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Analytical Reasoning

Lesson 5 of 113

1. Ye topic hai kya?

Analytical Reasoning ka matlab hai: diye gaye facts ko jod kar woh nikalna jo seedha likha nahi hai. Calculator nahi, formula nahi — sirf logic.

Exam mein yeh do badi families mein aata hai. Dono padhni hain, kyunki papers kabhi puzzle dete hain, kabhi statement.

Family A — Arrangement / Puzzle

Log ek line, circle, building, din-vaar baithe hain. Conditions di hain. Poochha: kaun kiske left, kaun kis floor, kaun sa rang.

Family B — Statement analysis

Ek ya do sentences. Pooche: assumption, conclusion, argument, course of action, cause–effect. Diagram kam, soch zyada.

Pro level tab hai jab tum:

  • jo zaroor sach hai, jo ho sakta hai, aur jo kah nahi sakte — teen alag rakh sako
  • left-right facing se ulta na ho
  • negative condition (“C, A ke immediate right nahi”) ko correctly use karo
  • 2–3 possible diagrams mein se options se kaat sako

2. Har question pe dimag kaise chale

4 dabbe yaad rakh:
1) Fact — jo sentence ne pakka keh diya
2) Deduction — fact se jo 100% nikalta hai
3) Possibility — jo contradict nahi karta, lekin pakka bhi nahi
4) Cannot say — data kam hai
Basic · Mini

Example. Diya hai: “A, B ke left mein hai.” (same row, same facing — abhi simple.) Fact, deduction, possibility aur cannot say alag-alag kya hain?

Solution shuru

  1. Step 1 — fact

    Jo sentence ne pakka kaha: A, B ke left taraf hai. B, A se right taraf hai — yahi drawing ka base.

  2. Step 2 — deduction

    Jo 100% nikalta hai: B, A ke right mein hai. Naya log nahi joda, sirf direction palat di.

  3. Step 3 — possibility

    “Immediate” nahi likha. Beech mein log ho sakte hain — contradict nahi, pakka bhi nahi.

  4. Step 4 — cannot say

    Beech mein kitne log? Data nahi. Answer mat ghadna.

Final answer — yahan khatamFact: A left of B. Deduction: B right of A. Possible: beech mein log. Cannot say: beech kitne.

Pehle padho, turant mat baithao

  1. Saari conditions ek baar padho. Kitne log? Line / circle / floor? Facing kya?
  2. Sabse sakht condition pehle lagao (jo jagah almost fix kar de).
  3. Diagram banao. Khali boxes. Names baad mein.
  4. Negative conditions last mein check (woh galat diagram hataati hain).
  5. Question padho — “who sits” alag, “how many between” alag, “which is true” alag.
Galati: pehli condition se poori line bhar dena. Analytical mein aksar 2 valid sketches banti hain. Options dekho, jo dono sketches mein common ho wahi answer.

3. Question ki bhasha — har shabd ka matlab

Likha haiMatlab
Immediate left / rightBilkul bagal. Beech mein koi nahi.
Left / right (bina immediate)Ussi taraf kahin. Beech allowed.
Second to the leftUs taraf do seats. Beech wali chhodkar doosri.
A sits third to the right of BB se start: right1, right2, right3 = A. B aur A ke beech 2 log.
Between A and BA aur B ke beech wale. Order A-?-B ya B-?-A dono possible jab tak side na kaha ho.
Only two betweenBeech mein exactly 2.
At one of the endsLine ke do siron mein se ek.
Neighbours / adjacentImmediate dono taraf.
Does not sit adjacentBagal nahi. Doosri conditions se jagah nikaalo.
Facing north / centre / each otherLeft-right badal sakta hai. Neeche alag section.
Who among the followingOptions mein se. Extra log ho sakte hain diagram mein.
All are facing the same direction exceptEk aadmi ulta — uske left-right invert.

“Second to the left” ko count kaise karein

Baithe 5: P Q R S T sab north face.

P1
Q2
R3
S4
T5

R se left ki taraf: pehli seat immediate left, doosri seat “second to the left”. North-face + west=left end ho to R=3 ka second left = seat 1. Left–right ka poora kanoon agle hisse mein hai — pehle shabd ka count samajh lo.

4. Left–right ka kanoon (yahi pe log atakte hain)

Rule ek hi hai: left aur right us vyakti ke hisaab se, jo baitha hai — tumhare hisaab se nahi. Paper par drawing us hisaab se ghumaani padti hai.

Case 1 — Ek row, sab NORTH face (oopar ki taraf)

Unka muh north. Unka daahina = east, baaya = west. Agar tum diagram mein left side west rakho:

Westunke LEFT
A
B
C
Eastunke RIGHT

A ke right = B. C ke left = B. Simple, kyunki muh oopar hai aur drawing bhi west← →east hai.

Case 2 — Ek row, sab SOUTH face (neeche, jaise tumhari taraf)

Ab unka left = east, right = west. Drawing mein names left-to-right rakho to unke left-right palat jaate hain.

South face row ko north wali drawing jaisa treat mat karo. Ya to diagram paltao, ya har baar “uske muh se” socho.

Do rows, aamne-saamne

Row-1 north face, Row-2 south face. Jo saamne baitha hai woh “opposite”.

  • Opposite = seedha saamne wali seat, left-right nahi.
  • Ek row ka left doosri row ke left se match nahi karega, kyunki facing ulta hai.
Basic · Opposite

Example. Row-1 muh north: west se A B C. Row-2 muh south, equal seats, aligned: west se D E F. A, B, C kiske opposite?

Solution shuru

  1. Step 1 — alignment

    West end saamne west end. Same column = opposite. Left-right yahan use mat karo.

  2. Step 2 — map

    A (west) ke saamne D. B ke saamne E. C (east) ke saamne F.

Final answer — yahan khatamA opposite D, B opposite E, C opposite F.

Row-2 ka left-right alag hoga kyunki muh south hai — opposite phir bhi column se.

Circle, sab CENTRE face

Centre face: us vyakti ka LEFT = clockwise, RIGHT = anti-clockwise (upar se dekho to).

Verify: circle ke south point par khade, muh centre (north). Tumhara left = west. Clockwise south se west ki taraf jaata hai. Isliye left = clockwise.

Circle, sab bahar face

Outside face: LEFT = anti-clockwise, RIGHT = clockwise — centre wale ka ulta.

Kuch andar, kuch bahar

Har aadmi ke liye alag rule. Neighbour ka left-right mix ho jayega. Pro question yahi hota hai.

5. Linear arrangement — ghar se pro

n log, ek seedhi line. Pehle n boxes banao. Ends mark karo.

Sakht vs dhili condition

ConditionKitni jagah
A end par2 jagah (left end ya right end)
A immediate right of BBlock B A (same facing north)
Exactly one between A and BA _ B ya B _ A
A left of B (not immediate)A kisi bhi left slot, B uske right kisi slot

Block banao: agar “P, Q ke immediate left” to unhe ek sticker [P Q] ki tarah move karo. Yeh pro trick hai — do log ko alag-alag nahi, unit ki tarah baithao.

Basic

Q1. 5 log A, B, C, D, E ek row mein, sab north. A end par. B, A ka immediate neighbour. C beech (middle) mein. D, E ke immediate left. Kaun left end par?

Solution shuru

  1. Step 1 — boxes

    5 seats. West = 1, east = 5. Beech = seat 3.

    1
    2
    3
    4
    5
  2. Step 2 — C

    C beech mein ⇒ C = 3.

  3. Step 3 — A aur B, do cases

    A end par ⇒ A = 1 ya A = 5. B, A ka bagal.

    Case 1: A = 1, B = 2.

    Case 2: A = 5, B = 4.

  4. Step 4 — D, E block

    Muh north ⇒ D, E ke immediate left = D, E se west. Sticker [D E].

  5. Step 5 — case 1 bharo

    A=1, B=2, C=3. Khali 4–5. [D E] ⇒ D=4, E=5.

    A1
    B2
    C3
    D4
    E5

    Left end = A. Yeh case chalta hai.

  6. Step 6 — case 2 bharo

    A=5, B=4, C=3. Khali 1–2. [D E] ⇒ D=1, E=2.

    D1
    E2
    C3
    B4
    A5

    Left end = D. Yeh case bhi chalta hai. Koi case yahi khatam nahi hua.

  7. Step 7 — sawaal

    Left end case 1 mein A, case 2 mein D. Dono alag. Unique answer nahi.

Final answer — yahan khatamCannot determine / data inadequate.

Seekh: basic sawaal mein bhi do diagrams. Unique nahi to options mein yahi tick.

Basic

Q2. Q1 jaisa hi, extra line: “E, C ke immediate right nahi.” Ab left end kaun?

Solution shuru

  1. Step 1 — Q1 ke dono cases lao

    Case 1: A B C D E (E=5). Case 2: D E C B A (E=2). C dono mein seat 3.

  2. Step 2 — extra line ka matlab

    C=3 ka immediate right (north) = seat 4. “E, C ke immediate right nahi” ⇒ E ≠ 4.

  3. Step 3 — case 1 check

    E=5, 5 ≠ 4. Condition true. Case 1 zinda. Left end = A.

  4. Step 4 — case 2 check

    E=2, 2 ≠ 4. Condition true. Case 2 bhi zinda. Left end = D.

  5. Step 5 — dono zinda

    Yeh extra line kisi case ko nahi maarti. Left end phir bhi A ya D.

Final answer — yahan khatamPhir bhi cannot determine.

Seekh: extra line ka kaam case maarna hota hai. Agar “E end par nahi” hota, case 1 mar jaata (E=5 end), tab left end = D.

Positions ko number do

Left end = 1, right = n (north face convention). “Third to the right of 2” = position 5. Formula:

  • North face: right = +1 index, left = −1 index (agar 1 left end)
  • “kth to the right of X” = index(X) + k
  • Range se bahar = impossible case

6. Do rows — aamne saamne

6+6 ya 5+5. Pehle opposite pairs fix karo — yeh sabse sakht hote hain.

Medium · Pattern

Example. 6 log. Row-1 (muh north): A, B, C. Row-2 (muh south): D, E, F. Har ke saamne ek. A ke opposite E nahi. B, C ke immediate left (C ki nazar se). Row-1 west se east kya-kya cases?

Solution shuru

  1. Step 1 — seats

    West se east: 1 2 3. Same number = opposite. Row-1 muh north ⇒ unka left = west.

  2. Step 2 — B, C block

    B, C ke immediate left ⇒ B, C se west. Sticker [B C].

    Case 1: B=1, C=2, A=3.

    Case 2: B=2, C=3, A=1.

    Block 3–4 jaisa kuch nahi — sirf 3 seats.

  3. Step 3 — “A opposite E nahi”

    Case 1: A=3 ⇒ F ya D opposite A ho sakte; E ≠ 3 (E seat 1 ya 2).

    Case 2: A=1 ⇒ E ≠ 1 (E seat 2 ya 3).

  4. Step 4 — yahan rukna

    Itne data se poori 6 names unique nahi. Row-1 ke 2 cases lock. Row-2 extra lines se kaato.

Final answer — yahan khatamRow-1: B C A ya A B C. E, A ke saamne nahi.

Pehle opposite pairs, phir har row ka left-right uske muh se. Arrow ↑ ↓ likho.

Row2 ko paper par upar likho ya neeche — farq nahi, par facing mark zaroor karo (↑ ↓). Bina arrow ke left-right galat hoga.

Pro: agar “X ke saamne wala, Y ka neighbour hai” to pehle X–opposite find, phir Y uske bagal. Yeh 3 logon ka chain banata hai.

7. Circular arrangement

n log circle. Pehla aadmi kahi bhi baith sakta hai — circle ka rotation same arrangement hai. Isliye ek log ko fix kar do (usually jo condition mein sabse zyada baar aaye, ya jo “right of” chain start kare).

Fix one person at the top of your sketch. Baaki relative. Time bachega, duplicate diagrams nahi.

8 log, centre face — standard

Har ke 2 neighbours. “Opposite” = n/2 seats door (8 mein 4 door, beech 3).

Clockwise names likho. Centre-face: clockwise = LEFT. To “immediate left of A” = clockwise next.

Medium

Q3. 6 log A–F circle, sab centre. A ke opposite D. B, A ke immediate right. F, A ke immediate left. C, D ke immediate right nahi. E, D ka neighbour. Clockwise order mein B ke turant baad kaun?

Solution shuru

  1. Step 1 — fix + opposite

    6 log. Opposite = 3 seats door. A ko seat 1. D opposite ⇒ D = 4.

  2. Step 2 — facing

    Centre: left = clockwise, right = anti-clockwise.

  3. Step 3 — B aur F

    B, A ke immediate right ⇒ anti-clockwise of 1 = 6. B = 6.

    F, A ke immediate left ⇒ clockwise of 1 = 2. F = 2.

  4. Step 4 — khali seats

    Bhari: 1=A, 2=F, 4=D, 6=B. Khali: 3 aur 5. Bache: C, E. Dono D ke neighbour hain (3 aur 5).

  5. Step 5 — “C, D ke immediate right nahi”

    D=4 ka immediate right = anti-clockwise = seat 3. C ≠ 3 ⇒ C = 5, E = 3.

  6. Step 6 — clockwise after B

    Clockwise: A → F → E → D → C → B → A.

    B=6 ke turant baad clockwise = seat 1 = A.

    A1
    F2
    E3
    D4
    C5
    B6
Final answer — yahan khatamB ke turant baad clockwise = A.

Seekh: ek log fix + opposite + facing se almost circle fill. “Nahi” last mein case kaatti hai.

Square / rectangle table

4 corners + kabhi mid-sides. Corner wale ke neighbours 2, lekin “left” facing andar/bahar se. Same circle logic: andar face ≈ centre. Corner vs side alag “who sits at corner” questions.

8. Floor puzzle aur din-vaar

Building: floor 1 neeche, highest upar (India papers usually 1 = ground/lowest). Kabhi ulta likha ho to pehle padhna.

  • “A, B se immediately above” = A ka floor = B + 1
  • “Two floors between A and B” = |floorA − floorB| = 3 (beech 2 floors)
  • Careful: between 2 floors vs 2 people between

Week: Mon–Sun. “A, B ke immediately after” = agla din. Gap wale din empty nahi chhodna jab 7 log 7 din hon.

Medium · Floor

Example. 7 floors, 7 log. “Only two floors between P and Q. P, Q se upar.” P aur Q ke kitne possible pairs?

Solution shuru

  1. Step 1 — formula

    Beech mein exactly 2 floors ⇒ |P − Q| = 3. P upar ⇒ P = Q + 3.

  2. Step 2 — Q chhota, list

    Q=1 → P=4. Q=2 → P=5. Q=3 → P=6. Q=4 → P=7.

    Q=5 → P=8 — floor nahi. Yahi khatam.

  3. Step 3 — exam mein aage

    Yahan 4 pairs. Doosri conditions se kaato jab tak ek pair bache.

Final answer — yahan khatam(Q,P) = (1,4), (2,5), (3,6), (4,7). Char cases.

9. Mix puzzle — log × city × rang × job

Yeh pro level ka core hai. Table banao, diagram akela kaafi nahi.

PersonCityColorDay
A
B

Har condition ek cell fill karti hai ya do cells ko jodti hai. “Jo Jaipur mein hai woh red nahi pehenta” = Jaipur row mein Color ≠ red.

Elimination grid: possible ticks, phir cross. Jahan ek tick bache, lock. Yeh CAT/Bank level bhi yahi hai, LDC mein chhota set (5×5).

Jo cheez sirf do logon mein se ek ke paas ho sakti hai, usse 50-50 case banao. Baaki conditions se ek case mar jaata hai. Dono live hon to question “which cannot be true” options se solve.
Medium · Mix

Example. 3 log A, B, C. Shehar: Jaipur, Kota, Ajmer. Rang: red, blue, green.

Conditions

  1. A Kota mein hai.
  2. Kota wala red nahi.
  3. B green pehenta hai.
  4. C Ajmer nahi.

Sawaal: Jaipur kaun, uska rang kya?

Solution shuru

  1. Step 1 — locks

    A = Kota. B = green. Kota ≠ red ⇒ A red nahi.

  2. Step 2 — A ka rang

    A green nahi (green B ke paas). A red nahi. Bachaa blue. A = Kota + blue.

  3. Step 3 — C, Jaipur

    Bache shehar: Jaipur, Ajmer. C Ajmer nahi ⇒ C = Jaipur, B = Ajmer.

  4. Step 4 — C ka rang

    Blue aur green nikal gaye. Bachaa red. C = Jaipur + red.

Final answer — yahan khatamJaipur: C. Rang: red.

10. Statement analysis — doosri family

Yahan seating nahi. Ek statement ko “todna” hai. LDC/SSC mein yeh bhi Analytical ke under aata hai.

10.1 Assumption (maan lena)

Assumption woh hidden fact hai jiske bina statement ka matlab adhura / bekaar ho jaye. Statement ke peeche ka vishwas.

  • Valid: bina iske speaker ka point toot jata
  • Invalid: naya moral, extreme (“hamesha”, “sirf”), ya statement se aage ki duniya
Basic

Q4. Statement: “Sarak par helmet lagao, dand se bachoge.”
I. Police helmet check karti hai.
II. Helmet se chot kam hoti hai.
Kaun si assumption valid?

Solution shuru

  1. Step 1 — statement ka daav

    Speaker ka point: helmet = dand se bacho. Safety/chot ka zikr nahi.

  2. Step 2 — I

    Dand tab lagta hai jab koi check kare. Bina check ke “dand se bacho” bekaar. I statement ke peeche chhupi hai. Valid.

  3. Step 3 — II

    Helmet se chot kam — acchi baat, lekin is statement ne yeh reason use hi nahi kiya. Dusra accha reason assumption nahi. II invalid.

Final answer — yahan khatamSirf I valid.

Seekh: assumption isi reason ko follow kare. Agar statement “suraksha ke liye helmet” hoti, to II valid hota.

10.2 Conclusion

Conclusion woh baat jo statements se zaroor nikalti ho. Naya fact mat ghusao.

  • “Some A are B. All B are C.” ⇒ Some A are C. (standard, overlapping)
  • Yahan full syllogism Venn se — woh bhi analytical deduction hai. Odd-one/series alag topic, lekin statement-conclusion yahan.
  • “Only” = “all reverse”. Only A are B ⇒ All B are A.
“Can be concluded” vs “probably true”. Exam chahta hai definite. News-jaisa guess mat maro.

10.3 Strong / weak argument

Question: “Should X happen?” Arguments for/against.

  • Strong: topic se directly, practical, facts/law/wide impact
  • Weak: emotion, example of one person, irrelevant comparison, extreme
Basic · Argument

Example. “Should plastic bag ban ho?”
A) Haan, naali band hoti hai aur pashu marte hain.
B) Haan, mujhe bags pasand nahi.
Kaun strong, kaun weak?

Solution shuru

  1. Step 1 — A

    Naali + pashu: topic se direct, public impact. Strong.

  2. Step 2 — B

    Personal pasand. Policy ke liye weak.

Final answer — yahan khatamA strong. B weak.

10.4 Course of action

Problem di, action socho.

  • Valid: practical, problem ke root se juda, legally theek, immediately useful
  • Invalid: overreact (poora shahar band), vague (“logon ko sochna chahiye”), problem se katin

Dono actions valid ho sakte hain independently. “Either” tab jab dono ek saath na chal saken.

10.5 Cause and effect

Do events. Pooche: 1 cause of 2, 2 cause of 1, independent, common cause.

Time order dekho. Correlation ko cause mat banao. “Barish aayi, sales giri” — ho sakta independent (festival off).

11. Eligibility / decision making

Criteria list + candidate data. Conditions: age ≥ 21, marks ≥ 60%, experience 2 yr, … Relaxation: SC age +5, etc.

  1. Har candidate ke liye table: har criterion Pass/Fail/Relax
  2. Agar ek bhi compulsory fail aur relaxation nahi → reject
  3. Agar data missing → “data inadequate”
  4. Agar sab pass → select; kuch refer to director — jab question aisa code de

Codes (a)(b)(c) yaad mat ratna — har set naya hota hai. Logic same: checklist.

Basic · Eligibility

Example. Criteria: umar ≥ 21, marks ≥ 60%. OBC ko umar +3. Candidate Ravi: 20 saal, 62%, OBC. Decision?

Solution shuru

  1. Step 1 — table

    Umar: 20. Marks: 62. Category: OBC (relaxation milegi).

  2. Step 2 — marks

    62 ≥ 60. Pass. Relaxation ki zaroorat nahi.

  3. Step 3 — umar

    20 ≥ 21? Nahi. OBC +3 ⇒ 23 ≥ 21. Pass.

  4. Step 4 — missing data?

    Koi field khali nahi. Dono compulsory pass.

Final answer — yahan khatamSelect.

Agar marks 58 hote aur relaxation sirf umar par, to reject. Data na ho to “inadequate”. Code (a)(b) is set par mat chipkao.

12. Input–output

Machine line rearrange — poora topic Input–output pe. Yahan arrangement + statement. Last sorted line se input wapas unique nahi milta — wahan trap.

13. Pro checklist (exam hall)

  1. Logon ki ginti = seats? Agar 8 log 8 seats, khaali nahi. Agar 6 log 8 seats, empty mark karo.
  2. Facing sentence skip mat karo — pehli line mein chhupa hota hai.
  3. “Not” wali lines last, lekin unhe likh ke rakho taaki case mar sako.
  4. Do cases parallel chalao, kaatna mat bhulo. Time: 8-log circle ~ 4–6 min LDC speed.
  5. Agar 3 questions ek set par hain: pehle poora diagram, phir 3 answers. Diagram galat to 3 galat — isliye diagram pe 70% time.
  6. Statement questions mein outside knowledge band. “Rajasthan ki sarkar…” current affair nahi, sirf diya text.
  7. Cannot determine / none of these — jab dono cases alag answers den.

14. Practice — basic se pro, poora solution

Pehle khud socho, phir neeche solution kholo. Upar Q1–Q4 theory ke saath; yahan Practice 1–10.

Basic · Linear

Practice 1. P, Q, R, S, T ek bench par baithe hain. Sabka muh north. R beech mein hai. P aur T dono ends par hain. Q, P ke immediate right mein hai. S kahan hai?

Solution shuru

  1. Step 1 — boxes

    5 seats. Left end = west = seat 1. Right end = east = seat 5. Beech = seat 3.

    1
    2
    3
    4
    5
  2. Step 2 — pakka fact

    R beech mein ⇒ R = seat 3.

  3. Step 3 — do cases

    P aur T ends par. Do possibilities:

    Case A: P = 1, T = 5

    Case B: T = 1, P = 5

  4. Step 4 — “Q, P ke immediate right”

    Muh north ⇒ unka right = east = bari seat number.

    Case B: P = 5 (right end). Seat 5 ke right mein koi seat nahi. Case B yahi khatam.

    Case A: P = 1 ⇒ Q = 2. T = 5. Bachi seat 4 = S.

  5. Step 5 — check
    P1
    Q2
    R3
    S4
    T5

    R beech. P, T ends. Q, P ke bagal right. Sab fit.

Final answer — yahan khatamS seat 4 par hai — R aur T ke beech.
Basic · Conclusion

Practice 2. Statements: All pens are books. Some books are files.
Conclusions: (i) Some pens are files. (ii) Some books are pens. Kaun sa follow karta hai?

Solution shuru

  1. Step 1 — drawing

    Saari pens books ke andar hain. Kuch books files se overlap karti hain. Yeh overlap pens wale hisse mein ho bhi sakta hai, nahi bhi.

  2. Step 2 — conclusion (i)

    “Some pens are files” tabhi pakka jab pens aur files ka cut zaroori ho. Yahan zaroori nahi. (i) follow nahi karta.

  3. Step 3 — conclusion (ii)

    “All pens are books” se wapas: kuch books pens hain. Exam nonempty maanta hai. (ii) follow karta hai.

Final answer — yahan khatamSirf (ii).
Medium · Circle

Practice 3. 6 log U, V, W, X, Y, Z circle mein, sabka muh centre. U ke immediate right V hai. W, V ke second left hai. X, W ke opposite hai. Y, X ka neighbour hai, Z nahi. Sawaal: Z kiske immediate left hai? (Matlab: Z kis ke left-bagal mein baitha hai?)

Solution shuru

  1. Step 1 — rules

    6 log. Opposite = 3 seats door.

    Centre face: left = clockwise, right = anti-clockwise.

  2. Step 2 — U aur V

    U ko seat 1 fix. V, U ke immediate right ⇒ anti-clockwise of 1 = seat 6. V = 6.

  3. Step 3 — W

    W = V ka second left. Left = clockwise.

    V=6 se clockwise pehli seat = 1 (U). Clockwise doosri = 2. W = 2.

  4. Step 4 — X

    X, W ke opposite. W=2 ka opposite = 5. X = 5.

    Ab bhari: 1=U, 2=W, 5=X, 6=V. Khali: 3 aur 4. Bache: Y, Z.

  5. Step 5 — Y, Z

    X=5 ke neighbours = 4 aur 6. 6 = V. To Y = 4. Bachi 3 = Z.

    U1
    W2
    Z3
    Y4
    X5
    V6

    Clockwise: U → W → Z → Y → X → V.

  6. Step 6 — sawaal

    “Z kiske immediate left hai?” = Z kis ke left mein hai.

    P ka left = Z hona chahiye ⇒ clockwise P ke baad Z. Z=3 to P=2 = W.

    Check: W=2 ka left = 3 = Z. Theek.

Final answer — yahan khatamZ, W ke immediate left hai. Answer: W.

Agar likha hota “Z ke immediate left kaun hai?” to answer Y hota.

Medium · Floor

Practice 4. 5 floors. Floor 1 sabse neeche, 5 sabse upar. A, B, C, D, E. A upar se doosre floor par. B aur C ke beech sirf ek floor. B, C se upar. D lowest nahi. C lowest nahi. E, A ke just neeche nahi. D kis floor par hai?

Solution shuru

  1. Step 1 — A

    Upar se doosra = floor 4. A = 4.

  2. Step 2 — B aur C

    B, C se upar + beech exactly 1 floor ⇒ B = C + 2.

    Possible: (B,C) = (5,3) ya (3,1). (4,2) nahi, kyunki 4 par A hai.

  3. Step 3 — C lowest nahi

    C ≠ 1. Pair (3,1) yahi khatam. Bachaa: B = 5, C = 3.

  4. Step 4 — D aur E

    Khali: 2 aur 1. D lowest nahi ⇒ D ≠ 1 ⇒ D = 2, E = 1.

  5. Step 5 — check

    A=4 ka just neeche = 3. Wahan C hai, E=1. “E, A ke just neeche nahi” OK.

    Upar se: 5=B, 4=A, 3=C, 2=D, 1=E.

Final answer — yahan khatamD floor 2 par hai.
Pro · Do rows

Practice 5. 8 log. North row: A, B, C, D — muh south. South row: E, F, G, H — muh north. Har ke saamne ek.

Conditions

  1. A ke opposite F nahi.
  2. B, A ke immediate left (A ki nazar se).
  3. H, B ke opposite.
  4. G, H ke immediate right (H ki nazar se).
  5. C kisi end par. D, C ka neighbour.

Sawaal: E kiske opposite hai?

Solution shuru

  1. Step 1 — seats

    West se east: seat 1 2 3 4. Same number = opposite.

    North row muh south: unka left = east (bari number).

    South row muh north: unka right = east (bari number).

  2. Step 2 — A, B

    B, A ke immediate left ⇒ B, A ke east. West-se-east: A phir B.

    Yeh block 1–2, 2–3, ya 3–4.

  3. Step 3 — H, G

    H, B ke opposite ⇒ H ka number = B.

    G, H ke immediate right ⇒ G, H ke east ⇒ H phir G.

    H seat 4 par nahi ho sakta. Isliye B ≠ 4. Block 3–4 yahi khatam.

  4. Step 4 — case 1–2

    A=1, B=2, H=2, G=3. C end: 1 already A, to C=4. D neighbour ⇒ D=3.

    North: A B D C. South khali 1 aur 4 = E, F.

    F, A ke opposite nahi ⇒ F ≠ 1 ⇒ F=4, E=1.

    E, A ke opposite. Yeh case chal gaya.

  5. Step 5 — case 2–3

    A=2, B=3, H=3, G=4. North khali 1 aur 4. C end par 1 ya 4.

    C=1 ⇒ D=2, lekin 2=A. Band. C=4 ⇒ D=3, lekin 3=B. Band.

    Yeh case yahi khatam.

  6. Step 6 — diagram

    North (muh south): A · B · D · C

    South (muh north): E · H · G · F

Final answer — yahan khatamE, A ke opposite hai.
Pro · Mix table

Practice 6. 5 log: Ram, Sham, Tom, Urvashi, Veena. Shehar: Jaipur, Kota, Ajmer, Udaipur, Bikaner. Rang: red, blue, green, yellow, white.

Conditions

  1. Ram Jaipur nahi, red nahi.
  2. Sham Kota, blue nahi.
  3. Tom green, Udaipur nahi.
  4. Urvashi yellow.
  5. Ajmer wala white nahi.
  6. Veena Bikaner nahi.
  7. Jaipur wala blue.
  8. Urvashi Udaipur nahi.

Sawaal: Udaipur kaun, uska rang kya?

Solution shuru

  1. Step 1 — seedhe locks

    Sham = Kota. Tom = green. Urvashi = yellow. Jaipur = blue.

  2. Step 2 — Jaipur-blue kaun

    Sham Kota mein. Tom green. Urvashi yellow. Ram Jaipur nahi.

    Bachaa Veena. Veena = Jaipur + blue.

  3. Step 3 — Ram ka rang

    Blue, green, yellow nikal gaye. Ram red nahi. Bachaa white. Ram = white.

    Ajmer white nahi ⇒ Ram Ajmer nahi.

  4. Step 4 — Udaipur

    Udaipur nahi ja sakte: Tom, Urvashi, Sham (Kota), Veena (Jaipur).

    Bachaa Ram. Ram = Udaipur + white.

Final answer — yahan khatamUdaipur: Ram. Rang: white.
Basic · Assumption

Practice 7. Statement: “Internet sasta karo taaki gaon mein padhai badhe.”
I. Gaon mein internet hai, lekin mehnga hai.
II. Sasta internet padhai badhata hai.
Kaun si assumption valid?

Solution shuru

  1. Step 1

    Statement ke do hisse: sasta karo + padhai badhe. Dono ke peeche hidden fact chahiye.

  2. Step 2 — I

    Agar net hi na ho ya pehle se sasta ho, to “sasta karo” bekaar. I valid.

  3. Step 3 — II

    “Taaki padhai” tabhi chalti hai jab sasta net padhai se juda ho. II valid.

  4. Step 4

    “Sirf internet se padhai” nahi kaha — itna extreme mat jodna.

Final answer — yahan khatamI aur II dono valid.
Medium · Course of action

Practice 8. Problem: Sheher mein dengue.
A) Saari flights band.
B) Larval inspection aur fogging.
C) Pamphlet se jagrukta.
Kaun se action valid?

Solution shuru

  1. Step 1 — A

    Dengue local machhar se. Flights band karna problem se kati hui. A invalid.

  2. Step 2 — B

    Larvae + fogging seedha control. Practical. B valid.

  3. Step 3 — C

    Jagrukta se pani jama kam. Problem se juda. C valid.

Final answer — yahan khatamB aur C. A nahi.
Pro · Eligibility

Practice 9. Criteria: umar ≥ 22, degree = graduate, Hindi pass. SC/ST ko umar −2. Meena: 21 saal, graduate, Hindi pass, ST. Decision?

Solution shuru

  1. Step 1 — table

    Umar 21. Degree haan. Hindi haan. ST ⇒ umar 22−2=20 tak chalegi.

  2. Step 2

    21 ≥ 20. Dono compulsory fields bhari. Select. Machine Input–output alag page.

Final answer — yahan khatamSelect.
Pro · Mixed facing

Practice 10. 6 log A, B, C, D, E, F circle. Muh alternate (ek centre, agla bahar…). A ka muh centre. Clockwise A ke baad B, phir C. B ka muh bahar. C ke opposite kaun hai, uska muh kahan?

Solution shuru

  1. Step 1 — order

    Clockwise: A=1, B=2, C=3, D=4, E=5, F=6.

  2. Step 2 — facing

    A centre, B bahar, alternate: 1 centre, 2 bahar, 3 centre, 4 bahar, 5 centre, 6 bahar.

  3. Step 3 — opposite

    6 log, opposite = +3. C=3 ka opposite = 6 = F. Seat 6 ka muh = bahar.

  4. Step 4 — yaad

    Agar “B ke left kaun” poochhein to B bahar hai: B ka left = anti-clockwise. A ka rule mat lagana.

Final answer — yahan khatamC ke opposite F. F ka muh bahar.

15. Itna kaafi hai kya?

Agar tum:

  • 3 dabbe (fact / deduction / possibility) alag kar sako
  • north-south aur centre-out left-right galti ke bina laga sako
  • do cases chala ke extra condition se ek maar sako
  • statement mein extra knowledge na ghusao

to LDC Pre/Mains ke analytical sawaal — seating, floor, mix table, assumption, course of action — basic se pro tak attempt ho sakte hain. Machine line Input–output pe.

Agle topic: Number series.

← Index · Reasoning · 1/Analytical Reasoning · RVUN JA/CA-II 2026