Q&A: practice questions with answers

Concept: Discount = discount rate × list price.

1. Discount $=25\%\text{ of }3000=\frac{25}{100}\times 3000$.
2. $=$ ₹750.

Trick: 25% = one-fourth: $3000÷4=750$.

Answer: (C) ₹750

Concept: For the same time period, profit is shared in the ratio of investments.

1. Aman's profit $=11600-2900=$ ₹8700; ratio Aman : Pawan $=8700:2900=3:1$.
2. Investments: Aman $=\frac{3}{4}\times 62800=$ ₹47100, Pawan $=$ ₹15700.
3. Difference $=47100-15700=$ ₹31400.

Trick: Difference $=\frac{3-1}{4}\times 62800=\frac{1}{2}\times 62800=$ ₹31400.

Answer: (C) ₹31,400

Concept: Product $n(n+1)$ is close to $n^2$; take $\sqrt{306}\approx 17.5$.

1. Try $17\times 18=306$ ✓.
2. The two numbers are 17 and 18; greater is 18.

Trick: Last digit 6 → numbers end in 2&3 or 7&8; near $\sqrt{306}$ gives 17, 18.

Answer: (B) $18$

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 87=7500$.
2. $r=\frac{7500\times 7}{44\times 87}=\frac{52500}{3828}$.
3. $r\approx 13.71\text{cm}$.

Trick: Quick estimate: $2\pi h\approx 547$, and $7500÷547\approx 13.7$.

Answer: (B) $13.71$

Concept: Share = (its ratio part ÷ total parts) × amount.

1. Total parts $=2+11+17=30$; 1 part $=\frac{976}{30}$.
2. Difference = $17-2=15$ parts.
3. $15\times \frac{976}{30}=\frac{976}{2}=$ ₹488.

Trick: 15 parts out of 30 is exactly half of ₹976.

Answer: (A) $488$

Concept: Divisible by 5 if the last digit is 0 or 5; by 3 if the digit sum is divisible by 3.

1. Last digit is 5 → divisible by 5 ✓.
2. Digit sum $=8+6+9+6+7+1+1+9+5=52$, not divisible by 3 → not by 15 or 12.
3. It is odd, so not divisible by 12 either.

Trick: Check the easiest rule first – the last digit.

Answer: (A) $5$

Concept: Total = average × count; split total in the given ratio.

1. Total price $=3\times 16035=$ ₹48105.
2. Total parts $=3+5+7=15$; 1 part $=\frac{48105}{15}=$ ₹3207.
3. Cheapest $=3\times 3207=$ ₹9621.

Trick: Average of the ratio terms is 5 parts = ₹16035, so 1 part = ₹3207 directly.

Answer: (D) $9,621$

Concept: $a^m\times b^m\times c^m=(abc{)}^m$ and $a^{-m}=\frac{1}{a^m}$.

1. ${14}^{-23}\times {25}^{-23}\times {15}^{-23}=(14\times 25\times 15{)}^{-23}$.
2. $14\times 25\times 15=5250$.
3. Value $={5250}^{-23}=\frac{1}{{5250}^{23}}$.

Trick: $14\times 15=210$, $210\times 25=5250$.

Answer: (D) $\frac{1}{{5250}^{23}}$

Concept: Relate the selling price (SP) to the cost price (CP) using the loss condition.

1. $\frac{2}{5}\text{SP}=0.76\text{CP}$ (loss 24%).
2. $\text{SP}=\frac{0.76\times 5}{2}\text{CP}=1.9\text{CP}$.
3. New price $=0.81\times 1.9\text{CP}=1.539\text{CP}$ → profit $53.9\%$.

Trick: Take CP = 100: $\frac{2}{5}\text{SP}=76$ → SP = 190; $81\%\text{ of }190=153.9$.

Answer: (D) 53.9%

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{52}^2}{26}=\frac{2704}{26}$.
2. $x=104$.

Trick: $52=2\times 26$, so $x=2\times 52=104$ (ratio doubles each time).

Answer: (B) $104$

Concept: Trace the path on axes (east = +x, north = +y) and compare the end point with the start.

1. Y(0, 0) → 12 km east → (12, 0).
2. Right turn = south 10 km → (12, −10); right = west 15 km → (−3, −10).
3. Right = north 10 km → Z(−3, 0).
4. Z is 3 km west of Y, so she must drive 3 km towards east.

Trick: North–south moves cancel (10 and 10); net east–west $=12-15=-3$ → 3 km west of Y.

Answer: (B) 3 km towards east

Concept: Check the differences between consecutive terms.

1. Differences: $1,5,9,13$ – increasing by 4.
2. Next difference $=17$.
3. $127+17=144$.

Trick: Second difference constant (4) → add next term of 1, 5, 9, 13, 17.

Answer: (C) $144$

Concept: Find the rule linking the three numbers: each term $=3\times \text{previous}+1$.

1. $23\times 3+1=70$, $70\times 3+1=211$ ✓; $12\times 3+1=37$, $37\times 3+1=112$ ✓.
2. Option (D): $33\times 3+1=100$, $100\times 3+1=301$ ✓.
3. (A): $5\times 3+1=16$ but $16\times 3+1=49\ne 48$ ✗.

Trick: Check both steps of every option – (A) passes the first step only.

Answer: (D) (33, 100, 301)

Concept: Track letter positions: word 1-2-3-4 → 2-3-1-4, and third = reverse of the first word.

1. LONG → O N L G (2, 3, 1, 4); GNOL = LONG reversed.
2. EARN → A R E N (2, 3, 1, 4); NRAE = EARN reversed ✓.
3. COIN → O I C N (2, 3, 1, 4); NIOC = COIN reversed ✓.
4. CUTE → UTCE would be needed (not UCTE); BACK, SKIP also fail.

Trick: Check the easy part first – the third term must be the exact reverse of the first.

Answer: (C) COIN - OICN - NIOC

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

1. V = 2; three boxes between V and W → W = 6.
2. X below V → X = 1. Only E between W and G → E = 5, G = 4.
3. U above W → U = 7; remaining F = 3.
4. Between F (3) and X (1): only V → one box.

Trick: Fix the 'second from bottom' clue first; everything else follows from it.

Answer: (C) One

Concept: Find the shift of each letter using place values (A = 1 … Z = 26).

1. YELP → BZOK: Y+3 = B, E−5 = Z, L+3 = O, P−5 = K.
2. EURF → HPUA: E+3 = H, U−5 = P, R+3 = U, F−5 = A ✓.
3. KKXV: K+3 = N, K−5 = F, X+3 = A, V−5 = Q → NFAQ.

Trick: Pattern +3, −5, +3, −5; wrap around after Z (X+3 = A).

Answer: (B) NFAQ

Concept: Arrange the digits in ascending order and pick positions.

1. 257146983 → 123456789.
2. Third from left = 3; second from right = 8.
3. Sum $=3+8=11$.

Trick: Digits 1–9 each once → position from left equals the digit itself.

Answer: (C) $11$

Concept: Compare each letter with its code letter.

1. JGCK → HEAI: each letter −2.
2. MJFN → KHDL: each letter −2 ✓.
3. PMIQ → N K G O.

Trick: Uniform shift: just move every letter two steps back.

Answer: (C) NKGO

Concept: Facing north, left/right are the viewer's left/right. Fix the end person first.

1. No one right of L → L = 7. Two between L and C → C = 4; G immediately right of C → G = 5.
2. Three between D and G → D = 1.
3. E–B pair (B immediately right of E) fits only 2–3 → E = 2, B = 3; F = 6.
4. Between B (3) and F (6): C, G → two.

Trick: Place the end person and the fixed pair; the rest drops into the gaps.

Answer: (A) Two

Concept: Facing north; number seats 1 (left) to 7 (right).

1. No one right of D → D = 7. Three between D and F → F = 3. Two between F and A → A = 6.
2. G third to the left of B and C immediately right of B → G = 1, B = 4, C = 5.
3. E takes the remaining seat 2.
4. Right of E: F, B, C, A, D → 5 people.

Trick: Seat 2 means $7-2=5$ people to the right.

Answer: (D) $5$