Q&A: practice questions with answers

Concept: Do the multiplication first, then the bracket, then add.

1. $(-\frac{5}{3})\times \frac{9}{25}=-\frac{3}{5}$.
2. $\frac{1}{5}-3=-\frac{14}{5}$.
3. $-\frac{3}{5}-\frac{14}{5}=-\frac{17}{5}$.

Trick: Use denominator 5 throughout: $-\frac{3}{5}+\frac{1}{5}-\frac{15}{5}=-\frac{17}{5}$.

Answer: (B) $-\frac{17}{5}$

Concept: An identity is true for all values of the variables.

1. $(a+b{)}^2=a^2+b^2+2ab$, so $(a+b{)}^2=a^2+b^2$ misses $2ab$.
2. Test $a=b=1$: LHS $=4$, RHS $=2$ – false.
3. Options (B), (C), (D) are standard identities.

Trick: Put $a=b=1$ in each option; the one that fails is not an identity.

Answer: (A) $(a+b{)}^2=a^2+b^2$

Concept: Equal work: (workers × days) must be equal in both cases.

1. $10(6m+8b)=2(26m+48b)\Rightarrow 60m+80b=52m+96b$.
2. $8m=16b\Rightarrow m=2b$.
3. Total work $=10(12b+8b)=200$ boy-days.
4. $40m+20b=100b$; time $=\frac{200}{100}=2$ days.

Trick: Convert all to boys: $1\text{man}=2\text{boys}$, then days $=\frac{\text{total work}}{\text{boys}}$.

Answer: (C) 2 days

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{P}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{D}\%}{100}$.

1. SP $=120\times 1.1=132$ (₹).
2. MP $\times 0.96=132$.
3. MP $=\frac{132}{0.96}=137.50$ (₹).

Trick: $\text{MP}=\text{CP}\times \frac{100+P}{100-D}=120\times \frac{110}{96}=137.5$.

Answer: (D) ₹137.50

Concept: To cross a bridge, the train covers (train length + bridge length).

1. Train length $=(24+1)\times 60=1500$ m.
2. Distance $=1500+1500=3000$ m $=3$ km.
3. Time $=\frac{3}{50}$ h $=\frac{3}{50}\times 60=3.6$ min.

Trick: Keep km and hours: $\frac{3\text{km}}{50\text{km/h}}\times 60$ gives minutes directly.

Answer: (D) 3.6 min

Concept: Find the shift applied to each letter of the first pair and apply it to NKSO.

1. J→H, G→E, O→M, K→I: each letter moves back 2 (−2).
2. Check: T→R, Q→O, Y→W, U→S – also −2.
3. N→L, K→I, S→Q, O→M gives LIQM.

Trick: Same position-wise shift in both example pairs → apply directly letter by letter.

Answer: (B) LIQM

Concept: Facing the centre, 'right' is anticlockwise and 'left' is clockwise. Fix M first.

1. Two people between M and N from M's right → N is 3rd seat to M's right.
2. Two between N and E from N's right → E is 3rd to N's right, i.e. immediate left of M.
3. Three between M and F from M's right → F is 4th to M's right.
4. Remaining seats go to D, G, L; D immediate left of G → order M, D, G, N, F, L, E (going right).
5. Third to the left of L: F, N, G → G.

Yaad rakhein: Draw the circle, place the fixed person, then count seats exactly – left/right flips when facing the centre.

Answer: (D) G

Concept: Remove all symbols, then count from the right end.

1. Numbers only: 3 9 2 4 6 8 7 1 5.
2. From the right: 5 (1st), 1 (2nd), 7 (3rd), 8 (4th), 6 (5th), 4 (6th), 2 (7th).

Trick: 7th from right among 9 numbers = $(9-7+1)=3$rd from left = 2.

Answer: (A) 2

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

1. Kevin from left $=45-16+1=30$th.
2. Facing north, left is the left end; Luther is 10 places to Kevin's left.
3. Luther from left $=30-10=20$th.

Trick: Convert to one end first, then move left (subtract) or right (add).

Answer: (A) 20^th

Concept: Replace the letters by signs, then apply BODMAS.

1. $85÷17+82-11\times 2$.
2. $=5+82-22$.
3. $=65$.

Yaad rakhein: Do ÷ and × first, then + and − from left to right.

Answer: (B) 65

Concept: 'Some' + 'All' gives a 'Some' conclusion.

1. Some books are trees + all trees are mats → some books are mats. (II) follows.
2. All mats are ponds → those books are also ponds → some ponds are books. (I) follows.

Yaad rakhein: Some A are B, all B are C ⇒ some A are C (and some C are A).

Answer: (B) Both conclusions (I) and (II) follow.

Concept: Decode each symbol from left to right.

1. M ÷ N: M is brother of N.
2. N × O: N is wife of O.
3. O + K: O is father of K.
4. K × L: K is wife of L; L − R: L is son of R.
5. So K is the wife of R's son.

Yaad rakhein: Read pair by pair and track gender: × makes K female, − makes L R's son.

Answer: (B) Son's wife

Concept: Test each option to find one rule that works on both sides.

1. Option (A): $81\times 2+1=163$ ✓.
2. Same rule: $74\times 2+1=149$ ✓.
3. Other options fail, e.g. $93\times 2+1=187\ne 163$.

Trick: 163 is odd and close to $2\times 81$ → try $2x+1$ first.

Answer: (A) X = 81, Y = 149

Concept: Common letters in two words share the common digits (order may be jumbled).

1. PILED = {P, I, L, E, D}, digits {6, 5, 4, 2, 8}.
2. PEDAL = {P, E, D, A, L}, digits {5, 7, 6, 2, 8}.
3. Common letters P, L, E, D ↔ common digits 6, 5, 2, 8.
4. Letter left in PILED is I, digit left is 4 → I = 4.

Trick: The letter present only in the first word takes the digit present only in its code.

Answer: (C) 4

Concept: Place the towns on a grid one statement at a time.

1. L — X — I lie on a west–east line (L west, I east of X).
2. N is north of L; P is east of N and north-east of L.
3. I is south-west of P, so P lies north-east of I.
4. X is west of I, so P is north and east of X → north-east.

Trick: If P is north-east of a town that is east of X, P is north-east of X too.

Answer: (B) North-east

Concept: Compare letter positions of the first cluster with the second.

1. IM-JL: I+1=J, M−1=L.
2. FJ-GI: F+1=G, J−1=I; QU-RT: Q+1=R, U−1=T.
3. LP-MQ: L+1=M but P+1=Q (not −1).
4. So LP-MQ breaks the (+1, −1) pattern.

Trick: Write the shifts as (+1, −1) for each pair – the pair with a different sign is the odd one.

Answer: (D) LP-MQ

Concept: Track each letter position separately.

1. 1st letters: R, U, X, A → +3 each → D.
2. 2nd letters: M, O, Q, S → +2 each → U.
3. 3rd letters: K, L, M, N → +1 each → O.
4. Next term: DUO.

Trick: After Z wrap to A: X + 3 = A, A + 3 = D.

Answer: (A) DUO

Concept: Vowels → next letter, consonants → previous letter.

1. M→L, A→B, J→I, O→P, R→Q, I→J, T→S, Y→X.
2. New group: L B I P Q J S X.
3. Letters also in MAJORITY: I and J → 2.

Trick: J becomes I and I becomes J – they swap, so both stay common.

Answer: (B) 2

Concept: A valid reason must directly explain why people migrate.

1. Reason I gives work, family, study, conflict and persecution – all genuine causes of migration.
2. Reason II gives climate change and natural disasters – also genuine causes.
3. Both explain the movement described in the statement.

Yaad rakhein: Push factors (conflict, disasters) and pull factors (jobs, study, family) both cause migration.

Answer: (D) Both I and II are possible reasons.

Concept: Find one operation that works on both pairs for an option.

1. Option (C): $23+52=75$.
2. Same rule: $29+52=81$ ✓.
3. Others fail: (A) $26\to 75$ is $+49$ but $29\to 76$ is $+47$; (D) $+54$ vs $+60$.

Trick: For an addition rule, $\text{Y}-29$ must equal $75-\text{X}$; only (C) gives $52=52$.

Answer: (C) X = 23, Y = 81