Q&A: practice questions with answers

Concept: Multiply the numerator first, then divide.

1. $1.5\times 0.2=0.30$.
2. $\frac{0.3}{0.3}=1$.

Trick: $1.5\times 0.2$: $15\times 2=30$ with two decimal places → $0.30$.

Answer: (B) 1

Concept: If CP of $a$ articles $=$ SP of $b$ articles, profit$\%=\frac{a-b}{b}\times 100$.

1. Let CP of each article be ₹1: $25=x\times \text{SP}$.
2. Profit $25\%\Rightarrow \text{SP}=1.25$.
3. $x=\frac{25}{1.25}=20$.

Trick: $\frac{25-x}{x}=\frac{1}{4}\Rightarrow 100-4x=x\Rightarrow x=20$.

Answer: (C) 20

Concept: Mode is the value that occurs most often.

1. Frequencies: 40 → 2, 41 → 1, 42 → 2, 43 → 2, 45 → 2, 46 → 2, 51 → 3, others → 1.
2. 51 appears 3 times, the maximum.

Trick: Tally only the values that repeat; one value with 3 occurrences stands out quickly.

Answer: (B) 51

Concept: Solve the innermost bracket first, then multiply, then add.

1. $(26-11+6)=21$.
2. $-9\times 21=-189$.
3. $30+(-189)=-159$.

Trick: Left-to-right inside the bracket: $26-11=15$, then $+6=21$ – do not subtract 17 from 26.

Answer: (C) -159

Concept: Leak's rate $=$ filling rate of A $-$ net filling rate.

1. A fills $\frac{1}{10}$ per hour; with leak net is $\frac{1}{12}$ per hour.
2. Leak empties $\frac{1}{10}-\frac{1}{12}=\frac{6-5}{60}=\frac{1}{60}$ per hour.
3. Leak alone empties the tank in 60 hours.

Trick: Leak time $=\frac{10\times 12}{12-10}=60$.

Answer: (A) 60 hours

Concept: Mean proportional $m$ between $a$ and $b$: $m^2=ab$.

1. $ab=900$ and $a-b=45$.
2. $(a+b{)}^2=(a-b{)}^2+4ab=2025+3600=5625\Rightarrow a+b=75$.
3. $b=\frac{75-45}{2}=15$ (and $a=60$; $60\times 15=900$ ✓).

Trick: Factor pairs of 900 with difference 45: $60\times 15$ – spot it directly.

Answer: (B) 15

Concept: $A=P{(1+\frac{r}{100})}^n$.

1. $\frac{74263}{70000}=1.0609$.
2. ${1.03}^2=1.0609$, so $n=2$.

Trick: ${1.03}^2=1.0609$, ${1.03}^3\approx 1.0927$ – the amount barely exceeds $6\%$, so 2 years.

Answer: (B) 2

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $SI=\frac{25200\times 6\times 4}{100}$.
2. $=252\times 24=6048$.

Trick: $6\%\times 4$ years $=24\%$ of 25200 $=\frac{24}{100}\times 25200=6048$.

Answer: (C) ₹6,048

Concept: Take total work $=$ LCM of days; work per day $=$ total $÷$ days.

1. Total work $=36$ units; A $=3$ units/day, B $=1$ unit/day.
2. In 4 days together: $4\times (3+1)=16$ units done; $20$ remain.
3. B alone: $\frac{20}{1}=20$ days.

Trick: Fraction left $=1-4(\frac{1}{12}+\frac{1}{36})=1-\frac{16}{36}=\frac{20}{36}$ → B takes $\frac{20}{36}\times 36=20$ days.

Answer: (A) 20

Concept: For numbers $3x,2x,13x$ with co-prime ratio terms, HCF $=x$ and LCM $=3\times 2\times 13\times x=78x$.

1. $78x=6318\Rightarrow x=\frac{6318}{78}=81$.
2. HCF $=x=81$.

Trick: $78\times 80=6240$, $6318-6240=78$ → one more, so $x=81$.

Answer: (D) 81

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

1. Only two boxes below G → G at 3. Only one box above B → B at 6. One box between B and A → A at 4 (8 is impossible).
2. Remaining places 1, 2, 5, 7 for C, D, E, F. F immediately above E needs consecutive places → E = 1, F = 2.
3. C and D take 5 and 7 with D below C → D = 5, C = 7.

Trick: Clues with 'only one/two' fix absolute positions – start with them, then fit the 'immediately' pair into the only consecutive gap.

Answer: (C) C

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

1. YI → BM: Y(25)+3 = 28 → B(2), I(9)+4 = M(13): shifts $+3,+4$.
2. GV → JZ: $+3,+4$; BU → EY: $+3,+4$.
3. OS → RV: O(15)+3 = R(18), S(19)+3 = V(22): shifts $+3,+3$ – different.

Trick: Write letter numbers (EJOTY: E=5, J=10, O=15, T=20, Y=25) and check both shifts; the odd one differs in the second shift.

Answer: (D) OS - RV

Concept: Facing north, 'right' is the higher position number. Number seats 1 to 8 from left.

1. Y is at an end and S is second to the left of Y → Y = 8, S = 6 (Y = 1 is impossible). S is third to the right of N → N = 3.
2. B–M are adjacent with Z two places right of M. Free seats 1, 2, 4, 5, 7: B = 4, M = 5, Z = 7 (B = 1, M = 2 leaves no adjacent pair for A–T).
3. T immediately right of A → A = 1, T = 2.

Trick: Fix the person at the extreme end first; every other clue then collapses to one option.

Answer: (B) A

Concept: Check the differences between consecutive terms for a repeating pattern.

1. Differences: $34-26=8$, $45-34=11$, $53-45=8$, $64-53=11$, $72-64=8$.
2. The pattern is $+8,+11$ alternating, so the next difference is $+11$.
3. $72+11=83$.

Trick: Two alternating differences → the series is really two interleaved series each growing by $19$: $26,45,64,83$.

Answer: (D) 83

Concept: Rearrange the letters alphabetically, then count positions from the required end.

1. NERVOUS in alphabetical order: E, N, O, R, S, U, V.
2. Fourth from the left = R; third from the right = S.
3. R(18) and S(19) are consecutive, so there are $0$ letters between them.

Trick: Letters between two positions $=|b-a|-1$; here $19-18-1=0$.

Answer: (C) 0

Concept: Swap the signs as instructed, then apply BODMAS.

1. After interchanging: $215+147\times 110-1604÷4$.
2. $147\times 110=16170$ and $1604÷4=401$.
3. $215+16170-401=15984$.

Trick: Rewrite the whole expression with new signs first – never swap and calculate at the same time.

Answer: (B) 15984

Concept: Number positions 1 (bottom) to 7 (top); fix the 'only' clues first.

1. Two boxes below A → A = 3. One box above E → E = 6. One box between E and B → B = 4.
2. Free places 1, 2, 5, 7 for C, D, F, G. C immediately above G needs consecutive places → G = 1, C = 2.
3. D below F → D = 5, F = 7. Between A (3) and D (5) there is only B → 1 box.

Trick: Boxes between positions $a$ and $b$ $=|a-b|-1=5-3-1=1$.

Answer: (B) 1

Concept: When the unknown is early in the series, work out the differences from the known terms at the end.

1. $110-80=30$, $80-56=24$, $56-38=18$ → differences fall by $6$ each step: next (earlier) differences are $12$ and $6$.
2. $?=38-12=26$; check: $26-20=6$ ✓.
3. Series: $20,26,38,56,80,110$ with differences $6,12,18,24,30$.

Trick: Differences that are multiples of one number ($6,12,18,\dots$) – spot the table of 6.

Answer: (D) 26

Concept: Find one operation on the whole first number that gives the second in both given pairs.

1. ${18}^2=324$, $324+1=325$ ✓; ${19}^2=361$, $361+1=362$ ✓. Rule: second $=$ first${}^2+1$.
2. Option C: ${22}^2+1=484+1=485$ ✓.
3. Others: ${14}^2+1=197\ne 196$; ${21}^2+1=442\ne 440$; ${20}^2+1=401\ne 399$.

Trick: When the second number is close to a perfect square, test $n^2\pm k$ first.

Answer: (C) 22, 485

Concept: Sort the digits in ascending order and compare position by position with the original.

1. Original: 3 1 4 7 8 5 6; ascending: 1 3 4 5 6 7 8.
2. Position-wise: 3→1, 1→3, 4→4, 7→5, 8→6, 5→7, 6→8.
3. Only the digit 4 (third place) stays where it was.

Trick: Write the two rows one below the other and tick the matching columns.

Answer: (A) One