Q&A: practice questions with answers

Concept: Compare the effective price paid per toy (take MP = ₹100 each).

1. (A) $100\times 0.79\times 0.92=$ ₹72.68 per toy
2. (B) $100\times 0.62\times 0.58\times 0.70\approx$ ₹25.17 per toy
3. (C) $8\times 83+2\times 80=824$ for 10, i.e. ₹82.40 per toy
4. (D) Pay for 10, get 17: $\frac{1000}{17}\approx$ ₹58.82 per toy
5. Highest effective price = scheme C → least beneficial.

Trick: Successive discounts always beat the plain sum; a scheme with only $17$–$20\%$ off can never compete with $21\%+8\%$ or three big discounts.

Answer: (C) C

Concept: Symmetric equations – add them to get $m+o$ directly.

1. $8m+5o=164$, $5m+8o=148$
2. Adding: $13(m+o)=312\Rightarrow m+o=24$
3. $3(m+o)=72$

Trick: When coefficients are swapped, adding gives the sum and subtracting gives the difference – no need to solve for each item.

Answer: (C) ₹72

Concept: In the right triangle, $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$.

1. $\cos {60}^{\circ }=\frac{15}{L}$
2. $\frac{1}{2}=\frac{15}{L}\Rightarrow L=30$ m

Trick: In a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle the hypotenuse is twice the side opposite ${30}^{\circ }$ (the base here).

Answer: (B) 30 m

Concept: $a^m\times a^n=a^{m+n}$, $(a^m{)}^n=a^{mn}$, $\frac{a^m}{a^n}=a^{m-n}$.

1. Numerator: $5^{3x}\times 5^{2(x-1)}=5^{3x+2x-2}=5^{5x-2}$
2. Divide: $5^{(5x-2)-(4x-1)}=5^{x-1}$

Trick: Put $x=1$: expression $=\frac{5^3\times 1}{5^3}=1=5^0$; only $5^{x-1}$ gives $5^0$.

Answer: (A) $5^{x-1}$

Concept: Surface area $=4\pi r^2$, so the ratio of areas $={\left(\frac{r_1}{r_2}\right)}^2$.

1. $\frac{r_A}{r_B}=\frac{6}{3}=2$
2. Area ratio $=2^2=4:1$

Trick: Double the radius → area $\times 4$, volume $\times 8$.

Answer: (A) 4:1

Concept: Draw the Venn diagram for two "All" statements – the sets nest inside each other.

1. All Chickens are Quails and all Quails are Pigeons → Chickens ⊂ Quails ⊂ Pigeons, so all Chickens are Pigeons.
2. Conclusion (I) "No Pigeons are Chickens" is the opposite of this – false.
3. Conclusion (II) "Some Pigeons are Chickens" is the converse of "All Chickens are Pigeons" – it follows.

Trick: All A are B ⇒ Some B are A (always true); "No" can never follow from two "All" statements.

Answer: (C) Only conclusion (II) follows

Concept: Treat the letter part and the number part as two separate series.

1. Letters: E→G→I→K (+2 each), I→K→M→O (+2), J→L→N→P (+2). Next: K+2 = M, O+2 = Q, P+2 = R → MQR.
2. Numbers: $33\to 27\to 21\to 15$, each $-6$ → $15-6=9$.
3. Required cluster: MQR 9.

Trick: Check the first letter and the number first – M and 9 together already point to option (D).

Answer: (D) MQR 9

Concept: In sentence coding, a word common to two sentences has the code common to both.

1. 'jaw bush oven' → ht vq xe; 'zebra egg jaw' → vq db ia.
2. Common word: jaw; common code: vq.
3. So jaw = vq. (ht/xe stand for bush/oven; db/ia for zebra/egg.)

Trick: Match the repeated word with the repeated code – no need to decode the rest.

Answer: (A) vq

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

1. Only four people to the right of Q → Q at seat 3. Only two to the right of E → E at seat 5.
2. K is a neighbour of both J and E. If K = 4, its neighbours are Q and E – no room for J. So K = 6, J = 7.
3. H is third to the right of D: remaining seats 1, 2, 4 → D = 1, H = 4, and C = 2.
4. Order: D C Q H E K J. Between C (2) and K (6) sit Q, H, E = three.

Trick: "Only n people to the right" fixes a seat instantly: seat = 7 − n.

Answer: (B) Three

Concept: Find the position shift of each letter separately.

1. GFK : JAE → G(7)+3 = J(10), F(6)−5 = A(1), K(11)−6 = E(5). Pattern: $+3,-5,-6$.
2. LGP : OBJ → L+3 = O, G−5 = B, P−6 = J. Same pattern confirmed.
3. CLW → C+3 = F, L(12)−5 = G(7), W(23)−6 = Q(17) → FGQ.
4. Check (B): G+3 = J, I−5 = D, M−6 = G ≠ H – fails.

Trick: Write the shift triplet $(+3,-5,-6)$ once, then test each option quickly on the middle letter.

Answer: (D) CLW : FGQ

Concept: Compare letters by position and numbers by difference.

1. FR 6 → JL 17: F(6)+4 = J(10), R(18)−6 = L(12), $6+11=17$.
2. EU 12 → IO 23: E+4 = I, U(21)−6 = O(15), $12+11=23$. Rule: $(+4,-6,+11)$.
3. SL 3 → S(19)+4 = W(23), L(12)−6 = F(6), $3+11=14$ → WF 14.

Trick: The number rule ($+11$) alone eliminates (A), (C), (D).

Answer: (B) WF 14

Concept: We need the pattern odd – odd – even, and count the middle digit.

1. Series: 2 8 7 5 6 9 7 6 8 4 8 2 1 8 2 8 6 8 1 3 1 1 3.
2. Scan each odd digit: 7 (before it 8, even – no); 5 (7 before, 6 after – yes); 7 (9 before, 6 after – yes); 1 (2 before – no); 3, 1, 1 at the end are followed by odd digits – no; the last 3 has nothing after it.
3. Count = 2.

Trick: Underline all odd digits first; then only check those whose left neighbour is also underlined.

Answer: (D) Two

Concept: Plot the path with coordinates; take A as origin, east = +x, north = +y. Every left turn rotates anticlockwise: S → E → N → W → S.

1. 21 km south → $(0,-21)$; left (east) 14 km → $(14,-21)$.
2. Left (north) 29 km → $(14,8)$; left (west) 21 km → $(-7,8)$.
3. Left (south) 8 km → P $=(-7,0)$.
4. P is 7 km due west of A, so he must drive 7 km east to reach A.

Trick: Net north–south: $-21+29-8=0$; net east–west: $14-21=-7$ → 7 km west of start.

Answer: (C) 7 km to the east

Concept: Apply the rule digit by digit: odd → $+1$, even → $-2$.

1. 5431786 → 5+1 = 6, 4−2 = 2, 3+1 = 4, 1+1 = 2, 7+1 = 8, 8−2 = 6, 6−2 = 4.
2. New number: 6242864.
3. Highest digit 8, lowest digit 2 → difference $8-2=6$.

Trick: Only the maximum and minimum are needed – the largest odd digit (7) becomes 8 and the smallest digits (1, 4) both become 2.

Answer: (A) 6

Concept: Arrange the letters alphabetically, locate the two required letters, then count letters between them in the alphabet.

1. MONKEYS → E K M N O S Y.
2. Second from left = K; fourth from right = N (Y S O N).
3. Between K(11) and N(14) in the alphabet: L, M → $14-11-1=2$.

Trick: Letters between two letters = difference of positions − 1.

Answer: (A) 2

Concept: Apply the rule to each digit: odd → $+2$, even → $-1$.

1. 2563174 → 2−1 = 1, 5+2 = 7, 6−1 = 5, 3+2 = 5, 1+2 = 3, 7+2 = 9, 4−1 = 3.
2. New number: 1755393.
3. Repeated digits: 5 (twice) and 3 (twice) → two digits appear more than once.

Trick: Write the new digits in a row and tick repeats – 1, 7, 5, 5, 3, 9, 3.

Answer: (A) Two

Concept: Replace every symbol by its real meaning, then apply BODMAS.

1. $17-3÷8+21\times 3$ becomes $17\times 3+8-21÷3$.
2. Division and multiplication first: $17\times 3=51$, $21÷3=7$.
3. $51+8-7=52$.

Trick: Rewrite the whole expression with new symbols before calculating – never substitute "on the fly".

Answer: (B) 52

Concept: Add the alphabetical positions of the letters and compare with the code.

1. WIAO: $23+9+1+15=48$; code 40 → $48-8$.
2. TNCR: $20+14+3+18=55$; code 47 → $55-8$. Rule: sum − 8.
3. FXAJ: $6+24+1+10=41$ → $41-8=33$.

Trick: When the code is close to the letter-sum, test "sum ± constant" first.

Answer: (B) 33

Concept: Find how the second cluster is obtained from the first in each pair.

1. NJ – LH: N−2 = L, J−2 = H. QM – OK: Q−2 = O, M−2 = K. WS – UQ: W−2 = U, S−2 = Q. All are $(-2,-2)$.
2. UQ – SP: U−2 = S, but Q−1 = P → $(-2,-1)$ – different.

Trick: Check the second letters only: J→H, M→K, S→Q are all −2; Q→P is −1.

Answer: (A) UQ - SP

Concept: Number seats 1–7 from the left; all face north, so right = higher number.

1. Y third from the right → seat 5. Only three between B and Y → B at seat 1.
2. N is a neighbour of both E and J, so E and J are on either side of N. N cannot be 6 (seat 5 is Y), hence N = 3 with E, J at 2 and 4.
3. F is second to the right of E: E = 2 gives F = 4 (clash with J), so E = 4, F = 6, J = 2. R takes seat 7.
4. Order: B J N E Y F R. Between J (2) and R (7): N, E, Y, F = four.

Trick: Place the fixed people (end/position clues) first, then the "neighbour of two" person – it has only one possible slot.

Answer: (C) Four