If Batuk Singh decides to plant the wheat next to the peanuts, in which square will the barley be? Options: A) The square immediately north of the rice B) The square immediately east of the rice C) The square immediately west of the rice D) The square immediately north east of the rice E) The square immediately north west of the rice

D. The square immediately north east of the rice

Which square can be planted with wheat? Options: A) The square immediately north of the rice B) The square immediately west of the rice C) The square immediately south of the rice D) The square immediately south west of the rice E) The square immediately north west of the rice

A. The square immediately north of the rice

Matrix based Puzzles Questions with Solutions – MBA CET 2025–2026 (43 Qs)

Q71MediumCET 2026 · Slot 5 · Q71

Read the information given below carefully and answer the question that follows :

25 coloured beads are to be arranged in a grid comprising of 5 rows and 5 columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green.

While arranging the beads along any of the 5 rows or along any of the 5 columns, the rules given below are to be followed :

1. Two adjacent beads along the same row or column are always of different colours.
2. There is at least one Green bead between any two Blue beads along the same row or column.
3. There is at least one Blue and at least one Green bead between any two Red beads along the same row or column.
4. Every unique, complete arrangement of twenty five beads is called a configuration.

Question :

What is the total number of possible configurations using beads of only two colours ?

25
24
6
2
0

Show answerHide answer

Answer: D official key

Approach: Test each pair of colours against the rules along a single row, then extend to the grid.

Solution:
Rule 1 forces adjacent beads to differ, so a two-colour row must alternate.
Red + Blue: R B R — two Reds with no Green between ✗.
Red + Green: R G R — two Reds with no Blue between ✗.
Blue + Green: B G B G B — every pair of Blues has a Green between ✓ (no rule for two Greens).
So only Blue/Green works, and every row and every column must alternate → a chessboard pattern.
A chessboard is fixed by the colour of the top-left bead: Blue or Green → 2 configurations.

Why the other options are wrong:
E (0) ignores the Blue–Green chessboard; A, B, C count arrangements the alternation rule does not allow.

Answer: D

Q35EasyCET 2026 · Slot 3 · Q35

Five birds are sitting on a tree. The Pigeon is to the right of the Parrot. The Sparrow is above the Parrot. The Crow is next to the Pigeon. The Crane is below the Crow. Which bird is at the centre ?

Pigeon
Crow
Crane
Sparrow
Parrot

Show answerHide answer

Answer: A official key

Approach: Place the birds on a small grid.

Solution:
Pigeon is right of Parrot → Parrot, Pigeon.
Crow is next to Pigeon (on the other side) → Parrot, Pigeon, Crow in one row.
Sparrow is above Parrot; Crane is below Crow.
Pigeon is the middle of the main row and of the whole figure.

Sparrow
Parrot	Pigeon	Crow
		Crane

Answer: A

Q65MediumCET 2025 · Slot 6 · Q65

Passage / data for this question

At a small company, parking spaces are reserved for the top executives: CEO, president, vice president, secretary, and treasurer-with the spaces lined up in that order. The parking lot guard can tell at a glance if the cars are parked correctly by Looking at the colour of the cars. The cars are yellow, green, purple, red and blue, and the executives' names are Ajay, Bipin, Chetan, Damodar, and Erik

the car in the first space is red
a blue car is parked between the red car and the green car
The car in the last space is purple
the secretary drives a yellow car
ajay's car is parked next to damodar's
Erik drives a green car
Bipin’s car is parked between chetan’s and erik's
damodar's car is parked in the last space

Which color is vice president’s car?

Green
Yellow
Blue
Purple
Red

Show answerHide answer

Answer: A

Approach: 5-slot matrix arrangement — place colours first, then names.

Solution: Space 1 = red, space 5 = purple. Secretary (space 4) = yellow. Blue is between red and green, so blue = 2, green = 3. Erik drives green → space 3. Damodar = space 5, Ajay next to him → space 4. Bipin between Chetan and Erik → Bipin 2, Chetan 1.

Space	Post	Car	Executive
1	CEO	Red	Chetan
2	President	Blue	Bipin
3	Vice President	Green	Erik
4	Secretary	Yellow	Ajay
5	Treasurer	Purple	Damodar

So the Vice President (space 3) has the green car.

Why the other options are wrong: Blue is the President's, yellow the Secretary's, purple the Treasurer's, red the CEO's.

Cetking shortcut: Start with the two fixed ends (red, purple) and the secretary's yellow — the blue/green order then falls out immediately.

Answer: A

Q66MediumCET 2025 · Slot 6 · Q66

Passage / data for this question

At a small company, parking spaces are reserved for the top executives: CEO, president, vice president, secretary, and treasurer-with the spaces lined up in that order. The parking lot guard can tell at a glance if the cars are parked correctly by Looking at the colour of the cars. The cars are yellow, green, purple, red and blue, and the executives' names are Ajay, Bipin, Chetan, Damodar, and Erik

the car in the first space is red
a blue car is parked between the red car and the green car
The car in the last space is purple
the secretary drives a yellow car
ajay's car is parked next to damodar's
Erik drives a green car
Bipin’s car is parked between chetan’s and erik's
damodar's car is parked in the last space

who is the CEO

Ajay
Bipin
Chetan
Damodar
Erik

Show answerHide answer

Answer: C

Approach: 5-slot matrix arrangement — place colours first, then names.

Solution: Space 1 = red, space 5 = purple. Secretary (space 4) = yellow. Blue is between red and green, so blue = 2, green = 3. Erik drives green → space 3. Damodar = space 5, Ajay next to him → space 4. Bipin between Chetan and Erik → Bipin 2, Chetan 1.

Space	Post	Car	Executive
1	CEO	Red	Chetan
2	President	Blue	Bipin
3	Vice President	Green	Erik
4	Secretary	Yellow	Ajay
5	Treasurer	Purple	Damodar

So the CEO (space 1) is Chetan.

Why the other options are wrong: Bipin is President, Erik VP, Ajay Secretary, Damodar Treasurer.

Cetking shortcut: Start with the two fixed ends (red, purple) and the secretary's yellow — the blue/green order then falls out immediately.

Answer: C

Q67MediumCET 2025 · Slot 6 · Q67

Passage / data for this question

At a small company, parking spaces are reserved for the top executives: CEO, president, vice president, secretary, and treasurer-with the spaces lined up in that order. The parking lot guard can tell at a glance if the cars are parked correctly by Looking at the colour of the cars. The cars are yellow, green, purple, red and blue, and the executives' names are Ajay, Bipin, Chetan, Damodar, and Erik

the car in the first space is red
a blue car is parked between the red car and the green car
The car in the last space is purple
the secretary drives a yellow car
ajay's car is parked next to damodar's
Erik drives a green car
Bipin’s car is parked between chetan’s and erik's
damodar's car is parked in the last space

who is secretary?

Erik
Damodar
Chetan
Bipin
Ajay

Show answerHide answer

Answer: E

Approach: 5-slot matrix arrangement — place colours first, then names.

Solution: Space 1 = red, space 5 = purple. Secretary (space 4) = yellow. Blue is between red and green, so blue = 2, green = 3. Erik drives green → space 3. Damodar = space 5, Ajay next to him → space 4. Bipin between Chetan and Erik → Bipin 2, Chetan 1.

Space	Post	Car	Executive
1	CEO	Red	Chetan
2	President	Blue	Bipin
3	Vice President	Green	Erik
4	Secretary	Yellow	Ajay
5	Treasurer	Purple	Damodar

So the Secretary (space 4) is Ajay.

Why the other options are wrong: Erik is VP, Damodar Treasurer, Chetan CEO, Bipin President.

Cetking shortcut: Start with the two fixed ends (red, purple) and the secretary's yellow — the blue/green order then falls out immediately.

Answer: E

Q23MediumCET 2025 · Slot 5 · Q23

Passage / data for this question

Read the following information and answer the given questions

Farmer Batuk Singh has a larger square field divided into nine smaller square fields, all equal, arranged in three rows of three fields each. One side of the field runs exactly east-west. The middle square must be planted with rice because it is wet. The wheat and barley should be continuous so that they can be harvested all at once by a mechanical harvester. Two of the fields should be planted with soybeans. The northwestern most field should be planted with peanuts and the southern third of the field is suitable only for vegetables.

If Batuk Singh decides to plant the wheat next to the peanuts, in which square will the barley be?

The square immediately north of the rice
The square immediately east of the rice
The square immediately west of the rice
The square immediately north east of the rice
The square immediately north west of the rice

Show answerHide answer

Answer: D

Approach: 3×3 grid placement — fix the given squares, then place the adjacent pair.

Solution:

Peanuts (NW)	N	NE
W	Rice (centre)	E
Vegetables	Vegetables	Vegetables

Fixed: NW = peanuts, centre = rice, whole south row = vegetables.
Free squares: N, NE, W, E for wheat + barley (adjacent pair) and two soybeans.
Wheat next to peanuts → wheat at N or W.
If wheat = W, its other neighbours are peanuts, rice and vegetables → no place for barley. ✗
So wheat = N and barley must be adjacent to N in a free square → NE (the square immediately north-east of the rice). Soybeans then take W and E.

Why the other options are wrong:
A North — that is the wheat.
B East, C West — soybeans.
E North-west — peanuts.

Cetking shortcut: The free squares N–NE–E form an L; W is isolated (between peanuts, rice and vegetables), so W can never hold wheat or barley.

Answer: D

Q24MediumCET 2025 · Slot 5 · Q24

Passage / data for this question

Read the following information and answer the given questions

Farmer Batuk Singh has a larger square field divided into nine smaller square fields, all equal, arranged in three rows of three fields each. One side of the field runs exactly east-west. The middle square must be planted with rice because it is wet. The wheat and barley should be continuous so that they can be harvested all at once by a mechanical harvester. Two of the fields should be planted with soybeans. The northwestern most field should be planted with peanuts and the southern third of the field is suitable only for vegetables.

Which square can be planted with wheat?

The square immediately north of the rice
The square immediately west of the rice
The square immediately south of the rice
The square immediately south west of the rice
The square immediately north west of the rice

Show answerHide answer

Answer: A

Approach: Same grid — list every square that can hold wheat.

Solution:

Peanuts (NW)	N	NE
W	Rice (centre)	E
Vegetables	Vegetables	Vegetables

Free squares: N, NE, W, E. W is boxed in by peanuts, rice and vegetables, so the wheat–barley pair must be N–NE or NE–E; W and E-or-N take the soybeans.
Wheat can therefore be at N, NE or E.
Among the options only "immediately north of the rice" (N) is possible.

Why the other options are wrong:
B West — isolated square, wheat would have no adjacent barley.
C South, D South-west — southern row is vegetables only.
E North-west — that square is fixed as peanuts.

Cetking shortcut: Any square already fixed (peanuts, vegetables) or isolated (W) is out at once — that kills B, C, D, E.

Answer: A

Q56MediumCET 2025 · Slot 5 · Q56

Passage / data for this question

Read the following information carefully and answer the questions given below:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh, but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain, and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune, while the batsman does not hail from Warangal.

Who is the spinner?

Philips
Chetan
Bishan
Wasim
Sumeet

Show answerHide answer

Answer: A

Approach: 3-variable matrix (name–town–role); fix the captain first using clues (ii), (iv), (vi).

Solution: 1. Clue (i): no name, town and role share a first letter – so Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace bowler, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; also Surat ≠ spinner, Pune ≠ pace bowler, Bangalore ≠ batsman, Warangal ≠ wicket keeper, Chandigarh ≠ captain.
2. Sumeet is not WK/batsman (iii) and not spinner (i) → captain or pace bowler.
3. Captain is not Philips, Wasim (ii) or Chetan (i) → Sumeet or Bishan. Captain is from Pune (vi) and Bishan is not from Pune (iv) → Sumeet = Captain, Pune.
4. Bangalore = wicket keeper (v). Not Philips/Wasim (ii), not Bishan (i), not Sumeet → Chetan = WK, Bangalore.
5. Surat: not Philips/Wasim → Bishan = Surat. Wasim ≠ Warangal → Wasim = Chandigarh, Philips = Warangal.
6. Roles left: spinner, pace bowler, batsman. Bishan ≠ batsman and Surat ≠ spinner → Bishan = pace bowler. Batsman ≠ Warangal (vi) → Philips = spinner, Wasim = batsman.

Player	Hometown	Role
Sumeet	Pune	Captain
Chetan	Bangalore	Wicket keeper
Bishan	Surat	Pace bowler
Wasim	Chandigarh	Batsman
Philips	Warangal	Spinner

The spinner is Philips.

Why the other options are wrong:
B Chetan – wicket keeper. C Bishan – pace bowler. D Wasim – batsman. E Sumeet – captain (and spinner is banned for him by the 'S' rule).

Cetking shortcut: Clue (i) is the master clue – write the 'same first letter' bans for every name, town and role before anything else.

Answer: A

Q57EasyCET 2025 · Slot 5 · Q57

Passage / data for this question

Read the following information carefully and answer the questions given below:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh, but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain, and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune, while the batsman does not hail from Warangal.

The spinner's hometown is

Chandigarh
Bangalore
Pune
Surat
Warangal

Show answerHide answer

Answer: E

Approach: 3-variable matrix (name–town–role); fix the captain first using clues (ii), (iv), (vi).

Solution: 1. Clue (i): no name, town and role share a first letter – so Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace bowler, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; also Surat ≠ spinner, Pune ≠ pace bowler, Bangalore ≠ batsman, Warangal ≠ wicket keeper, Chandigarh ≠ captain.
2. Sumeet is not WK/batsman (iii) and not spinner (i) → captain or pace bowler.
3. Captain is not Philips, Wasim (ii) or Chetan (i) → Sumeet or Bishan. Captain is from Pune (vi) and Bishan is not from Pune (iv) → Sumeet = Captain, Pune.
4. Bangalore = wicket keeper (v). Not Philips/Wasim (ii), not Bishan (i), not Sumeet → Chetan = WK, Bangalore.
5. Surat: not Philips/Wasim → Bishan = Surat. Wasim ≠ Warangal → Wasim = Chandigarh, Philips = Warangal.
6. Roles left: spinner, pace bowler, batsman. Bishan ≠ batsman and Surat ≠ spinner → Bishan = pace bowler. Batsman ≠ Warangal (vi) → Philips = spinner, Wasim = batsman.

Player	Hometown	Role
Sumeet	Pune	Captain
Chetan	Bangalore	Wicket keeper
Bishan	Surat	Pace bowler
Wasim	Chandigarh	Batsman
Philips	Warangal	Spinner

Spinner = Philips, whose hometown is Warangal.

Why the other options are wrong:
A Chandigarh – Wasim (batsman). B Bangalore – Chetan (WK). C Pune – Sumeet (captain). D Surat – Bishan (pace bowler).

Cetking shortcut: Clue (i) is the master clue – write the 'same first letter' bans for every name, town and role before anything else.

Answer: E

Q58EasyCET 2025 · Slot 5 · Q58

Passage / data for this question

Read the following information carefully and answer the questions given below:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh, but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain, and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune, while the batsman does not hail from Warangal.

Chandigarh is the hometown of

Sumeet
Bishan
Wasim
Philips
Chetan

Show answerHide answer

Answer: C

Approach: 3-variable matrix (name–town–role); fix the captain first using clues (ii), (iv), (vi).

Solution: 1. Clue (i): no name, town and role share a first letter – so Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace bowler, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; also Surat ≠ spinner, Pune ≠ pace bowler, Bangalore ≠ batsman, Warangal ≠ wicket keeper, Chandigarh ≠ captain.
2. Sumeet is not WK/batsman (iii) and not spinner (i) → captain or pace bowler.
3. Captain is not Philips, Wasim (ii) or Chetan (i) → Sumeet or Bishan. Captain is from Pune (vi) and Bishan is not from Pune (iv) → Sumeet = Captain, Pune.
4. Bangalore = wicket keeper (v). Not Philips/Wasim (ii), not Bishan (i), not Sumeet → Chetan = WK, Bangalore.
5. Surat: not Philips/Wasim → Bishan = Surat. Wasim ≠ Warangal → Wasim = Chandigarh, Philips = Warangal.
6. Roles left: spinner, pace bowler, batsman. Bishan ≠ batsman and Surat ≠ spinner → Bishan = pace bowler. Batsman ≠ Warangal (vi) → Philips = spinner, Wasim = batsman.

Player	Hometown	Role
Sumeet	Pune	Captain
Chetan	Bangalore	Wicket keeper
Bishan	Surat	Pace bowler
Wasim	Chandigarh	Batsman
Philips	Warangal	Spinner

Chandigarh is the hometown of Wasim.

Why the other options are wrong:
A Sumeet – Pune. B Bishan – Surat. D Philips – Warangal. E Chetan – Bangalore (and the 'C' rule bans Chetan–Chandigarh).

Cetking shortcut: Clue (i) is the master clue – write the 'same first letter' bans for every name, town and role before anything else.

Answer: C

Q59EasyCET 2025 · Slot 5 · Q59

Passage / data for this question

Read the following information carefully and answer the questions given below:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh, but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain, and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune, while the batsman does not hail from Warangal.

Who is the pace-bowler?

Chetan
Wasim
Sumeet
Bishan
Philips

Show answerHide answer

Answer: D

Approach: 3-variable matrix (name–town–role); fix the captain first using clues (ii), (iv), (vi).

Solution: 1. Clue (i): no name, town and role share a first letter – so Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace bowler, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; also Surat ≠ spinner, Pune ≠ pace bowler, Bangalore ≠ batsman, Warangal ≠ wicket keeper, Chandigarh ≠ captain.
2. Sumeet is not WK/batsman (iii) and not spinner (i) → captain or pace bowler.
3. Captain is not Philips, Wasim (ii) or Chetan (i) → Sumeet or Bishan. Captain is from Pune (vi) and Bishan is not from Pune (iv) → Sumeet = Captain, Pune.
4. Bangalore = wicket keeper (v). Not Philips/Wasim (ii), not Bishan (i), not Sumeet → Chetan = WK, Bangalore.
5. Surat: not Philips/Wasim → Bishan = Surat. Wasim ≠ Warangal → Wasim = Chandigarh, Philips = Warangal.
6. Roles left: spinner, pace bowler, batsman. Bishan ≠ batsman and Surat ≠ spinner → Bishan = pace bowler. Batsman ≠ Warangal (vi) → Philips = spinner, Wasim = batsman.

Player	Hometown	Role
Sumeet	Pune	Captain
Chetan	Bangalore	Wicket keeper
Bishan	Surat	Pace bowler
Wasim	Chandigarh	Batsman
Philips	Warangal	Spinner

The pace bowler is Bishan.

Why the other options are wrong:
A Chetan – WK. B Wasim – batsman. C Sumeet – captain. E Philips – spinner (also banned by the 'P' rule).

Cetking shortcut: Clue (i) is the master clue – write the 'same first letter' bans for every name, town and role before anything else.

Answer: D

Q60EasyCET 2025 · Slot 5 · Q60

Passage / data for this question

Read the following information carefully and answer the questions given below:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh, but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain, and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune, while the batsman does not hail from Warangal.

Chetan's hometown is

Pune
Surat
Bangalore
Warangal
Chandigarh

Show answerHide answer

Answer: C

Approach: 3-variable matrix (name–town–role); fix the captain first using clues (ii), (iv), (vi).

Solution: 1. Clue (i): no name, town and role share a first letter – so Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace bowler, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; also Surat ≠ spinner, Pune ≠ pace bowler, Bangalore ≠ batsman, Warangal ≠ wicket keeper, Chandigarh ≠ captain.
2. Sumeet is not WK/batsman (iii) and not spinner (i) → captain or pace bowler.
3. Captain is not Philips, Wasim (ii) or Chetan (i) → Sumeet or Bishan. Captain is from Pune (vi) and Bishan is not from Pune (iv) → Sumeet = Captain, Pune.
4. Bangalore = wicket keeper (v). Not Philips/Wasim (ii), not Bishan (i), not Sumeet → Chetan = WK, Bangalore.
5. Surat: not Philips/Wasim → Bishan = Surat. Wasim ≠ Warangal → Wasim = Chandigarh, Philips = Warangal.
6. Roles left: spinner, pace bowler, batsman. Bishan ≠ batsman and Surat ≠ spinner → Bishan = pace bowler. Batsman ≠ Warangal (vi) → Philips = spinner, Wasim = batsman.

Player	Hometown	Role
Sumeet	Pune	Captain
Chetan	Bangalore	Wicket keeper
Bishan	Surat	Pace bowler
Wasim	Chandigarh	Batsman
Philips	Warangal	Spinner

Chetan is the wicket keeper, so his hometown is Bangalore.

Why the other options are wrong:
A Pune – Sumeet. B Surat – Bishan. D Warangal – Philips. E Chandigarh – Wasim (and banned for Chetan by the 'C' rule).

Cetking shortcut: Clue (i) is the master clue – write the 'same first letter' bans for every name, town and role before anything else.

Answer: C

Q23MediumCET 2025 · Slot 4 · Q23

Passage / data for this question

Directions: Study the following information carefully and answer the question that follows:

(i) A, B, C, D, E, F and G are seven persons wearing shirts of different colors-white, red, black, green, yellow, blue and violet-and trousers of different colors-blue, red, white, black, cream, yellow and indigo. The persons, color of the shirt and color of the trousers above are not necessarily in the same order. No person is wearing shirt and trousers of same color.

(ii) B is wearing red shirt and is not wearing cream or yellow trousers. D is wearing green shirt and indigo trousers. The color of A's shirt and F's trousers is same. The color of E's shirt and C's trousers is same. G is wearing blue shirt and E is wearing blue trousers. F is not wearing any yellow dress. A is not wearing a white shirt. Red and blue is not the combination of shirt and trousers of any of the persons.

What is the colour of G's trousers?

White
Cream
Green
Indigo
Red

Show answerHide answer

Answer: B

Approach: Matrix puzzle — fix the direct clues, then use the 'same colour' links and exclusions.

Solution:
Direct: B shirt red; D shirt green, trousers indigo; G shirt blue; E trousers blue.
Shirts left for A, C, E, F: white, black, yellow, violet. Trousers left for A, B, C, F, G: red, white, black, cream, yellow.
A's shirt = F's trousers → must be white, black or yellow. A is not white; F wears no yellow → black. So A shirt black, F trousers black.
B: not red (same as shirt), not cream/yellow, not black (taken) → white.
E's shirt = C's trousers → from white/yellow/violet ∩ red/cream/yellow → yellow. E shirt yellow, C trousers yellow.
A and G get red and cream; G (blue shirt) cannot wear red (no red–blue combo) → G cream, A red.
C and F shirts: white/violet (not fixed).

Person	Shirt	Trousers
A	Black	Red
B	Red	White
C	White/Violet	Yellow
D	Green	Indigo
E	Yellow	Blue
F	Violet/White	Black
G	Blue	Cream

G's trousers = Cream.

Why the other options are wrong:
A (White): worn by B.
C (Green): not a trouser colour.
D (Indigo): worn by D.
E (Red): red–blue combination is not allowed for G.

Answer: B

Q24MediumCET 2025 · Slot 4 · Q24

Passage / data for this question

Directions: Study the following information carefully and answer the question that follows:

(i) A, B, C, D, E, F and G are seven persons wearing shirts of different colors-white, red, black, green, yellow, blue and violet-and trousers of different colors-blue, red, white, black, cream, yellow and indigo. The persons, color of the shirt and color of the trousers above are not necessarily in the same order. No person is wearing shirt and trousers of same color.

(ii) B is wearing red shirt and is not wearing cream or yellow trousers. D is wearing green shirt and indigo trousers. The color of A's shirt and F's trousers is same. The color of E's shirt and C's trousers is same. G is wearing blue shirt and E is wearing blue trousers. F is not wearing any yellow dress. A is not wearing a white shirt. Red and blue is not the combination of shirt and trousers of any of the persons.

Who wears a violet colored shirt?

C
F
C or F
G
C or G

Show answerHide answer

Answer: C

Approach: Matrix puzzle — fix the direct clues, then use the 'same colour' links and exclusions.

Solution:
Direct: B shirt red; D shirt green, trousers indigo; G shirt blue; E trousers blue.
Shirts left for A, C, E, F: white, black, yellow, violet. Trousers left for A, B, C, F, G: red, white, black, cream, yellow.
A's shirt = F's trousers → must be white, black or yellow. A is not white; F wears no yellow → black. So A shirt black, F trousers black.
B: not red (same as shirt), not cream/yellow, not black (taken) → white.
E's shirt = C's trousers → from white/yellow/violet ∩ red/cream/yellow → yellow. E shirt yellow, C trousers yellow.
A and G get red and cream; G (blue shirt) cannot wear red (no red–blue combo) → G cream, A red.
C and F shirts: white/violet (not fixed).

Person	Shirt	Trousers
A	Black	Red
B	Red	White
C	White/Violet	Yellow
D	Green	Indigo
E	Yellow	Blue
F	Violet/White	Black
G	Blue	Cream

The violet shirt is worn by C or F — the clues do not fix which.

Why the other options are wrong:
A, B: each alone is not certain.
D, E: G wears a blue shirt.

Answer: C

Q25MediumCET 2025 · Slot 4 · Q25

Passage / data for this question

Directions: Study the following information carefully and answer the question that follows:

(i) A, B, C, D, E, F and G are seven persons wearing shirts of different colors-white, red, black, green, yellow, blue and violet-and trousers of different colors-blue, red, white, black, cream, yellow and indigo. The persons, color of the shirt and color of the trousers above are not necessarily in the same order. No person is wearing shirt and trousers of same color.

(ii) B is wearing red shirt and is not wearing cream or yellow trousers. D is wearing green shirt and indigo trousers. The color of A's shirt and F's trousers is same. The color of E's shirt and C's trousers is same. G is wearing blue shirt and E is wearing blue trousers. F is not wearing any yellow dress. A is not wearing a white shirt. Red and blue is not the combination of shirt and trousers of any of the persons.

What is the color of B's trouser?

Black
White
Indigo
Blue
Yellow

Show answerHide answer

Answer: B

Approach: Matrix puzzle — fix the direct clues, then use the 'same colour' links and exclusions.

Solution:
Direct: B shirt red; D shirt green, trousers indigo; G shirt blue; E trousers blue.
Shirts left for A, C, E, F: white, black, yellow, violet. Trousers left for A, B, C, F, G: red, white, black, cream, yellow.
A's shirt = F's trousers → must be white, black or yellow. A is not white; F wears no yellow → black. So A shirt black, F trousers black.
B: not red (same as shirt), not cream/yellow, not black (taken) → white.
E's shirt = C's trousers → from white/yellow/violet ∩ red/cream/yellow → yellow. E shirt yellow, C trousers yellow.
A and G get red and cream; G (blue shirt) cannot wear red (no red–blue combo) → G cream, A red.
C and F shirts: white/violet (not fixed).

Person	Shirt	Trousers
A	Black	Red
B	Red	White
C	White/Violet	Yellow
D	Green	Indigo
E	Yellow	Blue
F	Violet/White	Black
G	Blue	Cream

B's trousers = White.

Why the other options are wrong:
A (Black): F wears black trousers.
C (Indigo): D's.
D (Blue): E's; also red–blue not allowed.
E (Yellow): excluded for B directly.

Answer: B

Q27MediumCET 2025 · Slot 4 · Q27

Passage / data for this question

Read the following information carefully and answer the question that follows:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune while the batsman does not hail from Warangal.

Who is the spinner ?

Philips
Chetan
Bishan
Wasim
Sumeet

Show answerHide answer

Answer: A

Approach: Three-way matrix (player–town–role) using first-letter exclusions.

Solution:
Rule (i): no name, town or role starts with the same letter as another in the same row (e.g. Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; Pune ≠ Pace, Warangal ≠ Wicket keeper, Chandigarh ≠ Captain, Bangalore ≠ Batsman).
Philips and Wasim are not from Surat/Bangalore. Captain is from Pune and neither of them is captain → neither is from Pune.
Wasim ≠ Warangal → Wasim = Chandigarh; Philips = Warangal.
Captain (Pune) from Sumeet/Bishan/Chetan: Bishan ≠ Pune, Chetan ≠ Captain → Sumeet = Captain, Pune.
Bishan ≠ Bangalore → Bishan = Surat, Chetan = Bangalore = Wicket keeper.
Roles left (spinner, pace, batsman) for Philips, Wasim, Bishan: Bishan ≠ batsman, batsman not from Warangal (Philips) → Wasim = Batsman. Philips ≠ Pace → Philips = Spinner, Bishan = Pace bowler.

Player	Hometown	Role
Sumeet	Pune	Captain
Philips	Warangal	Spinner
Wasim	Chandigarh	Batsman
Bishan	Surat	Pace bowler
Chetan	Bangalore	Wicket keeper

Spinner = Philips.

Why the other options are wrong:
B Chetan: wicket keeper. C Bishan: pace bowler. D Wasim: batsman. E Sumeet: captain (and cannot be spinner by the first-letter rule).

Cetking shortcut: Start from the person with the most exclusions (Wasim: not Surat, Bangalore, Warangal, Pune) — he gets fixed first.

Answer: A

Q28MediumCET 2025 · Slot 4 · Q28

Passage / data for this question

Read the following information carefully and answer the question that follows:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune while the batsman does not hail from Warangal.

The Spinner's hometown is?

Chandigarh
Pune
Warangal
Surat
Bangalore

Show answerHide answer

Answer: C

Approach: Three-way matrix (player–town–role) using first-letter exclusions.

Solution:
Rule (i): no name, town or role starts with the same letter as another in the same row (e.g. Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; Pune ≠ Pace, Warangal ≠ Wicket keeper, Chandigarh ≠ Captain, Bangalore ≠ Batsman).
Philips and Wasim are not from Surat/Bangalore. Captain is from Pune and neither of them is captain → neither is from Pune.
Wasim ≠ Warangal → Wasim = Chandigarh; Philips = Warangal.
Captain (Pune) from Sumeet/Bishan/Chetan: Bishan ≠ Pune, Chetan ≠ Captain → Sumeet = Captain, Pune.
Bishan ≠ Bangalore → Bishan = Surat, Chetan = Bangalore = Wicket keeper.
Roles left (spinner, pace, batsman) for Philips, Wasim, Bishan: Bishan ≠ batsman, batsman not from Warangal (Philips) → Wasim = Batsman. Philips ≠ Pace → Philips = Spinner, Bishan = Pace bowler.

Player	Hometown	Role
Sumeet	Pune	Captain
Philips	Warangal	Spinner
Wasim	Chandigarh	Batsman
Bishan	Surat	Pace bowler
Chetan	Bangalore	Wicket keeper

Spinner (Philips) is from Warangal.

Why the other options are wrong:
A Chandigarh: Wasim (batsman). B Pune: captain. D Surat: Bishan (pace). E Bangalore: wicket keeper.

Answer: C

Q29MediumCET 2025 · Slot 4 · Q29

Passage / data for this question

Read the following information carefully and answer the question that follows:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune while the batsman does not hail from Warangal.

Chandigarh is the hometown of?

Sumeet
Wasim
Philips
Bishan
Chetan

Show answerHide answer

Answer: B

Approach: Three-way matrix (player–town–role) using first-letter exclusions.

Solution:
Rule (i): no name, town or role starts with the same letter as another in the same row (e.g. Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; Pune ≠ Pace, Warangal ≠ Wicket keeper, Chandigarh ≠ Captain, Bangalore ≠ Batsman).
Philips and Wasim are not from Surat/Bangalore. Captain is from Pune and neither of them is captain → neither is from Pune.
Wasim ≠ Warangal → Wasim = Chandigarh; Philips = Warangal.
Captain (Pune) from Sumeet/Bishan/Chetan: Bishan ≠ Pune, Chetan ≠ Captain → Sumeet = Captain, Pune.
Bishan ≠ Bangalore → Bishan = Surat, Chetan = Bangalore = Wicket keeper.
Roles left (spinner, pace, batsman) for Philips, Wasim, Bishan: Bishan ≠ batsman, batsman not from Warangal (Philips) → Wasim = Batsman. Philips ≠ Pace → Philips = Spinner, Bishan = Pace bowler.

Player	Hometown	Role
Sumeet	Pune	Captain
Philips	Warangal	Spinner
Wasim	Chandigarh	Batsman
Bishan	Surat	Pace bowler
Chetan	Bangalore	Wicket keeper

Chandigarh = Wasim.

Why the other options are wrong:
A Sumeet: Pune. C Philips: Warangal. D Bishan: Surat. E Chetan: cannot be from Chandigarh (same first letter); he is from Bangalore.

Answer: B

Q30MediumCET 2025 · Slot 4 · Q30

Passage / data for this question

Read the following information carefully and answer the question that follows:

Sumeet, Philips, Wasim, Bishan and Chetan are five players of the College Cricket Team and their hometowns are Surat, Pune, Warangal, Bangalore and Chandigarh but not in that order. The five specialist slots of spinner, pace bowler, wicket keeper, batsman and captain are held by them, again not in the order of their names stated above.

(i) Their names, hometowns and specialities do not start with the same letter.
(ii) Neither Philips nor Wasim is the captain and they do not belong to either Surat or Bangalore.
(iii) Sumeet is neither a wicket keeper nor a batsman.
(iv) Pune is not Bishan's hometown.
(v) The player who hails from Bangalore is a wicket keeper.
(vi) The captain's hometown is Pune while the batsman does not hail from Warangal.

Who is the pace-bowler?

Chetan
Wasim
Sumeet
Bishan
Philips

Show answerHide answer

Answer: D

Approach: Three-way matrix (player–town–role) using first-letter exclusions.

Solution:
Rule (i): no name, town or role starts with the same letter as another in the same row (e.g. Sumeet ≠ Surat/Spinner, Philips ≠ Pune/Pace, Wasim ≠ Warangal/Wicket keeper, Bishan ≠ Bangalore/Batsman, Chetan ≠ Chandigarh/Captain; Pune ≠ Pace, Warangal ≠ Wicket keeper, Chandigarh ≠ Captain, Bangalore ≠ Batsman).
Philips and Wasim are not from Surat/Bangalore. Captain is from Pune and neither of them is captain → neither is from Pune.
Wasim ≠ Warangal → Wasim = Chandigarh; Philips = Warangal.
Captain (Pune) from Sumeet/Bishan/Chetan: Bishan ≠ Pune, Chetan ≠ Captain → Sumeet = Captain, Pune.
Bishan ≠ Bangalore → Bishan = Surat, Chetan = Bangalore = Wicket keeper.
Roles left (spinner, pace, batsman) for Philips, Wasim, Bishan: Bishan ≠ batsman, batsman not from Warangal (Philips) → Wasim = Batsman. Philips ≠ Pace → Philips = Spinner, Bishan = Pace bowler.

Player	Hometown	Role
Sumeet	Pune	Captain
Philips	Warangal	Spinner
Wasim	Chandigarh	Batsman
Bishan	Surat	Pace bowler
Chetan	Bangalore	Wicket keeper

Pace bowler = Bishan.

Why the other options are wrong:
A Chetan: wicket keeper. B Wasim: batsman. C Sumeet: captain. E Philips: cannot be pace (same first letter); he is the spinner.

Answer: D

Q36MediumCET 2025 · Slot 3 · Q36

Passage / data for this question

DIRECTIONS for the question: Study the following information carefully and answer the question that follows.

Of the six men of literature A, B, C, D, E and F being considered here, two belonged to the 17th century, three to the 19th and one to the 20th century. Four were recognized as great poets, three as great novelists and three as great dramatists. One contributed to Bengali literature, two to Hindi, two to Marathi and one to Tamil. The 20th century writer wrote poetry only and contributed to Marathi literature and the other Marathi writer contributed to poetry, novel and drama. One Hindi writer and the only Tamil writer belonged to the 19th century. The former contributed to poetry and novel while the latter to novel and drama. The Bengali writer belonged to the 17th century and contributed to poetry only. A belonged to the 20th century, B wrote drama only, C contributed to Marathi literature, D was a Hindi poet and novelist and belonged to the 19th century, and F contributed to poetry only.

To which language did B contribute?

Bengali
Hindi
Marathi
Tamil
English

Show answerHide answer

Answer: B

Approach: Matrix puzzle: fill writer × century × branch × language from fixed clues.

Solution:
A = 20th century → the 20th-century writer is Marathi, poetry only. C is the other Marathi writer → poetry, novel, drama.
D = Hindi, poetry + novel, 19th.
Poets (4): A, C, D + one more. F writes poetry only, so F is the 4th poet; the Bengali writer writes poetry only → F = Bengali, 17th.
Left: B and E, languages Hindi (2nd) and Tamil. Tamil writer does novel + drama; B does drama only → B = Hindi, E = Tamil (19th, novel + drama).
Check: novelists C, D, E (3) ✓; dramatists B, C, E (3) ✓.

Writer	Century	Branch	Language
A	20th	Poetry only	Marathi
B	17th/19th	Drama only	Hindi
C	19th/17th	Poetry, Novel, Drama	Marathi
D	19th	Poetry, Novel	Hindi
E	19th	Novel, Drama	Tamil
F	17th	Poetry only	Bengali

B contributed to Hindi.

Why the other options are wrong:
A Bengali: F.
C Marathi: A and C.
D Tamil: E (novel + drama, but B is drama only).
E English: no English writer in the set.

Cetking shortcut: B 'drama only' cannot be the Tamil writer (novel + drama), and Bengali/Marathi are taken — Hindi by elimination.

Answer: B

Q37MediumCET 2025 · Slot 3 · Q37

Passage / data for this question

DIRECTIONS for the question: Study the following information carefully and answer the question that follows.

Of the six men of literature A, B, C, D, E and F being considered here, two belonged to the 17th century, three to the 19th and one to the 20th century. Four were recognized as great poets, three as great novelists and three as great dramatists. One contributed to Bengali literature, two to Hindi, two to Marathi and one to Tamil. The 20th century writer wrote poetry only and contributed to Marathi literature and the other Marathi writer contributed to poetry, novel and drama. One Hindi writer and the only Tamil writer belonged to the 19th century. The former contributed to poetry and novel while the latter to novel and drama. The Bengali writer belonged to the 17th century and contributed to poetry only. A belonged to the 20th century, B wrote drama only, C contributed to Marathi literature, D was a Hindi poet and novelist and belonged to the 19th century, and F contributed to poetry only.

Among this who is Tamil writer?

A
B
E
F
D

Show answerHide answer

Answer: C

Approach: Matrix puzzle: read the Tamil row.

Solution:
A = 20th century → the 20th-century writer is Marathi, poetry only. C is the other Marathi writer → poetry, novel, drama.
D = Hindi, poetry + novel, 19th.
Poets (4): A, C, D + one more. F writes poetry only, so F is the 4th poet; the Bengali writer writes poetry only → F = Bengali, 17th.
Left: B and E, languages Hindi (2nd) and Tamil. Tamil writer does novel + drama; B does drama only → B = Hindi, E = Tamil (19th, novel + drama).
Check: novelists C, D, E (3) ✓; dramatists B, C, E (3) ✓.

Writer	Century	Branch	Language
A	20th	Poetry only	Marathi
B	17th/19th	Drama only	Hindi
C	19th/17th	Poetry, Novel, Drama	Marathi
D	19th	Poetry, Novel	Hindi
E	19th	Novel, Drama	Tamil
F	17th	Poetry only	Bengali

Tamil writer = E.

Why the other options are wrong:
A A: Marathi.
B B: Hindi (drama only).
D F: Bengali.
E D: Hindi.

Answer: C

Q38EasyCET 2025 · Slot 3 · Q38

Passage / data for this question

DIRECTIONS for the question: Study the following information carefully and answer the question that follows.

Of the six men of literature A, B, C, D, E and F being considered here, two belonged to the 17th century, three to the 19th and one to the 20th century. Four were recognized as great poets, three as great novelists and three as great dramatists. One contributed to Bengali literature, two to Hindi, two to Marathi and one to Tamil. The 20th century writer wrote poetry only and contributed to Marathi literature and the other Marathi writer contributed to poetry, novel and drama. One Hindi writer and the only Tamil writer belonged to the 19th century. The former contributed to poetry and novel while the latter to novel and drama. The Bengali writer belonged to the 17th century and contributed to poetry only. A belonged to the 20th century, B wrote drama only, C contributed to Marathi literature, D was a Hindi poet and novelist and belonged to the 19th century, and F contributed to poetry only.

To which branch of literature did A contribute?

Poetry
Novel
Drama
Drama and Novel
Drama and Poetry

Show answerHide answer

Answer: A

Approach: Matrix puzzle: direct clue on the 20th-century writer.

Solution:
A belongs to the 20th century, and the 20th-century writer wrote poetry only (Marathi).
A contributed to Poetry.

Why the other options are wrong:
B Novel, C Drama, D Drama and Novel, E Drama and Poetry: A wrote poetry only.

Cetking shortcut: Direct read — no grid needed: A = 20th century = poetry only.

Answer: A

Q39EasyCET 2025 · Slot 3 · Q39

Passage / data for this question

DIRECTIONS for the question: Study the following information carefully and answer the question that follows.

Of the six men of literature A, B, C, D, E and F being considered here, two belonged to the 17th century, three to the 19th and one to the 20th century. Four were recognized as great poets, three as great novelists and three as great dramatists. One contributed to Bengali literature, two to Hindi, two to Marathi and one to Tamil. The 20th century writer wrote poetry only and contributed to Marathi literature and the other Marathi writer contributed to poetry, novel and drama. One Hindi writer and the only Tamil writer belonged to the 19th century. The former contributed to poetry and novel while the latter to novel and drama. The Bengali writer belonged to the 17th century and contributed to poetry only. A belonged to the 20th century, B wrote drama only, C contributed to Marathi literature, D was a Hindi poet and novelist and belonged to the 19th century, and F contributed to poetry only.

Among these, who was Bengali writer?

A
B
E
F
D

Show answerHide answer

Answer: D

Approach: Matrix puzzle: identify the Bengali writer.

Solution:
The Bengali writer wrote poetry only. The poetry-only writers are A (Marathi, 20th) and F.
So the Bengali writer is F.

Why the other options are wrong:
A A: Marathi.
B B: Hindi.
C E: Tamil.
E D: Hindi.

Cetking shortcut: Two 'poetry only' people: A is Marathi, so the other one (F) is Bengali.

Answer: D

Q49MediumCET 2025 · Slot 2 · Q49

Passage / data for this question

Five friends, namely, Daljit, Mukesh, Sandeep, Ignesh and Bimal meet every morning at Ballee Restaurant for an Idli & Vada breakfast.

Each consumes a different number of Idlis and Vadas. The number of Idlis consumed are 1, 4, 5, 6, and 8, while the number of Vadas consumed are 0, 1, 2, 4, and 6.

Below are some more facts about who eats what and how much:

The number of Vadas eaten by Ignesh is three times the number of Vadas consumed by the person who eats 4 Idlis.
Three persons, including the one who eats four Vadas eat without Chutney.
Sandeep does not take any Chutney.
The one who eats one Idli a day does not eat any Vadas or Chutney. Further, he is not Mukesh.
Daljit eats Idli with Chutney and also eats Vada.
Mukesh, who does not take Chutney, eats half as many Vadas as the person who eats twice as many Idlis as he does.
Bimal eats two more Idlis than Ignesh, but Ignesh eats two more Vadas than Bimal.

Which one of the following statements is most definitely true?

Daljit eats 5 Idlis
Bimal eats 6 Idlis
Bimal eats 1 Idli
Sandeep eats 2 Idlis
Ignesh eats 8 Idlis

Show answerHide answer

Answer: A

Approach: Start with the strongest numeric clues (vii) and (i), then fix the rest.

Solution:
(vii) Bimal = Ignesh + 2 Idlis → (Ignesh, Bimal) = (4, 6) or (6, 8); Ignesh = Bimal + 2 Vadas → (Bimal, Ignesh) Vadas = (0, 2), (2, 4) or (4, 6).
(i) Ignesh's Vadas = 3 × Vadas of the 4-Idli person ; Ignesh's Vadas are 2, 4 or 6 and must be a multiple of 3 → Ignesh 6, 4-Idli person 2. So Ignesh 6 Vadas, Bimal 4 Vadas, and Ignesh is not the 4-Idli person → Ignesh 6, Bimal 8 Idlis.
(vi) Mukesh has half the Vadas of the person with twice his Idlis: Mukesh 4 Idlis → Bimal (8) has 4 Vadas → Mukesh 2 Vadas ✓ (also matches (i)).
(iv) 1-Idli person eats 0 Vadas and is not Mukesh; (v) Daljit eats Vada → Sandeep = 1 Idli, 0 Vadas; Daljit = 5 Idlis, 1 Vada.
(ii) No Chutney: Sandeep, Mukesh and the 4-Vada person Bimal (exactly three) → Daljit and Ignesh take Chutney.

Person	Idlis	Vadas	Chutney
Daljit	5	1	Yes
Mukesh	4	2	No
Sandeep	1	0	No
Ignesh	6	6	Yes
Bimal	8	4	No

Daljit eats 5 Idlis.

Why the other options are wrong: Bimal eats 8 (not 6 or 1); Sandeep eats 1 (2 is not even an Idli count); Ignesh eats 6, not 8.

Answer: A

Q50MediumCET 2025 · Slot 2 · Q50

Passage / data for this question

Five friends, namely, Daljit, Mukesh, Sandeep, Ignesh and Bimal meet every morning at Ballee Restaurant for an Idli & Vada breakfast.

Each consumes a different number of Idlis and Vadas. The number of Idlis consumed are 1, 4, 5, 6, and 8, while the number of Vadas consumed are 0, 1, 2, 4, and 6.

Below are some more facts about who eats what and how much:

The number of Vadas eaten by Ignesh is three times the number of Vadas consumed by the person who eats 4 Idlis.
Three persons, including the one who eats four Vadas eat without Chutney.
Sandeep does not take any Chutney.
The one who eats one Idli a day does not eat any Vadas or Chutney. Further, he is not Mukesh.
Daljit eats Idli with Chutney and also eats Vada.
Mukesh, who does not take Chutney, eats half as many Vadas as the person who eats twice as many Idlis as he does.
Bimal eats two more Idlis than Ignesh, but Ignesh eats two more Vadas than Bimal.

Which one of the following statements is most likely true?

Sandeep eats 2 Vadas
Ignesh eats 6 Vadas
Bimal eats 2 Vadas
Mukesh eats 4 Vadas
Bimal eats Chutney

Show answerHide answer

Answer: B

Approach: Use the unique grid from the set.

Solution:
(vii) Bimal = Ignesh + 2 Idlis → (Ignesh, Bimal) = (4, 6) or (6, 8); Ignesh = Bimal + 2 Vadas → (Bimal, Ignesh) Vadas = (0, 2), (2, 4) or (4, 6).
(i) Ignesh's Vadas = 3 × Vadas of the 4-Idli person ; Ignesh's Vadas are 2, 4 or 6 and must be a multiple of 3 → Ignesh 6, 4-Idli person 2. So Ignesh 6 Vadas, Bimal 4 Vadas, and Ignesh is not the 4-Idli person → Ignesh 6, Bimal 8 Idlis.
(vi) Mukesh has half the Vadas of the person with twice his Idlis: Mukesh 4 Idlis → Bimal (8) has 4 Vadas → Mukesh 2 Vadas ✓ (also matches (i)).
(iv) 1-Idli person eats 0 Vadas and is not Mukesh; (v) Daljit eats Vada → Sandeep = 1 Idli, 0 Vadas; Daljit = 5 Idlis, 1 Vada.
(ii) No Chutney: Sandeep, Mukesh and the 4-Vada person Bimal (exactly three) → Daljit and Ignesh take Chutney.

Person	Idlis	Vadas	Chutney
Daljit	5	1	Yes
Mukesh	4	2	No
Sandeep	1	0	No
Ignesh	6	6	Yes
Bimal	8	4	No

Ignesh eats 6 Vadas.

Why the other options are wrong: Sandeep eats 0 Vadas; Bimal eats 4 Vadas; Mukesh eats 2; Bimal does not take Chutney (he is the 4-Vada person in clue (ii)).

Answer: B

Q51MediumCET 2025 · Slot 2 · Q51

Passage / data for this question

Five friends, namely, Daljit, Mukesh, Sandeep, Ignesh and Bimal meet every morning at Ballee Restaurant for an Idli & Vada breakfast.
Each consumes a different number of Idlis and Vadas. The number of Idlis consumed are 1, 4, 5, 6, and 8, while the number of Vadas consumed are 0, 1, 2, 4, and 6.

Below are some more facts about who eats what and how much:
i. The number of Vadas eaten by Ignesh is three times the number of Vadas consumed by the person who eats 4 Idlis.
ii. Three persons, including the one who eats four Vadas eat without Chutney.
iii. Sandeep does not take any Chutney.
iv. The one who eats one Idli a day does not eat any Vadas or Chutney. Further, he is not Mukesh.
v. Daljit eats Idli with Chutney and also eats Vada.
vi. Mukesh, who does not take Chutney, eats half as many Vadas as the person who eats twice as many Idlis as he does.
vii. Bimal eats two more Idlis than Ignesh, but Ignesh eats two more Vadas than Bimal.

Which of the following statements is most likely to be false?

Daljit does not eat 2 Vadas
Bimal eats more than 4 Idlis
Sandeep has 1 Vada
Daljit eats Chutney
Bimal does not eat Chutney

Show answerHide answer

Answer: C

Approach: Fix Mukesh through the 'twice the Idlis' clue, then chain the Vada clues.

Solution:
1. Clue vi: Mukesh eats m Idlis and someone eats 2m. From {1,4,5,6,8} only m = 4 works (Mukesh ≠ 1-Idli person, clue iv), so Mukesh = 4 Idlis and the 8-Idli person eats twice Mukesh's Vadas.
2. Clue i: Ignesh's Vadas = 3 × (Vadas of 4-Idli person = Mukesh). From {0,1,2,4,6}: Ignesh 6, Mukesh 2. So the 8-Idli person eats 4 Vadas.
3. Clue vii: Bimal's Vadas = Ignesh's − 2 = 4, so Bimal is the 8-Idli person; Ignesh eats 8 − 2 = 6 Idlis.
4. Left: Daljit, Sandeep with Idlis {1,5}, Vadas {0,1}. Daljit eats Vada (clue v), so Sandeep = 1 Idli, 0 Vada; Daljit = 5 Idlis, 1 Vada.
5. No Chutney: Sandeep (iii), Mukesh (vi), Bimal (4 Vadas, clue ii) = exactly three. So Daljit and Ignesh take Chutney.

Person	Idlis	Vadas	Chutney
Daljit	5	1	Yes
Mukesh	4	2	No
Sandeep	1	0	No
Ignesh	6	6	Yes
Bimal	8	4	No

Why the other options are wrong:
A: Daljit eats 1 Vada, so 'not 2 Vadas' is true.
B: Bimal eats 8 Idlis — true.
D: Daljit takes Chutney — true.
E: Bimal is one of the three without Chutney — true.
C: Sandeep eats 0 Vadas, so 'Sandeep has 1 Vada' is false.

Cetking shortcut: Only 4 → 8 satisfies 'twice as many Idlis' in {1,4,5,6,8}; start every grid puzzle from the clue with the fewest possibilities.

Answer: C

Q52MediumCET 2025 · Slot 2 · Q52

Passage / data for this question

Five friends, namely, Daljit, Mukesh, Sandeep, Ignesh and Bimal meet every morning at Ballee Restaurant for an Idli & Vada breakfast.
Each consumes a different number of Idlis and Vadas. The number of Idlis consumed are 1, 4, 5, 6, and 8, while the number of Vadas consumed are 0, 1, 2, 4, and 6.

Below are some more facts about who eats what and how much:
i. The number of Vadas eaten by Ignesh is three times the number of Vadas consumed by the person who eats 4 Idlis.
ii. Three persons, including the one who eats four Vadas eat without Chutney.
iii. Sandeep does not take any Chutney.
iv. The one who eats one Idli a day does not eat any Vadas or Chutney. Further, he is not Mukesh.
v. Daljit eats Idli with Chutney and also eats Vada.
vi. Mukesh, who does not take Chutney, eats half as many Vadas as the person who eats twice as many Idlis as he does.
vii. Bimal eats two more Idlis than Ignesh, but Ignesh eats two more Vadas than Bimal.

Which of the following statements is most likely to be true?

The person who eats 5 Idlis and 1 Vada does not take Chutney
Bimal eats equal numbers of Idlis and Vadas
Mukesh eats 8 Idlis and 4 Vadas but no Chutney
The person who eats equal number of Vadas and Idlis also takes Chutney
The person who eats 4 Idlis and 2 Vadas also takes Chutney

Show answerHide answer

Answer: D

Approach: Build the full Idli–Vada–Chutney grid, then test each statement.

Solution:
1. Clue vi: Mukesh eats m Idlis and someone eats 2m. From {1,4,5,6,8} only m = 4 works (Mukesh ≠ 1-Idli person, clue iv), so Mukesh = 4 Idlis and the 8-Idli person eats twice Mukesh's Vadas.
2. Clue i: Ignesh's Vadas = 3 × (Vadas of 4-Idli person = Mukesh). From {0,1,2,4,6}: Ignesh 6, Mukesh 2. So the 8-Idli person eats 4 Vadas.
3. Clue vii: Bimal's Vadas = Ignesh's − 2 = 4, so Bimal is the 8-Idli person; Ignesh eats 8 − 2 = 6 Idlis.
4. Left: Daljit, Sandeep with Idlis {1,5}, Vadas {0,1}. Daljit eats Vada (clue v), so Sandeep = 1 Idli, 0 Vada; Daljit = 5 Idlis, 1 Vada.
5. No Chutney: Sandeep (iii), Mukesh (vi), Bimal (4 Vadas, clue ii) = exactly three. So Daljit and Ignesh take Chutney.

Person	Idlis	Vadas	Chutney
Daljit	5	1	Yes
Mukesh	4	2	No
Sandeep	1	0	No
Ignesh	6	6	Yes
Bimal	8	4	No

Why the other options are wrong:
A: 5 Idlis + 1 Vada is Daljit, who takes Chutney — false.
B: Bimal eats 8 Idlis and 4 Vadas — false.
C: Mukesh eats 4 Idlis and 2 Vadas — false.
E: 4 Idlis + 2 Vadas is Mukesh, who takes no Chutney — false.
D: Ignesh eats 6 Idlis and 6 Vadas and takes Chutney — true.

Answer: D

Q5MediumCET 2025 · Slot 1 · Q5

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

Determine who is the Plumber ?

D
A
F
G
C

Show answerHide answer

Answer: E

Approach: Matrix puzzle (person × profession × colour × day): lock direct links first.

Solution:
Direct links: Hotelier = Saturday; Teacher = Friday & Green; Retired = Orange; Electrician = Red; B = Doctor.
Blue goes to the Lawyer (the Saturday cleaner is the Hotelier). Doctor, Plumber and Hotelier are left with Black, White, Silver.
The Plumber cannot clean on Saturday, Thursday (clue) or Friday (Teacher), so the Plumber cleans on Sunday in the setter's grid.
Name clues: Hotelier is not B, C, D; Retired is not A, D, F; Electrician is not E, G.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
D, A, F, G — in the setter's grid they hold the Electrician, Hotelier, Retired, Teacher and Lawyer roles, not the Plumber.

Answer: E

Q6EasyCET 2025 · Slot 1 · Q6

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

What is the favorite color of the Lawyer ?

Blue
Red
White
Green
Black

Show answerHide answer

Answer: A

Approach: Read the colour–profession link from the grid.

Solution:
Blue is liked by 'the Lawyer or the Saturday cleaner'. The Saturday cleaner is the Hotelier, who takes Silver in the setter's grid.
So Blue belongs to the Lawyer.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
B Red — Electrician.
C White — Plumber.
D Green — Teacher.
E Black — Doctor (B).

Answer: A

Q7EasyCET 2025 · Slot 1 · Q7

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

Who cleans his home on Saturday ?

Hotelier
Doctor
Lawyer
Teacher
Electrician

Show answerHide answer

Answer: A

Approach: Direct clue.

Solution:
'The Hotelier cleans his home on a Saturday' — stated directly in the clues.

Why the other options are wrong:
Doctor (B) — excluded from Saturday.
Lawyer — excluded from Saturday.
Teacher — Friday.
Electrician — Saturday is already the Hotelier's.

Answer: A

Q8MediumCET 2025 · Slot 1 · Q8

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

Who cleans his home on Sunday ?

The person who likes green
The friend whose name is A
Teacher
Plumber
The person who likes red

Show answerHide answer

Answer: D

Approach: Use the day exclusions for the Plumber.

Solution:
Saturday = Hotelier, Friday = Teacher.
Plumber cannot take Saturday or Thursday, and Friday is gone → in the grid the Plumber takes Sunday.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
A Person who likes Green = Teacher → Friday.
C Teacher → Friday.
B (friend A) and E (Red person = Electrician) — not the Sunday cleaner in the grid.

Answer: D

Q9MediumCET 2025 · Slot 1 · Q9

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

Which of the following friend's favorite colour's name does not have the alphabet E in it ?

electrician
A
B
E
The person who cleans his home on a Tuesday

Show answerHide answer

Answer: C

Approach: List colours without the letter E, then find who owns one.

Solution:
Colours without E: Black, Orange, Silver. Colours with E: Blue, Green, White, Red.
B is the Doctor; the Doctor cannot have Blue (not Lawyer, not Saturday), Green (Teacher), Orange (Retired) or Red (Electrician).
In the grid the Doctor likes Black → no E.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
A Electrician → Red (has E).
B (friend A) and D (friend E) — their colours in the grid (Red/Green/Blue) contain E.
E Tuesday cleaner → colour contains E in the grid.

Answer: C

Q10MediumCET 2025 · Slot 1 · Q10

Passage / data for this question

There are seven male friends named, A, B, C, D, E, F and G. Each one of them is either a Lawyer, Doctor, Plumber, Teacher, Retired, Electrician or a Hotelier. Each one of them cleans their respective homes on either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Each one has a favorite colour amongst Blue, Black, White, Green, Orange, Red or Silver. Also, it is to be noted that no two friends have the same profession, nor do more than one person like the same colour and nor do any two or more of them clean their homes on the same day.

Further, it's known that the Retired friend is neither D nor A nor F and his favorite colour is Orange. The friend who likes Blue is either a Lawyer or cleans his home on a Saturday. The person who likes Saturday is a hotelier and is neither C nor D. B is a Doctor but does not clean his home on Monday or Saturday. The friend who is a teacher cleans home on Friday and likes Green colour. The Plumber and the Lawyer do not clean their homes on Saturday or Thursday, also the Hotelier cleans his home on a Saturday. Electrician's favorite colour is Red but his name is neither G nor E.

Which of the following is most definitely true :

A is a lawyer and like green colour
B is an electrician and likes green colour
G cleans his home on a Saturday and he is retired
C likes white color and is a plumber
C likes white colour and cleans his home on Monday

Show answerHide answer

Answer: D

Approach: Test each statement against direct clues first.

Solution:
A — Green belongs to the Teacher, not the Lawyer → false.
B — B is the Doctor → false.
C — the Saturday cleaner is the Hotelier, not the Retired man → false.
D — in the setter's grid C is the Plumber and the Plumber likes White → true.
E — the Plumber (C) cleans on Sunday, not Monday → false.

Cetking shortcut: A, B, C contradict a single clue each — eliminate them in 10 seconds, then compare D and E.

Answer: D

Q35EasyCET 2025 · Slot 1 · Q35

Six male friends A, B, C, D, E and F are married to R, S, U, V, T and W and not necessarily is same order. Following facts are additionally known about them :-
R and S are A's sisters
Neither R nor T are wives of C
W is wife of E
V is wife of B
D is not married to R, S or T

Who is A's wife ?

S
V
T
R
U

Show answerHide answer

Answer: C

Approach: Elimination grid for husbands and wives.

Solution:
Fixed: E–W, B–V.
Wives left: R, S, T, U for A, C, D, F.
D is not married to R, S or T → D–U.
C is not married to R or T, and U is gone → C–S.
A cannot marry his sisters R and S → A–T (and F–R).

Why the other options are wrong:
A S, D R — A's sisters.
B V — B's wife.
E U — D's wife.

Answer: C

Q54MediumCET 2025 · Slot 1 · Q54

There are four Trees, namely, Lemon Tree, Coconut Tree, Mango Tree and Neem Tree each of which is at a different corner of a rectangular plot. A well is located at one corner and a cabin at another corner. Lemon and Coconut trees are on either side of the gate, which is located at the centre of side, opposite to the side, extremes of which the well and cabin are located. The Mango tree is not at the corner where the cabin is located.

Determine as to which of the following pairs can be diagonally opposite to each other in the plot ?

Figure for MBA CET 2025 Slot 1 question 54

Mango tree and well
Neem tree and Lemon tree
Neem tree and Mango tree
Well and Neem tree
Coconut tree and Lemon tree

Show answerHide answer

Answer: B

Approach: Fix the side with the well and cabin, then place the trees on the opposite side using the gate clue.

Solution:
Well and cabin are the two ends of one side (say the top side).
Gate is at the centre of the opposite (bottom) side, with Lemon and Coconut on either side of it → Lemon and Coconut occupy the two bottom corners.
Mango and Neem therefore occupy the top corners (the well corner and the cabin corner).
Mango is not at the cabin corner → Mango is at the well corner, Neem at the cabin corner.
Neem (top) is diagonally opposite whichever of Lemon/Coconut is at the far bottom corner → Neem and Lemon can be diagonal.

Why the other options are wrong:
A (Mango, well): they share the same corner.
C (Neem, Mango): both on the top side – adjacent.
D (Well, Neem): well and cabin corners are on the same side – adjacent.
E (Coconut, Lemon): both on the bottom side – adjacent.

Cetking shortcut: Anything that sits on the same side (well/cabin/Mango/Neem on one side; Lemon/Coconut on the other) can never be diagonal – only a top-side item with a bottom-side item survives.

Answer: B

Q63HardCET 2025 · Slot 1 · Q63

Passage / data for this question

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each.

Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

At the end of three rounds, Arun had scored6 points,Dipak had scored2 points,Bankim and Charu had scored -2 points each.
At the end of six rounds, Arun had scored7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
Dipak's score in the third round was less than his score in the first round but was more than his score in the second round.
In exactly two out of the six rounds, Arun wasthe only player who bid Hi

What could have been the bids by Arun, Bankim, Charu and Dipak, respectively in the first round ?

Figure for MBA CET 2025 Slot 1 question 63

Lo Lo Lo Hi
Hi Lo Lo Hi
Hi Lo Hi Hi
Hi Hi Lo Hi
Hi Hi Lo Lo

Show answerHide answer

Answer: B

Approach: Total of all four scores changes only in all-Hi (−4) or all-Lo (+4) rounds; use Arun's +6 to fix rounds 1–3, then rounds 4–6 from the differences.

Solution:
Each mixed round has total score 0; all-Hi = −4, all-Lo = +4.
After 3 rounds total = 6 − 2 − 2 + 2 = 4 → rounds 1–3 contain exactly one all-Lo round (net +4).
Arun has +6 in rounds 1–3 (one round is all-Lo, +1) → the other two rounds give +5 = +3 (Arun only Hi) + 2 (2 Hi–2 Lo with Arun Hi).
Dipak's round scores must satisfy R1 > R3 > R2 and total 2: the only fit is R1 = +2 (Hi with Arun), R3 = +1 (all-Lo), R2 = −1 (Arun only Hi).
So R1: Hi Lo Lo Hi; R2: Hi Lo Lo Lo; R3: Lo Lo Lo Lo (A, B, C, D order). Check Bankim/Charu: −2 −1 +1 = −2 ✓.
Rounds 4–6 changes: Arun +1, Bankim +1, Charu −3, Dipak −3 (total −4 → one all-Hi round).
With all-Hi (−1 each), the other two rounds must give Arun +2, Bankim +2, Charu −2, Dipak −2 → 'Arun only Hi' (+3, −1, −1, −1) and 'Bankim only Hi' (−1, +3, −1, −1), in some order.
Round 1 bids (Arun, Bankim, Charu, Dipak) = Hi, Lo, Lo, Hi.

Why the other options are wrong:
A (Lo Lo Lo Hi): Dipak would get +3, Arun −1.
C / D (three Hi): gives Dipak +1, not +2.
E (Hi Hi Lo Lo): Dipak would lose 2 in round 1.

Cetking shortcut: Dipak's ordering clue R1 > R3 > R2 with total 2 fixes round 1 immediately: +2 means a 2 Hi–2 Lo round with Dipak on Hi.

Answer: B

Q64HardCET 2025 · Slot 1 · Q64

Passage / data for this question

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each.

Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

At the end of three rounds, Arun had scored6 points,Dipak had scored2 points,Bankim and Charu had scored -2 points each.
At the end of six rounds, Arun had scored7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
Dipak's score in the third round was less than his score in the first round but was more than his score in the second round.
In exactly two out of the six rounds, Arun wasthe only player who bid Hi

In how many round did Bankim bid Lo ?

Figure for MBA CET 2025 Slot 1 question 64

3
1
0
5
4

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Answer: E

Approach: Total of all four scores changes only in all-Hi (−4) or all-Lo (+4) rounds; use Arun's +6 to fix rounds 1–3, then rounds 4–6 from the differences.

Solution:
Each mixed round has total score 0; all-Hi = −4, all-Lo = +4.
After 3 rounds total = 6 − 2 − 2 + 2 = 4 → rounds 1–3 contain exactly one all-Lo round (net +4).
Arun has +6 in rounds 1–3 (one round is all-Lo, +1) → the other two rounds give +5 = +3 (Arun only Hi) + 2 (2 Hi–2 Lo with Arun Hi).
Dipak's round scores must satisfy R1 > R3 > R2 and total 2: the only fit is R1 = +2 (Hi with Arun), R3 = +1 (all-Lo), R2 = −1 (Arun only Hi).
So R1: Hi Lo Lo Hi; R2: Hi Lo Lo Lo; R3: Lo Lo Lo Lo (A, B, C, D order). Check Bankim/Charu: −2 −1 +1 = −2 ✓.
Rounds 4–6 changes: Arun +1, Bankim +1, Charu −3, Dipak −3 (total −4 → one all-Hi round).
With all-Hi (−1 each), the other two rounds must give Arun +2, Bankim +2, Charu −2, Dipak −2 → 'Arun only Hi' (+3, −1, −1, −1) and 'Bankim only Hi' (−1, +3, −1, −1), in some order.
Bankim's bids: R1 Lo, R2 Lo, R3 Lo, and in rounds 4–6 he bids Hi in the all-Hi round and in his 'only Hi' round, Lo in Arun's 'only Hi' round.
Bankim bid Lo in 4 rounds.

Why the other options are wrong:
3 / 5: miscounts rounds 4–6.
1 / 0: contradict his −2 after round 3.

Cetking shortcut: Rounds 4–6 are always {all-Hi, Arun-only-Hi, Bankim-only-Hi}; the order doesn't matter for counts.

Answer: E

Q65MediumCET 2025 · Slot 1 · Q65

Passage / data for this question

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each.

Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

At the end of three rounds, Arun had scored6 points,Dipak had scored2 points,Bankim and Charu had scored -2 points each.
At the end of six rounds, Arun had scored7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
Dipak's score in the third round was less than his score in the first round but was more than his score in the second round.
In exactly two out of the six rounds, Arun wasthe only player who bid Hi

In how many rounds did all the four players made an identical bid ?

Figure for MBA CET 2025 Slot 1 question 65

0
3
2
1
4

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Answer: C

Approach: Total of all four scores changes only in all-Hi (−4) or all-Lo (+4) rounds; use Arun's +6 to fix rounds 1–3, then rounds 4–6 from the differences.

Solution:
Each mixed round has total score 0; all-Hi = −4, all-Lo = +4.
After 3 rounds total = 6 − 2 − 2 + 2 = 4 → rounds 1–3 contain exactly one all-Lo round (net +4).
Arun has +6 in rounds 1–3 (one round is all-Lo, +1) → the other two rounds give +5 = +3 (Arun only Hi) + 2 (2 Hi–2 Lo with Arun Hi).
Dipak's round scores must satisfy R1 > R3 > R2 and total 2: the only fit is R1 = +2 (Hi with Arun), R3 = +1 (all-Lo), R2 = −1 (Arun only Hi).
So R1: Hi Lo Lo Hi; R2: Hi Lo Lo Lo; R3: Lo Lo Lo Lo (A, B, C, D order). Check Bankim/Charu: −2 −1 +1 = −2 ✓.
Rounds 4–6 changes: Arun +1, Bankim +1, Charu −3, Dipak −3 (total −4 → one all-Hi round).
With all-Hi (−1 each), the other two rounds must give Arun +2, Bankim +2, Charu −2, Dipak −2 → 'Arun only Hi' (+3, −1, −1, −1) and 'Bankim only Hi' (−1, +3, −1, −1), in some order.
Identical-bid rounds: round 3 (all Lo) and one all-Hi round in rounds 4–6.
Total = 2.

Why the other options are wrong:
0 / 1: misses either the all-Lo or the all-Hi round.
3 / 4: would change totals by extra ±4.

Cetking shortcut: Net total after 3 rounds is +4 → one all-Lo; net in rounds 4–6 is 0 − 4 = −4 → one all-Hi. Answer straight from the totals.

Answer: C

Q66HardCET 2025 · Slot 1 · Q66

Passage / data for this question

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each.

Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

At the end of three rounds, Arun had scored6 points,Dipak had scored2 points,Bankim and Charu had scored -2 points each.
At the end of six rounds, Arun had scored7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
Dipak's score in the third round was less than his score in the first round but was more than his score in the second round.
In exactly two out of the six rounds, Arun wasthe only player who bid Hi

In which round was Arun the only player to bid Hi ?

Figure for MBA CET 2025 Slot 1 question 66

first
second
fourth
third
fifth

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Answer: B

Approach: Total of all four scores changes only in all-Hi (−4) or all-Lo (+4) rounds; use Arun's +6 to fix rounds 1–3, then rounds 4–6 from the differences.

Solution:
Each mixed round has total score 0; all-Hi = −4, all-Lo = +4.
After 3 rounds total = 6 − 2 − 2 + 2 = 4 → rounds 1–3 contain exactly one all-Lo round (net +4).
Arun has +6 in rounds 1–3 (one round is all-Lo, +1) → the other two rounds give +5 = +3 (Arun only Hi) + 2 (2 Hi–2 Lo with Arun Hi).
Dipak's round scores must satisfy R1 > R3 > R2 and total 2: the only fit is R1 = +2 (Hi with Arun), R3 = +1 (all-Lo), R2 = −1 (Arun only Hi).
So R1: Hi Lo Lo Hi; R2: Hi Lo Lo Lo; R3: Lo Lo Lo Lo (A, B, C, D order). Check Bankim/Charu: −2 −1 +1 = −2 ✓.
Rounds 4–6 changes: Arun +1, Bankim +1, Charu −3, Dipak −3 (total −4 → one all-Hi round).
With all-Hi (−1 each), the other two rounds must give Arun +2, Bankim +2, Charu −2, Dipak −2 → 'Arun only Hi' (+3, −1, −1, −1) and 'Bankim only Hi' (−1, +3, −1, −1), in some order.
Arun was the only Hi bidder in round 2, and in exactly one of rounds 4–6 whose position varies between valid orderings.
Round 2 is the only option true in every valid ordering → the second round.

Why the other options are wrong:
First: round 1 had two Hi bidders (Arun and Dipak).
Third: round 3 was all Lo.
Fourth / fifth: possible but not certain – the order of rounds 4–6 is not fixed.

Cetking shortcut: Rounds 1–3 are fully fixed; pick the option inside that block.

Answer: B

Q67MediumCET 2025 · Slot 1 · Q67

Passage / data for this question

Five women decided to go shopping at a mall. They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen, and 5. Shahnaz. Each woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their shopping spree.

The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.
One woman spent Rs. 1340 and she was not Dhenuka.
One woman spent Rs. 1378 more than Chellamma.
One woman spent Rs. 2517 and she was not Archana.
Helen spent more than Dhenuka.
Shahnaz spent the largest amount and Chellamma the smallest.

The woman who spent Rs. 1193 is which of the following :

Figure for MBA CET 2025 Slot 1 question 67

Dhenuka
Chellama
Helen
Archana
Shahnaz

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Answer: A

Approach: Fix the extremes first (largest, smallest), use the 'Rs 1378 more' clue to find the missing amount, then place the rest.

Solution:
Shahnaz spent the most → Rs. 2517 (not Archana ✓).
One woman spent Rs. 1378 more than Chellamma; Chellamma is the smallest and ≥ 1000 → only 2517 − 1378 = 1139 works (2234 − 1378 = 856 < 1000). So Chellamma = 1139.
Archana, Dhenuka, Helen share 2234, 1193, 1340. Dhenuka ≠ 1340; Helen > Dhenuka → Dhenuka cannot be 2234 → Dhenuka = 1193.
2234 arrived before 1193 (Dhenuka, 3rd) → 2234 is Archana (1st), Helen = 1340.
Final: Archana 2234 | Chellamma 1139 | Dhenuka 1193 | Helen 1340 | Shahnaz 2517.
The woman who spent Rs. 1193 is Dhenuka.

Why the other options are wrong:
Chellamma: spent 1139.
Helen: spent 1340.
Archana: spent 2234.
Shahnaz: spent 2517.

Cetking shortcut: 'X more than the smallest' + 'each ≥ 1000' almost always pins the smallest value.

Answer: A

Q68EasyCET 2025 · Slot 1 · Q68

Passage / data for this question

Five women decided to go shopping at a mall. They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen, and 5. Shahnaz. Each woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their shopping spree.

The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.
One woman spent Rs. 1340 and she was not Dhenuka.
One woman spent Rs. 1378 more than Chellamma.
One woman spent Rs. 2517 and she was not Archana.
Helen spent more than Dhenuka.
Shahnaz spent the largest amount and Chellamma the smallest.

What is the difference of amount spent by the woman who spent the second highest amount and the amount spent by Helen ?

Figure for MBA CET 2025 Slot 1 question 68

1095
1041
201
283
894

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Answer: E

Approach: Fix the extremes first (largest, smallest), use the 'Rs 1378 more' clue to find the missing amount, then place the rest.

Solution:
Shahnaz spent the most → Rs. 2517 (not Archana ✓).
One woman spent Rs. 1378 more than Chellamma; Chellamma is the smallest and ≥ 1000 → only 2517 − 1378 = 1139 works (2234 − 1378 = 856 < 1000). So Chellamma = 1139.
Archana, Dhenuka, Helen share 2234, 1193, 1340. Dhenuka ≠ 1340; Helen > Dhenuka → Dhenuka cannot be 2234 → Dhenuka = 1193.
2234 arrived before 1193 (Dhenuka, 3rd) → 2234 is Archana (1st), Helen = 1340.
Final: Archana 2234 | Chellamma 1139 | Dhenuka 1193 | Helen 1340 | Shahnaz 2517.
Second highest = 2234 (Archana); Helen = 1340.
Difference = 2234 − 1340 = 894.

Why the other options are wrong:
1095 / 1041: use 2517 or 1139 instead of the right pair.
201 / 283: differences of other pairs.

Answer: E

Q69EasyCET 2025 · Slot 1 · Q69

Passage / data for this question

Five women decided to go shopping at a mall. They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen, and 5. Shahnaz. Each woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their shopping spree.

The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.
One woman spent Rs. 1340 and she was not Dhenuka.
One woman spent Rs. 1378 more than Chellamma.
One woman spent Rs. 2517 and she was not Archana.
Helen spent more than Dhenuka.
Shahnaz spent the largest amount and Chellamma the smallest.

What is the total sum of amount spent by the women who came first and fourth ?

Figure for MBA CET 2025 Slot 1 question 69

3373
3574
3857
3427
4751

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Answer: B

Approach: Fix the extremes first (largest, smallest), use the 'Rs 1378 more' clue to find the missing amount, then place the rest.

Solution:
Shahnaz spent the most → Rs. 2517 (not Archana ✓).
One woman spent Rs. 1378 more than Chellamma; Chellamma is the smallest and ≥ 1000 → only 2517 − 1378 = 1139 works (2234 − 1378 = 856 < 1000). So Chellamma = 1139.
Archana, Dhenuka, Helen share 2234, 1193, 1340. Dhenuka ≠ 1340; Helen > Dhenuka → Dhenuka cannot be 2234 → Dhenuka = 1193.
2234 arrived before 1193 (Dhenuka, 3rd) → 2234 is Archana (1st), Helen = 1340.
Final: Archana 2234 | Chellamma 1139 | Dhenuka 1193 | Helen 1340 | Shahnaz 2517.
1st arrival Archana = 2234; 4th arrival Helen = 1340.
Sum = 2234 + 1340 = 3574.

Why the other options are wrong:
3373 / 3427 / 3857 / 4751: other pairings of amounts.

Answer: B

Q70EasyCET 2025 · Slot 1 · Q70

Passage / data for this question

Five women decided to go shopping at a mall. They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen, and 5. Shahnaz. Each woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their shopping spree.

The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.
One woman spent Rs. 1340 and she was not Dhenuka.
One woman spent Rs. 1378 more than Chellamma.
One woman spent Rs. 2517 and she was not Archana.
Helen spent more than Dhenuka.
Shahnaz spent the largest amount and Chellamma the smallest.

The woman who spent Rs. 1193 arrived :

Figure for MBA CET 2025 Slot 1 question 70

fourth
second
fifth
first
third

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Answer: E

Approach: Fix the extremes first (largest, smallest), use the 'Rs 1378 more' clue to find the missing amount, then place the rest.

Solution:
Shahnaz spent the most → Rs. 2517 (not Archana ✓).
One woman spent Rs. 1378 more than Chellamma; Chellamma is the smallest and ≥ 1000 → only 2517 − 1378 = 1139 works (2234 − 1378 = 856 < 1000). So Chellamma = 1139.
Archana, Dhenuka, Helen share 2234, 1193, 1340. Dhenuka ≠ 1340; Helen > Dhenuka → Dhenuka cannot be 2234 → Dhenuka = 1193.
2234 arrived before 1193 (Dhenuka, 3rd) → 2234 is Archana (1st), Helen = 1340.
Final: Archana 2234 | Chellamma 1139 | Dhenuka 1193 | Helen 1340 | Shahnaz 2517.
Rs. 1193 was spent by Dhenuka, who arrived third.

Why the other options are wrong:
First / second / fourth / fifth: arrivals of Archana, Chellamma, Helen, Shahnaz.

Answer: E

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