Q&A: practice questions with answers

Concept: To undo an increase of $r\%$, reduce by $\frac{r}{100+r}\times 100\%$.

1. $\frac{25}{125}\times 100=20\%$.
2. Check: $125\times 0.8=100$ ✓.

Yaad rakhein: $25\%=\frac{1}{4}$ up → $\frac{1}{5}=20\%$ down; $20\%$ up → $16\frac{2}{3}\%$ down.

Answer: (B) 20%

Concept: Dividend $=$ Divisor $\times$ Quotient $+$ Remainder.

1. Divisor $=4\times 15=60$; quotient $=\frac{60}{5}=12$.
2. Dividend $=60\times 12+15=735$.

Trick: Remainder 15 must be less than the divisor 60 – a quick sanity check.

Answer: (D) 735

Concept: Take capacity = LCM of the times; add filling rates, subtract emptying rate.

1. Capacity $=\text{LCM}(4,10,10)=20$ units; rates $+5,+2,-2$ per hour → net $5$/h.
2. One-fourth $=5$ units → time $=\frac{5}{5}=1$ hour.

Trick: The 10-h filler and 10-h emptier cancel, leaving just the 4-h tap: quarter of 4 h = 1 h.

Answer: (C) 1

Concept: $x^m\times x^n=x^{m+n}$ and $\frac{x^m}{x^n}=x^{m-n}$.

1. $x^2\times x^3=x^5\Rightarrow a=5$.
2. $\frac{x^2}{x^b}=x^1\Rightarrow 2-b=1\Rightarrow b=1$.
3. $a+b=6$.

Yaad rakhein: Multiply → add powers; divide → subtract powers.

Answer: (B) 6

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$ (never the average of speeds).

1. Part 1: $40$ km at $20$ km/h → $2$ h.
2. Part 2: $3$ h at $40$ km/h → $120$ km.
3. Average $=\frac{40+120}{2+3}=\frac{160}{5}=32\text{km/h}$.

Trick: Time-weighted: $\frac{2\times 20+3\times 40}{5}=32$.

Answer: (A) 32 km/h

Concept: Compare corresponding letters of consecutive terms using positions A=1 … Z=26 (wrap around after A).

1. F(6) → Y(25) → R(18) → K(11): each step $-7$ (6−7 = −1 → 25).
2. C(3) → V(22) → O(15) → H(8): each $-7$. Z(26) → S(19) → L(12) → E(5): each $-7$.
3. Next: K(11)−7 = D(4), H(8)−7 = A(1), E(5)−7 = −2 → 24 = X.

Trick: Going back 7 is the same as going forward 19; when the number goes below 1, add 26.

Answer: (A) DAX

Concept: Find the rule applied to each letter and the number separately, then apply it to the new term.

1. BO 7 → EI −7: B(2)+3 = E(5); O(15)−6 = I(9); 7−14 = −7.
2. Check: CS 10 → FM −4: C+3 = F; S(19)−6 = M(13); 10−14 = −4. ✓
3. IJ 20: I(9)+3 = L(12); J(10)−6 = D(4); 20−14 = 6 → LD 6.

Trick: First letter +3, second letter −6, number −14 – verify with the second pair before applying.

Answer: (D) LD 6

Concept: Check the difference between corresponding letters of successive terms.

1. C(3) → F(6) → I(9) → L(12): $+3$ each time.
2. G(7) → J(10) → M(13) → P(16): $+3$. T(20) → W(23) → Z(26) → C(3): $+3$ (wrapping).
3. Next term: L+3 = O, P+3 = S, C+3 = F → OSF.

Trick: When every letter moves by the same +3, just add 3 to the last term.

Answer: (A) OSF

Concept: Number the positions 1 (bottom) to 7 (top) and place the definite clues first.

1. W is second from the bottom → W = 2; T is immediately above W → T = 3.
2. Three boxes between U and W → U = 6.
3. R lies between S and T, and V is neither at top nor bottom → R and V occupy 4 and 5 (in either order), S = 7, so X = 1 (bottom).
4. Between X (1) and T (3) there is only W → one box.

Trick: Start with fixed-position clues ("second from bottom"), then "exactly n between" clues; the leftover box falls into the last empty slot.

Answer: (C) ONE

Concept: Locate every vowel and check its immediate left (must be a consonant) and immediate right (must be a vowel).

1. Series: S Z E C A O Q M D P U I J L N X K D. Vowels: E, A, O, U, I.
2. E: Z before, C after (consonant) ✗. A: C before, O after ✓. O: A before (vowel) ✗.
3. U: P before, I after ✓. I: U before (vowel) ✗.
4. Pairs found: C-A-O and P-U-I → two.

Trick: Look only at adjacent vowel pairs (AO, UI); the first vowel of each such pair qualifies if a consonant precedes it.

Answer: (A) Two

Concept: Add the alphabetical positions of the letters and see how the sum relates to the code.

1. SFOR: S(19)+F(6)+O(15)+R(18) = 58; $58\times 2=116$ ✓.
2. WILA: W(23)+I(9)+L(12)+A(1) = 45; $45\times 2=90$ ✓.
3. GMVC: G(7)+M(13)+V(22)+C(3) = 45; $45\times 2=90$.

Trick: Same letter-sum as WILA (45) → same code 90; no need to multiply again.

Answer: (D) 90

Concept: Find one rule connecting the first two numbers to the third in both given sets.

1. (5, 3, 16): $5\times 3+1=16$. (6, 3, 19): $6\times 3+1=19$. Rule: $a\times b+1=c$.
2. (3, 5, 15): $16\ne 15$ ✗. (1, 7, 8): $1\times 7+1=8$ ✓.
3. (7, 3, 20): $22$ ✗. (4, 6, 21): $25$ ✗.

Trick: Third number is just one more than the product – test $a\times b+1$ on each option.

Answer: (B) (1, 7, 8)

Concept: Decode the chain from left to right and build a family tree.

1. G ÷ H: G is the sister of H. H + I: H is the husband of I.
2. I × T: I is the mother of T, so H is T's father. T − R: T is the brother of R.
3. Hence H is R's father and G, being H's sister, is R's father's sister.

Trick: Follow the chain: the person just before "mother/father" symbols is the parent; the sibling of a parent is an aunt/uncle.

Answer: (D) Father's sister

Concept: Number positions 1 (bottom) to 7 (top); use the fixed clues first.

1. No box below C → C = 1. Two boxes between E and C → E = 4.
2. Only Q is above D → Q = 7, D = 6.
3. Remaining 2, 3, 5 for B, S, R with B above S above R → B = 5, S = 3, R = 2.
4. Immediately above S (3) is position 4 = E.

Trick: "Only X is above Y" fixes two slots at once (top and second from top).

Answer: (B) E

Concept: Number positions 1 (bottom) to 7 (top) and fill the definite clues first.

1. C is second from the bottom → C = 2. Three boxes between C and G → G = 6.
2. F is below C → F = 1.
3. Only H between G and P → P = 4, H = 5 (P cannot be at 8).
4. E is above G → E = 7; the last box D = 3.
5. Between D (3) and E (7): P, H, G → three boxes.

Trick: Count boxes between positions as $|p_1-p_2|-1$; here $7-3-1=3$.

Answer: (A) Three

Concept: Decode each symbol step by step and draw the family tree.

1. G * K: G is the daughter of K. K & V: K is the husband of V → G is the daughter of K and V.
2. V @ P: V is the mother of P → P is G's brother (son of K and V).
3. P # W: P is the father of W. W % J: W is the son of J → J is W's mother, i.e. P's wife.
4. So J is the wife of G's brother → brother's wife.

Trick: Two parents named for the same child (P father of W, W son of J) ⇒ P and J are husband and wife.

Answer: (B) Brother's wife

Concept: Find the position-shift from the first cluster to the second, letter by letter, in each pair.

1. RL–IQ: R(18)→I(9) = $-9$; L(12)→Q(17) = $+5$.
2. VP–MU: V(22)→M(13) = $-9$; P(16)→U(21) = $+5$. CW–TB: C(3)→T(20) = $-9$ (3−9 = −6 → 20); W(23)→B(2) = $+5$ (28 → 2).
3. OI–EN: O(15)→E(5) = $-10$; I(9)→N(14) = $+5$. The first-letter shift differs.

Trick: Pattern (−9, +5) in three pairs; OI–EN breaks it with −10.

Answer: (B) OI – EN

Concept: Take successive differences and look for a pattern in them.

1. 217 → 190 → 165 → 142 → 121 → 102: differences $-27,-25,-23,-21,-19$.
2. Differences decrease by 2 each time, so the next difference is $-17$.
3. $102-17=85$.

Trick: Consecutive odd numbers subtracted (27, 25, 23 …) – a very common second-level pattern.

Answer: (D) 85

Concept: Within each term the letters are alternate (A, C, E), and each term shifts by a fixed amount.

1. A(1) → D(4) → G(7) → J(10): $+3$ each time.
2. C → F → I → L: $+3$; E → H → K → N: $+3$.
3. Next: J+3 = M, L+3 = O, N+3 = Q → MOQ.

Trick: The series is just ACE, DFH, GIK, JLN, MOQ – every letter +3; option MPQ is a trap (P breaks the +2 gap inside the term).

Answer: (C) MOQ

Concept: Find the shift applied to each letter position in the given pairs and apply the same shifts.

1. BGKM → HKMN: B+6 = H, G+4 = K, K+2 = M, M+1 = N → shifts (+6, +4, +2, +1).
2. Check FHNP → LLPQ: F+6 = L, H+4 = L, N+2 = P, P+1 = Q ✓.
3. NRTX: N(14)+6 = T(20), R(18)+4 = V(22), T(20)+2 = V(22), X(24)+1 = Y(25) → TVVY.

Trick: Shifts need not be equal – write them as a list (+6, +4, +2, +1) and reuse.

Answer: (B) TVVY