Q&A: practice questions with answers

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ for one sharpener.

1. CP of one $=\frac{150}{12}=$ ₹12.50
2. SP of one $=$ ₹41, profit $=$ ₹28.50
3. $\text{Profit}\%=\frac{28.5}{12.5}\times 100=228\%$

Trick: $\frac{41}{12.5}=3.28$ ⇒ profit $=3.28-1=2.28=228\%$.

Answer: (D) 228

Concept: Mode is the value that occurs most often.

1. Frequencies: 14 → 4 times, 17 → 2 times, the rest once.
2. Highest frequency is for 14.

Yaad rakhein: Mode = most frequent observation.

Answer: (B) 14

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (left to right).

1. ${55}^{-13}÷{55}^{14}={55}^{-13-14}={55}^{-27}$
2. ${55}^{-27}\times {55}^{-11}={55}^{-38}$

Trick: Same base: just add/subtract the exponents: $-13-14-11=-38$.

Answer: (C) ${55}^{-38}$

Concept: Express the actual SP in terms of CP using the loss case.

1. $\frac{6}{15}\text{SP}=0.4\text{SP}=0.86\text{CP}$
2. $\text{SP}=\frac{0.86}{0.4}\text{CP}=2.15\text{CP}$
3. $54\%$ of SP $=0.54\times 2.15\text{CP}=1.161\text{CP}$
4. $\text{Profit}\%=16.1\%$

Trick: Take CP $=100$: $0.4\text{SP}=86\Rightarrow \text{SP}=215$; $0.54\times 215=116.1$.

Answer: (C) 16.1%

Concept: A number is divisible by 3 if its digit sum is divisible by 3.

1. Digit sum $=3+2+7+0+8+7=27$, divisible by 3 ✓
2. It is odd, so not divisible by 8 or 12; last digit 7, so not by 5.

Trick: Odd numbers can never be divisible by 8 or 12.

Answer: (A) 3

Concept: Convert everything to man-days / woman-days.

1. Work $=4\times 18=72$ man-days $=9\times 8=72$ woman-days
2. So 1 man = 1 woman in efficiency.
3. $3$ women $+6$ men $=9$ men
4. Days $=\frac{72}{9}=8$

Trick: $M_1{D}_1=M_2{D}_2$ once the units are made equal.

Answer: (C) 8 days

Concept: Distance = speed × time (time in hours).

1. $2.4\text{min}=\frac{2.4}{60}=0.04\text{h}$
2. Length $=41\times 0.04=1.64\text{km}$

Trick: km/h × minutes ÷ 60 gives km directly.

Answer: (D) 1.64 km

Concept: Find the value of one ratio part from the given difference.

1. Largest − smallest $=16-10=6$ parts $=$ ₹29,547
2. 1 part $=\frac{29547}{6}=4924.5$
3. Total $=40$ parts $=40\times 4924.5=$ ₹1,96,980

Trick: Total $=\text{difference}\times \frac{40}{6}$.

Answer: (B) 1,96,980

Concept: Apply the percentages one after another on the remaining amount.

1. After debt: $70\%$ of salary remains.
2. Spends $20\%$ of this, saves $80\%$: $0.7\times 0.8S=0.56S$
3. $0.56S=602\Rightarrow S=\frac{602}{0.56}=1075$

Trick: Savings fraction $=\frac{7}{10}\times \frac{4}{5}=\frac{14}{25}$.

Answer: (A) 1,075

Concept: Profit is shared in the ratio capital × time.

1. $M+N=1494$, $M-N=498$ ⇒ $M=996$, $N=498$, ratio $2:1$
2. $\frac{3(N+48000)}{4N}=2\Rightarrow 3N+144000=8N$
3. $N=28800$, $M$'s capital $=28800+48000=76800$

Trick: Check: $3\times 76800=230400$, $4\times 28800=115200$ → $2:1$ ✓

Answer: (B) ₹76,800

Concept: Fix the definite positions first (ends and counts), then place the pairs.

1. No one left of V ⇒ V at position 1; four between V and I ⇒ I at 6.
2. Three to the right of K ⇒ K at 4.
3. J immediately left of W: only free pair is 2-3 ⇒ J2, W3.
4. L not next to K (4) ⇒ L at 7, U at 5.

Trick: In a 7-seat row, 'n people to the right' ⇒ position $7-n$.

Answer: (A) L

Concept: Check every number: the element just before it and just after it must both be symbols.

1. Series: 7 1 $ % 3 $ % 8 < 7 8 1 8 # < 7 @ * 5 5 +
2. % 3 $ ✓ ; % 8 < ✓ ; < 7 @ ✓
3. < 7 8 ✗ ; * 5 5 ✗ ; other numbers have a number beside them.
4. Such numbers: 3, 8, 7 ⇒ Three

Trick: Scan only the numbers and look at both neighbours.

Answer: (D) Three

Concept: Total $=$ rank from top $+$ rank from bottom $-1$.

1. Total $=27+34-1=60$

Yaad rakhein: Subtract 1 because the person is counted twice.

Answer: (B) 60

Concept: Common letters get common digits; the leftover letter takes the leftover digit.

1. MORE → 1432, MOVE → 3254
2. Common letters M, O, E ↔ common digits 2, 3, 4
3. R → 1 (left in first), V → 5 (left in second)

Trick: Eliminate common letters/digits, compare the remainders.

Answer: (B) 5

Concept: Draw nested circles: Kiwis inside Bananas inside Grapes.

1. All kiwis are bananas and all bananas are grapes ⇒ all kiwis are grapes. (I) follows.
2. All bananas are grapes ⇒ some grapes are bananas (conversion). (II) follows.

Yaad rakhein: 'All A are B' always gives 'Some B are A'.

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

Concept: Start from the definite clue 'only M to the right of K'.

1. K at 6, M at 7.
2. One person between I and K ⇒ I at 4.
3. Two between J and I ⇒ J at 1 (7 is taken).
4. L, O, N in 2, 3, 5 with L left of O left of N ⇒ L2, O3, N5.
5. Between N(5) and L(2): O, I ⇒ Two

Trick: Seats between positions $p$ and $q$ $=|p-q|-1=5-2-1=2$.

Answer: (A) Two

Concept: Build the family tree from S downwards.

1. T is the daughter of S and A is T's brother ⇒ A is S's son.
2. O is A's wife ⇒ O is S's son's wife.
3. B is O's father ⇒ B is S's son's wife's father.

Trick: Trace the path S → A → O → B and read each link.

Answer: (D) Son's wife's father

Concept: Round each value to the nearest convenient number, then follow BODMAS.

1. $\sqrt[3]{63.998}\approx \sqrt[3]{64}=4$
2. $\frac{5}{4}$ of $120\%$ of $400=\frac{5}{4}\times 480=600$
3. $29.99÷3.01\approx 30÷3=10$
4. $?\approx 4+600-10=594$

Trick: 'of' is multiplication; do it before $+$ and $-$.

Answer: (B) 594

Concept: Find the shift of each letter in the given pairs.

1. X→B, A→E, Y→C, A→E: each $+4$
2. G→K, J→N, H→L, J→N: each $+4$
3. M+4=Q, P+4=T, N+4=R, P+4=T ⇒ QTRT

Trick: Repeated letter (P, P) must give repeated letter (T, T).

Answer: (D) QTRT

Concept: Track net east-west and north-south displacement from A.

1. West 9, left → South 3, right → West 11, right → North 4, right → East 20.
2. East-West: $-9-11+20=0$
3. North-South: $-3+4=+1$ ⇒ P is 1 km north of A.
4. To return: 1 km towards south.

Trick: Left from West = South; right from South = West.

Answer: (A) 1 km to the south