Q&A: practice questions with answers

Concept: Use $(a\pm b{)}^2=a^2\pm 2ab+b^2$ and $(a+b)(a-b)=a^2-b^2$.

1. $(2x-5y{)}^2=4x^2-20xy+25y^2$
2. $(5x+2y{)}^2=25x^2+20xy+4y^2$
3. $(2x+5y)(2x-5y)=4x^2-25y^2$
4. Sum $=33x^2+4y^2$ (the $xy$ terms and $25y^2$ cancel)

Trick: Put $x=y=1$: $9+49-21=37$, and only $33+4=37$ matches.

Answer: (D) $33x^2+4y^2$

Concept: For two numbers, $\text{LCM}\times \text{HCF}=$ product of the numbers.

1. Numbers $=5k$ and $3k$ (5, 3 co-prime) ⇒ HCF $=k$, LCM $=15k$
2. $15k\times k=135$ ⇒ $k^2=9$ ⇒ $k=3$; LCM $=45$, HCF $=3$
3. $\frac{1}{45}+\frac{1}{3}=\frac{1+15}{45}=\frac{16}{45}$

Trick: Sum of reciprocals $=\frac{\text{LCM}+\text{HCF}}{\text{LCM}\times \text{HCF}}=\frac{48}{135}=\frac{16}{45}$.

Answer: (A) $\frac{16}{45}$

Concept: In the right triangle formed by ladder, wall and ground: $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Height on the wall $=30$ m, base $=x$
2. $\tan \theta =\frac{30}{x}=\frac{15}{8}$
3. $x=\frac{30\times 8}{15}=16$ m

Trick: $15\to 30$ means scale $\times 2$, so base $8\to 16$.

Answer: (B) $16$

Concept: Speed $=\frac{\text{distance}}{\text{time}}$ for the remaining part.

1. Covered $=\frac{2}{3}\times 519=346$ km in $\frac{5}{7}\times 7=5$ h
2. Remaining distance $=519-346=173$ km; time left $=7-5=2$ h
3. Speed $=\frac{173}{2}=86.5\text{km/h}$

Trick: Remaining $=\frac{1}{3}$ of 519 $=173$ km in 2 h.

Answer: (A) $86.5$

Concept: For $n$ (odd) observations in order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending order: 11, 18, 27, 30, 34, 35, 52, 76, 81, 83, 84
2. $n=11$ ⇒ median $=6^{\text{th}}$ value $=35$

Trick: Always sort first – the 6th value of the unsorted list (52) is the trap.

Answer: (B) $35$

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $28.4=3\times 88-2\text{Mean}=264-2\text{Mean}$
2. $2\text{Mean}=264-28.4=235.6$
3. $\text{Mean}=117.8$

Trick: Mean $=\frac{3\text{Median}-\text{Mode}}{2}$.

Answer: (C) $117.8$

Concept: Take CP $=100$; then MP and SP follow from the given percentages.

1. CP $=100$ ⇒ MP $=130$ (30% above CP)
2. Profit 2% ⇒ SP $=102$
3. Discount $=130-102=28$; $\text{Discount}\%=\frac{28}{130}\times 100\approx 21.54\%$

Trick: The ₹517 is not needed: $\frac{30-2}{130}\times 100$.

Answer: (D) $21.54\%$

Concept: Replace each sign by its meaning, then solve using BODMAS.

1. Given: $32÷14+16-8\times 4$
2. Replace (÷→+, +→−, −→×, ×→÷): $32+14-16\times 8÷4$
3. $16\times 8÷4=32$, so $32+14-32=14$

Trick: Write the new expression first, then apply BODMAS – never solve before replacing signs.

Answer: (A) 14

Concept: Check every digit: the term just before it AND the term just after it must both be symbols.

1. Series: # * £ 3 9 € # % £ 1 % £ 8 8 1 4 9 % 5 8 & 7
2. Pattern needed: symbol – number – symbol.
3. £ 1 % is the only such group (3 is followed by 9; 5 is followed by 8; 8 is preceded by 5; 7 is at the end).

Trick: Scan only the digits and look left/right of each – it is faster than scanning symbols.

Answer: (C) 1

Concept: Interchange + ↔ − and × ↔ ÷, then apply BODMAS.

1. Given: $38÷21+436\times 4-73$
2. After interchange: $38\times 21-436÷4+73$
3. $=798-109+73=762$

Trick: Do the swap carefully sign by sign; then BODMAS.

Answer: (B) 762

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

1. No box above F ⇒ F = 7. Three boxes between F and B ⇒ B = 3. D immediately above B ⇒ D = 4.
2. One box between G and D ⇒ G = 2 or 6. Four boxes between G and C: G = 2 gives C = 7 (taken), so G = 6, C = 1.
3. A above E in the remaining places 2 and 5 ⇒ A = 5, E = 2.
4. Order (bottom→top): C, E, B, D, A, G, F ⇒ boxes below D = C, E, B = 3.

Trick: Start with the "top/bottom" clue – it fixes one box exactly and the rest follow.

Answer: (C) 3

Concept: Take A as origin (0, 0); east = +x, north = +y. A left turn from east is north, from north is west, and so on.

1. 12 km east → (12, 0); left, 9 km north → (12, 9).
2. Left, 13 km west → (−1, 9); left, 11 km south → (−1, −2).
3. Left, 1 km east → P = (0, −2).
4. P is 2 km directly south of A, so he must drive 2 km north.

Trick: Net east-west: 12 − 13 + 1 = 0; net north-south: 9 − 11 = −2 (south) ⇒ go 2 km north.

Answer: (C) 2 km to the north

Concept: Chefs who are teachers but not lawyers = the region common to Chef and Teacher circles only, outside the Lawyer circle.

1. Chef ∩ Teacher has two parts: 7 (inside Lawyer too) and 9 (outside Lawyer).
2. Exclude the lawyer part (7) ⇒ required number = 9.

Trick: "But not X" ⇒ remove every region lying inside X's circle.

Answer: (B) 9

Concept: Find the letter-wise rule from # to VEC (test the options) and apply it to BNW.

1. Option (A): XHA → VEC: X−2=V, H−3=E, A+2=C (A wraps back: Y, Z, A … A+2 = C).
2. Apply −2, −3, +2 to BNW: B−2=Z, N−3=K, W+2=Y ⇒ ZKY.
3. This matches % = ZKY in option (A).

Trick: Use position numbers: X(24)→V(22), H(8)→E(5), A(1)→C(3); B(2)→Z(26), N(14)→K(11), W(23)→Y(25).

Answer: (A) # = XHA, % = ZKY

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

1. Sunita from left $=16+21=37$th.
2. From right $=73-37+1=37$th.

Trick: Sunita is exactly the middle student: 36 on each side of her in a row of 73.

Answer: (D) ${37}^{\text{th}}$

Concept: Check the gap between letters inside each cluster and between the clusters.

1. HL – JN: H+4=L, J+4=N (and H+2=J). NR – PT: N+4=R, P+4=T. SW – UY: S+4=W, U+4=Y.
2. MU – KH: M+8=U, K−3=H – the pattern breaks.
3. So MU – KH is the odd pair.

Trick: In three pairs both clusters have a +4 gap; spot the one that doesn't.

Answer: (B) MU - KH

Concept: Find the rule # → JLH by testing options, then apply it to IKG.

1. Option (C): EGC → JLH: E+5=J, G+5=L, C+5=H.
2. Apply +5 to IKG: I+5=N, K+5=P, G+5=L ⇒ NPL.
3. This matches % = NPL in option (C).

Trick: Position numbers: E(5)→J(10), G(7)→L(12), C(3)→H(8); I(9)→N(14), K(11)→P(16), G(7)→L(12).

Answer: (C) # = EGC, % = NPL

Concept: Every digit (odd or even) gets +2.

1. 5217643 → 7 4 3 9 8 6 5 ⇒ 7439865.
2. First digit from left = 7, first from right = 5.
3. Sum = 7 + 5 = 12.

Trick: Only the two end digits matter: 5+2 = 7 and 3+2 = 5.

Answer: (C) 12

Concept: Check the conditions on the first, last, second and third elements; code the rest from the table.

1. Group: 5 & 3 @ 7 % 2. First = 5 (odd number), last = 2 (even number) → condition (ii) applies: first and last become ©.
2. Condition (iii): second = &, third = 3 – not perfect squares, so not applied.
3. Middle codes: & = M, 3 = D, @ = Z, 7 = W, % = P.

Trick: Test condition (ii) first whenever both ends are numbers – it decides the two © at once.

Answer: (A) © M D Z W P ©

Concept: Look at the differences between consecutive terms.

1. $701-1=700,700-3=697,697-5=692,692-7=685$
2. Differences are odd numbers $1,3,5,7$; next is $9$.
3. $685-9=676$

Trick: Successive odd-number subtraction: also equals 701 minus a perfect square.

Answer: (D) $676$