Q&A: practice questions with answers

Concept: Total provision is fixed: $\text{men}\times \text{days}$ = constant (inverse proportion).

1. Provision $=100\times 26=2600$ man-days.
2. Remaining men $=100-50=50$.
3. Days $=\frac{2600}{50}=52$.

Trick: Half the men → double the days: $26\times 2=52$.

Answer: (C) 52

Concept: SP = MP × (1 − d₁)(1 − d₂); both articles have the same marked price, so find MP from X and use it for Y.

1. SP of X $=800+523=1323$. Also SP of X $=MP\times 0.55\times 0.27=0.1485MP$.
2. SP of Y $=MP\times 0.11$. So $\frac{{\text{SP}}_Y}{{\text{SP}}_X}=\frac{0.11}{0.1485}=\frac{20}{27}$.
3. SP of Y $=1323\times \frac{20}{27}=49\times 20=980$.
4. Profit% on Y $=\frac{980-800}{800}\times 100=22.5\%$.

Trick: Never compute MP (an ugly number); take the ratio of the two selling fractions instead – $1323=27\times 49$ cancels neatly.

Answer: (D) 22.5%

Concept: $\sqrt[3]{2744}=14$ (since ${14}^3=2744$) and $\sqrt[4]{256}=4$ (since $4^4=256$).

1. Expression $=\frac{14}{4}\times \frac{2}{42}\times 72$.
2. $\frac{14}{42}=\frac{1}{3}$, so it becomes $\frac{1}{3}\times \frac{2}{4}\times 72=\frac{1}{3}\times 36=12$.

Trick: Memorise cubes up to 20 (${14}^3=2744$) and powers of 2 ($2^8=256$).

Answer: (C) 12

Concept: The mode is the value that occurs most often in the data.

1. Count repeats: 4 → 2 times, 5 → 3, 7 → 4, 8 → 2, 9 → 2, 10 → 2, 11 → 2, 13 → 3.
2. 7 occurs the maximum (4) times, so mode = 7.

Trick: Data is already sorted – just look for the longest run of the same number.

Answer: (C) 7

Concept: If a sum becomes $n$ times in $T$ years at simple interest, the interest is $(n-1)P$, so $R=\frac{(n-1)\times 100}{T}$.

1. 5 times → SI $=4P$ in 16 years.
2. $R=\frac{4\times 100}{16}=25\%$ per annum.

Trick: $R\times T=(n-1)\times 100$ → $R=\frac{400}{16}=25$.

Answer: (B) 25

Concept: On a circular track, two runners in the same direction meet when the faster one gains the initial gap; gain per minute = difference of their speeds (in rounds per minute).

1. P: 5 rounds in 40 min → 8 min per round → speed $\frac{1}{8}$ round/min.
2. R does 7 rounds in the time P does 6 rounds (48 min) → speed $\frac{7}{48}$ round/min.
3. Relative speed $=\frac{7}{48}-\frac{6}{48}=\frac{1}{48}$ round/min.
4. They start at opposite ends of a diameter → gap $=\frac{1}{2}$ round; time $=\frac{1/2}{1/48}=24$ min.

Trick: R gains exactly 1 round on P in 48 min, so half a round in 24 min.

Answer: (C) 24

Concept: $A=P{(1+\frac{R}{100})}^n$; for 20% and 2 years the multiplier is ${1.2}^2=1.44$.

1. $A=28450\times 1.44$.
2. $28450\times 1.44=28450+28450\times 0.44=28450+12518=40968$.

Trick: 20% twice = $20+20+\frac{20\times 20}{100}=44\%$ effective; 44% of 28450 = 12518.

Answer: (B) ₹40,968

Concept: Convert the fractions to decimals and evaluate each bracket separately.

1. $\frac{4}{5}=0.8$, $\frac{8}{20}=0.4$.
2. First bracket: $3.75+0.8=4.55$; second bracket: $3.5-0.4=3.1$.
3. $4.55-3.1=1.45$.

Trick: Remove brackets: $3.75-3.5=0.25$ and $0.8+0.4=1.2$ → $1.45$.

Answer: (A) 1.45

Concept: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Note that the denominator uses $\frac{a}{10}$ and $\frac{b}{10}$.

1. Numerator $={732}^3+{268}^3=(732+268)({732}^2-732\times 268+{268}^2)=1000\times ({732}^2-732\times 268+{268}^2)$.
2. Denominator $={73.2}^2-73.2\times 26.8+{26.8}^2=\frac{{732}^2-732\times 268+{268}^2}{100}$.
3. Ratio $=1000\times 100=100000$.

Trick: When $a+b$ is a round number (1000) and the denominator is the "$a^2-ab+b^2$" pattern scaled by $\frac{1}{100}$, the answer is $(a+b)\times 100$.

Answer: (C) 100000

Concept: The string is the hypotenuse of a right triangle; height $=\text{string}\times \sin \theta$.

1. $\sin {30}^{\circ }=\frac{h}{24.4}$.
2. $h=24.4\times \frac{1}{2}=12.2$ m.

Trick: Opposite ${30}^{\circ }$, the side is always half the hypotenuse.

Answer: (C) 12.2 m

Concept: Take successive differences and look for a pattern in them.

1. Differences: $10-3=7$, $24-10=14$, $45-24=21$, $73-45=28$ → multiples of 7.
2. Next difference $=35$, so the next term $=73+35=108$.

Trick: Differences 7, 14, 21, 28 are the 7-times table – next is 35.

Answer: (D) 108

Concept: Find one rule linking the three numbers of each given set and test the options with it.

1. Try the form $c=4a-b$: $(19,21,55)$: $4\times 19-21=76-21=55$ ✓
2. $(8,9,23)$: $4\times 8-9=32-9=23$ ✓
3. Option A $(22,21,67)$: $4\times 22-21=88-21=67$ ✓
4. Others fail: $(1,21,19)$ gives $-17$; $(7,5,3)$ gives $23$; $(20,12,79)$ gives $68$.

Trick: When sums/products fail, try $ka\pm b$ with small $k$ (2, 3, 4) – both sets must obey the same $k$.

Answer: (A) (22,21,67)

Concept: Fix the extreme clues first ("only one above/below"), then fill the rest.

1. Only Z lives above U → Z = 6, U = 5.
2. T lives immediately below U → T = 4.
3. Only Y lives below S → Y = 1, S = 2; the remaining X = 3.
4. Floor 4 → T.

Trick: "Only X lives above Y" pins two floors at once – the top two. Start there.

Answer: (B) T

Concept: To name the third oldest of five, the full order (at least the top three) must be fixed.

1. Statement I: Lokesh > Ishan > Gaurav – Himanshu and Mukul unplaced → insufficient.
2. Statement II: Lokesh > Mukul > Ishan – two people unplaced → insufficient.
3. Together: Lokesh > Mukul > Ishan > Gaurav, but Himanshu can be anywhere, so the third oldest may be Ishan, Mukul or Himanshu.

Trick: If even one person is never mentioned in any statement, a "who is at position k" question is usually not answerable.

Answer: (D) Both statements I and II together are NOT sufficient to answer the question

Concept: Decode left to right and draw the family tree.

1. G + R: G is the daughter of R. R × W: R is the wife of W → G is the daughter of W and R.
2. W ÷ D: W is the father of D → D and G are sisters.
3. D − M: D is the mother of M.
4. So G is the sister of M's mother → mother's sister (maternal aunt).

Trick: Mark generations: G and D same level (siblings), M one level below → G = aunt.

Answer: (B) Mother's sister

Concept: Check the differences between consecutive terms.

1. $22-20=2,25-22=3,30-25=5,37-30=7$ – the differences are consecutive primes 2, 3, 5, 7.
2. Next prime = 11 → $37+11=48$.

Trick: Differences 2, 3, 5, 7, 11, 13 … = prime series; it appears very often in RRB papers.

Answer: (C) 48

Concept: Decode each symbol and build the tree from the person asked about.

1. G # A: G is the husband of A.
2. A − R: A is the daughter of R.
3. R + K: R is the father of K (K is just A's sibling).
4. G is married to R's daughter → G is R's daughter's husband (son-in-law).

Trick: The last relation (R + K) is a distractor – the answer only needs G–A–R.

Answer: (A) Daughter's husband

Concept: Scan each letter and check both neighbours.

1. Series: B * £ J E % $ * V € G * D H @ A £ O V $ T Y.
2. Letters with a symbol on both sides: * V €, € G *, @ A £.
3. J (£ J E), D (* D H), O (£ O V), T ($ T Y) have a letter on one side → not counted.
4. Count = 3.

Trick: Look only at symbol pairs with exactly one character between them – a letter there is a hit.

Answer: (D) 3

Concept: Track each letter position and the number separately.

1. Letters: Q(17)→U(21)→Y(25)→C(3)→G(7): each $+4$.
2. I(9)→M(13)→Q(17)→U(21)→Y(25); L(12)→P(16)→T(20)→X(24)→B(28−26=2): $+4$ each.
3. Numbers: 14, 18, 22, 26 → 30.
4. Next term = GYB 30.

Trick: All three letters jump $+4$ – pattern = $+4$ everywhere, so just add 4 (wrap after Z).

Answer: (D) GYB 30

Concept: Facing the centre, "right" is anticlockwise and "left" is clockwise.

1. Place N. Two people between N and Q from N's right → Q is 3rd to N's right.
2. Two between Q and M, and three between N and P: M is 3rd to the left of N (the other spot is N itself), P is opposite-ish, 4th from N.
3. Remaining three seats (1st, 2nd to N's right and 2nd to N's left) go to L, R, K with L immediately left of R → L 1st right of N, R 2nd right, K on the other side.
4. Counting to the left of K: P, Q, R → R is third to the left of K.

Trick: Facing centre: right = anticlockwise. Fix one person and draw – never solve seating in the head.

Answer: (D) R