. hence ignore dat 1 during dis last step of multiplication and just go wid 35 as d answer..
explanation is a bit lengthy but solving is by far a shorter process...
and so far as d 2nd qn is concerned., a simple trick - dont overlook d step between the data given in d qn and d data asked in d qn.. if hcf and lcm are given and u r askd to find d sum , it simply hints out to u dat - if u find d 2 numbers , u can add them to get their sum - dats d hidden intermediate step - so focus on determining d 2 numbers and dat z possible, indirectly finding d sum of d 2 numbers is possible if hcf and lcm are given.. there r som varities of qns based on dis concept.. but it will be better if u can post d exact qn as u did in case of qn no. 1
sorry... answer is 35 and not 45... evn if u multiply the entire thing u will get the answer as 2940521535 and so , ur solution is wrong..
ReplyDeletethank u prateek... d process is bit long... bt yeah i got it...
ReplyDeleteand for the 2nd ques.. hcf is 29 & lcm is 4147
anu , u r most welcome.. :)
ReplyDeleted process is actualy not dat long its a 2 step process and it wil b furder shortened if u can master vedic maths.. using vedic maths approach , other qns lyk solving quadratic eqns , or linear eqns in 2 variables ( where options lyk x > y or x >= y or x < y or x <=y or X = y or no reln possible ) , findng sqrs , cubes , sqr roots and cube roots etc etc become highly simplified and diz qns wil hardly tek som 30 seconds each to solve..
okk.. so d sum is eider 696 or 4176 :)
solution :
listen , HCF represents the largest number which is the factor common to both the numbers.. okk ??
so we can assume that if one number is 29a , thn d oder number is 29b
( note : a and b are co-primes to each other )
product of 2 nos is d product of der hcf and lcm.. ri8 ??
=> ( 29a ) ( 29b ) = 29 * 4147
=> ab = 143 :)
we can get 143 as ( 1 x 143 ) or ( 11 x 13 )
so d reqd numbers are :
either (29 x1 = ) 29 and (29 x 143 = ) 4147... case 1
or ( 29 x 11 = ) 319 and ( 29 x 13 = ) 377... case 2
considering case 1 : sum of d numbers is 4147 + 29 = 4176
considering case 2 : sum of d numbers is 319 + 377 = 696
in such qns , generally a condition is mentioned., lyk both the numbers are greater than 29 or atleast one of d numbers is not greater than 29.. so according to the condition mentioned in ur qn , u can use eider case 1 or case 2 :)
so, is my answer correct ?? what's d answer bdw ??
oh yes.. ur ans as well as ur explanation both r too gud... ans is 696.., no conditions mentioned but from the options givn its 696 only
ReplyDeletethanku very very much :) do u hav any tricks/techniqs for solving machine input and output qns in d fastest possible way ?? and whn z ur sbi po prelims exm ??
ReplyDeletehey evn i get stuck in those m/c i/p o/p probs..., sorry:(
ReplyDeletemy sbi exm on 21
:( okk
ReplyDeletemine is on 20th
all d best to u :)
do share ur exprnc aftr ur exm :)
ty n all d best to u too..!!:)
ReplyDeletelisten i got a shortcut to d profit and loss qn dat u askd first..
ReplyDeleteconsider cp = 100 , therefore sp = 105
now if cp = 95 , profit = 10% => sp = 95 + 9.5 = 104.5
difference between old sp and new sp = 0.5
but acc to qn, diference shud be 2 but we got it as 0.5
so 0.5 => 2
or 1 => 4
i.e 100 => 400
very short and simple.. :)
perfect..!! ok nw 1 more ques.... which of the following will have the maximum change in their values if 5 is added to both numerator nd denominator of all the fractions? 3/4, 2/3, 4/7,5/7? plz tell hw to solve it quickly
ReplyDeletewaah !! u always bring up some quality qns.. i lov solving such qns.. thanks for keeping my brain exercised.. but dear , r u preparing for CAT ??? morevr , i dont know any shortcut to sucha qn , if a simmilar qn coms in d exm , i wil eider leav it or evn if i solve it., i wil evaluate d decimal values :
ReplyDeletelike 0.75 to 0.88 : chang = 0.13
0.67 to 0.875 change = 0.205
0.57 to 0.75 change = 0.18
0.71 to 0.83 change = 0.12
so 2nd option shud b d correct answer
and diz calculations wil b easier if u remember d values lyk 1/2 =0.5 , 1/3 = 0.33...1/7 = 0.1428, 1/8 = 0.125...., 1/16 = 0.625
anyways, i will work upon it to find som shortcuts if possible.. but
are u preparing for CAT ?? watevr may be.. kip asking such qns irrespective of whedr u can solv or not..
and i will try my level best to find a shortcut to this qn and will comment back to ur next comment accordingly... okk ?? :)
ok..no prob.., m also going to leave these kind..., just wanted to know if there may be any shrtcut to such ques. no CAT.., jst preparing for bank exams..., Actually i prepare at home...So, basically i dont get any1 to clear my doubts
ReplyDeleteyeah ans is 2
ReplyDeleteheyy listen , i got d trick.. just experimenting it.. dont leav doz qns, evry qn fetches 1 mark and 1 mark means som thousands of competitors.. will let confirm u widin a day..
ReplyDeletedont worry., u can ask me ur doubts freely :) i am always der to help u and i wil b happy if u crack sbi po :) do post ur qns here.. i guess our teamwork wil help both of us equally :) nyways, ur comments are highly logical and qns are of gud quality.. kis me engaged.. :) thanku
anyways, i hav a doubt.. if i wanna b ur frnd on fb, thn suggest d shortcut :)
Hiii,it would be 400 Exactly,here m sharing u d detail solution-
ReplyDeletelet cp=x
1sp=x*105/100=21x/20
2nd sp=x*95/100=19x/20
now
for 10% Gain
=19x/20*110/100
=209x/200
so dat
1st sp- 2nd sp wid 10% Gain=2
21x/20-209x/200=2
=210x-209x/200=2
x=400
so cp = 400
the trick is :
ReplyDeletethe less the denominator is the more the change :)
subject to d condition :
1. fractions shud b proper fractions
2. the change in the denominator/numerator should be greater than the least denominator value.
heyy hi... plz explain d 2nd pt.. cudnt get it
ReplyDeleteokk...i'll try finding a shrtcut 4 ur doubt bt as of nw i dont think der is any use of solving such ques:)
ReplyDeleteokk.. so , in d gyvn qn : we hav 4 different fraction whose denominators are 3, 4 ,7..
ReplyDeleteleast denominator = 3
change in numr and/or denominator = 5
acc to d 2nd point dis change ( i.e. 5 ) must b greater dan d least denominator ( i.e. 3 )
an example to demonstrate :
case 1 : consider fractions :
(2/9) , (1/5) , (3/7) , (11/16)
decimal values : 0.22 0.2 0.43 0.68
Least denominator = 5
Add 6 to both numr and denominator ( 6 > 5 )
new fractions :
(8/15) , (7/11) , (9/13) , (17/22)
new decimal values : 0.53 0.63 0.69 0.77
changes = 0.31 0.43 0.26 0.09
here u c dat 1/5 having d least deciimal value shows d highest change...
now consider anoder set of 4 fractions :
(1/4) (3/6) (2/5) (3/4)
decimal
values : 0.25 0.5 0.4 0.75
add 3 to numerator and denominator
new fractions : (4/7) (6/9) (5/8) (6/7)
new decimal
values : 0.56 0.67 0.625 0.85
change : 0.31 0.17 0.22 0.10
here, least denominator was 4 but adding 3 to both numerators and denominators was observed in (1/4) and d least change was observed in (3/4)
dats wat.... now did u get it @Anu ??
yeah i think so...; dis way by just luking at d decimal vales we can say d ans...provided it sud follow d condition u mentioned in pt 2.. i.e d change in denomntr or numrtor sud b greater dan the least deno. value..., isnt it?/
ReplyDelete& 1 more thing in 2nd case the least deno value is i think 2 nt 4, so it follows d condition..as, 3>2.
no Anu no.. dont commit dat blunder... dont consider 3/6 as 1/2 in dis example.. if u do so, thn 3 > 2 is correct but answer wont be 1/2., it will be 1/4.. dont beliv ?? chk it urself
ReplyDeleted fact is.. dis trick is applicable only whn all d numerators are different.. if evn 2 numerators are same, thn u cant apply dis..
okk ?? now is it fine ??
i wil work furder on dis techniq to modify it so dat it can be used for any fraction.. but till thn go wid diz limitations.. may be i wil try doin so aftr saturday coz i hav my exm on saturday..
Hi prateek,
ReplyDeleteCan you share your logics and shortcuts to me via gmail. If possible... After ur SBI exam... It will be very useful for me as I am preparing for bank exams at home and it s for first time I am appearing so it s very difficult to me catch up all the things as I need to scratch my head for simple questions too..
Thank you. And all the best for ur SBI exam..
hey prateek.. how ws ur exam??hw ws d paper
ReplyDeletehi.. no intention to demotivate u.. but wat i say is my personal exprnc -
ReplyDeletequants :
a bit tidious calculations, u hav to tek care of units., 2 DIs wer der , one was easy and d oder was a bit dificult, 1 qn from si n it was bit cumbersome in calculations , 1 tym and work was der - moderate level , 5 inequality qns - moderate, 5 number series qns - too easy., profit and loss qn was easy, approximation involved less calculations.
reasoning :
1 seating arangement sample for 5 qns - easy 1 , 1 puzzle - left it , syllogism - 4 out of 5 wer easy 1s , no qn from machine input and output , blood reln qn was straight forward , inequality ( > , < etc ) was too direct and easy ,direction sensing qn was also very easy
english :
cloze test - 5 marks - easy words , error detection was easy , fill in d blanks was moderately tricky - involved some phrasal verbs , para jumbles was easy one , reading comprehension - was too too too long one to read., it was accompanied by 10 wns , left 6 of dem andswered d last 4 which askd synonym and antonym of words in BOLD and doz words wer easy ones.. so dats all
my attempt :-
quants - 18
reasoning - 19
english - 21
total - 58
all d best to u 4 2moro :)
4get not to share ur exprnc :)
seems quant was tough, resoning me no i/p o/p sounds gud..:) anyways..ur attempts r gud, dont worry.thanku.. it gave some idea for hw to proceed. sure will share my exp
ReplyDeletethanku.. all d best once agen :)
ReplyDeleteso far as my attempt is concerned , results wil decide abt dat , we cant say anything as of now..
awaiting 2 read ur exprnc :) all d best buddy..
kal dho daal :)