Q&A: practice questions with answers

Concept: Together, rates add: $\frac{1}{T}=\frac{1}{20}+\frac{1}{22}+\frac{1}{14}$.

1. LCM of 20, 22, 14 $=1540$
2. $\frac{77+70+110}{1540}=\frac{257}{1540}$
3. $T=\frac{1540}{257}=5\frac{255}{257}$ hours

Trick: Answer must be less than the fastest pipe (14 h) and close to $\frac{14}{3}$–$\frac{20}{3}$; only $5\frac{255}{257}$ fits.

Answer: (B) $5\frac{255}{257}$ hours

Concept: For equal time, profit is shared in the ratio of investments.

1. $1210:1650:1300=121:165:130$ (÷10)
2. Total $=416$
3. Avinash $=\frac{130}{416}\times 1200=375$

Trick: $\frac{1300}{4160}=\frac{5}{16}$, and $\frac{5}{16}\times 1200=375$.

Answer: (C) ₹375

Concept: Eliminate options by easy rules, then divide by the remaining one.

1. 769501 is odd → not divisible by 96 or 90
2. It ends in 1 → not divisible by 5, so not by 105
3. $769501÷97=7933$ exactly ($97\times 7933=769501$)

Trick: An odd number ending in 1 can't be divisible by any even number or multiple of 5 — only 97 is left.

Answer: (B) 97

Concept: Divisible by 22 ⇔ even AND divisible by 11 (alternate-digit sums differ by 0 or a multiple of 11).

1. 6580350: $(0+3+8+6)-(5+0+5)=17-10=7$ ✗; 6467614: $(4+6+6+6)-(1+7+4)=22-12=10$ ✗; 6136901 is odd ✗
2. 7132950: $(0+9+3+7)-(5+2+1)=19-8=11$ ✓ and even ✓

Trick: First drop odd numbers, then test 11 only on the rest.

Answer: (D) 71,32,950

Concept: Write both conditions as time equations and add them to eliminate Nikhil's time.

1. Let speeds be $a$ (Anmol), $n$ (Nikhil): $\frac{96}{a}-\frac{96}{n}=6$
2. Doubling: $\frac{96}{n}-\frac{96}{2a}=2$
3. Add: $\frac{96}{a}-\frac{48}{a}=8\Rightarrow \frac{48}{a}=8\Rightarrow a=6\text{km/h}$

Trick: Doubling speed halves time: the saving $\frac{T}{2}$ equals $6+2=8$ h, so $T=16$ h and speed $=\frac{96}{16}=6$.

Answer: (B) 6 km/hr

Concept: If $6A=9B=2C=k$, then $A:B:C=\frac{1}{6}:\frac{1}{9}:\frac{1}{2}$.

1. Multiply by LCM 18: $A:B:C=3:2:9$
2. Total parts $=14$
3. $A=\frac{3}{14}\times 73500=15750$

Trick: Total must be divisible by 14 and A's share by 3 → ₹15,750.

Answer: (C) ₹15,750

Concept: When a solid is melted and recast, volume is conserved: number of cubes $=\frac{\text{volume of cuboid}}{\text{volume of one cube}}$.

1. Cuboid volume $=24\times 16\times 10=3840{\text{cm}}^3$
2. Cube volume $=4^3=64{\text{cm}}^3$
3. Number $=\frac{3840}{64}=60$

Trick: Cancel first: $\frac{24\times 16\times 10}{4\times 4\times 4}=6\times 4\times 2.5=60$.

Answer: (A) 60

Concept: Two consecutive even numbers: $x$ and $x+2$, each square roughly half of 2740.

1. $\frac{2740}{2}=1370$, and $\sqrt{1370}\approx 37$ → try 36 and 38
2. ${36}^2+{38}^2=1296+1444=2740$ ✓
3. Sum $=36+38=74$

Trick: Sum $\approx 2\sqrt{\frac{2740}{2}}=2\times 37=74$.

Answer: (A) 74

Concept: Two 'All' statements chain: All A are B + All B are C ⇒ All A are C. 'All' cannot be reversed.

1. Lakes ⊂ Ponds ⊂ Rivers, so all lakes are rivers → I follows
2. Rivers may be outside ponds, so 'All rivers are ponds' is not certain → II does not follow

Trick: 'All A are B' never gives 'All B are A' — only 'Some B are A'.

Answer: (C) Only Conclusion I follows

Concept: Compare gaps inside each cluster using letter positions (A = 1 … Z = 26).

1. JM-DH: J(10)→M(13) +3, D(4)→H(8) +4
2. HK-BF: +3, +4; MP-GK: +3, +4
3. OR-IN: O(15)→R(18) +3, but I(9)→N(14) +5 ✗

Trick: Pattern '+3 then +4' (and first letters −6) breaks only in OR-IN.

Answer: (C) OR-IN

Concept: Find the position shift for each letter (A = 1 … Z = 26), wrapping around after A.

1. R(18)→K(11), B(2)→U(21), N(14)→G(7), H(8)→A(1): each −7
2. Check PZLF→ISEY: −7 each ✓
3. UEQK: U(21)→N(14), E(5)→X(24), Q(17)→J(10), K(11)→D(4)

Trick: −7 is the same as +19; for letters before H, go back around from Z.

Answer: (A) NXJD

Concept: Check differences between consecutive terms.

1. Differences: 10, 20, 30, 40
2. Next difference $=50$
3. $102+50=152$

Trick: Differences in an AP (+10 each time) → add the next multiple of 10.

Answer: (C) 152

Concept: Check differences between consecutive terms.

1. Differences: 0, 1, 2, 3
2. Next difference $=4$
3. $9+4=13$

Trick: Differences 0, 1, 2, 3, 4 — consecutive whole numbers.

Answer: (C) 13

Concept: Apply the rule digit by digit: odd → +2, even → −1.

1. 7 2 5 1 8 6 4 3 → 9 1 7 3 7 5 3 5
2. First digit = 9, last digit = 5
3. Difference $=9-5=4$

Trick: Only the first and last digits matter: 7 → 9 and 3 → 5.

Answer: (D) 4

Concept: Track each letter position separately.

1. 1st letters: C, E, G, I → K (+2)
2. 2nd letters: N, P, R, T → V (+2)
3. 3rd letters: Q, S, U, W → Y (+2)

Trick: Every letter moves +2: ITW → KVY.

Answer: (A) KVY

Concept: Each term is double the previous one.

1. $15\times 2=30$, $30\times 2=60$, $60\times 2=120$
2. $120\times 2=240$
3. Check: $240\times 2=480$ ✓

Trick: Always verify the missing term with the term after it.

Answer: (B) 240

Concept: Use letter positions (A = 1 … Z = 26) within each pair.

1. PT-WR: P(16)→T(20) +4, W(23)→R(18) −5
2. RV-YT and KO-RM: +4 and −5 as well
3. BF-IC: B(2)→F(6) +4, but I(9)→C(3) −6 ✗

Trick: Pattern '+4, then −5' (first letters +7) fails only in BF-IC.

Answer: (D) BF-IC

Concept: Plot the path on axes (East = +x, North = +y); each left turn rotates 90° anticlockwise.

1. A(0, 0) → 9 E → (9, 0) → left, 4 N → (9, 4)
2. Left, 11 W → (−2, 4) → left, 6 S → (−2, −2)
3. Left, 2 E → P(0, −2)
4. A is 2 km straight north of P

Trick: Net E–W: $9-11+2=0$; net N–S: $4-6=-2$ → P is 2 km south of A, so go 2 km North.

Answer: (B) 2 km to North

Concept: Find the shift for each letter position.

1. Y→U, U→Q, W→S, T→P: each −4
2. OKMJ→KGIF: each −4 ✓
3. LHJG: L→H, H→D, J→F, G→C

Trick: Uniform shift −4 on every letter.

Answer: (D) HDFC

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

1. Three boxes above G → G = 4; one box between F and G → F = 2 or 6
2. Three between F and N and N above G → F = 6, N = 2; O just below N → O = 3
3. Left: 1, 5, 7 for H, M, P; H not next to F (6) → H = 1
4. M above P → M = 5, P = 7

Trick: Start with the exact-position clue (G = 4) and test both F cases.

Answer: (B) P