Q&A: practice questions with answers

Concept: Speed in still water $=\frac{\text{downstream}+\text{upstream}}{2}$

1. Upstream speed $=\frac{176}{16}=11\text{km/h}$
2. Downstream speed $=\frac{85}{5}=17\text{km/h}$
3. Boat speed $=\frac{17+11}{2}=14\text{km/h}$ (stream $=3\text{km/h}$)

Trick: Boat = average of the two speeds; stream = half their difference.

Answer: (C) $14\text{km/h}$

Concept: $\text{CSA}=2\pi rh$, $\text{TSA}=2\pi r(h+r)$

1. $2\pi rh=\frac{1}{2}\times 2\pi r(h+r)$
2. $2h=h+r\Rightarrow r=h=12\text{cm}$
3. Diameter $=2r=24\text{cm}$

Trick: CSA = half of TSA means CSA = area of the two ends, i.e. $2\pi rh=2\pi r^2\Rightarrow h=r$.

Answer: (A) 24

Concept: BODMAS – solve innermost brackets first, then multiplication, then addition, then division.

1. $(5-7)=-2$
2. $9\times (-2)=-18$
3. $\{21-18\}=3$
4. $81÷3=27$

Trick: Watch the sign: the small bracket is negative, so the curly bracket becomes $21-18$.

Answer: (A) 27

Concept: After changing signs, solve using BODMAS.

1. Replace: ‘+’ → ‘−’, ‘×’ ↔ ‘÷’.
2. New expression: $20\times 5-8÷2-1$
3. $=100-4-1=95$

Trick: Rewrite the whole expression first, then apply BODMAS – never solve while changing signs.

Answer: (A) $95$

Concept: Find the highest and lowest of the given numbers, then use their first digits.

1. Numbers: 411, 559, 916, 247, 769.
2. Highest = 916 → first digit 9; Lowest = 247 → first digit 2.
3. $9-2=7$

Trick: Compare only the hundreds digit first – 9 (916) is highest, 2 (247) is lowest.

Answer: (B) $7$

Concept: Find the letter shifts from # to EPV and apply the same to RFS.

1. Try option (D): D → E (+1), N → P (+2), U → V (+1) gives EPV.
2. Apply +1, +2, +1 to RFS: R → S, F → H, S → T = SHT.
3. So # = DNU and % = SHT.

Trick: Check options quickly: only DNU gives EPV with a neat +1, +2, +1 pattern.

Answer: (D) # = DNU, % = SHT

Concept: A letter common to both words gets the digit common to both codes.

1. MICE → 6398 and PEAR → 3215.
2. Common letter: E. Common digit: 3.
3. So E = 3.

Trick: Common letter ↔ common digit (order of digits does not matter).

Answer: (C) $3$

Concept: Check each number: the element before it must be a number and the element after it must be a symbol.

1. 8 9 # → 9 (preceded by 8, followed by #) ✔
2. 1 9 © → 9 ✔; 3 5 © → 5 ✔
3. Other numbers fail (e.g. 5 after ©, 4 followed by 4).
4. Count = 3.

Trick: Scan for the pattern “number–number–symbol”; the middle number is counted.

Answer: (A) $3$

Concept: Fix the people with the most definite clues first (all face north, so left = left end side).

1. Five between G and D → they sit at the two ends. F is immediate left of D, so D = 7 (right end), F = 6, G = 1.
2. One person between F and M → M = 4.
3. N, O, E fill 2, 3, 5 with N left of O left of E → N = 2, O = 3, E = 5.
4. Third from left = O.

Trick: “Only five between” in a row of 7 always means the two ends.

Answer: (D) O

Concept: Interchange the signs, then solve by BODMAS.

1. ‘+’ ↔ ‘−’ and ‘×’ ↔ ‘÷’.
2. New expression: $215\times 4-822÷3+45$
3. $=860-274+45=631$

Trick: Do division/multiplication first: $822÷3=274$, $215\times 4=860$.

Answer: (B) $631$

Concept: Look at the differences between consecutive terms.

1. Differences: $3,9,15,21,27$ (increase by 6).
2. Next difference $=27+6=33$
3. $96+33=129$

Trick: Second difference constant (6) → next first difference is 33.

Answer: (D) $129$

Concept: Use letter positions: in each pair the 2nd letter = 1st + 3, and each pair starts 2 letters before the previous one.

1. QT-OR-MP: Q→T (+3), Q→O→M (−2 each).
2. XA-VY-TW: X→A (+3, wrapping), X→V→T (−2).
3. TW-RU-PS: T→W, R→U, P→S (+3) and T→R→P (−2) ✔
4. JM-HK-FJ fails (F→J is +4).

Trick: Check “+3 inside, −2 across” on each option.

Answer: (C) TW-RU-PS

Concept: Track the net east–west and north–south displacement.

1. South 37, right (west) 18, right (north) 58, left (west) 61, left (south) 21.
2. North–south: $-37+58-21=0$ → Z is level with Y.
3. East–west: $18+61=79$ km west of Y.
4. So he must drive 79 km towards the east.

Trick: When the vertical moves cancel, the answer is just the sum of the horizontal moves.

Answer: (D) 79 km towards the east

Concept: Write letter positions and find the gaps.

1. KNI: 11 → 14 → 9 (+3, −5)
2. MPK: 13 → 16 → 11 (+3, −5); ORM: 15 → 18 → 13 (+3, −5)
3. QTP: 17 → 20 → 16 (+3, −4) ✘

Trick: Here the 3rd letter = 1st letter − 2 (K→I, M→K, O→M); QTP breaks it.

Answer: (C) QTP

Concept: Facing the centre, “left” means clockwise and “right” means anticlockwise.

1. Place Amit; Yug is 2nd clockwise from Amit, Prabhat is 3rd anticlockwise.
2. Monu immediately right of Dinesh: the only free adjacent pair gives Dinesh and Monu the two seats after Prabhat (Monu next to Prabhat).
3. Shipra is not next to Prabhat → Shipra sits between Amit and Yug; Sakshi between Yug and Prabhat.
4. 2nd to the left of Dinesh = Shipra.

Trick: Draw the circle and mark the left (clockwise) arrow before placing anyone.

Answer: (A) Shipra

Concept: Replace each symbol by its meaning, then use BODMAS.

1. ÷ → +, × → ÷, − → ×, + → −
2. $109+76÷4\times 5-18$
3. $=109+19\times 5-18=109+95-18=186$

Trick: Write the decoded expression on one line before calculating.

Answer: (B) $186$

Concept: Track each letter position separately.

1. 1st letters: Q, O, M, K → I (−2)
2. 2nd letters: J, H, F, D → B (−2)
3. 3rd letters: L, J, H, F → D (−2)
4. Next term = IBD

Trick: All three positions move back by 2 every step.

Answer: (A) IBD

Concept: Find net north–south and east–west displacement from A.

1. Moves: N 16, W 8, S 7, W 4, S 12, E 18, N 3.
2. North–south: $16-7-12+3=0$
3. East–west: $-8-4+18=+6$ → P is 6 km east of A.
4. So he must drive 6 km to the west.

Trick: Vertical moves cancel → answer is the net horizontal shift.

Answer: (B) 6 km to the West

Concept: Total $=\text{rank from top}+\text{rank from bottom}-1$.

1. $16+55-1=70$
2. Rajat is counted in both ranks, so subtract 1 once.

Trick: Add the two ranks and subtract 1.

Answer: (A) $70$

Concept: Find the rule linking the middle number to the outer two.

1. $(2,20,4)$: $2^2+4^2=4+16=20$
2. $(5,61,6)$: $5^2+6^2=25+36=61$
3. $(3,25,4)$: $3^2+4^2=9+16=25$ ✔
4. Others: $7^2+2^2=53$, $4^2+6^2=52$, $1^2+3^2=10$ ✘

Trick: Middle = sum of squares of the end numbers.

Answer: (D) $(3,25,4)$