DELHI IAS ACADEMY T-4

STATE LEVEL OFFLINE/ONLINE

TEST SERIES 2023-24 CGPSC (PRELIMS)
(SUBJECT : CSAT - Maths, Reasoning, D.M., Hindi & Chhattisgarhi Language)
ijh{kkfFkZ; ksa }kjk dkys ckWy ikWb .V isu ls Hkjk tk;A mÙkj&iqfLrdk dk Øekad
ijh{kk dsUnzk /;{k dh eksgj To be filled by Candidates by Black Ball Point pen only Sr.No.of Answer Sheet
Seal of Superintendent of Examination Centre
vuqØekad
ROLL NO

?kks’k.kk % eSau s uhps fn;s x;s funsZ” k vPNh rjg i<+d j le> fy, gSaA
Declaration: I have read and understand the directions given below..

oh{kd ds gLrk{kj ijh{kkFkhZ ds gLrk{kj

(Signature of Candidates)......................................
(Signature of Invigilator)......................................
ijh{kkFkhZ dk uke
(Name of Candidates)...........................................
oh{kd dk uke
fnukad % l e;
(Name of Invigilator)........................................... (Date) .........../........./................. (Time) ..............................

iqfLrdk esa i`’ Bksa dh la[ ;k iqf Lrdk esa iz” uksa dh la[ ;k le; ?ka V s iw. kkZa d
28 100 2 200
Number of Pages in Booklet Number of Questions in Booklet Time Hours Total Marks

ijh{kkfFkZ;ksa ds fy, funsZ”k (Instructions for Candidates)

1- ¼d½ vH;fFkZ;ksa dks vks],e-vkj- mRrj”khV esa izfof’V;k¡ tSls uke] jksy 1. (a) Candidates are allowed time to fill up the basic information
ua- vkfn Hkjus ds fy, vfrfjDr le; fn;k x;k gSA about themselves in the OMR sheet such as Name, Roll No.
¼[k½ bl fn;s x;s le; ds i”pkr~ vH;fFkZ;ksa dks iz”u&iqfLrdk etc.
(b) After this, question booklet will be given to the candidates.
tk;sxhA iz”u iqfLrdk ds i`’Bksa rFkk iz”uksa dh la[;k dk feyku Tally the number of pages along with no. of questions printed
bl eq[k i`’B ij nh xbZ la[;kvksa ls dj ysAa ;fn blesa dksbZ on cover of the booklet in case of any discrepancy please get
fHkUurk gks rks d`i;k iz”u&iqfLrdk rRdky cny ysAa the booklet changed immediately.
¼x½ fn, x, vfrfjDr le; esa mRrj vafdr djus dh vuqefr ugha (c) Answering of question is not allowed in the given extra time.
gSA mRrj vafdr djus ds fy, nks ?kaVs dk le; fn;k tk,xkA Two hours time will be given for answering.
2- ¼d½ nh xbZ mRrj&”khV esa uhys ;k dkys cky ikW b.V isu ls 2. (a) On Answer Sheet, supplied to you write input Roll No.,
vuqØekad] iz”uiqfLrdk dk Øekad] iz”u iqfLrdk dk lsV Que. Booklet No., Set of Question Booklet (A,B,C or D).
Name of Candidate, Signature of candidate, Date of Exam, in
¼A, B, C vFkok D½] ijh{kkFkhZ dk uke] ijh{kkFkhZ ds gLrk{kj rFkk
Blue or Black Ball Point Pen.
ijh{kk dh frfFk] vafdr djsAa (b) On Answer Sheet till in your Roll No., Que. Booklet Set and
¼[k½ mRrj&”khV esa jksy ua-] iz”u iqfLrdk dk lsV] dsUnz dks vkfn Centre Code etc. by darkening corresponding circle O with
lacaf/kr xksys O dks uhys ;k dkys cky ikWb.V isu ls HkjsAa Blue or Black Ball Point Pen.
¼x½ mRrj&”khV esa iz”uksa ds mRrj vafdr djus gSaA bl lac/a k esa (c) On Answer Sheet only the answers to questions are to be
funs”Z k bl iz”u iqfLrdk ds ihNs fn;s x;s gSAa marked. The instructions for this are available on the back
cover page of the question booklet.
3- vkWfIVdy ekdZ jhMj (OMR) e”khu mRrj&”khV dh uhys ;k dkys 3. Optical Mark Reader (OMR) machine prepares the result by reading
cky ikWb.V isu ls Hkjsa xksys O dh izfof’V;ksa dks i<+dj ijh{kkQy the entries made in the circles O will the Blue or Black Ball Point Pen
rS;kj djrh gS] vr% ijh{kkfFkZ;ksa dks lpsr fd;k tkrk gS fd os on the Answer Sheet, hence the candidates must be extremely careful
mRrj&”khV esa izfof’V;ksa dks Hkjrs le; iwjh&iwjh lko/kkuh cjrsa ,oa dksbZ in marking these entries and must not commit errors.
=qfV u djsAa 4. Please do not write mark on Answer Sheet anything extra except
4- mRrj&”khV ij fu/kkZfjr LFkkuksa ij pkgh xbZ izfof’V;k¡ Hkjus ds vykok what is asked for.
dqN u fy[ks@ a vafdr djsAa 5. USE OF ANY CALCULATOR, LOG TABLES, MOBILE
PHONE ETC IS PROHIBITED.
5- fdlh Hkh izdkj ds dSydqyVs j] ykWx Vscy ,oa eksckby Qksu 6. Rough work should be done on the blank pages or in the space
vkfn dk iz;ksx oftZr gSA provided for this on each page of this question booklet. Extra
6- jQ dk;Z bl iz”u&iqfLrdk ds fu/kkZfjr [kkyh i`’Bksa esa vFkok vU; paper will not be supplied.
i`’Bksa esa fu/kkZfjr txgksa ij djsa vfrfjDr i`’B ugha fn;s tk;saxAs 7. If there is any sort of mistakes discrepancy (in Hindi or English
Version of the question) either of printing or of factual nature then
7- ;fn fdlh iz”u esa fdlh izdkj dh dksbZ eqn.z k ;k rF;kRed izdkj regarding this Institute no action will be taken on any type of
dh =qfV gks] rks bl lEcU/k esa laLFkk dk fu.kZ; vfUre gksxkA representation complaint as stated above.
8- ijh{kk lekfIr ds i”pkr~ mDr ds lac/a k esa fdlh Hkh izdkj dk 8. After the completion of the examination, no action will be taken on
vH;kosnu@f”kdk;r ij dksbZ dk;Zokgh ugha dh tk,xhA any representation/complaint regarding the above.
(Note: For Instructions regarding marking the answer please
¼uksV% mRrj vafdr djus ds fy, d`i;k iz”u iqfLrdk ds see the Last page of this question Booklet.)
ihNs doj ist ij fn, x, funs”Z kksa dks ns[ ksa½

DELHI IAS ACADMEY T-4 Page No.: 1

 iz”uksa ds mRrj nsus lEcU/kh funs”Z k
(Instruction Regarding Method of Answering Questions)

¼d`i;k dkys cky ikWb.V isu dk gh iz;ksx djs½a (Please Use Black Ball Point Pen Only)
1 - mRrj nsus dk rjhdk% 1. Method of Marking Answer
mRrj nsus ds fy;s vks-,e-vkj- mRrj iqfLrdk esa lEcfU/kr iz”u To give an answer, please darken one bubble out of the
ds cktw esas fn;s x;s pkj xksyksa esa ls dsoy ,d xksys dks iwjk given four, in the OMR Answer Sheet against that ques-
dkyk dhft,A tion.

2 - ewY;kadu i)fr% 2. Valuation Procedure

izR;sd iz”u ds pkj lEHkkfor mRrj gSa] muesa ls ,d mRrj There are four answers to a question, only one of
lgh@yxHkx lgh gSA iz”u dk lgh@yxHkx lgh mRrj them is correct/ nearly correct. 02 marks will be
vafdr djus ls 02 vad izkIr gksx
a s rFkk xyr mRrj vafdr awarded for each correct/ nearly correct answer and
djus ij 1@3 vad dkVk tk;sxkA ;fn ,d ls T;knk xksys 1/3 mark will be deducted for each wrong answer.
dkys fd;s tkrs gSa rks ;g xyr mRrj ekuk tk;sxkA If more than one bubble are darkened for a ques-
tion, it will be treated as wrong answer.

3 - mRrj “khV oh{kd dks lkSaiuk%

(i) oh{kd dks mRrj”khV lkSaius ls igys lqfuf”pr dj ysa 3. Handing over of Answer Sheet to Invigilator:
fd mRrj&”khV ds nksuksa i`’Bksa ij lHkh iwfrZ;k¡ tSls uke] (i) Please ensure that all the entries in the answer
jksy&uEcj] gLrk{kj] iz”u&iqfLrdk dk uEcj] vkfn sheet are filled up properly i.e. Name, Roll No.,
fu/kkZfjr LFkku ij Bhd&Bhd Hkjs x;s gSaA Signature, Question Booklet No. etc.
(ii) 2 ?k.Vs dh le; lhek ds iwoZ ijh{kk gky NksM+rs le; (ii) Before the limitation of 2 Hours duration while
iz”u&i= oh{kd ds ikl tek djsaA ;fn vko”;d gqvk leaving the examination hall, hand over the ques-
rks le; lhek ds i”pkr~ dsUnzk/;{k ls viuk iz”u&i= tion paper to the invigilator. If needed after du-
okil izkIr dj ldrs gSaA ration of period, you can ask for your question
paper from the Centre Superintendent.

4 - mRrj”khV ds mi;ksx esa lko/kkuh % 4. Care in Handing the Answer Sheet.

mRrj”khV dk iz;ksx djrs le; iwjh rjg ls lko/kkuh While using answer sheet adequate care should be
cjrsaA bls QVus] eksM+us ;k lyoV iM+us ls [kjkc u gksus taken about tear or spoil due to folds or wrinkles.
nsaA

DELHI IAS ACADMEY T-4 Page No.: 2

Q.1- daiuh C 1200 #i;s ds vkSlr [kqnjk ewY; ds lkFk 25 mRiknksa Q.1. Company C sells a line of 25 products with an average
dh J`a[kyk csprh gSA ;fn buesa ls dksbZ Hkh mRikn 420 #i;s retail price of Rs.1200. If none of these products sells for
ls de esa ugha fcdrk gS vkSj Bhd 10 mRikn 1000 #i;s ls de less than Rs.420 and exactly 10 of the products sell for less
esa fcdrs gSa] rks lcls egaxs mRikn dk vfèkdre laHko foØ; than Rs.1000, then what is the greatest possible selling
ewY; D;k gS\ price of the most expensive product?
(a) 2600 #i;s (a) Rs.2600
(b) 3900 #i;s (b) Rs.3900
(c) 7800 #i;s (c) Rs.7800
(d) 11800 #i;s (d) Rs.11800
Ans: d Ans: d
Sol: lcls egaxs mRikn dk mPpre laHkkfor foØ; ewY; [kkstus ds Sol: To find the greatest possible S.P. of the most expensive
fy,] gesa 'ks"k 24 mRiknksa ds U;wure foØ; ewY; ij fopkj product, we need to consider the minimum S.P. of the re-
djus dh vko';drk gS tks fd 10 mRiknksa ds fy, 420 #i;s maining 24 products which is Rs.420 each for 10 products
vkSj vU; 14 mRiknksa ds fy, 1000 #i;s gSA and Rs.1000 each for other 14 products.
24 mRiknksa dk U;wure foØ; ewY; Minimum S.P. of 24 products
¾ #-¼420×10 $1000 ×14½ = Rs.(420 × 10 + 1000 × 14)
¾ #-¼4200 $ 14000½ ¾ #-18200- = Rs.(4200 + 14000) = Rs.18200.
25 mRiknksa dk dqy foØ; ewY; ¾ #-¼1200×25½ ¾ #-30000- Total S.P. of 25 products = Rs.(1200 × 25) = Rs.30000.
 lcls egaxs mRikn dk vfèkdre laHko foØ; ewY;  Greatest possible S.P. of the most expensive product
¾ #-¼30000&18200½ ¾ #-11800- = Rs.(30000 – 18200) = Rs.11800.
Q.2- ,d Nk= }kjk HkkSfrdh] jlk;u foKku vkSj xf.kr esa çkIr dqy Q.2. The total marks obtained by a student in Physics, Chemis-
vad mlds }kjk jlk;u foKku esa çkIr vadksa ls 120 vfèkd gSA try and Mathematics together is 120 more than the marks
HkkSfrdh vkSj xf.kr esa feykdj mlds }kjk çkIr vkSlr vad obtained by him in Chemistry. What is the average marks
fdrus gSa\ obtained by him in Physics and Mathematics together?
(a) 40 (a) 40
(b) 60 (b) 60
(c) 120 (c) 120
(d) buesa ls dksbZ ugha (d) None of these
Ans: b
Ans: b
Sol: P + C + M = C + 120 P + M = 120
Sol: P + C + M = C + 120 P + M = 120
PM 120
PM 120  Required average = = = 60.
 vHkh’V vkSlr = = = 60 2 2
2 2
Q.3. The average height of 35 girls in a class was calculated as
Q.3- ,d d{kk esa 35 yM+fd;ksa dh vkSlr Å¡pkbZ 160 lseh vkadh xbZA
160 cm. It was later found that the height of one of the girls
ckn esa irk pyk fd d{kk dh ,d yM+dh dh ÅapkbZ xyrh ls
in the class was wrongly written as 144cm, whereas her
144 lseh fy[k nh xbZ Fkh] tcfd mldh okLrfod ÅapkbZ 104
actual height was 104 cm. What is the actual average height
lseh FkhA d{kk esa yM+fd;ksa dh okLrfod vkSlr Å¡pkbZ D;k gS\
of the girls in the class? (rounded off to 2 digits after deci-
¼n'keyo ds ckn 2 vadksa rd iw.kkZafdr½
mal)
(a) 158-54 lseh
(a) 158.54cm
(b) 158-74 lseh
(b) 158.74cm
(c) 159-56 lseh
(c) 159.56cm
(d) buesa ls dksbZ ugha
(d) None of these
Ans: d Ans: d
Sol: lgh ;ksx ¾ ¼160 × 35 $ 104 & 144½ lseh ¾ 5560 lsehA Sol: Correct sum = (160 × 35 + 104 – 144) cm = 5560cm.
 5560   5560 
 okLrfod vkSlr ÅapkbZ ¾  35  lseh  Actual average height =   cm
   35 

DELHI IAS ACADMEY T-4 Page No.: 3

 ¾ 158-857 lseh  158-86 lseh- = 158.857cm  158.86cm.

Q.4- ,d bathfu;fjax d‚yst esa eSdsfudy VªsM ls lHkh bathfu;fjax Q.4. In an engineering college the average salary of all engi-
Lukrdksa dk vkSlr osru 2-45 yk[k #i;s çfr o"kZ gS vkSj neering graduates from mechanical trade is Rs.2.45 lacs per
bysDVª‚fuDl VªsM ls bathfu;fjax Lukrdksa dk vkSlr osru 3-56 annum and that of the engineering graduates from Elec-
yk[k #i;s çfr o"kZ gSA lHkh eSdsfudy vkSj bysDVª‚fuDl tronics trade is Rs.3.56 lacs per annum. The average salary
Lukrdksa dk vkSlr osru 3-12 yk[k #i;s çfr o"kZ gSA bl of all Mechanical and Electronics graduates is Rs.3.12 lacs
laLFkku ls mÙkh.kZ gksus okys bysDVª‚fuDl Lukrdksa dh U;wure per annum. Find the least number of Electronics graduates
la[;k Kkr dhft,A passing out from this institute.
(a) 43 (a) 43
(b) 59 (b) 59
(c) 67 (c) 67
(d) fuèkkZfjr ugha fd;k tk ldrk (d) Cannot be determined
Ans: c
Ans: c
Sol: Let the number of mechanical Engineering graduates be M
Sol: ekuk fd eSdfs udy bathfu;fjax Lukrdksa dh la[;k M gS vkSj
and the number of Electronics Engineering graduates be E.
bysDVª‚fuDl bathfu;fjax Lukrdksa dh la[;k E gSA
Then, 2.45M + 3.56E = 3.12(M + E)
fQj] 2.45M + 3.56E = 3.12(M + E)
 2.45M + 3.56E = 3.12M + 3.12E
 2.45M + 3.56E = 3.12M + 3.12E
 0.44E = 0.67M
 0.44E = 0.67M
M 0.44 44
M 0.44 44   
   E 0.67 67
E 0.67 67
 Since the ratio 44 : 67 is in the simplest from, so least
 pw¡fd vuqikr 44 % 67 vHkkT; gS] blfy, bysDVª‚fuDl
number of Electronics graduates = 67.
Lukrdksa dh U;wure la[;k ¾ 67 gSA
0.032  0.212  0.0652 0.032  0.212  0.0652
Q.5. The value of is
Q.5-
0.0032  0.0212  0.00652 dk eku gS 0.0032  0.0212  0.00652
(a) 0.1
(a) 0-1
(b) 10
(b) 10
(c) 102
(c) 102
(d) 103
(d) 103
Ans: b
Ans: b
0.032  0.212  0.0652
0.032  0.212  0.0652 2 2 2
2 2 2 Sol: Given exp. =  0.03   0.21   0.065 
Sol: fn;k x;k gS ¾  0.03   0.21   0.065       
       10   10   10 
 10   10   10 

2 2 2
100[(0.03) 2  (0.21) 2  (0.065) 2 ]
100[(0.03)  (0.21)  (0.065) ]
0.032  0.212  (0.065) 2 = 100 = 10
0.032  0.212  (0.065) 2 = 100 = 10

12
12 Q.6. is equal to
Q.6- ds cjkcj gS 3 5  2 2
3 5  2 2
(a) 1  5  2  10
(a) 1  5  2  10
(b) 1  5  2  10
(b) 1  5  2  10
(c) 1  5  2  10
(c) 1  5  2  10
(d) 1  5  2  10
(d) 1  5  2  10

DELHI IAS ACADMEY T-4 Page No.: 4

Ans: b Ans: b

12 3 ( 5  2 2) 12 3 ( 5  2 2)
 
Sol: fn;k x;k gS& Sol: Given exp.
3 ( 5  2 2) 3 ( 5  2 2) 3 ( 5  2 2) 3 ( 5  2 2)

12(3  5  2 2 ) 12(3  5  2 2 ) 12(3  5  2 2 ) 12(3  5  2 2 )

= = = 2 2 =
2
3  ( 5  2 2) 2
9  (5  8  4 10 ) 3  ( 5  2 2) 9  (5  8  4 10 )

12(3  5  2 2 ) 3( 5  2 2  3) 12(3  5  2 2 ) 3( 5  2 2  3)
 = 
= (4  4 10 ) 10  1
(4  4 10 ) 10  1

3( 5  2 2  3) 10  1 3( 5  2 2  3) 10  1
 = 
= 10  1 10  1
10  1 10  1

3 50  3 5  6 20  6 2  9 10  9 3 50  3 5  6 20  6 2  9 10  9
= =
10  1 10  1

15 2  3 5  12 5  6 2  9 10  9 15 2  3 5  12 5  6 2  9 10  9
= =
9 9

9 2  9 5  9 10  9 9 2  9 5  9 10  9
= =
9 9

= 1  2  5  10 = 1  2  5  10
64 25
Q.7- la[;k 2564 × 6425 ,d çk—r la[;k n dk oxZ gSA n ds vadksa Q.7. The number 25 × 64 is square of a natural number n. The
dk ;ksx gS& sum of the digits of n is :
(a) 7 (a) 7
(b) 14 (b) 14
(c) 21 (c) 21
(d) 28 (d) 28
Ans: b
Ans: b
Sol: n 2  (25) 64  (64) 25  (52 ) 64  (2 6 ) 25
Sol: n 2  (25) 64  (64) 25  (52 ) 64  (2 6 ) 25
= 5128  2150  5128  2128  2 22
= 5128  2150  5128  2128  2 22
n = 5 64  2 64  211  (5  2) 64  211
n = 5 64  2 64  211  (5  2) 64  211
= 10 64  2048
= 10 64  2048
 Sum of digits of n = 2 + 0 + 4 + 8 = 14
 n ds vadksa dk ;ksx ¾ 2 $ 0 $ 4 $ 8 ¾ 14
Q.8. The average of 25 results is 18. The average of first 12 of
Q.8- 25 ifj.kkeksa dk vkSlr 18 gSA muesa ls igys 12 dk vkSlr 14
those is 14 and the average of last 12 is 17. What is the 13th
gS vkSj vafre 12 dk vkSlr 17 gSA 13oka ifj.kke D;k gksxk\
result?
(a) 74
(a) 74
(a) 75
(b) 75
(a) 69
(c) 69
(a) 78
(d) 78
Ans: d
Ans: d
Sol: çFke 12 ds ifj.kkeksa dk ;ksx ¾ 12 × 14
Sol: Sum of 1st 12 results = 12 × 14
fiNys 12 ifj.kkeksa dk ;ksx ¾ 12 × 17 Sum of last 12 results = 12 × 17
13ok¡ ifj.kke ¾ x ¼ekuk½ 13th result = x (let)
vc] Now,
12 × 14 + 12 × 17 + x = 25 × 18
12 × 14 + 12 × 17 + x = 25 × 18
;k x = 78 or x = 78
DELHI IAS ACADMEY T-4 Page No.: 5
Q.9- ;fn 3x  7 = x 2  P  7 x  5 , gS] rks P dk eku D;k gS\ Q.9. If 3x  7 = x 2  P  7 x  5 , what is the value of P?
1 1
(a) (a)
2 2
1 1
(b) 8 (b) 8
4 4
1 1
(c) 8 (c) 8
2 2
(d) fuèkkZfjr ugha fd;k tk ldrk (d) Cannot be determined
Ans: b Ans: b
Sol: 3 x  7  7 x  5  7 x  3x  2  4 x  2 Sol: 3 x  7  7 x  5  7 x  3x  2  4 x  2

1 1
 x .  x .
2 2

3 1 3 1
vc] 3x  7  x 2  P  7  P Now, 3x  7  x 2  P  7  P
2 4 2 4

17 1 33 1 17 1 33 1
P  = =8 . P  = =8 .
2 4 4 4 2 4 4 4
Q.10- ,d flad esa Bhd 12 yhVj ikuh gksrk gSA ;fn flad ls ikuh Q.10. A sink contains exactly 12 litres of water. If water is drained
rc rd fudkyk tkrk gS tc rd fd mlesa fudkys x, ikuh from the sink until it holds exactly 6 litres of water less than
dh ek=k ls Bhd 6 yhVj de ikuh u jg tk,] rks fdrus yhVj the quantity drained away, then how many litres of water
ikuh cg x;k\ were drained away?
(a) 2 (a) 2
(b) 3 (b) 3
(c) 6 (c) 6
(d) 9 (d) 9
Ans: d
Ans: d
Sol: Let the quantity of water drained away be x litres.
Sol: eku yhft, fd cgk, x, ikuh dh ek=k x yhVj gSA
Then, 12 – x = x – 6 2x = 18 x = 9
fQj] 12 – x = x – 6 2x = 18 x = 9
m 4 r 9 3mr  nt
m 4 r 9 3mr  nt Q.11. If  and  , the value of is
Q.11- ;fn  rFkk  gks] rks dk eku gS n 3 t 14 4nt  7 mr
n 3 t 14 4nt  7 mr
1
1 (a)  5
(a)  5 2
2
11
11 (b) 
(b)  14
14
1
1 (c)  1
(c)  1 4
4
11
11 (d)
(d) 14
14 Ans: b
Ans: b
m 4 r 9 mr 4 9 6
Sol:  and     
m 4 r 9 mr 4 9 6 n 3 t 14 nt 3 14 7
Sol:  vkSj     
n 3 t 14 nt 3 14 7
mr 6
3 1 3 1
mr 6 3mr  nt nt 7
3 1 3 1  
3mr  nt  mr 6
 nt  7 4nt  7mr
47 4 7
 4nt  7mr mr 6 nt 7
47 4 7
nt 7

DELHI IAS ACADMEY T-4 Page No.: 6

 18 18
1 1
7 11  1  11 7 11  1  11
=      . =      .
46 7  2 14 46 7  2 14
Q.12- 1 lsaVhehVj dk lkSoka fgLlk tc 1 fdyksehVj ds va'kksa esa fy[kk Q.12. One hundredth of centimetre when written in fractions of 1
tkrk gS] rks cjkcj gksrk gS kilometer, is equal to
(a) 0-0000001 (a) 0.0000001
(b) 0-000001 (b) 0.000001
(c) 0-0001 (c) 0.0001
(d) 0-001 (d) 0.001
Ans: a Ans: a

1  1  1  1 
cm  cm cm  cm
 100  Sol: Required fraction = 100 =  100 
Sol: vHkh"V fHkUu ¾ 100 =
1km (1000 100)cm 1km (1000 100)cm

1 1
= =
100  1000  100 100  1000  100

1 1
= =
10000000 10000000
= 0.0000001 = 0.0000001
Q.13- oSKkfud ladsru esa O;ä djus ij la[;k 518]000]000 ds Q.13. The number 518,000,000 when expressed in scientific nota-
cjkcj gksrh gS tion, equals
(a) 51.8 × 1010 (a) 51.8 × 1010
(b) 51.8 × 109 (b) 51.8 × 109
(c) 51.8 × 107 (c) 51.8 × 107
(d) 51.8 × 1011 (d) 51.8 × 1011
Ans: c
Ans: c
Sol: 518,000,000 = 5.18 × 100000000 = 5.18 × 108
Sol: 518,000,000 = 5.18 × 100000000 = 5.18 × 108
= 51.8 × 107
= 51.8 × 107
Q.14. 534.596 + 61.472 – 496.708 = ? + 27.271
Q.14- 534.596 + 61.472 – 496.708 = ? + 27.271
(a) 62.069
(a) 62-069
(b) 72.089
(b) 72-089
(c) 126.631
(c) 126-631
(d) 132.788
(d) 132-788
Ans: b
Ans: b Sol: Let 534.596 + 61.472 – 496.708 = x + 27.271
Sol: eku yhft, 534.596 + 61.472 – 496.708 = x + 27.271 Then,
rc] x = (534.596 + 61.472) – (496.708 +27.271)
x = (534.596 + 61.472) – (496.708 +27.271) = 596.068 – 523.979 = 72.089
= 596.068 – 523.979 = 72.089 Q.15. The traffic lights at three different signal points change
Q.15- rhu vyx&vyx flXuy fcanqvksa ij VªSfQd ykbVsa Øe'k% gj after every 45 seconds, 75 seconds and 90 seconds re-
45 lsdaM] 75 lsdaM vkSj 90 lsdaM ds ckn cny tkrh gSaA ;fn spectively. If all change simultaneously at 7:20 : 15 hours,
lHkh ,d lkFk 7 % 20 % 15 ?kaVs ij cnyrs gSa] rks os iqu% ,d then they will change again simultaneously at :
lkFk dc cnysxh& (a) 7 : 27 : 45 hours
(a) 7 % 27 % 45 ?kaVs (b) 7 : 28 : 00 hours
(b) 7 % 28 % 00 ?kaVs (c) 7 : 27 : 30 hours
(c) 7 % 27 % 30 ?kaVs (d) 7 : 27 : 50 hours
(d) 7 % 27 % 50 ?kaVs

DELHI IAS ACADMEY T-4 Page No.: 7

Ans: a Ans: a

5 45 , 75 , 90 5 45 , 75 , 90
3 9 , 15 , 18 3 9 , 15 , 18
Sol: 3 3 , 5, 6 Sol: 3 3 , 5, 6
1, 5, 2 1, 5, 2

 5 × 3 × 3 × 5 × 2 = 450  5 × 3 × 3 × 5 × 2 = 450
45] 75] 90 dk y?kqÙke lekiorZd 450 gS LCM of 45, 75, 90 is 450
 VªSfQd ykbVsa 7 feuV 30 lsdaM ds ckn ;kuh 7% 27 % 45  Traffic lights will change simultaneously after 7 min-
?kaVs ij ,d lkFk cny tk,axhA utes 30 seconds i.e. at 7 : 27 : 45 hours
Q.16- rhu ifg;s Øe'k% 40] 24 vkSj 16 çfr feuV pDdj yxk ldrs Q.16. Three wheels can complete 40, 24 and 16 revolutions per
gSaA çR;sd ifg;s ij ,d yky èkCck gksrk gS tks ,d fuf”pr minute respectively. There is a red spot on each wheel that
le; esa tehu dks Nwrk gSA fdrus le; ckn ;s lHkh èkCcs ,d touches the ground at time zero. After how much time, all
lkFk iqu% tehu dks Li'kZ djsaxs\ these spots will simultaneously touch the ground again?

1 1
(a) 7 lsd
a M
a (a) 7 sec
2 2
(b) 18 lsdaM (b) 18 sec

1 1
(c) 7 feuV (c) 7 min
2 2
(d) 18 feuV (d) 18 min
Ans: a
Ans: a
Sol: For one complete revolution, the first second and third
Sol: ,d iw.kZ pDdj yxkus esa] igys nwljs vkSj rhljs ifg;s dks
60 60 60 3 5 15
60 60 60 3 5 15 wheels take , , seconds i.e. , , seconds
Øe'k% , , lsdaM ;kuh , , lsdaM yxrs gSaA 40 24 16 2 2 4
40 24 16 2 2 4
respectively.
 lHkh yky èkCcksa dks ,d lkFk iqu% tehu dks Nwus esa yxus  Time taken for all red spots to touch the ground again
okyk le;A simultaneously.
 3 5 15 
 , ,  lsdaM dk y?kqRre lekiorZd ysus ij  3 5 15   L.C.M . of 3,5,15 
2 2 4  L.C.M of  , ,  sec =  H .C.F . of 2,2,4  sec
 2 2 4   
 3] 5] 1] 5 dk y?kqRre lekiorZd 
¾  2] 2] 4 dk egRre lekiorZd  lsdaM =
15 1
sec = 7 Sec.
 
2 2
15 1 Q.17. A gardener has to plant trees in rows containing equal
¾ lsdaM ¾ 7 lsdaM-
2 2 number of trees. If he plants in rows of 6,8, 10 and 12, then
Q.17- ,d ekyh dks leku la[;k esa isM+ksa okyh iafä;ksa esa isM+ yxkus five tree are left unplanted. But if he plants in rows of 13
gksrs gSaA ;fn og 6]8] 10 vkSj 12 dh iafä;ksa esa ikSèks yxkrk gS] trees each, then no tree is left. What is the number of trees
rks ikap isM+ fcuk yxk, jg tkrs gSAa ysfdu ;fn og 13&13 isMk+ as that the gardener plants?
dh iafä;ksa esa ikSèks yxkrk gS] rks dksbZ Hkh isM+ ugha cprkA ekyh (a) 485
}kjk yxk, x, isM+ksa dh la[;k fdruh gS\ (b) 725
(a) 485 (c) 845
(b) 725 (d) None of these
(c) 845 Ans: c
(d) buesa ls dksbZ ugha Sol: L.C.M of 6, 8, 10, 12 = 120.
Ans: c  Required number is of the form 120k + 5.
Sol: 6] 8] 10] 12 dk y?kqRre lekiorZd ¾ 120-
 vko';d la[;k 120k $ 5 ds :i esa gSA

DELHI IAS ACADMEY T-4 Page No.: 8

  dk U;wure eku ftlds fy, ¼120k $ 5½ 13 ls foHkkT; gS] Least value of k for which (120k + 5) is divisible by 13 is k =
k ¾ 7 gSA 7.
 vHkh"V la[;k ¾ ¼120 × 7 $ 5½ ¾ 845-  Required number = (120 ×7 + 5) = 845.
Q.18- 21 vke ds isM+] 42 lsc ds isM+ vkSj 56 larjs ds isM+ iafä;ksa esa Q.18. 21 mango trees, 42 apple trees and 56 orange trees have to
bl çdkj yxk, tkus gSa fd çR;sd iafä esa dsoy ,d gh fdLe be planted in rows such that each row contains the same
ds isM+ksa dh leku la[;k gksA iafä;ksa dh U;wure la[;k ftlesa number of trees of one variety only. Minimum number of
isM+ yxk, tk ldrs gSa& rows in which the trees may be planted is :
(a) 3 (a) 3
(b) 15 (b) 15
(c) 17 (c) 17
(d) 20 (d) 20
Ans: c Ans: c
Sol: iafä;ksa dh U;wure la[;k ds fy,] çR;sd iafä esa isMk+ as dh la[;k Sol: For the minimum number of rows, the number of trees in
vfèkdre gksuh pkfg,A each row must be the maximum.
 çR;sd iafä esa isM+ksa dh la[;k ¾ 21] 42] 56 dk egRre  Number of trees in each row = H.C.F. of 21, 42, 56 = 7.
lekiorZd ¾ 7- Hence, number of rows
vr% iafä;ksa dh la[;k  21  42  56  119
=  = = 17.
 21  42  56  119  7  7
=  = = 17.
 7  7 Q.19. The H.C.F. and L.C.M. of two numbers are 50 and 250 re-
Q.19- nks la[;kvksa dk egRre lekiorZd rFkk y?kqRre lekiorZd spectively. If the first number is divided by 2, the quotient
Øe'k% 50 vkSj 250 gSaA ;fn igyh la[;k dks 2 ls foHkkftr is 50. The second number is :
fd;k tk,] rks HkkxQy 50 gksrk gSA nwljh la[;k gS& (a) 50
(a) 50 (b) 100
(b) 100 (c) 125
(c) 125 (d) 250
Ans: c
(d) 250
Sol: Dividend = Divisor × Quotient + Reminder
Ans: c
First number = (50 × 2) = 100.
Sol: HkkT; ¾ Hkktd × HkkxQy $ “ks’k
çFke la[;k ¾ ¼50 × 2½ ¾ 100-
 50  250 
 50  250  Second number = 
= 125. 
nwljh la[;k =  100  = 125.  100 
 
{HCF × LCM = First Number × Second Number}
¿egRre lekiorZd × y?kqRre lekiorZd ¾ igyh la[;k ×
Q.20. What mathematical operation should come at the place of
nwljh la[;kÀ
‘?’ in the equation :
Q.20- lehdj.k esa ^\* ds LFkku ij dkSu lh xf.krh; lafØ;k dk fpUg
2 ? 6 – 12 ÷ 4 + 2 = 11
vkuk pkfg,A
(a) +
2 ? 6 – 12 ÷ 4 + 2 = 11
(b) –
(a) +
(c) ×
(b) –
(d) ÷
(c) ×
Ans: c
(d) ÷
Sol: Let 2 x 6 – 12 ÷ 4 + 2 = 11
Ans: c
Then, 2 x 6 – 3 + 2 = 11  2 x 6 = 11 + 3 – 2 = 12
Sol: eku yhft, 2 x 6 – 12 ÷ 4 + 2 = 11
So, ‘x’ must be replaced by ‘×’.
fQj] 2 x 6 – 3 + 2 = 11  2 x 6 = 11 + 3 – 2 = 12
blfy,] ^x* dks ^×* ls çfrLFkkfir fd;k tkuk pkfg,A

DELHI IAS ACADMEY T-4 Page No.: 9

Q.21- (? – 968) ÷ 79 × 4 = 512 Q.21. (? – 968) ÷ 79 × 4 = 512
(a) 10185 (a) 10185
(b) 10190 (b) 10190
(c) 11075 (c) 11075
(d) 11080 (d) 11080
Ans: d Ans: d
Sol: eku yhft, (x – 968) ÷ 79 × 4 = 512. Sol: Let (x – 968) ÷ 79 × 4 = 512.

x  968 512  79 x  968 512  79

rc]  4 = 512  x  968   10112 Then,  4 = 512  x  968   10112
79 4 79 4
 x  10112  968  11080  x  10112  968  11080

3 4 5 21 3 4 5 21
Q.22- 504 dk dk 7 dk 9 dk ¾\ Q.22. of of of of 504= ?
5 24 5 7 9 24
(a) 63 (a) 63
(b) 69 (b) 69
(c) 96 (c) 96
(d) buesa ls dksbZ ugha (d) None of these
Ans: d
Ans: d
 3 4 5 21 
 3 4 5 21  Sol: Given expression =      504  = 84.
Sol: fn;k x;k O;atd ¾      504  ¾ 84-  5 7 9 24 
 5 7 9 24 
5 6  5 1
5 6  5 1 Q.23. The value of  of 1  ÷ 2  3  is
Q.23-  dk 1  ÷ 2  3  dk eku gS  7 13   7 4
7 13   7 4
20
20 (a)
(a) 169
169
(b) 1
(b) 1
5
5 (c)
(c) 4
4
119
119 (d) 1
(d) 1 180
180
Ans: c
Ans: c
 5 19   19 4  5  19 7  13 5
 5 19   19 4  5  19 7  13 5 Sol: Given exp. =        =   .
Sol: fn;k x;k gS&       =   .  7 13   7 13  7  13 19  4 4
 7 13   7 13  7  13 19  4 4
5 7
7 5 65
6 5 8 11 8  3 13  3
11 8 3 8  3 13  Q.24. The expression 3 of 1 ÷  2   of equals
Q.24- O;atd 1 dk 3 ÷ dk  2   cjkcj gS& 6 9 9  11 22  5
9 6 5 9  11 22  7 8
8 7
(a) 1
(a) 1
1
1 (b)
(b) 2
2
7
7 (c)
(c) 9
9
5
5 (d)
(d) 12
12 Ans: d
Ans: d

DELHI IAS ACADMEY T-4 Page No.: 10

  45 7   73 8  8  25 13  3  45 7   73 8  8  25 13  3
Sol: fn;k x;k gS&    of    ÷    of Sol: Given exp.    of    ÷    of
 8 45   11 73  9  11 22  5  8 45   11 73  9  11 22  5

7 8  8 63  3 7 8  8 63  3
= of ÷    of = of ÷    of
8 11  9 22  5 8 11  9 22  5

7 28 3 7 84 7 55 5 7 28 3 7 84 7 55 5
=  of =  =  = =  of =  =  =
11 11 5 11 55 11 84 12 11 11 5 11 55 11 84 12
Q.25- 2a 3  [3a 3  4b 3  {2a 3  (7a 3 )}  5a 3  7b 3 ] dk eku gS Q.25. The value of 2a 3  [3a 3  4b 3  {2a 3  (7a 3 )}
(a)  11a 3  3b 3  5a 3  7b 3 ] is
(b) 7b 3  3a 3 (a)  11a 3  3b 3
(c) 11a  3b
3 3
(b) 7b 3  3a 3
(d)  (11a 3  3b 3 ) (c) 11a 3  3b 3
Ans: a (d)  (11a 3  3b 3 )
Sol: fn;k x;k gS& Ans: a
= 2a 3  [3a 3  4b 3  {5a 3 }  5a 3  7b 3 ] Sol: Given exp.

= 2a 3  [3a 3  4b 3  5a 3  5a 3  7b 3 ] = 2a 3  [3a 3  4b 3  {5a 3 }  5a 3  7b 3 ]

= 2a 3  [13a 3  3b 3 ] = 2a 3  13a 3  3b 3 = 2a 3  [3a 3  4b 3  5a 3  5a 3  7b 3 ]

=  11a 3  3b 3 = 2a 3  [13a 3  3b 3 ] = 2a 3  13a 3  3b 3

=  11a 3  3b 3
Q.26- ,d fuf'pr dksM esa 'DUCK' dks 'GXFN' fy[kk tkrk gSA vki
'TIGER' 'kCn dks dSls dksfMr djsx
a s\ Q.26. In a certain code ‘DUCK’ is written as ‘GXFN’. How will
(a) WLJHU you code the word ‘TIGER’
(b) WLHJU (a) WLJHU
(c) WLJIU (b) WLHJU
(d) WLKHU (c) WLJIU
Ans: a (d) WLKHU
D U C K Ans: a
+3 +3 +3 +3
Sol: D U C K
G X F N Sol:
+3 +3 +3 +3

T I G E R G X F N
+3 +3 +3 +3 +3
T I G E R
W L J H U +3 +3 +3 +3 +3

Q.27- ;fn 'TIGER' dks '27136' vkSj 'TRAIN' dks '26879' ds :i esa W L J H U
dksfMr fd;k x;k gSA vki 'GREAT' dks dSls dksfMr djsaxs\ Q.27. If ‘TIGER’ is coded as ‘27136’ and ‘TRAIN’ as ‘26879. How
(a) 16283 will you code ‘GREAT’?
(b) 16823 (a) 16283
(c) 16223 (b) 16823
(d) 16382 (c) 16223
Ans: d (d) 16382
Sol: bl ç'u esa T dks 2 ds :i esa n'kkZ;k x;k gS] I dks 7 ds :i Ans: d
esa n'kkZ;k x;k gS] G dks 1 ds :i esa n'kkZ;k x;k gS bR;kfnA Sol: In this question T indicate as 2, I indicates as 7, G indicates
rks] 'GREAT' dk dksM ¾ 16382 as 1 and so on,
So, Code of ‘GREAT’ = 16382

DELHI IAS ACADMEY T-4 Page No.: 11

Q.28- ,d fuf'pr dwV Hkk"kk esa "te me se" dk vFkZ gS “grapes are Q.28. In a certain code language “te me se” means “grapes are
sweet”] "de se ro" dk vFkZ gS “strawberry and grapes” vkSj sweet”, “de se ro” means “strawberry and grapes” and “ye
"ye me re" dk vFkZ gS “apples are red” rks Hkk"kk esa fdl 'kCn me re” means “apples are red”. Which word in the lan-
dk vFkZ sweet gksrk gS\ guage means sweet?
(a) se (a) se
(b) te (b) te
(c) ye (c) ye
(d) buesa ls dksbZ ugha (d) None of these
Ans: b Ans: b
Sol: ç'u ds vuqlkj] Sol: According to question,
te me se“grapes are sweet” te me se“grapes are sweet”
de se ro “strawberry and grapes” de se ro “strawberry and grapes”
ye me re  “apples are red” ye me re  “apples are red”
mijksä dksM ls from the above code
grapes = se, are = me, sweet = te grapes = se, are = me, sweet = te
Q.29- ;fn 'PEAR' dks 'GFDN' fy[kk tkrk gS] rks bl dksM esa 'REAP' Q.29. If ‘PEAR’ is written as ‘GFDN’, how is ‘REAP’ written in
dSls fy[kk tk,xk\ this code ?
(a) D N G F (a) D N G F
(b) N F D G (b) N F D G
(c) N D F G (c) N D F G
(d) F D N G (d) F D N G
Ans: b Ans: b
Sol: P E A R Sol: P E A R
       
G F D N G F D N
PEAR dks myVus ds ckn After reversing PEAR
R E A P R E A P
       
N F D G N F D G

Q.30- ;fn fdlh fuf'pr dksM es]a "ADVENTURE" dks "BFYISZBZN" Q.30. If in a certain code, “ ADVENTURE” is coded as
ds :i esa dksfMr fd;k tkrk gS] rks ml dksM esa "COUNTRY" “BFYISZBZN”, how is “COUNTRY” coded in that code?
dks dSls dksfMr fd;k tk,xk\ (a) E Q W P V T A
(a) E Q WPVTA (b) D Q X R Y X F
(b) DQ XRYXF (c) B N T M S Q X
(c) BN TMSQX (d) D P V O U S Z
Ans: b
(d) DPVOUSZ
Ans: b A D V E N T U R E
+1 +2 +3 +4 +5 +6 +7 +8 +9
A D V E N T U R E Sol:
+1 +2 +3 +4 +5 +6 +7 +8 +9 B F Y I S Z B Z N
Sol:
B F Y I S Z B Z N C O U N T R Y
+1 +2 +3 +4 +5 +6 +7
C O U N T R Y
+1 +2 +3 +4 +5 +6 +7 D Q X R Y X F
D Q X R Y X F

DELHI IAS ACADMEY T-4 Page No.: 12

 fn'kkfunZs'k ¼31&35½ % nh xbZ tkudkjh dks è;kuiwoZd i<+sa Directions (31-35): Read the given information carefully
vkSj vkxs fn, x, ç'uksa ds mÙkj nsa& and answer the questions given beside:
vkB yksx vk#"k] fcanw] pqUuw] fnO;k] ,êh] i`Foh] xksiky vkSj Eight people Aarush, Bindu, Chunnu, Divya, Etti, Prithvi,
gjh'k ,d vkB eaftyk bekjr esa jgrs gSa] ysfdu t:jh ugha Gopal and Harish live in an eight storey building, but not
fd blh Øe esa gksaA lcls fupyh eafty dh la[;k 1 gS vkSj necessarily in the same order. The lowermost floor is num-
lcls Åijh eafty dh la[;k 8 gSA bered 1 and the topmost floor is numbered 8.
vk#"k le la[;k okyh eafty ij jgrk gS ysfdu nwljs ;k Aarush lives on an even-numbered floor but not on
pkSFks eafty ij ughaA vk#"k vkSj fcanq ds chp dsoy rhu eaftysa the floor numbered second or fourth. Only three floors are
gSaA pqUuw vkSj ,êh ds chp dsoy nks O;fä jgrs gSaA i`Foh] fnO;k there between Aarush and Bindu. Only two people live
ls Åij okyh eafty ij jgrk gSA ftu eaftyksa ij ,êh vkSj between Chunnu and Etti. Prithvi lives on a floor above
fcanw jgrs gSa vkSj ftu eaftyksa ij vk#"k vkSj ,êh jgrs gS]a muds Divya. There are equal numbers of floors between the floors
chp eaftyksa dh la[;k leku gSA gjh'k] fnO;k dh eafty ds on which Etti and Bindu live and between the floors on
Bhd uhps jgrk gSA xksiky] vk#"k dh eafty ds Bhd uhps jgrk which Aarush and Etti live. Harish lives immediately below
gSA Divya’s floor. Gopal lives immediately below Aarush’s floor.
Q.31- i`Foh fuEufyf[kr esa ls fdl eafty ij jgrk gS\ Q.31. Prithvi lives on which of the following floors?
(a) rhljh (a) Third
(b) ikapoh (b) Fifth
(c) NBh (c) Sixth
(d) nwljh (d) Second
Ans: b Ans: b
Sol: eaf ty l a[ ; k O; f ä Sol: Floor Number Person
8 vk#"k 8 Aarush
7 xksiky 7 Gopal
6 ,êh 6 Etti
5 i`Foh 5 Prithvi
4 fcUnq 4 Bindu
3 pqUuw 3 Chunnu
2 fnO;k 2 Divya
1 gjh'k 1 Harish

i`Foh ik¡poha eafty ij jgrk gSA Prithvi lives on fifth floor.

blfy, fodYi b] lgh mÙkj gSA Option B, is hence the correct answer.
Q.32- fcanq vkSj fnO;k ds chp fdruh eaftysa gSa\ Q.32. How many floors are there between Bindu and Divya?
(a) ikap (a) Five
(b) ,d (b) One
(c) nks (c) Two
(d) rhu (d) Three
Ans: b Ans: b
Sol: eaf ty l a[ ; k O; f ä Sol: Floor Number Person
8 vk#"k 8 Aarush
7 xksiky 7 Gopal
6 ,êh 6 Etti
5 i`Foh 5 Prithvi
4 fcUnq 4 Bindu
3 pqUuw 3 Chunnu
2 fnO;k 2 Divya
1 gjh'k 1 Harish

DELHI IAS ACADMEY T-4 Page No.: 13

 mijksä rkfydk ls fcanq vkSj fnO;k ds chp dsoy ,d eafty From the above table There is only one floor between Bindu
gSA and Divya.
Q.33- fuEufyf[kr esa ls dkSu igyh eafty ij jgrk gS\ Q.33. Who among the following lives on the first floor?
(a) gjh'k (a) Harish
(b) pqUuw (b) Chunnu
(c) i`Foh (c) Prithvi
(d) ,êh (d) Etti
Ans: a Ans: a
Sol: eaf ty l a[ ; k O; f ä Sol: Floor Number Person
8 vk#"k 8 Aarush
7 xksiky 7 Gopal
6 ,êh 6 Etti
5 i`Foh 5 Prithvi
4 fcUnq 4 Bindu
3 pqUuw 3 Chunnu
2 fnO;k 2 Divya
1 gjh'k 1 Harish

mijksä rkfydk ls ge ns[k ldrs gSa fd gjh'k igyh eafty From the above table we can see Harish Lives on the first
ij jgrk gSA floor.
Q.34- fuEufyf[kr esa ls dkSu lkroha eafty ij jgrk gS\ Q.34. Who among the following lives on the seventh floor?
(a) ,êh (a) Etti
(b) i`Foh (b) Prithvi
(c) xksiky (c) Gopal
(d) pqUuw (d) Chunnu
Ans: c Ans: c
Sol: eaf ty l a[ ; k O; f ä Sol: Floor Number Person
8 vk#"k 8 Aarush
7 xksiky 7 Gopal
6 ,êh 6 Etti
5 i`Foh 5 Prithvi
4 fcUnq 4 Bindu
3 pqUuw 3 Chunnu
2 fnO;k 2 Divya
1 gjh'k 1 Harish

mijksä rkfydk ls ge ns[k ldrs gSa fd xksiky lkroha eafty From the above table we can see Gopal lives on the sev-
ij jgrk gSA enth floor.
Q.35- fuEufyf[kr esa ls dkSu lk dFku lR; gS@gSa\ Q.35. Which of the following statements is/are true?
(a) vk#"k NBh eafty ij jgrk gSA (a) Aarush lives on the sixth floor.
(b) fnO;k lcls Åijh eafty ij jgrh gSA (b) Divya lives on the topmost floor.
(c) i`Foh vkSj pqUuw ds chp nks yksx gSaA (c) There are two people between Prithvi and Chunnu.
(d) pqUuw rhljh eafty ij jgrk gSA (d) Chunnu lives on the third floor.
Ans: d Ans: d
Sol: Sol:

DELHI IAS ACADMEY T-4 Page No.: 14

 eaf ty l a[ ; k O; f ä Floor Number Person
8 vk#"k 8 Aarush
7 xksiky 7 Gopal
6 ,êh 6 Etti
5 i`Foh 5 Prithvi
4 fcUnq 4 Bindu
3 pqUuw 3 Chunnu
2 fnO;k 2 Divya
1 gjh'k 1 Harish

mijksä rkfydk ls ge ns[k ldrs gSa fd pqUuw rhljh eafty ij From the above table we can see Chunnu lives on the third
jgrk gS & ;g ,dek= lR; dFku gSA floor - is the only true statement.
Q.36- uhps ,d ikls dh nks fLFkfr;k¡ fn[kkbZ xbZ gSaA 5 vad okys Qyd Q.36. Two positions of a dice are shown below. Which number
ds foijhr Qyd ij dkSu lk vad fn[kkbZ nsxk\ will appear on the face opposite to the face with the num-
ber 5?

(a) 2@6 (a) 2/6

(b) 2 (b) 2
(c) 6 (c) 6
(d) 4 (d) 4
Ans: c Ans: c
Sol: bl ç'u esa ikls esa la[;k 3 okys mHk;fu"B Qyd leku fLFkfr Sol: In this question dice common faces with number 3, are in
esa gSaA vr: vad 5 okys vkeus&lkeus dh la[;k 6 gksxhA same positions. Hence the number of the opposite face to
Q.37- 6 ds foijhr Qyd ij dkSu lh la[;k gS\ face with number 5 will be 6.
Q.37. Which number is on the face opposite to 6?

(a) 4
(b) 1 (a) 4 (b) 1
(c) 2 (c) 2 (d) 3
Ans: b
(d) 3
Sol: As the numbers 2, 3, 4 and 5 are adjacent to 6. Hence the
Ans: b
number on the face opposite to 6 is 1.
Sol: pw¡fd la[;k,¡ 2] 3] 4 vkSj 5] 6 ds vklUu gSaA blfy, 6 ds
Q.38. Here two positions of dice are shown. If there are two dots
foijhr Qyd ij la[;k 1 gSA
in the bottom, then how many dots will be on the top?
Q.38- ;gka iklksa dh nks fLFkfr;k¡ fn[kkbZ xbZ gSaA ;fn uhps nks fcanq gSa
rks Åij fdrus fcanq gksaxs\

(a) 2
(a) 2
(b) 3
(b) 3

DELHI IAS ACADMEY T-4 Page No.: 15

 (c) 5 (c) 5
(d) 6 (d) 6
Ans: c Ans: c
Sol: ;gka 4 fcanqvksa okys lkekU; Qyd leku fLFkfr esa gSaA vr% 2] Sol: Here the common faces with 4 dots are in same positions.
5 ds foijhr gksxkA Hence 2 will be opposite to 5.
Q.39- ,d ?kukdkj Cy‚d dh nks fLFkfr;k¡ fn[kkbZ xbZ gSAa tc 5 lcls Q.39. Two positions of a cubical block are shown. When 5 is at
Åij gS rks dkSu lh la[;k lcls uhps gksxh\ the top which number will be at bottom?

(a) 1 (a) 1
(b) 2 (b) 2
(c) 3 (c) 3
(d) 4 (d) 4
Ans: c Ans: c
Sol: In these 2 positions one common face with number 3, is in
Sol: bu 2 fLFkfr;ksa esa la[;k 3 okyk ,d mHk;fu"B Qyd ,d gh
same position. 1 is opposite to 6 and 4 is opposite to 2.
fLFkfr esa gSA 1] 6 ds foijhr gS vkSj 4] 2 ds foijhr gSA blfy,
Therefore 5 is opposite to 3.
5] 3 ds foijhr gSA
Q.40. You are given three positions of dice then which face is
Q.40- vkidks iklksa dh rhu fLFkfr;k¡ nh xbZ gSa rks dkSu lk Qyd v{kj
opposite to the face with alphabet B?
B okys Qyd ds foijhr gS\

(a) E
(b) F
(a) E
(c) D
(b) F
(d) A
(c) D
Ans: b
(d) A
Sol: From figures (i) and (ii), we can conclude that the alpha-
Ans: b
bets C, D, A and E lie adjacent to the alphabet F. So, the
Sol: vk—fr;ksa ¼i½ vkSj ¼ii½ ls] ge ;g fu"d"kZ fudky ldrs gSa fd
alphabet B lies opposite F and conversely F lies opposite
v{kj C, D, A vkSj E v{kj F ds vklUu gSaA blfy,] v{kj B] F B.
ds foijhr gS vkSj blds foijhr F, B ds foijhr gSA Q.41. An accurate clock shows 7 o'clock in the morning. Through
Q.41- ,d ?kM+h lqcg 7 cts lVhd le; fn[kkrh gSA tc ?kM+h esa
how may degrees will the hour hand rotate when the clock
nksigj ds 3 cts gksx
a s rks ?kaVs dh lqbZ fdrus fMxzh rd ?kwe pwdh shows 3 o'clock in the afternoon?
gksxh\ (a) 244º
(a) 244º
(b) 250º
(b) 250º
(c) 268º
(c) 268º
(d) 240º
(d) 240º Ans: d
Ans: d Sol: In 1 hour hand moves 30º
Sol: 1 ?kaVs esa lqbZ 30º pyrh gS Angle traced by the hour hand in 8 hours
?kaVs dh lqbZ }kjk 8 ?kaVs esa dks.k dk irk yxk;k x;k

DELHI IAS ACADMEY T-4 Page No.: 16

 º º
 360   360 
=  8   240º =  8   240º
 12   12 
Q.42- ,d fnu esa ?kM+h dh lqb;k¡ fdruh ckj feyrh gSa\ Q.42. How many times do the hands of a clock coincide in a day?
(a) 20 (a) 20
(b) 21 (b) 21
(c) 22 (c) 22
(d) 24 (d) 24
Ans: c Ans: c
Sol: ,d ?kM+h dh lwb;k¡ çR;sd 12 ?kaVs esa 11 ckj feyrh gS]a blfy,] Sol: The hands of a clock coincide 11 times in every 12 hours
24 ?kaVs esa ,d ?kM+h dh lqb;k¡ 24 ckj feyrh gSaA So, in 24hours the hands of a clock coinside 24times.
Q.43- tc ?kM+h esa le; 8-30 cts gS rks feuV dh lqbZ vkSj ?kaVs dh Q.43. The angle between the minute hand and the hour hand of
lqbZ ds chp dk dks.k gS& a clock when the time is 8.30, is:
(a) 80º (a) 80º
(b) 75º (b) 75º
(c) 60º (c) 60º
(d) 105º (d) 105º
Ans: b Ans: b
Sol: We know that
Sol: Kkr gS&
11
11 30 H  M
30 H  M 2
2
11
11 30  8   30
30  8   30 2
2
240 – 165 = 75º
240 – 165 = 75º
Q.44. A watch is 1 minute slow at 1 p.m. on Tuesday and 2 min-
Q.44- ,d ?kM+h eaxyokj dks nksigj 1 cts 1 feuV /kheh gks tkrh gS
utes fast at 1p.m. on Thursday. When did it show the cor-
vkSj xq:okj nksigj 1 cts 2 feuV rst gks tkrh gSA ?kM+h }kjk
rect time?
lgh le; dc fn[kk;k x;k\
(a) 1.00 a.m. on Wednesday
(a) cqèkokj dks jkr 1-00 cts
(b) 5.00 a.m. on Wednesday
(b) cqèkokj lqcg 5-00 cts
(c) 1.00 p.m. on Wednesday
(c) nksigj 1-00 cts cqèkokj dks
(d) 5.00 p.m. on Wednesday
(d) 'kke 5-00 cts cqèkokj dks
Ans: b
Ans: b
le; nksigj 1 cts ls- eaxyokj dks nksigj 1 cts rd xq#okj Sol: Time from 1 p.m. on Tuesday to 1 p.m. on Thursday = 48
Sol:
hours
¾ 48 ?kaVs
So, the watch gains ( 1 + 2) minute or 3 min in 48 hrs.
rks] ?kM+h 48 ?kaVksa esa ¼1 $ 2½ feuV ;k 3 feuV c<+ tkrh gSA
Now, 3 min are gained in 48 hrs.
vc] 48 ?kaVksa esa 3 feuV c<+ tkrs gSaA
So, 1 min is gained in (48/3) = 16 hrs.
rks] ¼48@3½ ¾ 16 ?kaVs esa 1 feuV çkIr gksrk gSA
Thus, the watch showed the correct time 16 hrs. after 1 p.m.
bl çdkj] ?kM+h us eaxyokj nksigj 1 cts ds 16 ?kaVs ckn vFkkZr~
on Tuesday, i.e. 5 a.m. on Wednesday
cq/kokj dks lqcg 5 cts lgh le; fn[kk;kA
Q.45. A bus leaves at 12 : 25 noon and reaches destination at 10
Q.45- ,d cl nksigj 12 % 25 cts jokuk gksrh gS vkSj lqcg 10 % 45
: 45 am. The duration of the journey is :
cts xarO; ij igqaprh gSA ;k=k dh vofèk gS&
(a) 22 hrs 20 min
(a) 22 ?kaVs 20 feuV
(b) 22 hrs 40 min
(b) 22 ?kaVs 40 feuV
(c) 24 hrs 20 min
(c) 24 ?kaVs 20 feuV
(d) 24 hrs 40 min
(d) 24 ?kaVs 40 feuV
Ans: a
Ans: a

DELHI IAS ACADMEY T-4 Page No.: 17

Sol: ;k=k dh vofèk Sol: Duration of the journey
¾ ¼nksigj 12 % 25 cts ls vkèkh jkr rd dh vofèk½ $ ¼jkr = (duration from 12 : 25 noon to midnight) + (Duration
12 % 00 cts ls lqcg 10 % 45 cts rd dh vofèk½ from 12 : 00 midnight to 10 : 45am)
¾ 11 ?kaVs 35 feuV $ 10 ?kaVs 45 feuV ¾ 22 ?kaVs 20 feuVA = 11 hrs 35 min + 10 hrs 45 min = 22 hrs 20 min.
Q.46- vkidks vHkh&vHkh vkids fe= us lwfpr fd;k gS fd og vxys
fnu gksus okys fØdsV eSp ds fVdVksa dk çcaèku dj ldrk gSA Q.46. You have just been informed by your friend that he can
;g ,d Hkkjr&ikfdLrku eSp gS tks vkids 'kgj esa nqyZHk gS vkSj manage the tickets for a cricket match that is being held
vki okLro esa bls ns[kus ds fy, mRlqd gSaA gkyk¡fd] vkidk the next day. It is an India-Pakistan match which is rare in
c‚l dHkh Hkh flQZ ,d fnu dk uksfVl nsdj Nqêh ugha nsrk gSA your city and you are really keen to watch it. However,
vki D;k djsaxs\ your boss does not ever grant leave by just giving one-
(a) vius c‚l dks lwfpr djsaxs fd vki [kjkc LokLF; ds day notice. What should you do?
dkj.k dk;kZy; ugha vk ldrsA (a) Inform your boss that you cannot attend office due
(b) i;Zos{kd dks lwfpr djsaxs fd ,d vkikrdkyhu fLFkfr to ill health
mRiUu gks xbZ gS vkSj vki Nqêh dk ykHk ysuk pkgrs gSaA (b) Notify the supervisor that an emergency has occurred
(c) vius ofj"B vfèkdkjh ls iwNsaxs fd D;k og vkidks dksbZ and you want avail of leave.
jkLrk lq>k ldrk gS rkfd vki eSp ns[kus tk ldsaxsaA (c) Ask your senior officer if he can suggest a way for
(d) ,d dgkuh cuk;sx a s vkSj vius ofj"B vfèkdkjh ls Nqêh ekaxx
as As you to be excused so you can attend the match
Ans: c (d) Cook up a story and ask your senior officer for leave.
Sol: vius ofj"B vfèkdkjh ls iwNsaxs fd D;k og vkidks dksbZ jkLrk Ans: c
lq>k ldrk gS rkfd vki eSp ns[kus tk ldsaxsaA Sol: Ask your senior officer if he can suggest a way for you to
Q.47- x'rh MîwVh ds nkSjku] vki NsM[
+ kkuh vkSj NsMN
+ kM+ dh ?kVukvksa be excused so you can attend the match
dks ns[krs gSaA NsM+NkM+ djus okyk ,d jktusrk dk csVk gS tks Q.47. While on patrol duty, you come across eve teasing and
dkuwu dk ikyu djus okyk ukxfjd vkSj esgurh foèkk;d gSA molestation taking place. The molester is the son of a poli-
vki D;k djsaxs\ tician who is a law abiding citizen and a hard working
(a) tkdj iwNrkN djsaxs vkSj ml O;fä ls iwNsaxs fd D;k py MLA. What should you do?
jgk gS lkFk gh yM+dh dh lqj{kk lqfuf'pr djus ds ckn (a) Go and inquire and ask the person that what is going
vkxs dh tkap djsaxsaA on and carry further investigation after ensuring the
(b) pwafd NsM+NkM+ djus okyk LFkkuh; foèkk;d dk csVk gS] safety of the girl
blfy, vki ekeys dks utjvankt djus dk QSlyk ysaxsA (b) As the molester is the son of local MLA you decided
to ignore the matter.
(c) pwfa d foèkk;d dkuwu dk ikyu djus okys gS]a blfy, vkius
(c) Since the MLA is law abiding, you decided to let his
muds csVs dks psrkouh ds ckn tkus nsus dk QSlyk djsx
a As
son go after warning.
(d) vki vius ofj"B dks Qksu djsaxs vkSj fLFkfr crk;saxs rFkk
(d) You call up your senior and explain the situation and
muls iwNsaxs fd D;k fd;k tkuk pkfg,A
ask him that what should be done
Ans: a
Ans: a
Sol: tkdj iwNrkN djsaxs vkSj ml O;fä ls iwNsaxs fd D;k py jgk
Sol: Go and inquire and ask the person that what is going on
gS lkFk gh yM+dh dh lqj{kk lqfuf'pr djus ds ckn vkxs dh and carry further investigation after ensuring the safety
tkap djsaxsaA of the girl
Q.48- vki Hkkjrh; jsyos esa Mh-,e- gSaA Vªsuksa esa ,lh Dykl vkèkh [kkyh Q.48. You are a D.M. in the Indian Railways. The AC class in
tk jgh gS- ,slk ,;jykbu }kjk fn, tkus okys de fdjk, ds trains are going half empty. This is caused by low fares
dkj.k gksrk gSA ;k=h gokbZ ;k=k djuk ilan djrs gSaA lhVksa ds offered by airline. Passengers prefer to travel by air. To
mi;ksx dh nj c<+kus ds fy, vki D;k djsaxs\ increase the rate of occupancy. You would:
(a) ;kf=;ksa dks okil ykus ds fy, fdjk;k de djsaA (a) reduce the fare to get back passengers.
(b) mPp oxZ dks feyus okyh {kerk dks de djrs gq, fuEu (b) reduce the capacity of upper classes to match lower
oxZ dh ekax vuqlkj lhVsa iznku djsaxsA demand.
(c) ;g lkspdj leL;k dks utjvankt djsaxsa fd ;g vius (c) ignore the problem thinking that it will solve itself.
vki gy gks tk,xhA (d) eliminate the upper class facility.
(d) mPp oxZ dh lqfoèkk lekIr djukA Ans: b
Ans: b

DELHI IAS ACADMEY T-4 Page No.: 18

Sol: fuEu ekax ls esy [kkus ds fy, mPp oxksZa dh {kerk de djsaA Sol: reduce the capacity of upper classes to match lower de-
Q.49- ,d usrk esa vPNs O;fäRo fo'ks"kKrk] Hkk"kk ij idM+ vkSj vkilh mand.
lEeku iSnk djus tSls xq.kksa ds ekè;e ls nwljksa dks --------------- Q.49.A leader has the ability to ..................... others through
qualities such as a good personality expertise, command
-------- djus dh {kerk gksrh gS( bu lHkh ds fy, etcwr
of language, and the creation of mutual respect; all of
ikjLifjd dkS'ky dh vko';drk gksrh gSA which require strong interpersonal skills.
(a) vlj (a) impact
(b) çHkko (b) effect
(c) control
(c) fu;a=.k
(d) dominate
(d) gkoh gksuk Ans: c
Ans: c Sol: A leader has the ability to control others through qualities
Sol: ,d usrk esa vPNs O;fäRo fo'ks"kKrk] Hkk"kk ij idM+ vkSj vkilh such as a good personality expertise, command of lan-
guage, and the creation of mutual respect; all of which
lEeku ds fuekZ.k tSls xq.kksa ds ekè;e ls nwljksa dks fu;af=r
require strong interpersonal skills.
djus dh {kerk gksrh gS( bu lHkh ds fy, etcwr ikjLifjd Q.50. They manage conflict effectively by devising win-win
dkS'ky dh vko';drk gksrh gSA solutions, constructively influencing the behavior of oth-
Q.50- nksuksa i{kksa ds ykHk ls tqM+k ¼win-win solution½ lekèkku rS;kj ers, and using effective communication ...............
(a) targets (b) weapons
djds] nwljksa ds O;ogkj dks jpukRed :i ls çHkkfor djds
(c) goals (d) strategies
vkSj çHkkoh lapkj dk mi;ksx djds la?k"kZ dks çHkkoh <ax ls Ans: d
çcafèkr djrs gSa----------------- Sol: They manage conflict effectively by devising win-win
(a) mn~n s”; (b) gfFk;kj solutions, constructively influencing the behavior of oth-
(c) y{; (d) j.kuhfr;k¡ ers, and using effective communication strategies.
Ans: d
Sol: nksuksa i{kksa ds ykHk ls tqMk+ lekèkku dh j.kuhfr;k¡ rS;kj djds]
nwljksa ds O;ogkj dks jpukRed :i ls çHkkfor djds vkSj çHkkoh
lapkj dk mi;ksx djds la?k"kZ dks çHkkoh <ax ls çcafèkr djrs gSAa

Q.51. fuEufyf[kr mica/kksa dks lqesfyr dhft, & [Ans] c

lwph&1 ¼miHkk’kk½ lwph&2 ¼cksfy;ka½ [SOL] N% Hkk’kkvksa dks “kkL=h; Hkk’kk dk ntkZ iznku gS &
A. if”peh fgUnh 1- esokrh 1- rfey] 2- rsyxq] 3- ey;kye] 4- dUuM+] 5- laLd`r] 6-
B. jktLFkkuh fgUnh 2- vo/kh vksfM+;kA
C. iwohZ fgUnh 3- x<+okyh Q.53. fuEu esa ls dkSu fgUnh fodkl Øe ds lgh izk:i dks lanfHkZr
D. igkM+h fgUnh 4- caqnsyh djrk gSa \
dwV% (a) oSfnd laLd`r → ykSfdd laLd`r → ikfy → izkd`r
A B C D A B C D → viHkz”a kA
(a) 4 1 2 3 (b) 1 2 3 4 (b) ykSfdd laLd`r → oSfnd laLd`r → ikfy → izkd`r
(c) 2 3 1 4 (d) 1 3 4 2 → viHkz”a kA
[Ans] a (c) ykSfdd laLd`r → oSfnd laLd`r → viHkza”k →
[SOL] 4] 1] 2] 3 lgh lqesfyr gSA izkd`r → ikfy
Q.52. orZeku esa fdrus Hkk’kkvksa dks “kkL=h; Hkk’kk dk ntkZ iznku fd;k (d) oSfnd laLd`r → ykSfdd laLd`r → izkd`r → ikfy →
x;k gS \ viHkz”a kA
(a) 5 (b) 4 [Ans] a
(c) 6 (d) 8 [SOL] oSfnd laLd`r → ykSfdd laLd`r → ikfy → izkd`r →
viHkz”a kA

DELHI IAS ACADMEY T-4 Page No.: 19

Q.54. dFku 1 % vkBoha vuqlp
w h esa 22 Hkk’kkvksa dks lfEefyr fd;k x;k (c) m (d) buesa ls dksbZ ugha
gSA [Ans] a
dFku 2 % mnwZ Hkk’kk 8oha vuqlwph esa “kkfey ugha gSA [SOL] ftu Lojksa ds mPpkj.k esa ftºok dk vkxs dk Hkkx lfØ; jgrk
dFku 3 % vaxzsth Hkk’kk 8oha vuqlwph esa “kkfey ugha gSA gS] mUgsa ^vxz Loj* dgrs gSaA tSls& v] b] bZ] ,] ,sA
dwV % Q.59. fuEufyf[kr esa ls dkSu&lk O;atu var%LFk O;atu gS\
(a) dsoy dFku 1 lR; gS (a) ;
(b) dFku 1 vkSj 2 lR; gS (b) j
(c) dFku 1] 2] 3 lR; gS (c) y
(d) 1 vkSj 3 lR; gS (d) mijksDr lHkh
[Ans] d [Ans] d
[SOL] mnqZ Hkk’kk vkBoha vuqlwph esa “kkfey gS tcfd vaxzsth Hkk’kk bl [SOL] var%LFk O;atu 4 izdkj ds gksrs gSa& ;]j]y]o
vuqlwph esa “kkfey ugha gSA Q.60. fuEufyf[kr esa ls dkSu lk mica/k lR; gS \
Q.55. fyfi fodklØe dh lkFkZd O;k[;k fuEu esa ls dkSu lk fodYi (a) mfR{kIr O;atu & i<+uk
djrk gS \ (b) vuqLokj & canj
(a) fp=fyfi → izrhdkRed fyfi → lw=fyfi → Hkkoewyd (c) g¡luk & vuqukfld
fyfi → Hkko/ofuewyd fyfi (d) mi;qZDr lHkh dFku lR; gSA
(b) lw=fyfi → fp=fyfi → izr hdkRed fyfi → [Ans] d
Hkkoewyd fyfi → Hkko/ofuewyd fyfi [SOL] lHkh dFku lR; gSA
(c) fp=fyfi → lw=fyfi → izr hdkRed fyfi → Q.61. lgh lqesfyr dhft, &
Hkkoewyd fyfi → Hkko/ofuewyd fyfi lwph&1 ¼O;atu ds izdkj½ lwph&2 ¼O;atu½
(d) fp=fyfi → lw=fyfi → Hkkoewy d fyfi → A. la?k’khZ 1- g
izrhdkRed fyfi → Hkko/ofuewyd fyfiA B. vuqukfld 2- j
[Ans] c C. ikf”oZd 3- .k
[SOL] fp=fyfi → lw=fyfi → izrhdkRed fyfi → Hkkoewyd D. yqafBr 4- y
fyfi → Hkko/ofuewyd fyfi dwV%
Q.56. fuEufyf[kr dFkuksa esa dkSu lk dFku vlR; gS \ A B C D A B C D
(a) /ofu;k¡ JO; :i esa O;Dr gksrh gSA (a) 1 4 3 2 (b) 1 3 4 2
(b) fyfi;k¡ n`”; :i esa O;Dr gksrh gSA (c) 4 1 2 3 (d) 4 2 1 3
(c) nsoukxjh fyfi dk ewy czkãh fyfi gSA [Ans] b
(d) Hkk’kk dh y?kqRre bdkbZ “kCn gSA [SOL] 1] 3] 4] 2 lgh gSA
[Ans] d Q.62. fuEukafdr esa ls “kq) “kCn dk p;u dhft, \
[SOL] Hkk’kk dh y?kqRre bdkbZ /ofu gSA (a) dSyk”k (b) ,jkor
Q.57. lqesfyr dhft, & (c) mTtoy (d) }kjdk
lwph&1 lwph&2 [Ans] d
A. gLo Loj 1- v [SOL] dSyk”k & dSykl] ,jkor & ,sjkor] mTtoy & mTToy]
B. la;qDr Loj 2- , Q.63. ifjPNsn “kCn dk lgh laf/k foPNsn D;k gS \
C. nh?kZ Loj 3- bZ (a) ij $ bPNsn (b) ifj $ Nsn
D. Iyqr Loj 4- vksme~ (c) ifj% $ Nsn (d) ijh $ Nsn
dwV% [Ans] b
A B C D A B C D [SOL] ifjPNsn ¾ ifj $ Nsn
(a) 2 1 4 3 (b) 1 2 4 3 Q.64. iou “kCn dk laf/k foPNsn D;k gS\
(c) 1 2 3 4 (d) 4 3 2 1 (a) iks $ vu
[Ans] c (b) ikS $ vu
[SOL] 1] 2] 3] 4 lgh gSA (c) iks $ ou
Q.58. fuEufyf[kr esa ls dkSu&lk vxz Loj gS& (d) buesa ls dksbZ ugha
(a) b (b) vk [Ans] a

DELHI IAS ACADMEY T-4 Page No.: 20

[SOL] iou ¾ iks $ vu ¼v;kfn Loj laf/k½ Q.72. la;qDr O;atu fdrus izdkj ds gksrs gSa\
Q.65. Li”kZ O;atuksa dh la[;k fdruh gksrh gS\ (a) 4 (b) 5
(a) 20 (b) 22 (c) 3 (d) 6
(c) 28 (d) buesa ls dksbZ ugha [Ans] a
[Ans] d [SOL] fganh o.kZekyk esa la;qä O;atuksa dh dqy la[;k 4 gS&
[SOL] Li”kZ O;atuksa dh la[;k 25 gksrh gS rFkk M+ vkSj <+ dks feykdj 1- K 2- =
27 gks tkrh gSA 3- {k 4- J
Q.66. lgh lqesfyr dhft, Q.73. fuEu esa ls dkSu lk o.kZ narks’B~; o.kZ gS \
lwph&1 ¼”kCn½ lwph&2 ¼lfU/k½ (a) ; (b) o
A. rnzwi 1- O;atu laf/k (c) j (d) y
B. i;ksn 2- folxZ laf/k [Ans] b
C. vUo; 3- ;.k Loj laf/k [SOL] narks’B~; o.kZ & o
D. ,dSd 4- o`f) Loj laf/k Q.74. fuEu esa ls dkSu iwohZ fgUnh dh miHkk’kk gS \
dwV% 1- vof/k 2- c?ksyh
A B C D A B C D 3- NÙkhlx<+h 4- dukSth
(a) 4 3 2 1 (b) 3 2 1 4 dwV %
(c) 4 2 3 1 (d) 1 2 3 4 (a) 1] 2 lR; gS (b) 1] 2] 3 lR; gS
[Ans] d (c) 1 lR; gS (d) 1] 2] 3] 4 lR; gS
[SOL] 1] 2] 3] 4 lgh lqesfyr gSA [Ans] b
Q.67. fuEufyf[kr “kCnksa esa ls lgh “kCn pqfu, [SOL] iwohZ fgUnh & vof/k] c?ksyh] NÙkhlx<+hA
(a) mfUeyhr (b) mUehyhu Q.75. dkSu lk “kCn “kq) gS \
(c) mUeyhr (d) mUehfyr (a) v{kksfg.kh (b) v{kkSfg.kh
[Ans] d (c) v{kksgh.kh (d) v{kkSghuh
[SOL] mUehfyr “kCn orZuh dh n`f’V ls “kq) gSA [Ans] b
Q.68. orZuh dh n`f’V ls “kq) “kCn gS \ [SOL] “kq) orZuh & v{kkSfg.khA
(a) fuekZsgh (b) u`eksgh
(c) fueksgh (d) fueksghZ Q.76. NÙkhlx<+ jktHkk’kk vk;ksx }kjk dk;kZy; LFkkiuk fnol dc
[Ans] a euk;k tkrk gS\
[SOL] fueksZgh dh orZuh “kq) gSA (a) 25 uoEcj
Q.69. fuEufyf[kr o.kksZ esa ls dkSu vYiizk.k ugha gS \ (b) 25 vxLr
(a) d (b) <+ (c) 14 vxLr
(c) M (d) buesa ls dksbZ ugha (d) 28 uoEcj
[Ans] b Ans: c
[SOL] <+ & egkizk.k Sol: vxLr] 2008 dks NÙkhlx<+ jktHkk’kk vk;ksx dk xBu gqvk vkSj
Q.70. fuEufyf[kr o.kZ esa ?kks’k o.kZ dkSu lk gS \ 14 vxLr] 2008 dks vk;ksx dh izFke dkjksckjh cSBd dh
(a) t “kq:vkr gSA blfy, bl fnol dks dk;kZy; LFkkiuk fnol ds
(b) M :i eas euk;k tkrk gSA
(c) x Q.77. fuEufyf[kr esa ls dkSu lk fodYi NÙkhlx<+h ds fodkl Øe dks
(d) mijksDr lHkh n”kkZrk gS \
[Ans] d (a) v/kZekx/kh  iwohZ fgUnh  NÙkhlx<+h
[SOL] rhuksa o.kZ ?kks’k o.kZ gSaA (b) iwohZ fgUnh  v/kZekx/kh  NÙkhlx<+h
Q.71. fuEufyf[kr esa ls dkSu lk o.kZ oRL;Z O;atu ugh gS \ (c) v/kZekx/kh  NÙkhlx<+  iwohZ fgUnh
(a) e~ (b) l~ (d) buesa ls dksbZ ugha
(c) u~ (d) y~ [Ans] a
[Ans] a [SOL] jktHkk’kk NÙkhlx<+h dk fodkl Øe &
[SOL] oRL;Z O;atu & u~] y~] l~] t~A oSfnd laLd`r → ykSfdd laLd`r → ikfy → izkd`r →

DELHI IAS ACADMEY T-4 Page No.: 21

 viHkza”k → v/kZekx/kh → iwohZ fgUnh → NÙkhlx<+h B. fodkl 2- jk;x<+
Q.78. NÙkhlx<+h Hkk’kk dkSu lh fyfi esa fy[kh tkrh gS \ C. NÙkhlx<+ lgdkj 3- jk;iqj
(a) czkãh (b) nsoukxjh D. NÙkhlx<+ lg;ksx 4- nqxZ
(c) [kjks’Bh (d) buesa ls dksbZ ugha dwV%
[Ans] b A B C D A B C D
[SOL] NÙkhlx<+h Hkk’kk nsoukxjh fyfi esa fy[kh tkrh gSA (a) 1 2 4 3 (b) 1 2 3 4
Q.79. lqesfyr djsa & (c) 2 1 3 4 (d) 2 1 4 3
lwph&1 ¼{ks=½ lwph&2 ¼Hkk’kk½ [Ans] d
A. mÙkjh NÙkhlx<+h 1- /kkdM+ [SOL] 2] 1] 4] 3 lgh lqesfyr gSA
B. if”peh NÙkhlx<+h 2- iaMks Q.85. Þviu ns”kß fdl lekpkj i= ls lacfa /kr gS \
C. iwohZ NÙkhlx<+h 3- ejkjh (a) ve`r lans”k (b) nSfud HkkLdj
D. nf{k.kh NÙkhlx<+h 4- dyaxk (c) uoHkkjr (d) gfjHkwfe
dwV% [Ans] a
A B C D A B C D [SOL] Þviu ns”kß ,d lkIrkfgd NÙkhlx<+h LraHk gS tks ve`r lans”k
(a) 3 1 2 4 (b) 3 2 4 1 uked lekpkj i= esa Nirk gSA blds lEiknd xksfoan yky
(c) 2 3 4 1 (d) 2 3 1 4 oksjk gSA
[Ans] c Q.86. NÙkhlx<+h cksyh] O;kdj.k vksj dks”k uked iqLrd fdlds }kjk
[SOL] 2] 3] 4] 1 lgh lqesfyr gSA fy[kh x;h gS \
Q.80. NÙkhlx<+h Hkk’kk dk izFke f”kykys[k fdl LFkku ls feyk gS \ (a) jatu yky ikBd (b) fouksn dqekj oekZ
(a) fcykliqj (b) tkatxhj pkaik (c) fou; dqekj ikBd(d) buesa ls dksbZ ugha
(c) lqdek (d) narsokM+k [Ans] d
[Ans] d [SOL] NÙkhlx<+h cksyh] O;kdj.k vksj dks”k uked iqLrd MkW- dkafrdqekj
[SOL] NÙkhlx<+h Hkk’kk dk lcls izkphu ,ao fyf[kr :i narsokM+k ds tSu }kjk fy[kh x;h gSA
f”kykys[k ij feyrk gSA Q.87. MkW- dkfUr dqekj tSu us NÙkhlx<+h dks HkkSxksfyd vk/kkj ij
Q.81. ÞNÙkhlx<+h ds milxZß uked iqLrd fdlus fy[kh gS \ fdrus Hkkxksa esa foHkkftr fd;k gS \
(a) ujsUnz dqekj lqn”kZu (a) 5 (b) 6
(b) MkW- lq/khu oekZ (c) 3 (d) 4
(c) MkW- jes”kpanz esgjks=k [Ans] c
(d) MkW- fouksn dqekj oekZ [SOL] MkW- dkfUr dqekj tSu us NÙkhlx<+h dks HkkSxksfyd vk/kkj ij ^3*
[Ans] a Hkkxksa esa foHkkftr fd;k gSA
[SOL] NÙkhlx<+h ds milxZ uked iqLrd ujsUnz dqekj lqn”kZu us Q.88. [kYVkgh cksyh fuEufyf[kr esa fdl ftys esa cksyh tkrh gS\
fy[kh gSA (a) fcykliqj (b) eaqxsyh
Q.82. Þfca>okjhß cksyh fdlds varxZr vkrh gS \ (c) dchj/kke (d) mijksDr lHkh
(a) dsUnzh; NÙkhlx<h+ (b) iwohZ NÙkhlx<+h [Ans] d
(c) mÙkjh NÙkhlx<+h (d) if”peh NÙkhlx<+ [SOL] ;g NÙkhlx<+ ds if”peh lhekar {ks=ksa esa cksyh tkus okyh ize[
q k
[Ans] b cksyh gS blesa fcykliqj] eax
q y
s h] dchj/kke] jktukanxkao ftys ds
[SOL] fca>okjh cksyh iwohZ NÙkhlx<+h ds varxZr vkrh gSA if”peh {ks= “kkfey gSA
Q.83. ÞmRFkkuß uked if=dk dc izdkf”kr dh xbZ Fkh \ Q.89. Þvkslkjhß “kCn NÙkhlx<+h esa dkSu&lh Hkk’kk ls vk;k gS\
(a) 1923 (b) 1925 (a) Hkkstiqjh (b) rsyxw
(c) 1935 (d) 1940 (c) vksfM+;k (d) caqnsyh
[Ans] c [Ans] c
[SOL] mRFkku uked if=dk dk izdk”ku jk;iqj ls 1935 bZ- esa lqn
a jyky [SOL] NÙkhlx<+h esa vksfM+;k ls izHkkfor “kCn & xksaM+k ¼xksaM½] vkslkjh]
f=ikBh }kjk fd;k x;k FkkA NsM+h ¼Nsjh½A
Q.84. lqesfyr dhft,%& Q.90. ejkBh izHkkfor NÙkhlx<+h :i dks D;k dgk tkrk gS \
lwph&1¼lekpkj i=½ lwph&2 ¼izdk”ku LFkku½ (a) gYch (b) cSxkuh
A. NÙkhlx<+ 1- fcykliqj (c) dyaxk (d) buesa ls dksbZ ugha

DELHI IAS ACADMEY T-4 Page No.: 22

[Ans] a Q.96. fuEufyf[kr esa ls dkSu lh cksyh dsUnzh; NÙkhlx<+h esa “kkfey
[SOL] ejkBh izHkkfor NÙkhlx<+h :i dks gYch dgk tkrk gSA ugha gS \
Q.91. fuEufyf[kr esa ls dkSu lh cksyh nzfoM+ Hkk’kk ifjokj esa “kkfey (a) [kSjkx<+h (b) jk;iqjh
ugha gS \ (c) dkadjs h (d) buesa ls dksbZ ugha
(a) ijth (b) xnck [Ans] d
(c) mjkao (d) nksjyh [SOL] mi;qZDr lHkh cksfy;k¡ dsUnzh; NRrhlx<+ esa “kkfey gSaA
[Ans] b Q.97. iafM+r ykspu izlkn ik.Ms; }kjk , xzkej vkWQ NÙkhlx<+h
[SOL] xnck vkfLVªd@nf{k.k@fu’kkn Hkk’kk ifjokj dh cksyh gSA Mk;yDV vkWQ fgUnh fdl o’kZ fy[kk x;k Fkk \
Q.92. Þe;k: ekVhÞ ds laiknd dkSu Fks \ (a) 1931 (b) 1941
(a) ujsUnz oekZ (b) lqanjyky “kekZ (c) 1911 (d) buesa ls dksbZ ugha
(c) lq”khy oekZ (d) uanfd”kksj frokjh [Ans] d
[Ans] c [SOL] iafMr ykspu izlkn ik.Ms; }kjk , xzkej vkWQ NÙkhlx<+h
[SOL] e;k: ekVh ds laiknd lq”khy oekZ FksA Mk;yDV vkWQ fgUnh 1921 esa fy[kk x;k FkkA
Q.93. izkphudky esa NÙkhlx<+h cksyh dks D;k dgk tkrk Fkk \ Q.98. Þyksdk{kjß uked NÙkhlx<+h if=dk ds lEiknd dkSu Fks\
(a) dkslyh (b) dksly (a) uUn fd”kksj frokjh
(c) vf/k’Bh (d) buesa ls dksbZ ugha (b) MkW- ikys”oj “kekZ
[Ans] a (c) fgeka”kq f}osnh
[SOL] NÙkhlx<+ {ks= igys nf{k.k dksly ds uke ls tkuk tkrk Fkk (d) ek;kjke lqjtu
vkSj NÙkhlx<+h cksyh dks ^dkslyh* dgk tkrk FkkA [Ans] a
Q.94. NÙkhlx<+h Hkk’kk ds fodkl esa ;ksxnku ds fy, fofHkUu lkfgR;dkjksa [SOL] yksdk{kj uked NÙkhlx<+h if=dk ds lEiknd uUn fd”kksj
dks lEekfur djus ds mís”; ls fdl o’kZ ls NÙkhlx<+h jktHkk’kk frokjh FksA
}kjk lEeku fn;k tk jgk gS \ Q.99. ^izdV* “kCn ds fy, NÙkhlx<+h esa D;k mi;ksx fd;k tkrk gS\
(a) 2012 (b) 2013 (a) ijxV (b) izfdV
(c) 2015 (d) buesa ls dksbZ ugha (c) fijxV (d) mijksDr lHkh
[Ans] d [Ans] a
[SOL] NÙkhlx<+h Hkk’kk ds fodkl esa ;ksxnku ds fy, fofHkUu lkfgR;dkjksa [SOL] ^izdV* “kCn ds fy, NÙkhlx<+h esa ijxV “kCn dk mi;ksx fd;k
dks lEekfur djus ds mís”; ls 2010 ls NÙkhlx<+h jktHkk’kk tkrk gSA
}kjk lEeku fn;k tk jgk gSA Q.100. NÙkhlx<+ jktHkk’kk vk;ksx ds v/;{kksa dks muds dk;Zdky ds
Q.95. lqesfyr djsa %& vuqlkj Øe esa tek,¡ &
lwph&1 lwph&2 1- nkus”oj “kekZ
¼NÙklhx<+h jktHkk’kk vk;ksx ds izkarh; ¼vk;ksftr LFkku½ 2- “;keyky prqosZnh
lEesyu½ 3- MkW- fou; dqekj ikBd
A. izFke 1- jk;iqj (a) 1] 2] 3 (b) 2] 1] 3
B. r`rh; 2- fHkykbZ (c) 3] 2] 1 (d) 2] 3] 1
C. iape 3- fcykliqj [Ans] b
D. f}rh; 4- jkfte [SOL] NÙkhlx<+ jktHkk’kk vk;ksx ds v/;{kksa dk Øe&
dwV% 1- “;keyky prqosZnh
A B C D A B C D 2- nkus”oj “kekZ
(a) 3 2 4 1 (b) 2 3 4 1 3- MkW- fou; dqekj ikBd
(c) 2 3 1 4 (d) 3 2 1 4
[Ans] b
[SOL] 2] 3] 4] 1 lgh lqesfyr gSA

DELHI IAS ACADMEY T-4 Page No.: 23

 DELHI IAS ACADEMY
STATE LEVEL EXAM - 2023
(Subject : Test-4, CSAT)
TIME - 2 hours TOTAL QSTN. -100 NEG. MARKING: 1/3

ANSWER KEY (16-11-23)

1 D 21 D 41 D 61 B 81 A
2 B 22 D 42 C 62 D 82 B
3 D 23 C 43 B 63 B 83 C
4 C 24 D 44 B 64 A 84 D
5 B 25 A 45 A 65 D 85 A
6 B 26 A 46 C 66 D 86 D
7 B 27 D 47 A 67 D 87 C
8 D 28 B 48 B 68 A 88 D
9 B 29 B 49 C 69 B 89 C
10 D 30 B 50 D 70 D 90 A
11 B 31 B 51 A 71 A 91 B
12 A 32 B 52 C 72 A 92 C
13 C 33 A 53 A 73 B 93 A
14 B 34 C 54 D 74 B 94 D
15 A 35 D 55 C 75 B 95 B
16 A 36 C 56 D 76 C 96 A
17 C 37 B 57 C 77 A 97 D
18 C 38 C 58 A 78 B 98 A
19 C 39 C 59 D 79 C 99 A
20 C 40 B 60 D 80 D 100 B
DELHI IAS ACADMEY T-4 Page No.: 24
 Rough Work

ekWMy vkalj osclkbV ij viyksM fd;k tk,xkA VsLV iqfLrdk % ewY; 30@&
Website : www.delhiias.com
DELHI IAS ACADMEY T-4 Page No.: 25