19th July Shift 1RRB JE PYQ

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

Sample Questions from this Test

Question 1:

The base and the corresponding height of a parallelogram are 9393 cm and 2929 cm, respectively. The area of the parallelogram (in cm2^2) is:
A.2701
B.2697
C.2714
D.2696

1. Area of a parallelogram is given by the formula: Area=Base×Height\text{Area} = \text{Base} \times \text{Height}.
2. Substituting the given values:
Area=93×29\text{Area} = 93 \times 29.
Shortcut/Exam Trick: Use the unit digit method. The unit digit of
3×93 \times 9 is 77. Only option 'b' has a unit digit of 77, making 26972697 the correct answer.

Question 2:

Simplify the following: 112÷[16÷8×{18+12(3+10(4+7))}]112 \div [16 \div 8 \times \{18 + 12 - (3 + 10 - (4 + 7))\}]
A.3
B.4
C.11
D.2

1. Follow BODMAS (Brackets, Orders, Div, Mult, Add, Sub). Solve innermost bracket: (4+7)=11(4 + 7) = 11.
2. Next bracket:
(3+1011)=1311=2(3 + 10 - 11) = 13 - 11 = 2.
3. Curly bracket:
{18+122}=302=28\{18 + 12 - 2\} = 30 - 2 = 28.
4. Square bracket:
[16÷8×28]=[2×28]=56[16 \div 8 \times 28] = [2 \times 28] = 56.
5. Final step:
112÷56=2112 \div 56 = 2.
Shortcut/Exam Trick: Keep calculations mental and process brackets from inside out sequentially to save writing time.

Question 3:

The quadratic equation whose roots are 12\frac{1}{\sqrt{2}} and 12\frac{1}{\sqrt{2}} is:
A.2x2(22)x+1=02x^2 - (2\sqrt{2})x + 1 = 0
B.3x2(22)x1=03x^2 - (2\sqrt{2})x - 1 = 0
C.2x2(42)x1=02x^2 - (4\sqrt{2})x - 1 = 0
D.2x2+(25)x+1=02x^2 + (2\sqrt{5})x + 1 = 0

1. Let the roots be α=12\alpha = \frac{1}{\sqrt{2}} and β=12\beta = \frac{1}{\sqrt{2}}.
2. Sum of roots:
α+β=12+12=22=2\alpha + \beta = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.
3. Product of roots:
αβ=(12)×(12)=12\alpha \beta = (\frac{1}{\sqrt{2}}) \times (\frac{1}{\sqrt{2}}) = \frac{1}{2}.
4. A quadratic equation is given by
x2(Sum of roots)x+(Product of roots)=0x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0.
5. Substituting the values:
x2(2)x+12=0x^2 - (\sqrt{2})x + \frac{1}{2} = 0.
6. Multiply the entire equation by
22 to remove the fraction: 2x222x+1=02x^2 - 2\sqrt{2}x + 1 = 0.
Shortcut/Exam Trick: If roots are equal and positive, the sum is positive, making the middle term's coefficient negative (
22x-2\sqrt{2}x), and the product is positive (+1/2+1/2), so the constant term is positive (+1+1). Option 1 is the only one matching these exact signs and values.

Question 4:

Sonora-64 and Lerma Rojo are high-yielding varieties (HYVs) of which crop introduced during the Green Revolution in India?
A.Rice
B.Barley
C.Wheat
D.Maize

  • Key Fact: Sonora-64 and Lerma Rojo are semi-dwarf, high-yielding varieties of wheat.
  • Supporting Detail: They were introduced in India from Mexico during the Green Revolution to boost agricultural production.
  • Related Concept: IR8 and Jaya are famous high-yielding varieties of Rice introduced during the same period.
  • Why wrong options are wrong: Rice involves IR8, not Sonora; Barley and Maize were not the primary focus crops for these specific Mexican HYV seeds.
  • Exam Trick: Remember 'Rojo' means red in Spanish, linking to Mexican wheat varieties brought by Norman Borlaug.

Question 5:

When methane (CH4CH_4) burns completely in the presence of sufficient oxygen, which products are formed?
A.Carbon dioxide and carbon
B.Carbon and hydrogen
C.Carbon dioxide and water
D.Carbon monoxide and water

  • Key Fact: Complete combustion of hydrocarbons like methane always produces carbon dioxide (CO2CO_2) and water (H2OH_2O).
  • Supporting Detail: The balanced chemical equation is CH4+2O2CO2+2H2O+HeatCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{Heat}.
  • Related Concept: Incomplete combustion (lack of oxygen) produces carbon monoxide (COCO) or soot (Carbon).
  • Why wrong options are wrong: Carbon monoxide forms only in limited oxygen; solid carbon and hydrogen are not final products of complete combustion.
  • Exam Trick: 'Complete' combustion = max oxidation state of Carbon (CO2CO_2) + max oxidation state of Hydrogen (H2OH_2O).

Question 6:

A question is followed by two statements numbered (I) and (II). You have to decide whether the data provided in the statements are sufficient to answer the question.
Question: Five people, A, B, C, D and E, are sitting in a straight line facing north. Who sits to the immediate left of C?
I. B sits second from one of the extreme ends of the line. Only one person sits between B and C. D sits third to the right of C.
II. D sits to the immediate right of B. Only three people sit between D and A. E sits to the immediate right of C.
A.Data in statement I alone is sufficient
B.Data in both statements I and II together is NOT sufficient
C.Data in both statements I and II together is sufficient
D.Data in statement II alone is sufficient

  • Key Fact: Check statement II: D is right of B (BD). 3 people between D and A forces A at position 1 and D at 5, meaning B is at 4. E sits immediate right of C (CE), which fills positions 2 and 3. The arrangement is A C E B D.
  • Supporting Detail: Statement I gives two possibilities for B (position 2 or 4), but when you place D third to the right of C, it forces B to position 4. So A/E are at 1. We don't know who is immediate left of C.
  • Related Concept: Data Sufficiency requires a unique, definitive answer.
  • Why wrong options are wrong: Statement I doesn't uniquely define the person on the left.
  • Exam Trick: Always plot out the 'forced' positions (like 3 people between D and A in a 5-person line means they must be at extreme ends) to quickly test sufficiency.

Question 7:

Which of the following is NOT true regarding the Poona Pact of 1932?
A.Seats reserved for depressed classes increased in the legislatures.
B.Ambedkar refused to sign the pact.
C.It was accepted by the government.
D.It dropped the idea of separate electorates for depressed classes.

  • Key Fact: Dr. B.R. Ambedkar actively negotiated and signed the Poona Pact on behalf of the depressed classes. Therefore, the statement that he refused to sign it is false.
  • Supporting Detail: The pact was signed on September 24, 1932, at Yerwada Central Jail in Pune, agreeing to joint electorates instead of separate ones.
  • Related Concept: The pact increased the reserved seats for depressed classes in provincial legislatures from 71 to 147.
  • Why wrong options are wrong: Options 1, 3, and 4 are historically true events of the Poona Pact.
  • Exam Trick: Remember 'Poona Pact = Compromise'. Ambedkar gave up separate electorates to save Gandhi's life, but gained more reserved seats in return.

Question 8:

How does the magnetic field strength at the center of a circular loop change if the current through the loop is doubled?
A.It becomes twice as large
B.It becomes four times as large
C.It becomes half as large
D.It remains unchanged

  • Key Fact: The magnetic field BB at the center of a circular current-carrying loop is directly proportional to the current II. Formula: B=μ0I2RB = \frac{\mu_0 I}{2R}.
  • Supporting Detail: If the current II is doubled, the magnetic field BB also strictly doubles, assuming the radius RR remains constant.
  • Related Concept: If the radius RR is doubled, the magnetic field strength becomes half.
  • Why wrong options are wrong: Squaring or halving effects only occur if parameters like power or distance squared are involved, which isn't the case here.
  • Exam Trick: BIB \propto I and B1RB \propto \frac{1}{R}. Direct relationship means whatever happens to current happens exactly to the magnetic field.

Question 9:

When two aqueous solutions react inside a sealed flask and form products, what happens to the total mass of the system?
A.It fluctuates during reaction
B.It increases due to gas formation
C.It decreases due to precipitation
D.It remains unchanged throughout

  • Key Fact: According to the Law of Conservation of Mass, mass can neither be created nor destroyed in a chemical reaction.
  • Supporting Detail: Because the flask is sealed, no matter (gas or liquid) can escape or enter. Hence, the total mass remains exactly the same.
  • Related Concept: Antoine Lavoisier discovered this principle in 1789.
  • Why wrong options are wrong: Mass never fluctuates randomly. Gas formation or precipitation only changes the state of matter, not the total mass in a closed system.
  • Exam Trick: Sealed flask = Closed system = Mass is always constant.

Question 10:

Which of the following metals is NOT refined by electrolytic refining?
A.sodium
B.copper
C.tin
D.zinc

  • Key Fact: Sodium is a highly reactive metal and is extracted through the electrolytic reduction of its molten salts (like molten NaCl), but not 'refined' via electrolytic refining involving aqueous solutions.
  • Supporting Detail: Electrolytic refining is commonly used for less reactive metals where impure metal forms the anode and pure metal forms the cathode in an aqueous electrolyte.
  • Related Concept: Copper, tin, lead, zinc, and silver are widely purified using the electrolytic refining method.
  • Why wrong options are wrong: Copper, tin, and zinc all undergo electrolytic refining commercially to achieve high purity.
  • Exam Trick: Highly reactive metals (K, Na, Ca, Mg, Al) are extracted by electrolytic reduction of molten salts, not aqueous refining.

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