Q&A: practice questions with answers

Concept: Make all numbers have the same number of decimal places, take LCM of the integers, then restore the decimal.

1. $5=5.00$, $3.2=3.20$, $0.03$ → integers 500, 320, 3.
2. $\text{LCM}(500,320)=8000$ ($500=2^2{5}^3$, $320=2^{6}5$ → $2^6{5}^3$); with 3 → 24000.
3. Divide by 100: LCM $=240$.

Trick: LCM of decimals $=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$ when written as fractions over 100.

Answer: (D) 240

Concept: Net income $=$ income $-$ tax. The extra tax equals the fall in net income.

1. Let income $=100$ and tax rate $=t\%$; net $=100-t$.
2. Tax rises by $18\%$ of $t$: extra tax $=0.18t$. Net falls by $2\%$ of $(100-t)$.
3. $0.18t=0.02(100-t)\Rightarrow 0.2t=2\Rightarrow t=10\%$.

Trick: Test the options: at $10\%$, tax goes $10\to 11.8$; net goes $90\to 88.2$, a fall of $1.8=2\%$ of 90 ✓.

Answer: (A) 10%

Concept: Write SP in terms of CP for both situations and equate.

1. Let CP $=x$. Original SP $=1.58x$.
2. New: CP $=x+20$, SP $=1.58x-74$, profit $52.5\%$: $1.58x-74=1.525(x+20)$.
3. $1.58x-1.525x=30.5+74\Rightarrow 0.055x=104.5\Rightarrow x=1900$.

Trick: Check: CP 1900, SP 3002; new CP 1920, SP 2928; $\frac{1008}{1920}=52.5\%$ ✓.

Answer: (A) 1,900

Concept: $P=\frac{SI\times 100}{R\times T}$.

1. $P=\frac{1440\times 100}{8\times 3}=\frac{144000}{24}$.
2. $P=$ ₹6,000.

Trick: $8\%\times 3=24\%$ of the sum $=1440\Rightarrow$ sum $=1440÷0.24=6000$.

Answer: (A) 6,000

Concept: Solve brackets first, treat "of" as multiplication before $÷$, then $÷$, then $+$.

1. $\frac{20}{33}÷\frac{20}{7}=\frac{20}{33}\times \frac{7}{20}=\frac{7}{33}$.
2. $\frac{3}{5}\times \frac{10}{9}+\frac{2}{11}=\frac{2}{3}+\frac{2}{11}=\frac{22+6}{33}=\frac{28}{33}$; so $\frac{7}{33}÷\frac{28}{33}=\frac{1}{4}$.
3. $\frac{28}{23}\text{ of }\frac{23}{4}=7$; $\frac{7}{4}÷7=\frac{1}{4}$.
4. Total $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$.

Trick: "of" binds tighter than $÷$: compute $\frac{28}{23}\text{ of }\frac{23}{4}$ first – it collapses to 7.

Answer: (C) $\frac{1}{2}$

Concept: Scan the series left to right and test each even digit: the digit before it must be a perfect square (1, 4 or 9) and the digit after it must be odd.

1. Series: 9 6 3 1 8 7 2 9 5 6 1 4 7 5 3 4 5 6 2 6 8 7 2 3 7 6 5 8.
2. Matches: 9 6 3 (6 between 9 and 3), 1 8 7 (8 between 1 and 7), 1 4 7 (4 between 1 and 7).
3. Check the rest: 4 5 6 → 6 is followed by 2 (even) ✗; 9 5 → 5 is odd ✗. So exactly three.

Trick: Underline every 1, 4, 9 first and look only at the digit right after each – the search shrinks to a few spots.

Answer: (C) प

Concept: Write the place values (A = 1 … Z = 26) and compare how the second cluster is obtained from the first.

1. OR–KM: O(15)→K(11) is $-4$, R(18)→M(13) is $-5$.
2. QT–MO: Q(17)→M(13) $-4$, T(20)→O(15) $-5$. SV–OQ: S(19)→O(15) $-4$, V(22)→Q(17) $-5$.
3. UX–QR: U(21)→Q(17) $-4$, but X(24)→R(18) is $-6$ – breaks the $(-4,-5)$ pattern.

Trick: Also see the gap inside each cluster: OR, QT, SV, UX all $+3$; KM, MO, OQ are $+2$ but QR is $+1$ – the odd one jumps out.

Answer: (A) ह

Concept: Decode each symbol step by step and build the family tree, then read L’s relation to T.

1. L # I → L is the wife of I; I $ G → I is the father of G. So L is the mother of G.
2. G @ H → G is the sister of H; H & T → H is the brother of T. So G, H and T are all children of I and L.
3. Hence L is the mother of T.

Trick: In a chain, fix the generation of the first and last person only – here L is one level above G, H and T, and she is female, so “mother”.

Answer: (D) Mother

Concept: Look at the differences: $+4,+9,+4,+22,+4,?$ – the $+4$ steps alternate with a second rule.

1. $7+4=11,11\times 2-2=20$
2. $20+4=24,24\times 2-2=46$
3. $46+4=50,50\times 2-2=98$

Trick: When differences alternate small–big–small, test the big jumps as “$\times 2\pm k$”: $11\to 20$ and $24\to 46$ are both $\times 2-2$.

Answer: (A) ज

Concept: Position from the left $=$ Total $-$ position from the right $+1$. Moving to the left decreases the left-end position.

1. Ccd from the left $=122-54+1=69$th.
2. Vks is 17th to the left of Ccd: $69-17=52$nd from the left.

Trick: “Left of” → subtract from the left-position; “right of” → add. First convert everyone to the same end.

Answer: (A) “

Concept: Each term is obtained by multiplying by an increasing number and subtracting the next number: $\times 1-2,\times 2-3,\times 3-4,\dots$

1. $8\times 1-2=6,6\times 2-3=9$
2. $9\times 3-4=23,23\times 4-5=87$
3. $87\times 5-6=435-6=429$

Trick: A series that grows faster and faster (6, 9, 23, 87) usually hides a multiplier that increases by 1 each step.

Answer: (B) त

Concept: For people facing the centre, “right” means anticlockwise and “left” means clockwise. Fix Av and place the others.

1. Bg is 2nd to the right (anticlockwise) of Av; Gp is 2nd to the left (clockwise) of Av.
2. Dq is 3rd to the right of Gp – going anticlockwise from Gp past Av, Dq lands just before Bg. Three seats remain: one between Av and Gp, and two between Gp and Bg.
3. Ck is immediately right of Fx, so Fx and Ck take the two adjacent seats between Gp and Bg (Ck next to Gp); Et takes the seat between Av and Gp.
4. Counting from the left (clockwise) of Et: Gp, then Ck – one person between.

Trick: Draw the circle first and mark arrows “L ↻, R ↺” – half the mistakes in such questions are direction slips.

Answer: (D) प

Concept: Track the first letters of the clusters and the internal pattern of each cluster.

1. First letters: U(21) → G(7) → S(19) → E(5) → ? : alternately $-14$ and $+12$, so next $=5+12=17=$ Q.
2. Inside every cluster: 2nd letter $=$ 1st $-3$, 3rd letter $=$ 1st $+2$ (U R W: 21, 18, 23).
3. For Q(17): $17-3=14=$ N, $17+2=19=$ S → QNS.

Trick: Check the internal pattern against the options: only QNS has the shape (x, x−3, x+2).

Answer: (A) भ

Concept: Number the positions from the top (1) to the bottom (7) and place the boxes using the definite clues first.

1. Only one box above E → E is 2nd; B is immediately above E → B is 1st (top).
2. Two boxes between C and E → C is 5th.
3. “Only Q is below S” → S is 6th and Q is 7th (bottom). The remaining D and R go to 3rd and 4th with D above R.
4. Order (top→bottom): B, E, D, R, C, S, Q. Between B (1st) and S (6th): E, D, R, C = four boxes.

Trick: Start from clues that fix an exact slot (“only one above”, “immediately above”) – they anchor the stack in seconds.

Answer: (B) ठ

Concept: Find how each letter and the number change in the given pairs, then apply the same change.

1. MJ 8 → OM $-6$: M$+2=$O, J$+3=$M, $8-14=-6$.
2. RW 12 → TZ $-2$: R$+2=$T, W$+3=$Z, $12-14=-2$ ✓ same rule.
3. UR 1 → U$+2=$W, R$+3=$U, $1-14=-13$ → WU $-13$.

Trick: Numbers going negative almost always mean a fixed subtraction – find it from the first pair ($8\to -6$ is $-14$) and confirm with the second.

Answer: (C) ऋ

Concept: Compare the shift of the first letters and of the second letters (use place values, wrapping past Z back to A).

1. SU–KA: S(19)→K(11) $-8$; U(21)→A(1) $-20$.
2. UW–MC: U(21)→M(13) $-8$; W(23)→C(3) $-20$. MO–EU: M(13)→E(5) $-8$; O(15)→U(21) is $-20$ after wrapping ($15-20=-5\Rightarrow -5+26=21$).
3. ZB–PH: Z(26)→P(16) is $-10$, not $-8$ – the odd pair (B→H is $-20$ with wrap, consistent, but the first letter breaks).

Trick: Check first letters alone first: 19→11, 21→13, 13→5 are all $-8$; 26→16 is $-10$ – answer found without touching the second letters.

Answer: (A) प

Concept: When two words share letters, the common letters share the common digits; the leftover letter gets the leftover digit.

1. DOLT → 4 9 6 3 and LOTS → 6 4 3 5.
2. Common letters O, L, T ↔ common digits 4, 6, 3.
3. The extra letter in LOTS is S and the extra digit is 5, so S = 5 (and D = 9).

Trick: Don’t try to match positions – just cancel the common letters and common digits; what is left pairs up.

Answer: (B) स

Concept: Find the letter-wise shift in the given pairs and apply it to the options.

1. QVI : KZJ → Q(17)→K(11) $-6$, V(22)→Z(26) $+4$, I(9)→J(10) $+1$.
2. USH : OWI → U(21)→O(15) $-6$, S(19)→W(23) $+4$, H(8)→I(9) $+1$ ✓ pattern $(-6,+4,+1)$.
3. LQG : FUH → L(12)→F(6) $-6$, Q(17)→U(21) $+4$, G(7)→H(8) $+1$ ✓.

Trick: Test only the third letter first ($+1$): LQG→FUH (G→H ✓), CNK→WRJ (K→J ✗), BRH→VVF ✗, AHL→UMK ✗ – one check settles it.

Answer: (A) प

Concept: Compare corresponding letters of consecutive clusters.

1. First letters F, G, H, I → next J; second letters H, I, J, K → next L; third letters D, E, F, G → next H.
2. Every letter moves $+1$ each step, so the next cluster is JLH.

Trick: When all three positions step uniformly, just add 1 to each letter of the last cluster: IKG → JLH.

Answer: (B) ज

Concept: Facing the centre: right = anticlockwise, left = clockwise. Begin with the pair that is fixed relative to each other (J and Z).

1. Z is immediately right (anticlockwise) of J. J is 3rd to the left (clockwise) of K, so K is 3 seats anticlockwise from J (one seat beyond Z... i.e. Z, then a gap, then K). Z is 3rd to the right of I, so I is 3 seats clockwise from Z, i.e. 2 seats clockwise from J.
2. Going clockwise from J: J, _, I, _, K, _, Z. The gaps are for H, W, Y.
3. Y is not a neighbour of J, so Y is not in the seat between J and I; one person between H and W counted from the left (clockwise) of W → W between J and I, H between I and K, Y between K and Z.
4. From the right (anticlockwise) of Y: K, H, I, then W → three people between Y and W.

Trick: Fill the circle with blanks first, then use “not a neighbour” clues to kill the wrong blanks.

Answer: (C) प