Q&A: practice questions with answers

Concept: $MP\times (1-d)=CP\times (1+p)$; find MP in terms of CP, then apply the new discount.

1. $0.75MP=1.1CP\Rightarrow MP=\frac{1.1}{0.75}CP=\frac{22}{15}CP$.
2. With $10\%$ discount: SP $=0.9\times \frac{22}{15}CP=1.32CP$.
3. Profit $=32\%$.

Trick: Take CP $=75$: MP $=110$; SP at 10% off $=99$; profit $=24$ on 75 $=32\%$.

Answer: (D) 32%

Concept: Translate the two statements into equations.

1. Keeran $=k$, Ramu $=k+8$.
2. Seven years ago: $k+1=3(k-7)\Rightarrow k+1=3k-21\Rightarrow k=11$.
3. Ramu $=11+8=19$ years.

Trick: Seven years ago the difference (8) was twice Keeran's age → Keeran was 4, Ramu 12; add 7 → 19.

Answer: (C) 19 years

Concept: Discount of ₹2 on ₹30 is a $\frac{2}{30}=\frac{1}{15}$ discount on the marked price.

1. Let CP $=100$ → MP $=120$.
2. SP $=120\times \frac{14}{15}=112$.
3. Gain $=12\%$.

Trick: Take CP $=25$: MP $=30$, pay $28$ → gain $3$ on $25$ $=12\%$.

Answer: (D) 12% gain

Concept: Convert the mixed fraction to a decimal and add.

1. $1\frac{1}{4}=1.25$.
2. $0.75+1.25=2.00$ cups.

Yaad rakhein: $\frac{1}{4}=0.25$, $\frac{3}{4}=0.75$ – they pair up to make 1.

Answer: (C) 2.0

Concept: $169={13}^2$ and $2.56={1.6}^2$, so numerator and denominator cancel neatly.

1. $(169{)}^2={13}^4$, $(2.56{)}^2={1.6}^4$ → these cancel with the denominator.
2. Left: $\sqrt{\frac{(0.25{)}^2}{(0.5{)}^2}}=\frac{0.25}{0.5}=0.5$.

Trick: Spot perfect squares ($169$, $2.56$) before multiplying anything.

Answer: (D) 0.5

Concept: Find the letter gaps inside each cluster (A=1 … Z=26, wrap-around allowed).

1. AWY: $A-4=W$, $W+2=Y$
2. CYA: $C-4=Y$, $Y+2=A$; EAC: $E-4=A$, $A+2=C$
3. OLM: $O-4=K$, but the second letter is L ($O-3$) → breaks the pattern.

Trick: Check the first gap of every option first – OLM fails at once.

Answer: (C) OLM

Concept: Track each letter position separately.

1. 1st letters: C, I, O, U → $+6$ each → next $U+6=A$.
2. 2nd letters: I, O, U, A → $+6$ → next $G$.
3. 3rd letters: F, L, R, X → $+6$ → next $D$. So the term is AGD.

Trick: All three positions move by the same $+6$ – find one, apply to all.

Answer: (B) AGD

Concept: Write the word and its alphabetical arrangement one below the other and compare position by position.

1. COMBINE: C O M B I N E
2. Alphabetical: B C E I M N O
3. Only position 6 (N) matches; all other positions differ. So 1 letter stays unchanged.

Trick: Alphabetical order of COMBINE: B C E I M N O – compare column-wise.

Answer: (A) 1

Concept: Find the shift applied to each letter.

1. DVUK → GZZQ: $+3,+4,+5,+6$
2. PLOI → SPTO: $+3,+4,+5,+6$ (same rule)
3. YXDA: $Y+3=B$, $X+4=B$, $D+5=I$, $A+6=G$ → BBIG

Trick: Increasing shifts $+3,+4,+5,+6$; remember wrap-around after Z.

Answer: (B) BBIG

Concept: Fix boxes with exact positions first (1 = top, 7 = bottom).

1. Two boxes above F → F = 3. Three boxes below C → C = 4.
2. E immediately above F → E = 2; E below A → A = 1.
3. B above G, below D, in 5–7 → D = 5, B = 6, G = 7.
4. Between A (1) and C (4): E and F → two boxes.

Trick: Count 'above' clues from the top and 'below' clues from the bottom.

Answer: (B) Two

Concept: Convert both positions to the same end, then subtract.

1. Kl from the left $=195-36+1=160$th.
2. Rt is 143rd from the left.
3. People between $=160-143-1=16$.

Trick: Between two ranks from the same end $=$ difference $-1$.

Answer: (D) 16

Concept: Compare each letter of the first cluster with the matching letter of the second.

1. GY–FV: $G-1=F$, $Y-3=V$; TL–SI: $T-1=S$, $L-3=I$
2. AI–ZF: $A-1=Z$, $I-3=F$ (wrap-around)
3. QC–PY: $Q-1=P$, but $C-3=Z$, not Y ($C-4$) → odd one.

Trick: Pattern $(-1,-3)$; only QC–PY breaks it at the second letter.

Answer: (D) QC - PY

Concept: Letters common to both words share the codes common to both codes.

1. MOPS → 5863; POEM → 8765.
2. Common letters M, O, P ↔ common digits 5, 6, 8.
3. Leftover: S ↔ 3 in the first word, E ↔ 7 in the second.

Trick: Remove common letters/digits – the leftover pairs are fixed.

Answer: (C) 7

Concept: Vowels → $+2$, consonants → $-1$.

1. V O L U N T A R Y
2. → U Q K W M S C Q X ($O+2=Q$, $R-1=Q$)
3. Q appears twice; M, W, K appear once each.

Trick: Watch for a vowel and a consonant landing on the same letter (O and R both → Q).

Answer: (B) Q

Concept: Each letter position follows its own constant jump.

1. 1st: M, P, S, V → $+3$ → Y
2. 2nd: A, E, I, M → $+4$ → Q
3. 3rd: I, N, S, X → $+5$ → $X+5=C$ (wrap-around). Term: YQC.

Trick: Jumps $+3$, $+4$, $+5$ for the three positions.

Answer: (A) YQC

Concept: Replace the symbols first, then apply BODMAS.

1. $9-4+45\times 5÷3$ becomes $9\times 4-45÷5+3$
2. $=36-9+3$
3. $=30$

Trick: Write the new expression fully before calculating – avoids mixing old and new signs.

Answer: (B) $30$

Concept: Common letters ↔ common digits.

1. FAME → 8341; FEAL → 6438.
2. Common letters F, A, E ↔ common digits 3, 4, 8.
3. Leftover: M ↔ 1 and L ↔ 6.

Trick: Strike out the common part; the remaining letter takes the remaining digit.

Answer: (A) 6

Concept: Decode each symbol and draw the family tree.

1. H – I: H is the brother of I. I + O: I is the wife of O.
2. O * D: O is the father of D, so I is D's mother. D × J: D is the mother of J.
3. So I is J's mother's mother, and H (her brother) is J's mother's mother's brother.

Trick: Go from J upwards: mother (D) → her mother (I) → her brother (H).

Answer: (A) Mother's mother's brother

Concept: Find the backward shift from each letter of the first cluster to the second.

1. OP–EA: $O-10=E$, $P-15=A$; UV–KG: $U-10=K$, $V-15=G$
2. MN–CY: $M-10=C$, $N-15=Y$ (wrap-around)
3. JK–WV: $J+13=W$, $K+11=V$ → does not follow $(-10,-15)$.

Trick: First clusters are consecutive letters; the shifts $-10$ and $-15$ are fixed.

Answer: (C) JK - WV

Concept: Interchange symbols as told, then use BODMAS.

1. $38-25÷133+703\times 19$ becomes $38\times 25+133-703÷19$
2. $703÷19=37$, $38\times 25=950$
3. $950+133-37=1046$

Trick: $38\times 25=\frac{3800}{4}=950$ – quick mental shortcut.

Answer: (A) $1046$