16th March 2025 Shift 1 — RRB NTPC 2025

100 Questions • 90 Minutes • 2025 Mock Test in Hindi and English

Sample Questions from this Test

Question 1:

Find the area of a quadrilateral ABCD whose area is twice the area of a triangle PQR. The sides of triangle PQR are in the ratio 4:5:6 and the perimeter of the triangle is 90 cm (use 7=2.6\sqrt{7} = 2.6).
A.656 cm2
B.702 cm2
C.763 cm2
D.698 cm2

• Given: Sides of △\trianglePQR in ratio 4:5:6, perimeter = 90 cm; Area(ABCD) = 2 ×\times Area(PQR); 7=2.6\sqrt{7}=2.6.

• Basic Method (Solution):
Step 1: Let sides be
4x4x, 5x5x, 6x6x. Then 4x+5x+6x=90⇒15x=90⇒x=64x+5x+6x=90 \Rightarrow 15x=90 \Rightarrow x=6.
Step 2: Sides = 24, 30, 36 cm.
Step 3: Semi-perimeter
s=45s = 45 cm.
Step 4: Area by Heron's formula
=s(s−a)(s−b)(s−c)=45(45−36)(45−30)(45−24)=45×9×15×21=127575= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{45(45-36)(45-30)(45-24)} = \sqrt{45 \times 9 \times 15 \times 21} = \sqrt{127575}.
Step 5:
127575=25×5103=25×9×567=25×9×81×7127575 = 25 \times 5103 = 25 \times 9 \times 567 = 25 \times 9 \times 81 \times 7. Thus 127575=5×3×9×7=135×2.6=351\sqrt{127575} = 5 \times 3 \times 9 \times \sqrt{7} = 135 \times 2.6 = 351 cm2^2.
Step 6: Area(ABCD)
=2×351=702= 2 \times 351 = 702 cm2^2.

• Shortcut Method: After finding sides 24, 30, 36, note that they form a scaled 4-5-6 triangle. Area of 4-5-6 triangle =7.5(7.5−6)(7.5−5)(7.5−4)= \sqrt{7.5(7.5-6)(7.5-5)(7.5-4)}. Scaling factor x=6x=6, area scales by x2=36x^2=36, so area =36×(15/2)(3/2)(5/2)(7/2)=36×(1575/4)= 36 \times \sqrt{(15/2)(3/2)(5/2)(7/2)} = 36 \times (\sqrt{1575}/4). But direct Heron with s=45s=45 quickly yields 1357=351135\sqrt{7}=351, then double to 702.

• Final Answer: 702 cm2

Question 2:

What should come in place of the question mark (?) in the given series?
89 87 84 79 72 ?
A.57
B.59
C.63
D.61

• Given conditions: Series 89, 87, 84, 79, 72, ?

• Deduction process: Differences: 89−87=2, 87−84=3, 84−79=5, 79−72=7. The differences are consecutive prime numbers 2, 3, 5, 7. Next prime is 11. So 72−11=61.

• Exam Trick: Look for prime-number difference pattern; primes increase by successive primes.

• Final Answer: 61

Question 3:

20% of a number is more than 20% of 620 by 210. The number is:
A.1520
B.1720
C.1670
D.1770

• Given: 20% of number = 20% of 620 + 210; find the number.

• Basic Method (Solution):
Step 1:
0.2x=0.2×620+2100.2x = 0.2 \times 620 + 210
Step 2:
0.2x=124+210=3340.2x = 124 + 210 = 334
Step 3:
x=334/0.2=1670x = 334 / 0.2 = 1670

• Shortcut Method: 20% of number − 20% of 620 = 210 ⇒\Rightarrow 20% of (number − 620) = 210 ⇒\Rightarrow number − 620 = 210 ×\times 5 = 1050 ⇒\Rightarrow number = 1050 + 620 = 1670.

• Final Answer: 1670

Question 4:

Working of hydraulic lift and hydraulic brakes is based on:
A.Faraday's law
B.Gay-Lussac's law
C.Pascal's law
D.Coulomb's law

• Key Fact: Hydraulic lift and hydraulic brakes work on Pascal's law, which states that pressure applied to an enclosed fluid is transmitted equally in all directions.

• Supporting Detail: Pascal's law is the principle behind hydraulic systems where a small force on a small piston creates a large force on a larger piston.

• Related Concept: Hydraulic machines multiply force using the relation F1/A1=F2/A2F_1/A_1 = F_2/A_2.

• Why others wrong: Faraday's law relates to electromagnetic induction; Gay-Lussac's law relates pressure and temperature of gases; Coulomb's law is about electrostatic force.

• Exam Trick: Remember “Pascal → Pressure → Hydraulic”.

• Additional Info: Blaise Pascal formulated this law in the 17th century.

Question 5:

P, Q, R, S, T, U and V sit around a circular table facing the centre. Only one person sits between U and P when counted from the right of U. Only one person sits between P and R. Only one person sits between U and Q. Only two people sit between V and R. Only one person sits between T and Q. How many people sit between T and V when counted from the left of T?
A.3
B.2
C.4
D.1

• Given conditions: 7 people circular, facing centre. 1 between U-P (right of U); 1 between P-R; 1 between U-Q; 2 between V-R; 1 between T-Q.

• Deduction process: From U, right side one person then P. From P one then R. From U one then Q (so Q is on the other side of U). Two between V and R. One between T and Q. Possible arrangement (clockwise): U, S, P, V, R, T, Q. Counting from left of T to V gives 1 person (R).

• Exam Trick: Fix U and place others relative to it using “from the right of” as clockwise or anticlockwise consistently; draw both possible directions and eliminate.

• Final Answer: 1

Question 6:

What is the name of the powerful non-nuclear bomb unveiled by Turkey at the 17th International Defence Industry Fair (IDEF) in July 2025?
A.Nebula
B.Phantom-X
C.Gazap
D.Titan

• Key Fact: Turkey unveiled the powerful non-nuclear bomb named Gazap at the 17th IDEF in July 2025.

• Supporting Detail: IDEF is Turkey’s major international defence exhibition held in Istanbul.

• Related Concept: Non-nuclear high-explosive bombs are developed for precision strike capabilities without nuclear fallout.

• Why others wrong: Nebula, Phantom-X and Titan are not the names associated with the Turkish bomb unveiled at IDEF 2025.

• Exam Trick: Link “Turkey + IDEF 2025 → Gazap”.

• Additional Info: Gazap means “wrath” in Turkish, reflecting its destructive power.

Question 7:

Anand has ₹1407 with him. He divides this amount between his two sons, Anil and Ashok, and asks them to invest their respective shares at a compound interest rate of 10% per annum, compounded annually. It is observed that Anil's investment amounts to the same value after 15 years as Ashok's investment does after 16 years. How much (in ₹) did Anand give to Ashok?
A.737
B.837
C.520
D.670

• Given: Total = 1407; rate 10% CI; Anil’s amount after 15 yr = Ashok’s amount after 16 yr; find Ashok’s share.

• Basic Method (Solution):
Step 1: Let Ashok get
xx, Anil get 1407−x1407-x.
Step 2:
(1407−x)(1.1)15=x(1.1)16(1407-x)(1.1)^{15} = x(1.1)^{16}
Step 3:
(1407−x)/x=1.1⇒1407−x=1.1x⇒1407=2.1x⇒x=1407/2.1=670(1407-x)/x = 1.1 \Rightarrow 1407-x = 1.1x \Rightarrow 1407 = 2.1x \Rightarrow x = 1407/2.1 = 670.

• Shortcut Method: Since amounts are equal after 15 and 16 years, Ashok’s principal is Anil’s principal multiplied by 1.1. Ratio Anil : Ashok = 1 : 1.1 = 10 : 11. Total parts 21; Ashok = (11/21)×1407=670(11/21)\times1407 = 670.

• Final Answer: 670

Question 8:

Which of the following Articles of the Indian Constitution directs the State to ensure equal pay for equal work for both men and women?
A.Article 42(b)
B.Article 15
C.Article 39(d)
D.Article 14

• Key Fact: Article 39(d) of the Directive Principles of State Policy directs the State to secure equal pay for equal work for both men and women.

• Supporting Detail: Article 39 is part of Part IV (DPSP) and contains principles of policy to be followed by the State.

• Related Concept: Equal Remuneration Act, 1976 gives statutory force to this principle.

• Why others wrong: Article 42 deals with just and humane conditions of work and maternity relief; Article 15 prohibits discrimination; Article 14 is equality before law.

• Exam Trick: Remember “39(d) = equal pay for equal work”.

• Additional Info: The principle was also recognised by the Supreme Court as a fundamental right under Article 14 in certain cases.

Question 9:

Which of the following letter-number clusters will replace the question mark “?” in the given series to make it logically complete?
VPG 7, ZTK 11, DXO 15, HBS 19, ?
A.LEU 23
B.MGX 23
C.LGY 24
D.LFW 23

• Given conditions: Series VPG 7, ZTK 11, DXO 15, HBS 19, ?

• Deduction process: Letters: V→Z (+4), Z→D (+4, wrap), D→H (+4); P→T (+4), T→X (+4), X→B (+4); G→K (+4), K→O (+4), O→S (+4). Numbers increase by +4 each time: 7, 11, 15, 19 → next 23. Next letters: H+4=L, B+4=F, S+4=W → LFW 23.

• Exam Trick: Each letter advances +4 in alphabet; number +4; check wrap-around from Z to A.

• Final Answer: LFW 23

Question 10:

EGDH is related to FIFI in a certain way based on the English alphabetical order. In the same way, JMHL is related to KOJM. To which of the following is NPRS related, following the same logic?
A.ORTT
B.ORST
C.OSST
D.OOPT

• Given conditions: EGDH → FIFI; JMHL → KOJM; find for NPRS.

• Deduction process: Pattern: each letter +1, +2, +3, +4 respectively. E+1=F, G+2=I, D+3=G→I? Wait: actually E→F (+1), G→I (+2), D→F (+2)? Recheck: EGDH to FIFI: E+1=F, G+2=I, D+5? Better: reverse pairs or alternate. Observed: first letter +1, second +2, third +5? Consistent pattern: positions advance by +1, +2, +5, +2? Actual observed: E→F (+1), G→I (+2), D→F (+2), H→I (+1) no. Looking at both examples: EGDH → F I F I; JMHL → K O J M. Pattern is +1, +2, +2, +1? E+1=F, G+2=I, D+2=F, H+1=I. Yes! J+1=K, M+2=O, H+2=J, L+1=M. Perfect. So N+1=O, P+2=R, R+2=T, S+1=T → ORTT.

• Exam Trick: Check successive +1,+2,+2,+1 pattern on consecutive letters.

• Final Answer: ORTT

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