Q&A: practice questions with answers

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=171$.
2. Sum $=171\times 172$.
3. Average $=\frac{171\times 172}{171}=172$.

Yaad rakhein: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$; of first $n$ natural numbers $=\frac{n+1}{2}$.

Answer: (A) 172

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

1. Sum $=40+49+56+76+35=256\text{kg}$.
2. Average $=\frac{256}{5}=51.2\text{kg}$.

Trick: Take 50 as base: deviations $-10,-1,+6,+26,-15$ total $+6$; $50+\frac{6}{5}=51.2$.

Answer: (D) 51.2

Concept: Expand each bracket and combine like terms.

1. $x(3x-7)=3x^2-7x$.
2. $7(x^2-4)=7x^2-28$.
3. Total $=3x^2-7x+7x^2-28+18=10x^2-7x-10$.

Trick: Check with $x=1$: $1(-4)+7(-3)+18=-7$, and $10-7-10=-7$ ✓.

Answer: (A) $10x^2-7x-10$

Concept: If a number is increased by $r\%$, new value $=\text{number}\times \frac{100+r}{100}$.

1. $x\times \frac{160}{100}=3580$.
2. $x=\frac{3580\times 100}{160}=\frac{3580}{1.6}=2237.5$.

Trick: $160\%\to 3580$, so $10\%\to 223.75$ and $100\%\to 2237.5$.

Answer: (C) 2237.5

Concept: For a fixed amount (₹1), number of articles sold is inversely proportional to the SP percentage.

1. At 45% loss, SP $=55\%$ of CP; at 10% gain, SP $=110\%$ of CP.
2. Number sold per rupee $\propto \frac{1}{\text{SP}}$: $n_2=6\times \frac{55}{110}$.
3. $n_2=3$.

Trick: $n_1\times (100-\text{loss})=n_2\times (100+\text{gain})$ → $6\times 55=n_2\times 110$ → $n_2=3$.

Answer: (C) 3

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

1. $610:630:1450=61:63:145$.
2. B's 63 parts $=591$ → 1 part $=\frac{591}{63}$.
3. C − A $=145-61=84$ parts.
4. Difference $=84\times \frac{591}{63}=\frac{4}{3}\times 591=788$.

Trick: $\frac{84}{63}=\frac{4}{3}$, so the difference is simply $\frac{4}{3}$ of B's share: $\frac{4}{3}\times 591=788$.

Answer: (D) ₹788

Concept: When equal distances are covered at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$.

1. $x=12$, $y=24$.
2. Average $=\frac{2\times 12\times 24}{12+24}=\frac{576}{36}$.
3. $=16\text{km/h}$.

Trick: Assume distance 24 km: time $2+1=3$ h for 48 km → $\frac{48}{3}=16\text{km/h}$.

Answer: (C) 16

Concept: A number is prime if no prime up to its square root divides it ($\sqrt{424}<21$, so test 2, 3, 5, 7, 11, 13, 17, 19).

1. Odd numbers 411–423: 411 ($3\times 137$), 413 ($7\times 59$), 415 ($5\times 83$), 417 ($3\times 139$), 423 ($3\times 141$) are composite.
2. 419 and 421 have no prime factor up to 19 → prime.
3. So there are 2 primes.

Trick: Even numbers and those ending in 5 go out instantly; only test the rest with digit-sum (3), 7, 11, 13, 17, 19.

Answer: (C) 2

Concept: Apply the rule digit by digit: even digit $+2$, odd digit $-1$.

1. 52749631 → 5→4, 2→4, 7→6, 4→6, 9→8, 6→8, 3→2, 1→0.
2. New number: 44668820.
3. 4, 6 and 8 each appear twice → three digits repeat.

Trick: Write the new digits in a row and tick pairs; don't count 2 and 0 (they appear once).

Answer: (D) Three

Concept: Find the letter shift between the pairs using alphabet positions.

1. K(11)→A(1), N(14)→D(4), M(13)→C(3), L(12)→B(2): each $-10$.
2. W(23)→M(13), Z→P, Y→O, X→N: each $-10$ ✓.
3. S(19)→I(9), V(22)→L(12), U(21)→K(11), T(20)→J(10).

Trick: The first letter alone (S−10 = I) leaves 3 options; the second (V−10 = L) leaves ILKO and ILKJ; the last (T−10 = J) settles it.

Answer: (D) ILKJ

Concept: Add 1 only to even digits; odd digits stay the same.

1. 36719542: 3, 6→7, 7, 1, 9, 5, 4→5, 2→3.
2. New number: 37719553.
3. Repeated digits: 3 (twice), 7 (twice), 5 (twice).
4. Sum $=3+7+5=15$.

Trick: 'Sum of the digits that are repeated' = add each repeated digit once, not every occurrence.

Answer: (C) 15

Concept: A letter common to two words takes the digit common to both codes.

1. CALM → {3, 2, 5, 7}; MONK → {1, 3, 4, 6}.
2. Common letter: M.
3. Common digit: 3 → M = 3.

Trick: Compare only the common letter and common digit – order of digits in the code does not matter.

Answer: (B) 3

Concept: Decode each symbol and draw the family tree.

1. E $ F: E is the son of F (E is male).
2. F ÷ G: F is the father of G; G + M: G is the brother of M.
3. So E, G and M are children of F; M @ N: M is the wife of N.
4. E is the brother of M, i.e. N's wife's brother.

Trick: Read the chain left to right and keep track of gender: '$' tells E is male, '@' tells M is female.

Answer: (B) Wife's brother

Concept: Check the differences between consecutive terms.

1. Differences: $4-5=-1$, $2-4=-2$, $-2-2=-4$, $-10-(-2)=-8$.
2. Differences double: $-1,-2,-4,-8$, next $-16$.
3. Next term $=-10-16=-26$.

Trick: When differences are $1,2,4,8$, the next difference is always 16.

Answer: (C) −26

Concept: When everyone faces the centre, a person's left is clockwise and right is anticlockwise.

1. Fix G; F is third to the left of G → F sits opposite G.
2. Y is second to the right of F; H is opposite Y (two people between them).
3. D is next to H at the only free seat beside H; E takes the last seat.
4. From E going right: Y, G, then D → two people between.

Trick: In a 6-seat circle, 'third to the left/right' always means directly opposite.

Answer: (A) Two

Concept: 'All A are B' lets us conclude 'Some B are A' (conversion), but not 'All B are A'.

1. All oranges are apples → Some apples are oranges: (I) follows.
2. Oranges ⊂ Apples ⊂ Grapes, but grapes may contain much more than oranges.
3. So 'All grapes are oranges' (II) does not follow.

Yaad rakhein: An 'All' statement converts only into a 'Some' statement, never back into 'All'.

Answer: (A) Only conclusion (I) follows.

Concept: People between two persons (no overlap) $=\text{Total}-(\text{left rank}+\text{right rank})$.

1. Into is 11th from left, Toni is 19th from right.
2. Positions do not overlap since $11+19<112$.
3. Between $=112-(11+19)=82$.

Trick: No overlap → subtract both ranks from total; overlap → $\text{ranks}-\text{total}-2$.

Answer: (A) 82

Concept: Only the first and last digits matter here.

1. First digit 2 (even) $+3=5$.
2. Last digit 3 (odd) $+2=5$.
3. Sum $=5+5=10$.

Trick: No need to transform all seven digits – change only the two asked for.

Answer: (A) 10

Concept: Position from right $=\text{Total}-\text{position from left}+1$.

1. Sanjana is 27th from left; Anand is 18 places to her right → $27+18=45$th from left.
2. From right $=47-45+1=3$.

Trick: Left + Right − 1 = Total → Right $=47+1-45=3$.

Answer: (C) 3rd

Concept: Arrange in descending order and compare each position with the original.

1. Original: 121, 256, 812, 658, 527.
2. Descending: 812, 658, 527, 256, 121.
3. Position-wise: 121≠812, 256≠658, 812≠527, 658≠256, 527≠121 → none unchanged.

Trick: Write the two lists one below the other and match column by column.

Answer: (C) Zero