A pen is used to write; a knife is used to cut.

A is B’s mother, and C is A’s daughter, so C is B’s sister.

The sequence follows the pattern of triangular numbers: n(n+1)/2. For n=4, 45/2 = 10; for n=5, 56/2 = 15. Next is 15.

Work rate of 2 workers = 1 job/6 days = 1/6 job/day. Per worker = 1/12 job/day. For 3 workers, rate = 3/12 = 1/4 job/day. Time = 1/(1/4) = 4 days.

Distance = 120 km, speed = 60 km/h, so time = 120/60 = 2 hours. Departure at 9 AM + 2 hours = 11 AM.

Days are numbered sequentially: Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5). Days are numbered sequentially: Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5). Days are numbered sequentially: Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5).

A is not next to B, so A and B can be at the ends (e.g., A-C-D-B). C and D are next to each other, so they cannot both be at the ends.

Y is next to Z, so they are adjacent (e.g., Y-Z or Z-Y). Since X is not next to Y, Y must be in the middle (e.g., X-Y-Z or Z-Y-X).

A is to the left of B, and C is to the right of B, so the order is A, B, C. B is in the middle of this sequence.

Some pens are pencils, and since all pencils are erasers, those pens that are pencils are also erasers. Thus, some pens are erasers.

Each letter in “APPLE” is shifted forward by 2 positions (A→C, P→R, L→N, E→G). Applying the same to “ORANGE” (O→Q, R→T, A→C, N→P, G→I, E→F) gives “QTCPF.” Each letter in “APPLE” is shifted forward by 2 positions (A→C, P→R, L→N, E→G). Applying the same to “ORANGE” (O→Q, R→T, A→C, N→P, G→I, E→F) gives “QTCPF.” Each letter in “APPLE” is shifted forward by 2 positions (A→C, P→R, L→N, E→G). Applying the same to “ORANGE” (O→Q, R→T, A→C, N→P, G→I, E→F) gives “QTCPF.”

All numbers are even and form an arithmetic sequence with a common difference of 2. No number is an outlier. (Note: If an odd number were present, it would be the answer.)

The pattern is (a + b)². For 3 + 5 = 8, 8² = 64, but the result is given as 16 (possibly a typo or specific coding). Assuming the pattern holds, 5 + 7 = 12, so 12² = 144, but the options suggest a simpler pattern. Correcting for consistency, (5 + 7)² = 36.

Each letter in “DOG” is replaced by the next letter in the alphabet (D→E, O→P, G→H). Applying the same rule to “CAT” (C→D, A→B, T→U) gives “DBU.”

P is Q’s brother (male sibling), and Q is R’s sister (female sibling). Thus, P is R’s brother.

Since all roses are flowers, some flowers must be roses. The color red does not affect this conclusion.

If the day after tomorrow is Wednesday, tomorrow must be Tuesday, so today is Monday.

From the given statements, we know all cats are mammals, so some mammals must be cats. The color black is irrelevant to the conclusion.

The pattern is placing vowels at the start in the order they appear and consonants at the end in reversed order. “SARASWATI” → vowels: A, A, A, I → AASS, consonants: S, R, S, W, T → IRWAT Thus, it becomes AASSIRWAT.

Each letter in MONDAY is assigned numbers 1 to 6. The same letters in WEDNESDAY get the same numbers, and new letters are added in sequence. It’s a logical mapping of unique alphabets in a sequence. It tests memory mapping and reasoning.

Placing B as the central point, A is directly south, and C is directly east. From C's point, A lies to the southwest. This is a directional logic problem requiring diagrammatic visualization. Such questions are common in spatial reasoning.

It is a geometric progression: each number is multiplied by 2. 64×2 = 128. It tests pattern recognition under multiplication.