RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: $5^2+{12}^2={13}^2$, so the triangle is right-angled. Altitude to the hypotenuse $=\frac{\text{product of legs}}{\text{hypotenuse}}$.

1. Area $=\frac{1}{2}\times 5\times 12=30$
2. Also area $=\frac{1}{2}\times 13\times h$
3. $h=\frac{60}{13}\approx 4.615\approx 4.62$ cm

Trick: Altitude on hypotenuse $=\frac{ab}{c}$; it is always less than the shorter leg (5).

Answer: (D) 4.62 cm

Concept: A regular hexagon = 6 equilateral triangles: Area $=\frac{3\sqrt{3}}{2}{a}^2$.

1. $a=10$, $a^2=100$
2. Area $=\frac{3\times 1.732}{2}\times 100=2.598\times 100$
3. $\approx 259.8$ m²

Trick: 173.2 is the area of just 4 such triangles — a common trap.

Answer: (B) 259.8 square meters

Concept: $\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100$.

1. Profit $=575-500=75$
2. $\frac{75}{500}\times 100=15\%$

Trick: Profit % is always on cost price, not selling price.

Answer: (B) 15%

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $SI=\frac{3200\times 12\times 5}{100}$
2. $=32\times 60=1920$

Trick: Total rate $=12\times 5=60\%$; $60\%$ of 3200 $=1920$.

Answer: (C) ₹1920

Concept: $68=4\times 17$; a number must be divisible by both 4 and 17.

1. Test by 4 (last two digits): 17350 → 50 ✗; 35350 → 50 ✗.
2. 23800 $=68\times 350$ ✓; 28560 $=68\times 420$ ✓.
3. Not divisible: (i) and (iv).

Trick: The quick test by 4 eliminates (i) and (iv) at once.

Answer: (C) ( i ) and ( iv )

Concept: Greatest number leaving the same remainder = HCF of the differences of the numbers.

1. $321-186=135$, $402-321=81$, $402-186=216$
2. $135=27\times 5$, $81=27\times 3$, $216=27\times 8$
3. HCF $=27$

Trick: Same remainder → take differences, then HCF.

Answer: (D) 27

Concept: Milk does not change; only water increases.

1. Milk $=5x$, water $=3x$.
2. $\frac{5x}{3x+48}=\frac{5}{5}$ ⇒ $3x+48=5x$ ⇒ $x=24$
3. Milk $=5\times 24=120$ L

Trick: Water parts go from 3 to 5: 2 parts = 48 L.

Answer: (B) 120 liters

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$ with base $=25$ m for both angles.

1. Building: $25\tan {30}^{\circ }=\frac{25}{\sqrt{3}}$
2. Building + pole: $25\tan {60}^{\circ }=25\sqrt{3}$
3. Pole $=25\sqrt{3}-\frac{25}{\sqrt{3}}=\frac{50}{\sqrt{3}}\approx 28.9\approx 29$ m

Trick: For angles ${30}^{\circ }$ and ${60}^{\circ }$ from the same point, pole $=2\times$ building.

Answer: (C) 29 m

Concept: ${\left(\sqrt{a}\right)}^2=a$; evaluate each root first.

1. $\sqrt{196}=14$, $\sqrt{81}=9$
2. ${\left(\sqrt{16}\right)}^2=16$
3. $14-9+16=21$

Trick: Squaring a square root just gives back the number.

Answer: (A) 21

Concept: Write the ratio in lowest terms: numbers $=\text{HCF}\times$ co-prime ratio terms.

1. $6:4=3:2$
2. Numbers $=3\times 49=147$ and $2\times 49=98$
3. LCM $=49\times 3\times 2=294$

Trick: LCM $=\text{HCF}\times$ product of co-prime ratio terms.

Answer: (A) 294

Concept: Position from left $=$ Total $-$ position from right $+1$.

1. Gagan from left $=54-13+1=42$nd.
2. Nitin is 17 places to the left of Gagan: $42-17=25$.

Trick: Left + Right $=$ Total $+1$.

Answer: (D) 25^th

Concept: Draw the Venn diagram: Drawings inside Mountains, Mountains inside Planes.

1. (I) Every mountain is a plane, so some planes are certainly mountains → follows.
2. (II) Drawings lie inside Mountains which lie inside Planes → all drawings are planes → follows.

Trick: All A are B + All B are C ⇒ All A are C; and 'All B are C' always gives 'Some C are B'.

Answer: (C) Both conclusions (I) and (II) follow

Concept: Fix the boxes from top (1) to bottom (7) using the definite clues first.

1. Only Q above I → Q = 1, I = 2. R immediately below I → R = 3.
2. Two boxes between I and S → S = 5. Only K below L → K = 7, L = 6.
3. Remaining J = 4. Second above J (4) is position 2 = I.

Trick: 'Only X above Y' means Y is 2nd from top – start with such fixed clues.

Answer: (A) I

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

1. No one to the right of E → E = 7. Two between E and T → T = 4; T immediately left of I → I = 5.
2. Three between N and I → N = 1.
3. A immediately right of R in seats 2, 3, 6 → R = 2, A = 3; G = 6.
4. Between G (6) and A (3): T and I → two people.

Trick: Seats between positions $a$ and $b$ $=|a-b|-1=6-3-1=2$.

Answer: (A) Two

Concept: Look at the differences between consecutive terms.

1. $152-12=140,140-6=134,134-12=122,122-6=116,116-12=104$
2. The pattern alternates $-12,-6$, so next $=104-6=98$.

Trick: Alternating differences → the next step repeats the pair.

Answer: (A) 98

Concept: Replace the symbols (+ → −, − → ×, × → ÷, ÷ → +) and then apply BODMAS.

1. $62-9÷94+480\times 15\Rightarrow 62\times 9+94-480÷15$
2. $=558+94-32=620$

Trick: Rewrite the whole expression first, then solve ÷, × before +, −.

Answer: (A) 620

Concept: We need the pattern Vowel–Consonant–Vowel. Vowels here: U, U, E, U, E, U.

1. Check the letter right after each vowel: U U K G, E W Y, U K Q, E Q Q, U S V.
2. In every case the consonant after a vowel is followed by another consonant, so no consonant sits between two vowels.

Trick: Scan only around the vowels – fewer letters to check.

Answer: (C) 0

Concept: Replace A → ÷, B → ×, C → +, D → − and apply BODMAS.

1. $41-44+9÷21\times 7$
2. $9÷21\times 7=\frac{9}{21}\times 7=3$
3. $41-44+3=0$

Trick: Divide and multiply left to right: $\frac{9\times 7}{21}=3$.

Answer: (D) 0

Concept: Relate the 2nd and 3rd numbers to the 1st.

1. $158-17=141,158+19=177$; $164-17=147,164+19=183$.
2. Check (C): $119-17=102,119+19=138$ ✓.
3. (A) $146+19=165\ne 163$; (B) $122+19=141\ne 139$; (D) $131+19=150\ne 151$.

Trick: Test the easier rule ($-17$) first, then confirm with $+19$.

Answer: (C) (119, 102, 138)

Concept: Track each letter position separately (A = 1 … Z = 26).

1. 1st letter: G → B → W → R → M (each $-5$).
2. 2nd letter: T → X → B → F → J (each $+4$).
3. 3rd letter: X → U → R → O → L (each $-3$).
4. 4th letter: I → K → M → O → Q (each $+2$).

Trick: Wrap around: $2-5=-3\to 26-3=23=$ W.

Answer: (D) MJLQ