Q&A: practice questions with answers

Concept: Equivalent single change $=a+b+\frac{ab}{100}$.

1. $a=b=15$.
2. $15+15+\frac{15\times 15}{100}=30+2.25=32.25\%$

Trick: ${1.15}^2=1.3225$ → $32.25\%$ increase.

Answer: (C) $32.25\%$

Concept: Under CI, if money becomes $x$ times in $n$ years, it becomes $x^k$ times in $kn$ years.

1. $3$ times in 6 years.
2. $27=3^3$ → time $=3\times 6=18$ years.

Trick: CI multiplies: $\times 3$ every 6 years → $1\to 3\to 9\to 27$ (3 steps = 18 years).

Answer: (D) $18$ years

Concept: If numbers are in ratio $a:b$ (co-prime) with HCF $h$, numbers are $ha$, $hb$ and LCM $=h\times a\times b$.

1. Numbers $=6\times 3=18$ and $6\times 8=48$.
2. LCM $=6\times 3\times 8=144$

Trick: LCM must be a multiple of 6 and 8 (i.e. of 24) → only 144 fits.

Answer: (D) $144$

Concept: Number $=10a+b$, reversed $=10b+a$.

1. $4(a+b)=10b+a\Rightarrow 3a=6b\Rightarrow a=2b$.
2. $a-b=3\Rightarrow b=3$, $a=6$.
3. Number $=63$; check: $4\times 9=36$ ✔

Trick: Test options: only 63 gives $4\times (6+3)=36$ = its reverse.

Answer: (A) $63$

Concept: $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$.

1. CP per kg $=\frac{1000}{20}=50$.
2. Profit per kg $=55-50=5$.
3. $\text{Profit}\%=\frac{5}{50}\times 100=10\%$

Trick: Total SP $=20\times 55=1100$ → ₹100 profit on ₹1000 = 10%.

Answer: (B) $10\%$

Concept: Track each letter position and the number separately.

1. 1st letters: D, J, P, V, B, H – each $+6$ (V → B wraps round).
2. 2nd letters: I, N, S, X, C, H – each $+5$.
3. 3rd letters: X, B, F, J, N, R – each $+4$; 4th letters: C, F, I, L, O, R – each $+3$.
4. Numbers: 6, 12, 18, 24, 30, 36 – each $+6$. So the term is HHRR 36.

Trick: Letter steps $+6,+5,+4,+3$ fall by one from left to right.

Answer: (C) HHRR 36

Concept: Fix the definite positions first (seats 1–7 from left, all facing north).

1. Three people left of B → B is 4th.
2. Only Z is right of T → T = 6th, Z = 7th.
3. Three people between T and O → O = 2nd.
4. D, J, A fill seats 1, 3, 5 with D left of J left of A → D = 1, J = 3, A = 5.
5. Immediately right of B (4th) is A (5th).

Trick: Start with ‘only’ clues – they lock seats at the ends.

Answer: (D) A

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

1. No box below C → C = 7 (bottom).
2. Two boxes between E and C → E = 4.
3. Only Q above D → Q = 1, D = 2.
4. S below B and above R: positions 3, 5, 6 left → B = 3, S = 5, R = 6.
5. Box immediately above S (5) is E (4).

Trick: ‘No box below X’ = X at bottom; ‘only Q above D’ = Q top, D second.

Answer: (D) E

Concept: Number positions 1 (top) to 7 (bottom).

1. No box above M → M = 1.
2. Three boxes between M and H → H = 5; N immediately above H → N = 4.
3. Two boxes between L and H → L = 2 (8 is not possible).
4. Four boxes between L and I → I = 7.
5. G above O in the remaining 3 and 6 → G = 3, O = 6. Second above G (3) is position 1 = M.

Trick: Chain the ‘between’ clues from the fixed box M: M → H → L → I.

Answer: (C) M

Concept: Compare 1st letters of the two parts and 2nd letters of the two parts.

1. HV – KT: $\text{H}+3=\text{K}$, $\text{V}-2=\text{T}$.
2. OA – RY: $\text{O}+3=\text{R}$, $\text{A}-2=\text{Y}$; YT – BR: $\text{Y}+3=\text{B}$, $\text{T}-2=\text{R}$.
3. SZ – VW: $\text{S}+3=\text{V}$ but $\text{Z}-3=\text{W}$ (not $-2$).
4. So SZ – VW is the odd one.

Trick: Pattern $(+3,-2)$; the alphabet wraps (Y → B, A → Y).

Answer: (C) SZ - VW

Concept: Interchange the signs first, then apply BODMAS.

1. Given: $14\times 2-12÷4+8$.
2. After interchange: $14÷2+12\times 4-8$.
3. $=7+48-8=47$.

Trick: Rewrite the whole expression with new signs before calculating.

Answer: (B) 47

Concept: Decode each symbol and draw the family tree.

1. B = C → B is mother of C; C ? O → C is daughter of O.
2. So O is C’s father and B’s husband.
3. O % P → O is father of P, so B is mother of P.
4. P € X → P is wife of X. B is mother of X’s wife.

Trick: Mother of wife = mother-in-law; here read it as ‘Wife’s mother’.

Answer: (D) Wife’s mother

Concept: Arrange the digits in ascending order and pick the positions.

1. 4317856 in ascending order → 1345678.
2. Second from left = 3; second from right = 7.
3. Sum $=3+7=10$.

Trick: With 7 digits, 2nd from right = 6th from left.

Answer: (A) 10

Concept: Check the gap inside each pair and the gap between pairs.

1. JM – NQ – RU: inside each pair $+3$ (J→M, N→Q, R→U); first letters J→N→R are $+4$.
2. BE – FI – JM: same, $+3$ inside and $+4$ between.
3. MP – QT – UX: M→P, Q→T, U→X ($+3$) and M→Q→U ($+4$) ✓.
4. Others break the pattern, e.g. NE has $-9$.

Trick: Pattern: pair gap $+3$, triad step $+4$.

Answer: (C) MP - QT - UX

Concept: Track the 1st, 2nd and 3rd letters separately.

1. 1st: G, J, M, P → $+3$ → S.
2. 2nd: N, R, V, Z → $+4$ → D (Z wraps to D).
3. 3rd: I, L, O, R → $+3$ → U.
4. Next term = SDU.

Trick: Z $+4$ = D because after Z we restart at A.

Answer: (B) SDU

Concept: Find the differences between consecutive terms.

1. Differences: $372-321=51$, $321-277=44$, $277-240=37$, $240-210=30$, $210-187=23$.
2. They decrease by 7 each: 51, 44, 37, 30, 23, next 16.
3. $187-16=171$.

Trick: Second difference is constant ($7$).

Answer: (B) 171

Concept: Find the operation that turns the first number into the second.

1. $123-7=116$ and $102-7=95$ → rule: subtract 7.
2. Check options: $365-357=8$, $413-402=11$, $604-596=8$, $541-534=7$ ✓.
3. So (541, 534) follows the rule.

Trick: Just compute first minus second for each option.

Answer: (D) 541, 534

Concept: Draw the least-overlap Venn diagram and test each conclusion.

1. Brooms lie inside Rings; Lamps are completely outside Brooms.
2. Lamps may or may not touch Rings, so ‘Some lamps are rings’ is not certain → (I) does not follow.
3. ‘All rings are lamps’ is impossible, because rings that are brooms are not lamps → (II) does not follow.

Trick: A conclusion follows only if it is true in every possible diagram.

Answer: (A) Neither conclusion (I) nor (II) follows

Concept: Compare first letters and second letters of the two clusters.

1. SA – LI: $\text{S}-7=\text{L}$, $\text{A}+8=\text{I}$.
2. YG – RO: $\text{Y}-7=\text{R}$, $\text{G}+8=\text{O}$; AI – TQ: $\text{A}-7=\text{T}$, $\text{I}+8=\text{Q}$.
3. OW – HF: $\text{O}-7=\text{H}$ but $\text{W}+8=\text{E}$, not F.
4. So OW – HF is the odd one.

Trick: Rule $(-7,+8)$; W $+8$ wraps: X, Y, Z, A, B, C, D, E.

Answer: (C) OW – HF

Concept: Two ‘All’ statements give a chain: Kings ⊂ Mountains ⊂ Arrows.

1. All kings are mountains → some mountains are kings (conversion of ‘All’) → (I) follows.
2. Kings lie inside Mountains, which lie inside Arrows → all kings are arrows → (II) follows.
3. Both conclusions are true in the diagram.

Trick: All A are B + All B are C ⇒ All A are C, and Some B are A.

Answer: (D) Both conclusions (I) and (II) follow