Q&A: practice questions with answers

Concept: If $a:x::x:b$, then $x^2=a\times b$ (x is the mean proportional).

1. $x^2=3.2\times 16.2=51.84$
2. $x=\sqrt{51.84}=7.2$ (taking $x>0$)

Trick: $3.2\times 16.2=\frac{32\times 162}{100}=\frac{5184}{100}$ and $\sqrt{5184}=72$.

Answer: (B) 7.2

Concept: $a^2+2ab+b^2=(a+b{)}^2$.

1. Here $a=75$, $b=25$
2. Expression $=(75+25{)}^2={100}^2$
3. $=10000$

Trick: Spot the pattern $a\times a+2\times a\times b+b\times b$ – no long multiplication needed.

Answer: (D) 10000

Concept: $\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$.

1. Time 1 $=\frac{60}{40}=1.5$ h, time 2 $=\frac{120}{80}=1.5$ h
2. Total distance $=180$ km, total time $=3$ h
3. Average $=\frac{180}{3}=60$ km/h

Trick: Equal times ⇒ average speed $=\frac{40+80}{2}=60$ km/h.

Answer: (C) 60 km/h

Concept: Adding the same number to every observation increases the average by that number.

1. Old average $=7$
2. Each number $+21$ ⇒ average $+21$
3. New average $=7+21=28$

Yaad rakhein: Add/subtract $k$ to all → average changes by $k$; multiply all by $k$ → average $\times k$.

Answer: (C) 28

Concept: Diagonals of a rhombus bisect each other at ${90}^{\circ }$; area $=\frac{1}{2}{d}_1{d}_2$.

1. Side $=\frac{100}{4}=25$ cm; half of given diagonal $=20$ cm
2. Other half $=\sqrt{{25}^2-{20}^2}=\sqrt{225}=15$ ⇒ $d_2=30$ cm
3. Area $=\frac{1}{2}\times 40\times 30=600{\text{ cm}}^2$

Trick: $15,20,25$ is the triplet $3,4,5$ scaled by $5$.

Answer: (B) 600 ${\text{cm}}^2$

Concept: Half-diagonals and a side form a right triangle: $a=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$.

1. Half-diagonals $=5$ cm and $12$ cm
2. $a=\sqrt{5^2+{12}^2}=\sqrt{169}$
3. $a=13$ cm

Trick: $5,12,13$ is a Pythagorean triplet.

Answer: (A) 13 cm

Concept: A word common to two sentences gets the code common to both coded forms.

1. 'happy children play' → 'ha ch pl'
2. 'children love toys' → 'ch lo tk'
3. Common word: children; common code: ch

Trick: Here each code is also the first two letters of the word – children → ch.

Answer: (D) ch

Concept: 'All A are B' always gives 'Some B are A'. 'Some B are C' does not connect A with C definitely.

1. All onions are tomatoes ⇒ some tomatoes are onions: I follows
2. Potatoes overlap tomatoes, but may lie outside the onion circle
3. So 'some onions are potatoes' is not certain: II does not follow

Yaad rakhein: 'All' converts to 'Some' (reverse); two 'Some'/'All–Some' links via the bigger set give no definite conclusion.

Answer: (C) Only conclusion I follows

Concept: With only 'Some' statements, conclusions follow only between directly linked pairs.

1. Plates–Glasses, Glasses–Bowls, Bowls–Cups are the only definite links
2. Cups and Glasses are not directly linked: I does not follow
3. Bowls and Plates are not directly linked: II does not follow

Yaad rakhein: Some + Some = no conclusion.

Answer: (D) Neither conclusion (I) nor (II) follows.

Concept: Swap the signs first, then solve using BODMAS.

1. $-\leftrightarrow ÷$ and $\times \leftrightarrow +$
2. $16÷2\times 5-12+2$
3. $=8\times 5-12+2=40-12+2$
4. $=30$

Trick: Write the new expression fully before calculating – avoids sign slips.

Answer: (D) 30

Concept: In a series of $n$ digits, $k^{\text{th}}$ from right $=(n-k+1{)}^{\text{th}}$ from left.

1. Total digits $=30$
2. ${15}^{\text{th}}$ from left $=9$
3. ${11}^{\text{th}}$ from right $=(30-11+1)={20}^{\text{th}}$ from left $=1$
4. Sum $=9+1=10$

Trick: Count in blocks of $5$: 7 6 3 9 5 | 4 4 8 6 4 | 8 8 9 6 9 | 2 2 6 3 1.

Answer: (D) 10

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

1. ARID → 9, 0, 7, 8; DIRE → 8, 0, 9, 5
2. Common letters D, I, R ↔ common digits 0, 8, 9
3. So A ↔ 7 (left in ARID) and E ↔ 5 (left in DIRE)

Trick: Odd letter out ↔ odd digit out.

Answer: (C) 5

Concept: A valid course of action tackles the root cause practically and is the duty of the right body.

1. Cause: industrial emissions → strict emission norms by authorities removes the cause: I follows
2. Distributing masks is not the industries' duty and does not solve the problem: II does not follow

Yaad rakhein: Prefer actions that remove the cause over temporary relief by the wrong agency.

Answer: (A) Only I follows

Concept: Check differences between consecutive terms.

1. $472-329=143$, $615-472=143$, $758-615=143$, $901-758=143$
2. Constant difference $=143$
3. Next $=901+143=1044$

Trick: Unit digits go $9,2,5,8,1$ (+3 each) ⇒ next ends in $4$ → only $1044$.

Answer: (C) 1044

Concept: In a series of $n$ digits, $k^{\text{th}}$ from right $=(n-k+1{)}^{\text{th}}$ from left.

1. Total digits $=30$
2. ${10}^{\text{th}}$ from left $=1$
3. ${12}^{\text{th}}$ from right $=(30-12+1)={19}^{\text{th}}$ from left $=2$
4. Sum $=1+2=3$

Trick: Blocks of $5$: 5 1 9 4 7 | 7 1 4 4 1 | 2 8 5 4 2 | 6 0 3 2 2.

Answer: (B) 3

Concept: Decode each symbol and build the family tree step by step.

1. Q × R: Q is mother of R
2. R – S: R is husband of S
3. S + T: S is sister of T; T ÷ U: T is son of U
4. Q is the mother of S's husband ⇒ Q is S's husband's mother (mother-in-law)

Trick: Only the chain Q → R → S is needed; T and U do not affect the answer.

Answer: (C) Husband's mother

Concept: Find the rule linking the three numbers in the given sets.

1. $(18,6,24)$: $18=3\times 6$, $24=18+6$
2. $(27,9,36)$: $27=3\times 9$, $36=27+9$
3. $(33,11,44)$: $33=3\times 11$, $44=33+11$ ✓

Trick: Pattern is $(3x,x,4x)$: $x=11$ gives $(33,11,44)$.

Answer: (A) (33, 11, 44)

Concept: Compare the gap inside each cluster using letter positions.

1. FI-DG: F(6)→I(9) $+3$, D(4)→G(7) $+3$
2. DG-BE: $+3$, $+3$; IL-GJ: $+3$, $+3$
3. NQ-LP: N(14)→Q(17) $+3$ but L(12)→P(16) $+4$ ✗

Trick: Both clusters should have the same gap; NQ-LP breaks it.

Answer: (A) NQ-LP

Concept: Find the letter-wise shift in the example pairs.

1. MILH → KGJF: each letter $-2$
2. QMPL → OKNJ: each letter $-2$ ✓
3. TPSO $-2$: T→R, P→N, S→Q, O→M = RNQM

Trick: Same shift for all four letters – apply it in the same order.

Answer: (B) RNQM

Concept: Inside each term letters jump $+3$; first letters of terms jump $+8$ (cyclic).

1. CFI, KNQ, SVY, ADG: each $+3$ inside
2. First letters: C→K→S→A, each $+8$ (S $+8$ wraps to A)
3. Next first letter $=$ A $+8=$ I ⇒ I, L, O

Trick: Last letter $+2$ gives the next first letter: I→K, Q→S, Y→A, so G→I.

Answer: (B) ILO