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JEE Advanced Practice Questions

Original practice questions for the Joint Entrance Examination (Advanced) (IITs (organising IIT rotates each year)), mixed across 5 topics with instant answers and explanations. Practise by topic below or take the timed mock test.

JEE Advanced mixed practice

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  1. Q1.AlgebraSolve: 5x + 4y = 43 and 5x + y = 52

    Show answer & explanation

    Answer: B — x = 11, y = -3

    Elimination/substitution gives x = 11, y = -3. Check: 5(11) + 4(-3) = 43 ✓ and 5(11) + 1(-3) = 52 ✓.

  2. Q2.Probability & CombinatoricsIn how many ways can the letters of the word ORANGE be arranged?

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    Answer: C — 720

    All 6 letters are different, so arrangements = 6! = 720.

  3. Q3.Geometry & MensurationFind the distance between (-4, -4) and (4, 11).

    Show answer & explanation

    Answer: D — 17

    Distance = √((Δx)² + (Δy)²) = √(8² + 15²) = 17.

  4. Q4.PhysicsThe speed of light in vacuum is approximately:

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    Answer: A — 3 × 10⁸ m/s

    c ≈ 2.998 × 10⁸ m/s.

  5. Q5.ChemistryThe periodic table is arranged in order of increasing:

    Show answer & explanation

    Answer: B — Atomic number

    The modern periodic table orders elements by atomic number.

  6. Q6.AlgebraSolve: x² − 2x − 48 = 0

    Show answer & explanation

    Answer: D — -6, 8

    Find two roots with sum 2 and product -48: 8 and -6. So x² − 2x − 48 = (x − 8)(x + 6) = 0.

  7. Q7.Probability & CombinatoricsA card is drawn from a well-shuffled deck of 52. What is the probability that it is a red card?

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    Answer: A — 1/2

    There are 26 such cards in 52, so P = 26/52 = 1/2.

  8. Q8.Geometry & MensurationA right triangle has legs 6 cm and 8 cm. Find the hypotenuse.

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    Answer: D — 10 cm

    By Pythagoras, hypotenuse = √(6² + 8²) = √100 = 10.

  9. Q9.PhysicsA body with initial velocity 14 m/s accelerates at 5 m/s² for 7 s. Find the distance covered.

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    Answer: A — 220.5 m

    s = ut + ½at² = 14×7 + ½×5×7² = 220.5 m.

  10. Q10.ChemistryWhat is the pH of a solution with [OH⁻] = 10⁻¹ mol/L at 25 °C?

    Show answer & explanation

    Answer: D — 13

    [OH⁻] = 10⁻¹ → pOH = 1; pH = 14 − 1 = 13 (at 25 °C).

Free JEE Advanced Practice Questions by pGento

About JEE Advanced

Joint Entrance Examination (Advanced) — conducted by IITs (organising IIT rotates each year). The entrance exam for undergraduate programmes at the Indian Institutes of Technology.

Original practice questions written for this site in the style and difficulty of each exam’s syllabus. They are not past papers and the site is not affiliated with any exam body. Exam patterns, marking schemes and syllabi change — always check the latest official notification.

Practise by topic

JEE Advanced practice topics

TopicSectionQuestion source
AlgebraMathematicsCalculated (new numbers every set)
Probability & CombinatoricsMathematicsCalculated (new numbers every set)
Geometry & MensurationMathematicsCalculated (new numbers every set)
PhysicsScienceCalculated + question bank
ChemistryScienceCalculated + question bank

Frequently asked questions

Solve: 5x + 4y = 43 and 5x + y = 52
Answer: B — x = 11, y = -3. Elimination/substitution gives x = 11, y = -3. Check: 5(11) + 4(-3) = 43 ✓ and 5(11) + 1(-3) = 52 ✓.
In how many ways can the letters of the word ORANGE be arranged?
Answer: C — 720. All 6 letters are different, so arrangements = 6! = 720.
Find the distance between (-4, -4) and (4, 11).
Answer: D — 17. Distance = √((Δx)² + (Δy)²) = √(8² + 15²) = 17.
Are these real past exam papers?
No. Every question is an original practice question written in the style, syllabus and difficulty of the exam, with a worked explanation. The site is not affiliated with any exam body.
Can I get new questions?
Yes. Click “New questions” for a fresh set. Calculation questions are generated with new numbers every time, so you never run out of practice.
JEE Advanced Practice Questions

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