Q&A: practice questions with answers

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=40+49+56+75+39=259$.
2. Average $=\frac{259}{5}=51.8$ kg.

Trick: Assume 50: deviations $-10,-1,+6,+25,-11$ sum to $+9$; $50+\frac{9}{5}=51.8$.

Answer: (B) 51.8

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$.

1. After 10%: $1200\times 0.9=1080$.
2. After 5%: $1080\times 0.95=1026$.

Trick: Equivalent single discount $=10+5-\frac{10\times 5}{100}=14.5\%$; $1200\times 0.855=1026$.

Answer: (C) ₹1,026

Concept: Perimeter of an equilateral triangle $=3\times \text{side}$.

1. Side $=13$ cm.
2. Perimeter $=3\times 13=39$ cm.

Yaad rakhein: Equilateral triangle: perimeter $=3a$, area $=\frac{\sqrt{3}}{4}{a}^2$.

Answer: (B) 39 cm

Concept: One-day work $=\frac{1}{\text{days}}$; subtract combined rates to get an individual rate.

1. Anjali + Ayushi $=\frac{1}{18}$; all three $=\frac{1}{9}$.
2. Ankita $=\frac{1}{9}-\frac{1}{18}=\frac{1}{18}$.
3. Anjali + Ankita $=\frac{1}{30}+\frac{1}{18}=\frac{3+5}{90}=\frac{8}{90}$.
4. Days $=\frac{90}{8}=\frac{45}{4}$.

Trick: Take total work $=90$ units: Anjali 3/day, Ankita $10-5=5$/day; together $\frac{90}{8}=\frac{45}{4}$ days.

Answer: (A) $\frac{45}{4}$ days

Concept: C’s one-day work $=(\text{A+B+C})-\text{A}-\text{B}$.

1. A+B+C $=\frac{1}{5}$, A $=\frac{1}{19}$, B $=\frac{1}{16}$.
2. C $=\frac{1}{5}-\frac{1}{19}-\frac{1}{16}=\frac{304-80-95}{1520}=\frac{129}{1520}$.
3. C alone $=\frac{1520}{129}=11\frac{101}{129}$ days.

Trick: C must take more than 5 days but less than A or B alone (16 days); $1520÷129\approx 11.8$, so option A.

Answer: (A) $11\frac{101}{129}$ days

Concept: Profit is shared in the ratio of investments (same time period).

1. $63:35:42=9:5:6$ (divide by 7); total $=20$ parts.
2. One part $=\frac{5120}{20}=256$.
3. B − C difference $=6-5=1$ part $=256$.

Trick: Simplify the ratio first — a difference of 1 part means the answer is just one part.

Answer: (D) ₹256

Concept: Divisible by 5 if the last digit is 0 or 5; by 15 only if divisible by both 3 and 5.

1. 245015 ends in 5 → divisible by 5.
2. Digit sum $=2+4+5+0+1+5=17$, not divisible by 3 → not by 15.
3. Odd → not by 2; for 11: alternate digit sums $2+5+1=8$ and $4+0+5=9$ differ by 1 → not by 11.

Trick: 15 is the tempting trap — always check the 3-rule too.

Answer: (C) 5

Concept: The largest circle inside a rectangle has diameter equal to the shorter side.

1. Shorter side $=126$ m, so $d=126$ m.
2. Circumference $=\pi d=\frac{22}{7}\times 126$.
3. $=22\times 18=396$ m.

Trick: 126 is a multiple of 7 → $126÷7=18$, then $\times 22$.

Answer: (B) 396

Concept: For equal distances at speeds $a$ and $b$: $\text{Average speed}=\frac{2ab}{a+b}$.

1. $a=72$, $b=90$.
2. Average $=\frac{2\times 72\times 90}{72+90}=\frac{12960}{162}$.
3. $=80\text{km/h}$.

Trick: Average speed is always a little less than the simple mean $\frac{72+90}{2}=81$ → 80.

Answer: (A) 80

Concept: ${\text{cosec}}^2\beta -1={\cot }^2\beta$ and ${\sec }^2\beta -{\tan }^2\beta =1$.

1. $\frac{1}{{\text{cosec}}^2\beta -1}=\frac{1}{{\cot }^2\beta }={\tan }^2\beta$.
2. Expression $={\sec }^2\beta -{\tan }^2\beta$.
3. $=1$.

Trick: Put $\beta ={45}^{\circ }$: $2-\frac{1}{2-1}=1$.

Answer: (C) 1

Concept: Check each symbol: the element just before it must be a number and the one just after it must be a symbol (pattern: number–SYMBOL–symbol).

1. Scan left to right for “number, symbol, symbol”.
2. 2 $ © ✓; 2 £ @ ✓; 5 £ % ✓.
3. 5 # 6 fails (followed by a number); others are preceded by symbols.
4. Total $=3$.

Trick: First mark every symbol followed by a symbol, then keep only those with a number before them.

Answer: (B) 3

Concept: Facing the centre, a person’s left is clockwise and right is anticlockwise.

1. Fix D; L is immediately right (anticlockwise) of D.
2. Going left (clockwise) from L, three people sit before C: D and two more seats, then C.
3. Three between D and J places J next to C (other side is C’s seat).
4. A is immediate right of K → only possible in the two seats after D: D, A, K, C; B fills the last seat.
5. From the right of K (anticlockwise): K → A → D, so 1 person.

Trick: In circular puzzles, place the most fixed pair (D–L) first and read “from the right/left” as anticlockwise/clockwise.

Answer: (D) One

Concept: Compare the letter gaps in each pair using positions (A = 1 … Z = 26).

1. KP-LI: $K+1=L$, $P(16)-7=I(9)$.
2. MR-NK: $M+1=N$, $R(18)-7=K(11)$; IN-JG: $I+1=J$, $N(14)-7=G(7)$.
3. GL-HF: $G+1=H$ but $L(12)-6=F(6)$ — gap 6, not 7.

Trick: First letters all move $+1$, so the odd one hides in the second letter.

Answer: (A) GL-HF

Concept: Apply the rule digit-wise: odd digit $+3$, even digit $-1$.

1. 6 2 4 3 5 1 4 → 5 1 3 6 8 4 3.
2. Highest digit $=8$, lowest $=1$.
3. Difference $=8-1=7$.

Trick: Only the extremes matter: largest odd (5) gives 8, smallest result comes from 2 → 1.

Answer: (B) 7

Concept: Differences doubling: $+1,+2,+4,+8,\dots$

1. $84+1=85$, $85+2=87$, $87+4=91$, $91+8=99$.
2. Next: $99+16=115$.

Trick: Differences are powers of 2: $2^0,2^1,2^2,2^3,2^4$.

Answer: (B) 115

Concept: Replace letters by signs (A = $÷$, B = $\times$, C = $+$, D = $-$) and solve using BODMAS.

1. $(117÷3)+(119÷7)+24-6+(117÷13)\times 2$.
2. $=39+17+24-6+9\times 2$.
3. $=39+17+24-6+18=92$.

Trick: Finish all brackets and $\times$ first: $39,17,18$; then add/subtract.

Answer: (B) 92

Concept: Swap $+\leftrightarrow -$ and $\times \leftrightarrow ÷$, then apply BODMAS.

1. $25÷6-8+20\times 4$ becomes $25\times 6+8-20÷4$.
2. $=150+8-5$.
3. $=153$.

Trick: Rewrite the whole expression first, then calculate — avoid swapping mid-way.

Answer: (B) 153

Concept: A letter common to both words takes the digit common to both codes (codes may be jumbled).

1. LEFT → 5427, EARS → 6153.
2. Common letter: E.
3. Common digit: 5. So E = 5.

Trick: Intersect letters and intersect digits — one common each means a direct match.

Answer: (A) 5

Concept: Track the position with coordinates (east and north positive).

1. A $(0,0)$ → east 8: $(8,0)$ → left (north) 4: $(8,4)$.
2. Left (west) 4: $(4,4)$ → left (south) 11: $(4,-7)$.
3. Right while facing south = west, 4 km: P $(0,-7)$.
4. A is 7 km straight north of P.

Trick: Net east $=8-4-4=0$, net north $=4-11=-7$ → go 7 km north.

Answer: (C) 7 km towards north

Concept: Treat each letter position and the number as separate series.

1. 1st letters: C, E, G, I → $+2$ → K.
2. 2nd letters: O, Q, S, U → $+2$ → W; 3rd: B, D, F, H → $+2$ → J.
3. Numbers: 5, 10, 15, 20 → $+5$ → 25.

Trick: All three letters jump $+2$ — option with K…W…J and 25.

Answer: (C) KWJ 25