Q&A: practice questions with answers

Concept: Mode is the value that occurs most often.

1. 7 occurs twice; 8, 9, 10 occur once each.
2. Mode $=7$.

Yaad rakhein: Mode = most frequent, median = middle value, mean = average.

Answer: (A) 7

Concept: Opposite directions: relative speed = sum; distance = sum of lengths.

1. Relative speed $=40+60=100\text{km/h}=100\times \frac{5}{18}=\frac{250}{9}\text{m/s}$.
2. Distance $=100+150=250\text{m}$.
3. Time $=250÷\frac{250}{9}=9\text{s}$.

Trick: Time (s) $=\frac{\text{total length (m)}\times 18}{\text{speed (km/h)}\times 5}=\frac{250\times 18}{500}=9$.

Answer: (C) 9 seconds

Concept: After $n$ years, the sum of two ages increases by $2n$.

1. Present sum $=60-2\times 6=48$.
2. $5x+7x=12x=48\Rightarrow x=4$.
3. A $=5\times 4=20$ years.

Trick: A $=\frac{5}{12}\times 48=20$.

Answer: (B) 20 years

Concept: Net price $=MP\times \frac{100-d}{100}$.

1. $2400\times \frac{75}{100}=1800$.

Trick: $25\%=\frac{1}{4}$: $2400-600=1800$.

Answer: (D) ₹1,800

Concept: $A=P(1+\frac{RT}{100})$, $SI=\frac{PRT}{100}$.

1. $5040=P(1+0.6)=1.6P\Rightarrow P=3150$.
2. SI for 2 years $=3150\times \frac{20\times 2}{100}=1260$.

Trick: 3 years' SI $=60\%$ of P, amount $=160\%$; 2 years $=40\%$: $\frac{40}{160}\times 5040=1260$.

Answer: (D) 1,260

Concept: Check the gaps between consecutive letters of each cluster.

1. BCDF: $+1,+1,+2$
2. HIJL: $+1,+1,+2$; OPQS: $+1,+1,+2$
3. UVWX: $+1,+1,+1$ – different pattern.

Trick: Three clusters skip one letter at the end (E, K, R missing); UVWX skips nothing.

Answer: (B) UVWX

Concept: Look at the first letters and at the link inside each term.

1. First letters: F, J, N, R → each $+4$, so next is V.
2. Inside each term: 2nd = 1st $+4$, 3rd = 2nd $-1$ (F→J→I, J→N→M, R→V→U).
3. V $+4$ = Z, Z $-1$ = Y → VZY.

Trick: Every letter of each term moves $+4$: FJI → JNM → NRQ → RVU → VZY.

Answer: (B) VZY

Concept: Letters common to both words take the digits common to both codes.

1. CONE = 1536, NOTE = 3146.
2. Common letters: O, N, E ↔ common digits: 1, 3, 6.
3. Left over: CONE has C ↔ 5; NOTE has T ↔ 4.

Trick: Remove the common part; the unmatched letter gets the unmatched digit.

Answer: (D) 4

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

1. A = $÷$, B = $\times$, C = $+$, D = $-$.
2. $43-28+35÷7\times 11$
3. $=43-28+5\times 11=43-28+55$
4. $=70$

Trick: Do $÷$ and $\times$ left to right first: $35÷7\times 11=55$.

Answer: (C) $70$

Concept: Number positions 1 (top) to 7 (bottom) and fix the definite clues first.

1. Three boxes above G → G = 4. One box between J and G → J = 2 or 6.
2. Three between J and K, with K above G: J = 6, K = 2 (J = 2 would put K at 6, below G). S just below K → S = 3.
3. F, H, I take 1, 5, 7. H is not next to J (5, 7) → H = 1; F above I → F = 5, I = 7.
4. Order: H, K, S, G, F, J, I. Between H(1) and J(6): K, S, G, F = 4.

Trick: Boxes between = $6-1-1=4$.

Answer: (B) Four

Concept: Decode each symbol one link at a time and draw the tree.

1. S ÷ D: S is the son of D (S is male).
2. D & R: D is the father of R. R + Y: R is the brother of Y, so S, R, Y are siblings.
3. Y # H: Y is the wife of H.
4. S is the brother of Y, i.e. brother of H's wife.

Trick: Brother of wife = brother-in-law (wife's brother).

Answer: (D) Wife's brother

Concept: Number seats 1–7 from the left (all face north, so their right = our right).

1. Two people right of E → E = 5; S immediately left of E → S = 4.
2. Two between E and C → C = 2. Two between T and S → T = 1 or 7.
3. R, K must be adjacent (K right of R). If T = 7, seats 1, 3, 6 are left – no adjacent pair. So T = 1, R = 6, K = 7, I = 3.
4. Row: T C I S E R K → 3rd from left is I.

Trick: Test the two cases of T; the one that leaves a free adjacent pair for R–K wins.

Answer: (C) I

Concept: Find the moves from 1st→2nd and 2nd→3rd letter.

1. YBX: Y $+3$ = B (Z, A, B), B $-4$ = X.
2. NQM: N $+3$ = Q, Q $-4$ = M; JMI: J $+3$ = M, M $-4$ = I.
3. UYT: U $+4$ = Y, Y $-5$ = T – different.

Trick: In the group, 3rd letter = 1st letter $-1$ (Y→X, N→M, J→I); U→T also, so check the middle: only UYT has $+4$.

Answer: (C) UYT

Concept: Each term is the previous term multiplied by a fixed number.

1. $4\times 3=12$, $12\times 3=36$, $36\times 3=108$
2. $108\times 3=324$
3. Check: $324\times 3=972$ ✓

Trick: $\frac{972}{108}=9=3\times 3$, so the missing term is $108\times 3$.

Answer: (A) $324$

Concept: Write the word and its alphabetical rearrangement one below the other and compare position by position.

1. Original: D E P A R T S
2. Sorted: A D E P R S T
3. Positions 1–7: D/A, E/D, P/E, A/P, R/R, T/S, S/T → only R (5th) is unchanged.

Trick: R is the 5th smallest letter and is already 5th in the word.

Answer: (C) 1

Concept: Replace each sign as given, then use BODMAS.

1. '+' → $-$, '−' → $\times$, '×' → $÷$, '÷' → $+$.
2. $54\times 2+120÷2-85$
3. $=108+60-85$
4. $=83$

Trick: Rewrite the whole expression first, then calculate – never mix the old and new signs.

Answer: (B) $83$

Concept: Fix genders from the clues and draw the family tree.

1. P is the sister of Q; Q is the brother of S → P, Q, S are siblings.
2. X is the son of S and T is the father of X → S is X's mother (female), T her husband.
3. So X is the son of P's sister S.

Trick: When the father is named separately, the other parent (S) must be the mother.

Answer: (D) Sister's son

Concept: Number positions 1 (top) to 7 (bottom) and place the fixed boxes first.

1. O = 1. Only one box below P → P = 6. Two boxes between M and P → M = 3.
2. Left: 2, 4, 5, 7 for L, N, Q, R. Q just above N, L three places above N, R below N.
3. Only N = 5 works: Q = 4, L = 2, R = 7.
4. Order: O, L, M, Q, N, P, R. Between Q(4) and R(7): N, P = 2.

Trick: Boxes between = $7-4-1=2$.

Answer: (C) Two

Concept: Find how the third number is made from the first two.

1. $(27,12,322)$: $27\times 12=324$, $324-2=322$ ✓
2. $(24,13,310)$: $24\times 13=312$, $312-2=310$ ✓
3. (A) $31\times 8=248$, $248-2=246$ ✓; (B) $290-2=288\ne 292$; (C) $306-2=304\ne 308$; (D) $224-2=222\ne 224$.

Trick: Rule: third = first × second $-2$.

Answer: (A) $(31,8,246)$

Concept: Track the east–west and north–south displacements separately.

1. North 22, right → East 24, right → South 28, right → West 30, right → North 6.
2. North–south: $22-28+6=0$ → P is on the same level as A.
3. East–west: $24-30=-6$ → P is 6 km west of A.
4. So to return, drive 6 km towards the east.

Trick: Net vertical $=0$, so the answer is just the net horizontal gap: $30-24=6$.

Answer: (B) 6 km to the east