Q&A: practice questions with answers

Concept: For in-centre O of triangle PQR, $\angle QOR={90}^{\circ }+\frac{\angle P}{2}$.

1. ${130}^{\circ }={90}^{\circ }+\frac{\angle P}{2}$
2. $\frac{\angle P}{2}={40}^{\circ }\Rightarrow \angle QPR={80}^{\circ }$

Yaad rakhein: In-centre: ${90}^{\circ }+\frac{A}{2}$; Circumcentre: $2A$; Orthocentre: ${180}^{\circ }-A$.

Answer: (B) ${80}^{\circ }$

Concept: Take total work $=\text{LCM}(8,16,24)=48$ units; P, Q, R do $6,3,2$ units/hour.

1. One P-Q-R round (3 h) $=6+3+2=11$ units; 9 h = 3 rounds $=33$ units.
2. Remaining $=48-33=15$ units.
3. Now Q and R alternate: every 2 h they do $3+2=5$ units $\Rightarrow 15$ units in $\frac{15}{5}\times 2=6$ h (Q–R–Q–R–Q–R exactly).

Trick: In alternate-work problems, first find work per full cycle, then handle the leftover cycle by cycle.

Answer: (B) 6

Concept: Cylinders are cut whole, so count how many fit along each edge; then TSA of cylinder $=2\pi r(r+h)$.

1. Base $8\times 8$ m, diameter 1 m $\Rightarrow 8\times 8=64$ cylinders per layer; height $8÷4=2$ layers $\Rightarrow 128$ cylinders.
2. TSA of one cylinder $=2\times \frac{22}{7}\times \frac{1}{2}\times (\frac{1}{2}+4)=\frac{22}{7}\times \frac{9}{2}=\frac{99}{7}{\text{m}}^2$
3. Total area $=128\times \frac{99}{7}$; cost $=128\times \frac{99}{7}\times 14=128\times 198=25344$

Trick: Keep $\frac{99}{7}$ unsimplified — the rate ₹14 cancels the 7 neatly.

Answer: (C) ₹25,344

Concept: Find the two numbers from sum and difference, then their LCM.

1. Difference $=\frac{1}{7}\times 56=8$
2. Numbers $=\frac{56+8}{2}=32$ and $\frac{56-8}{2}=24$
3. $\text{LCM}(32,24)=96$

Trick: Numbers $=\frac{S+D}{2}$ and $\frac{S-D}{2}$; LCM $=\frac{32\times 24}{\text{HCF }8}=96$.

Answer: (A) 96

Concept: Evaluate the roots first: ${19}^3=6859$ and $4^4=256$.

1. $\sqrt[3]{6859}=19,\sqrt[4]{256}=4$
2. $\frac{19}{4}\times \frac{2}{57}\times 168=\frac{19\times 2\times 168}{4\times 57}$
3. $57=19\times 3$, so $=\frac{2\times 168}{4\times 3}=\frac{336}{12}=28$

Trick: Cube ending in 9 → root ends in 9; ${19}^3=6859$ is worth memorising.

Answer: (A) 28

Concept: Money doubles on SI when interest $=$ principal, i.e. $R\times T=100$.

1. $R\times 16=100$
2. $R=\frac{100}{16}=6.25\%$

Trick: Double → $RT=100$; triple → $RT=200$; $n$ times → $RT=100(n-1)$.

Answer: (D) $6.25$

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$.

1. Total $=78\times 136+22\times 383+34\times 457=10608+8426+15538=34572$
2. Members $=78+22+34=134$
3. Average $=\frac{34572}{134}=258$

Trick: Check divisibility: $134\times 258=34572$ exactly, confirming the option.

Answer: (C) 258

Concept: Work backwards from R's price: each step is a multiplier.

1. Q's selling price (60% loss) $=0.4\times$ Q's cost $=3234\Rightarrow$ Q's cost $=\frac{3234}{0.4}=8085$
2. P sold at 50% profit: $1.5\times$ P's cost $=8085\Rightarrow$ P's cost $=\frac{8085}{1.5}=5390$

Trick: ${\text{CP}}_P=\frac{3234}{1.5\times 0.4}=\frac{3234}{0.6}=5390$ in one step.

Answer: (A) ₹5390

Concept: Find the common marked price from article X, then apply the 77% discount to Y.

1. SP of X $=M\times 0.5\times 0.72=0.36M$; profit 958 $\Rightarrow 0.36M=500+958=1458\Rightarrow M=4050$
2. SP of Y $=4050\times 0.23=931.5$
3. Profit on Y $=431.5$; $\%=\frac{431.5}{500}\times 100=86.3\%$

Trick: Successive discounts 50% and 28% $\equiv$ single discount $64\%$ (pay 36%).

Answer: (A) $86.3\%$

Concept: Third proportional to $a$ and $b$ is $c$ such that $a:b=b:c\Rightarrow c=\frac{b^2}{a}$.

1. $c=\frac{{24}^2}{16}=\frac{576}{16}=36$

Yaad rakhein: Third proportional $=\frac{b^2}{a}$; mean proportional $=\sqrt{ab}$; fourth proportional $=\frac{bc}{a}$.

Answer: (B) 36

Concept: Look at the differences between consecutive terms – here they form the prime numbers.

1. $89-87=2,87-84=3,84-79=5,79-72=7$.
2. Differences $2,3,5,7$ are consecutive primes, so the next difference is $11$.
3. $?=72-11=61$.

Trick: When differences are irregular (2, 3, 5, 7…), check primes first – it is the most common pattern in NTPC series.

Answer: (D) 61

Concept: Facing the centre, "right" means anticlockwise and "left" means clockwise. Fix one person and place the rest.

1. Fix P. One person between U and P from U's right ⇒ going clockwise from P: P, (V), U.
2. One between P and R and two between V and R ⇒ R is 2nd anticlockwise from P. One between U and Q ⇒ Q is next to R.
3. One between T and Q ⇒ T sits between R and P. S takes the last seat. Clockwise order: P, V, U, S, Q, R, T.
4. From T's left (clockwise): T → P → V. Only one person (P) sits between T and V.

Trick: Always draw the circle and mark "left = clockwise, right = anticlockwise" at the top before reading the clues.

Answer: (D) 1

Concept: Check each letter position and the number separately.

1. 1st letters V, Z, D, H: $+4$ each ⇒ next L. 2nd letters P, T, X, B: $+4$ ⇒ F. 3rd letters G, K, O, S: $+4$ ⇒ W.
2. Numbers $7,11,15,19$: $+4$ ⇒ $23$.
3. Required cluster: LFW 23.

Trick: Write A=1 … Z=26 once on the rough sheet; $+4$ with wrap-around (Z → D) becomes instant.

Answer: (D) LFW 23

Concept: Find the position-wise shift from the first word to the second and apply the same shift.

1. EGDH → FIFI: E$+1=$F, G$+2=$I, D$+2=$F, H$+1=$I. Pattern $+1,+2,+2,+1$.
2. Check: JMHL → K, O, J, M = KOJM ✓.
3. NPRS: N$+1=$O, P$+2=$R, R$+2=$T, S$+1=$T ⇒ ORTT.

Trick: Verify the pattern on the second given pair before applying – it rules out wrong guesses in seconds.

Answer: (A) ORTT

Concept: Seven days, seven people – start with the most restrictive clue.

1. Five people between V and T ⇒ they occupy Monday and Sunday. S is after T, so T = Monday, V = Sunday.
2. Two between R and T ⇒ R = Thursday; Q immediately after R ⇒ Q = Friday.
3. S immediately before P and after T ⇒ S = Tuesday, P = Wednesday; U gets Saturday.
4. Between P (Wed) and U (Sat): R and Q ⇒ two people.

Trick: "Five between two people" in a 7-slot row always means the two ends – place them first.

Answer: (B) Two

Concept: Scan the series once, checking each number's left and right neighbours.

1. Series: © 3 @ 8 5 1 £ 1 6 1 5 7 2 8 # € £ © € © 1 3.
2. Numbers with a symbol on both sides: "© 3 @" ✓. The 8 before # has 2 on its left; the 1 after © has 3 on its right; the 1 after £ has 6 on its right – all fail.
3. Count $=1$.

Trick: Underline every symbol first; then look only for a lone number sandwiched between two underlined marks.

Answer: (A) 1

Concept: Letters and numbers follow separate rules – check both.

1. Letters: F→X→P→H each is $-8$ (F=6, X=24, P=16, H=8). Similarly Z→R→J→B ($-8$) and N→F→X→P ($-8$).
2. Next: H$-8=$Z, B$-8=$T, P$-8=$H ⇒ ZTH.
3. Numbers $4,6,9,13$: differences $2,3,4$ ⇒ next $+5$ ⇒ $18$.

Trick: A drop of 8 is the same as $+18$; whichever is easier with wrap-around (Z = 0 or 26).

Answer: (C) ZTH 18

Concept: Track the position on a grid; a left turn while facing South means facing East, and so on.

1. A → 25 km South; left ⇒ East 17 km; left ⇒ North 28 km (now 3 km north of A's level, 17 km east).
2. Left ⇒ West 20 km (now 3 km west of A); left ⇒ South 3 km ⇒ P is exactly 3 km due West of A.
3. To return he must drive 3 km towards East.

Trick: Net North–South: $-25+28-3=0$; net East–West: $+17-20=-3$ (west). Net values give the answer without the full diagram.

Answer: (D) 3 km to the east

Concept: Decode from the right end (P) backwards and draw the family tree.

1. V + P: V is P's mother. T × V: T is V's sister. K ÷ T: K is father of T (hence of V too).
2. S − K: S is K's wife ⇒ S is the mother of V.
3. V is P's mother, so S is P's mother's mother.

Trick: In a long chain, the person next to P decides the final link (here V = mother); everyone else just tells you which side of the family.

Answer: (D) Mother's mother

Concept: Compare the two words: letters common to both must share the common digits.

1. BELI → 6 3 1 5 and EVIL → 9 6 5 3. Common letters: E, L, I. Common digits: 6, 3, 5.
2. The leftover letter in BELI is B ⇒ code 1; the leftover letter in EVIL is V ⇒ code 9.

Trick: Do not try to match positions – just cancel the common letters and common digits; what remains pairs up.

Answer: (C) 9