Q&A: practice questions with answers

Concept: Average of the first $n$ natural numbers $=\frac{n+1}{2}$ (average of an AP = mean of first and last term).

1. Here $n=63$.
2. Average $=\frac{63+1}{2}=\frac{64}{2}=32$.

Trick: For consecutive numbers, average = middle term. The middle of 1…63 is the 32nd term = 32.

Answer: (C) 32

Concept: Successive increases multiply: final $=P(1+\frac{a}{100})(1+\frac{b}{100})$.

1. After 1st year: $2,90,000\times 1.10=3,19,000$.
2. After 2nd year: $3,19,000\times 1.20=3,82,800$.

Trick: Net change $=a+b+\frac{ab}{100}=10+20+2=32\%$; $2,90,000\times 1.32=3,82,800$.

Answer: (C) 3,82,800

Concept: With parallel lines, corresponding angles are equal, co-interior angles add to ${180}^{\circ }$ and vertically opposite angles are equal.

1. $t_1\parallel t_2$ cut by line $u$: $\angle C$ is corresponding to $\angle A$ → $\angle C={92}^{\circ }$.
2. $u\parallel v$ cut by $t_2$: $\angle B$ and $\angle C$ are co-interior → $\angle B={180}^{\circ }-{92}^{\circ }={88}^{\circ }$.
3. $\angle D$ is vertically opposite to $\angle B$ → $\angle D={88}^{\circ }$.

Trick: Two pairs of parallel lines form a parallelogram; $\angle A$ and $\angle D$ are at opposite corners but $\angle D$ is the exterior angle there, so $\angle D={180}^{\circ }-\angle A$.

Answer: (A) ${88}^{\circ }$

Concept: With a false weight, the dealer charges for 1 kg but hands over only $w$ kg; his real cost is for $w$ kg.

1. Let cost price $=$ ₹100 per kg. Professed 25% loss → selling price of "1 kg" $=75$.
2. Actual goods given $=w$ kg, costing $100w$. Gain 70% → $75=1.7\times 100w$.
3. $w=\frac{75}{170}=0.4412$ kg $=441.2$ g.

Trick: $\frac{\text{True weight}}{\text{False weight}}=\frac{100+\text{gain}\%}{100-\text{loss}\%}$ → $\frac{1000}{w}=\frac{170}{75}$, $w=441.2$ g.

Answer: (B) 441.2 gm

Concept: Add the per-hour work of filling taps and subtract that of the emptying tap.

1. Net work per hour $=\frac{1}{4}+\frac{1}{27}-\frac{1}{27}=\frac{1}{4}$.
2. So the full cistern fills in 4 hours.
3. One-fourth of the cistern $=\frac{1}{4}\times 4=1$ hour.

Trick: The 27-hour filler and the 27-hour emptier cancel each other – only the 4-hour tap matters.

Answer: (C) 1

Concept: For a discrete frequency distribution, median is the value of $x$ whose cumulative frequency first reaches or exceeds $\frac{N}{2}$.

1. $N=8+10+11+16+20+25+15+9+6=120$, so $\frac{N}{2}=60$.
2. Cumulative frequencies: 8, 18, 29, 45, 65, 90, 105, 114, 120.
3. The first cumulative frequency $\ge 60$ is 65, at $x=5$ → median $=5$.

Trick: Both the 60th and 61st observations lie in the $x=5$ group (46th to 65th), so median = 5 without any averaging.

Answer: (D) 5

Concept: A percentage discount on the marked price is a fixed rupee amount; the second discount here is that same rupee amount again.

1. First discount $=d\%$ of 880 $=8.8d$ rupees.
2. Second discount $=$ the same amount $=8.8d$. Total discount $=17.6d$.
3. $880-17.6d=132\Rightarrow 17.6d=748\Rightarrow d=42.5$.

Trick: Total discount $=880-132=748$ = two equal discounts → each $=374$ = $\frac{374}{880}\times 100=42.5\%$.

Answer: (A) 42.5

Concept: Write the ratio terms with a common multiplier $k$ and substitute in the equation.

1. Let $a=3k$, $b=4k$, $c=5k$.
2. $2a-3b+4c=6k-12k+20k=14k=56\Rightarrow k=4$.
3. $c=5k=20$.

Trick: Compute the "ratio value" of the expression first: $2(3)-3(4)+4(5)=14$; then $k=\frac{56}{14}=4$.

Answer: (A) 20

Concept: Scan each symbol and check its left neighbour (must be a letter) and right neighbour (must be a symbol).

1. Series: T O @ $ J $ S € M L I * N G © Y & T P £ Y T.
2. @ : left O (letter), right $ (symbol) ✓.
3. Other symbols ($, €, *, ©, &, £) are each followed by a letter, so they fail.

Trick: Look only for the pattern Letter–Symbol–Symbol; the middle one counts.

Answer: (C) 1

Concept: Find the operation linking the three numbers of each given set.

1. $336-216=120$, $216-156=60$; $379-259=120$, $259-199=60$.
2. Rule: 1st − 120 = 2nd, 2nd − 60 = 3rd.
3. Check (C): $412-120=292$, $292-60=232$ ✓. Others fail (e.g. $432-332=100$).

Trick: Check the first difference (120) – only option (C) has it.

Answer: (C) (412, 292, 232)

Concept: Find the chain of operations in the given sets and apply it to the options.

1. $157\stackrel{+5}{\to }162\stackrel{+10}{\to }172\stackrel{÷2}{\to }86$; $93\stackrel{+5}{\to }98\stackrel{+10}{\to }108\stackrel{÷2}{\to }54$.
2. (C): $185+5=190$, $190+10=200$, $200÷2=100$ ✓.
3. (A) ends with $+20$, (B) starts with $+4$, (D) ends with $\times 2$.

Trick: Check the last step first – only (C) halves the third number.

Answer: (C) 185 – 190 – 200 – 100

Concept: Draw the family tree generation by generation.

1. A is B's wife; B is father of C and D (D is C's brother) → A is mother of D.
2. E is D's son → A is the mother of E's father.

Trick: Trace E upward: E → D (father) → A (D's mother) = paternal grandmother.

Answer: (A) Father's mother

Concept: Look for the pattern Letter–Letter–Symbol; the middle letter is counted.

1. Series: O N & # K € S * P # £ * G I & * @ I A * @ ©.
2. N (O before, & after) ✓; I (G before, & after) ✓; A (I before, * after) ✓.
3. S and P are preceded by symbols, so they do not count. Total 3.

Trick: Underline every letter whose right neighbour is a symbol, then check its left neighbour.

Answer: (C) 3

Concept: The same letter-shift must convert # to NMH and VUP to %.

1. Test (D): S→N, R→M, M→H – each letter moves back 5 (S=19→N=14, R=18→M=13, M=13→H=8).
2. Apply −5 to VUP: V(22)→Q(17), U(21)→P(16), P(16)→K(11) = QPK ✓.
3. Other options do not give a uniform shift on both sides.

Trick: Options give # directly – check which one differs from NMH by a constant.

Answer: (D) # = SRM, % = QPK

Concept: Letters common to both words share the common digits; the leftover letter takes the leftover digit.

1. ROAD → 9416, DORM → 3691. Common letters: R, O, D; common digits: 9, 1, 6.
2. In DORM, the extra letter M takes the extra digit 3.
3. (Similarly A = 4 in ROAD.)

Trick: Subtract the common digits from the second code – what remains is the new letter's code.

Answer: (B) 3

Concept: Change every digit by the rule, then pick the greatest and smallest.

1. 21673458: odd digits +1, even digits −1 → 2→1, 1→2, 6→5, 7→8, 3→4, 4→3, 5→6, 8→7.
2. New number: 12584367.
3. Greatest 8, smallest 1 → product $=8\times 1=8$.

Trick: Odd digits become even and even become odd; 7→8 is the largest, 2→1 the smallest.

Answer: (C) 8

Concept: Decode each symbol pair from right to left starting at B.

1. N @ B: N is son of B. D + N: D is wife of N → D is B's son's wife.
2. R # D: R is sister of D. E % R: E is father of R → E is also father of D.
3. So E is the father of B's son's wife.

Trick: Build the tree from the person asked about (B) and move link by link.

Answer: (C) Son's wife's father

Concept: Look at the differences between consecutive terms.

1. $64\to 62\to 59\to 54\to 47$: differences $2,3,5,7$ – consecutive prime numbers.
2. Next prime is 11: $47-11=36$.

Trick: Differences 2, 3, 5, 7, 11, 13 … = primes; a very common series pattern.

Answer: (D) 36

Concept: Test each statement alone, then together; the answer must be unique to be "sufficient".

1. I alone: C sits between A and E, but which of them is on C's right is unknown → not sufficient.
2. II alone: fixes F, B (2nd left) and C (2nd right); C's neighbours can be any of A, D, E → not sufficient.
3. Together: D is F's immediate left neighbour, so A and E occupy the two seats beside C – but they can swap. Still two possible answers (A or E).

Trick: If two people can interchange without breaking any condition, the data is insufficient.

Answer: (A) Both statements I and II together are not sufficient to answer the question.

Concept: Place the fixed people first (Q, V, S), then fit the pair P–T, then the rest.

1. Two before Q → Q = Wed; three after V → V = Thu; three between S and Q → S = Sun.
2. P immediately before T, P not Monday: only Fri–Sat is free as a pair → P = Fri, T = Sat.
3. R not Tuesday → R = Mon, U = Tue. Between R (Mon) and T (Sat): Tue, Wed, Thu, Fri = 4 people.

Trick: "Only n people before/after X" fixes X's position at once.

Answer: (C) Four