Q&A: practice questions with answers

Concept: Overall factor = product of individual factors.

1. After 6% loss: factor $0.94$; after 65% gain: factor $1.65$
2. Overall $=0.94\times 1.65=1.551$
3. Gain $=55.1\%$ (₹347 → ₹538.20)

Trick: $-6+65-\frac{6\times 65}{100}=59-3.9=55.1\%$.

Answer: (D) $55.1\%$

Concept: The two cases differ only on the extra 24 km, which is run at full speed in case 2 instead of $\frac{3}{4}$ speed.

1. Time saved $=35-25=10$ min $=\frac{1}{6}$ h
2. $\frac{24}{\frac{3}{4}v}-\frac{24}{v}=\frac{32}{v}-\frac{24}{v}=\frac{8}{v}$
3. $\frac{8}{v}=\frac{1}{6}\Rightarrow v=48\text{km/h}$

Trick: At $\frac{3}{4}$ speed time becomes $\frac{4}{3}$, i.e. extra $\frac{1}{3}$ of normal time: $\frac{1}{3}\times \frac{24}{v}=\frac{1}{6}$.

Answer: (D) 48 km/hr

Concept: Multiply the factors of all changes.

1. Factor $=0.55\times 0.50\times 0.75\times 1.65$
2. $=0.275\times 0.75\times 1.65=0.20625\times 1.65=0.3403$
3. Decrease $=100-34.03=65.97\%$

Trick: Take original = 100: 55 → 27.5 → 20.625 → 34.03.

Answer: (B) $65.97\%$

Concept: Work done = rate $\times$ time; filling pipe positive, emptying pipe negative.

1. Both for 10 h: $10(\frac{1}{15}-\frac{1}{20})=10\times \frac{1}{60}=\frac{1}{6}$
2. Remaining $\frac{5}{6}$ by X: $\frac{5}{6}\times 15=12.5$ h
3. Total $=10+12.5=22.5$ h

Trick: LCM capacity 60 units: X $=+4$, Y $=-3$; 10 h → 10 units; remaining 50 units $÷4=12.5$ h.

Answer: (D) 22.5 hours

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

1. Numbers: $47\times 4=188$ and $47\times 1=47$
2. LCM $=47\times 4\times 1=188$

Trick: When one number divides the other, LCM is the bigger number.

Answer: (B) 188

Concept: Letters common to two words share the common codes (order is jumbled).

1. CHIN → {9, 4, 1, 3}; NICE → {2, 3, 1, 9}.
2. Common letters C, I, N ↔ common codes 9, 1, 3.
3. Left-over in NICE: E ↔ 2 (and H ↔ 4).

Trick: Remove common codes; the leftover code belongs to the leftover letter.

Answer: (D) 2

Concept: Check each letter position separately across the terms.

1. 1st letters: Z, U, P, K, F (each $-5$).
2. 2nd letters: E, Z, U, P, K (each $-5$).
3. 3rd letters: K, F, A, V, Q (each $-5$, wrapping A → V).

Trick: Each term's 1st letter = previous term's 2nd letter, so the answer starts F, K.

Answer: (C) FKQ

Concept: Replace letters with signs (A $÷$, B $\times$, C $+$, D $-$) and apply BODMAS.

1. (C): $85÷5-7\times 5+22$
2. $=17-35+22=4$ ✓
3. (A) $=425-7+\frac{5}{22}\ne 4$; (B) $=85+5-\frac{7}{5}\times 22\ne 4$; (D) $=85-\frac{5}{7}\times 5+22\ne 4$.

Trick: Look for an option where division gives a whole number ($85÷5=17$).

Answer: (C) $85÷5-7\times 5+22=4$

Concept: Compare letter positions within and between the clusters.

1. GJ – FI: $G+3=J$, $F+3=I$, and $G-1=F$ ✓
2. PS – OR and OR – NQ follow the same $+3$ / $-1$ pattern ✓
3. KN – HU: $K+3=N$ but $H+13=U$ (and $K-3=H$) ✗

Trick: Check the gap inside the second cluster first – only HU breaks the $+3$ gap.

Answer: (B) KN - HU

Concept: A word common to two sentences takes the code common to both.

1. ‘uncle steak gum’ → xl wc iv; ‘sir gum barn’ → wc mn ut.
2. Common word: gum; common code: wc.
3. So gum = wc.

Trick: Common word ↔ common code.

Answer: (D) wc

Concept: Facing the centre, a person's left is the clockwise direction.

1. Fix Z; three people between Z and T counting from Z's right → T is fourth to the right of Z (third to the left).
2. B is second to the left of T.
3. Empty seats: two between Z and T (on Z's left), one between T and B, one between B and Z. A immediately left of J needs two adjacent seats → J, A sit between Z and T.
4. D cannot be next to T → D is beside Z on the right; O sits between T and B.
5. From A going left: T, O, then B → 2 people.

Trick: In a circle of 7, “3 between on one side” means 2 between on the other side.

Answer: (B) Two

Concept: Find the shift applied to each letter.

1. CZAZ → VSTS: $C-7=V$, $Z-7=S$, $A-7=T$, $Z-7=S$.
2. EBCB → XUVU: every letter $-7$ again ✓
3. GDED: $G-7=Z$, $D-7=W$, $E-7=X$, $D-7=W$ → ZWXW.

Trick: Same letter in the word (Z, B, D) → same letter in the code; only ZWXW keeps 2nd = 4th with a different 3rd.

Answer: (D) ZWXW

Concept: Test the product of the first two numbers against the third.

1. $17\times 4\times 3=204$ ✓; $19\times 5\times 3=285$ ✓ → third $=$ first $\times$ second $\times 3$.
2. (A) $16\times 8\times 3=384$ ✓
3. (B) $21\times 4\times 3=252$, (C) $15\times 7\times 3=315$, (D) $18\times 6\times 3=324$ ✗

Trick: Third $÷$ (first $\times$ second) is constant: $\frac{204}{68}=\frac{285}{95}=3$.

Answer: (A) (16, 8, 384)

Concept: Start with the clues that fix exact positions (count from the top).

1. Only K above H → K = 1, H = 2.
2. Only two boxes above E → E = 3; E immediately above J → J = 4.
3. I immediately above L → I = 5, L = 6.
4. Between K (1) and I (5): H, E, J → 3 boxes.

Trick: Boxes between positions $p$ and $q$ $=|p-q|-1=5-1-1=3$.

Answer: (D) Three

Concept: Fix top and bottom first, then place the rest by the order clues.

1. Only T above I → T = 1, I = 2. No box below J → J = 7.
2. Two boxes between L and J → L = 4.
3. G above H above K in the free places 3, 5, 6 → G = 3, H = 5, K = 6.
4. Between T (1) and H (5): I, G, L → 3.

Trick: Count gap: $5-1-1=3$.

Answer: (C) Three

Concept: Draw the least-overlap Venn diagram satisfying the statements.

1. All ribbons are goats → Ribbons circle inside Goats.
2. Some crows are ribbons → those crows are inside Goats too → Some crows are goats: (I) follows.
3. Goats may have members that are not crows → (II) ‘All goats are crows’ does not follow.

Trick: “Some A are B” + “All B are C” ⇒ “Some A are C” always.

Answer: (B) Only conclusion (I) follows

Concept: Find odd digits whose both neighbours are odd: pattern odd–odd–odd.

1. Odd digits: 5, 3, 1, 9, 9, 9, 9 (positions 2, 7, 8, 11, 18, 19, 23).
2. Only two places have adjacent odd digits: ‘3 1’ (preceded by 2, followed by 8) and ‘9 9’ (preceded by 2, followed by 2).
3. No odd digit has odd digits on both sides → none.

Trick: You need three odd digits in a row; scan for that block – there is none.

Answer: (C) None

Concept: Find both letters by counting from the stated ends, then count the alphabet gap.

1. 4th from left: T, Y, E, V.
2. 6th from right: S, M, L, O, Z, R.
3. Between R (18) and V (22) in the alphabet: S, T, U → $22-18-1=3$.

Trick: Letters between positions $p$ and $q$ $=|p-q|-1$.

Answer: (A) Three

Concept: Replace letters (A $+$, B $-$, C $\times$, D $÷$) and use BODMAS.

1. $126-13+42÷7\times 15$
2. $42÷7=6$, $6\times 15=90$
3. $126-13+90=203$

Trick: Do $÷$ and $\times$ first, then $+$ and $-$ from left to right.

Answer: (A) 203

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

1. UQL → QMH: each letter $-4$. CAT → YWP: each letter $-4$ (C → Y wraps).
2. ZBM → VXI: $Z-4=V$, $B-4=X$, $M-4=I$ ✓
3. NOP → OPQ is $+1$; UTW → RQT is $-3$; LJH → NLJ is $+2$ ✗

Trick: Check only the first letter of each option: $-4$ from Z gives V – unique.

Answer: (B) ZBM : VXI