Daily Practice — today's mixed quiz

Question 1

A car moving at 36 km/h is brought to rest by applying brakes over a distance of 10 m. Find the deceleration produced.

Accepted Answer

Step 1 — Convert speed.
  36 km/h = 36 × 5/18 = 10 m/s.

Step 2 — Givens.
  u = 10 m/s, v = 0, s = 10 m. Use v² = u² + 2as.

Step 3 — Substitute.
  0² = 10² + 2 × a × 10
  0 = 100 + 20a
  a = −5 m/s².

✦ Answer: Deceleration = 5 m/s² (negative sign indicates motion is decelerating).

Real-world: braking decelerations on dry road are typically 3–7 m/s². Wet road or worn tyres reduce this.

Question 2

Plot P(3, 0), Q(0, 4), R(−3, 0), S(0, −4). What figure do they form? Find its area.

Accepted Answer

Step 1 — Recognise where each point lies.
  P(3, 0) and R(−3, 0): on the x-axis, 3 units on either side of the origin.
  Q(0, 4) and S(0, −4): on the y-axis, 4 units on either side of the origin.

Step 2 — Join them in order PQRS.
  This produces a quadrilateral whose vertices are on the axes — a 'kite'-like shape sitting symmetrically about both axes.

Step 3 — Use the diagonal lengths.
  Diagonal PR (along x-axis): length = 3 − (−3) = 6.
  Diagonal QS (along y-axis): length = 4 − (−4) = 8.
  The diagonals are PERPENDICULAR (one is x-axis, the other is y-axis) and BISECT each other (both pass through the origin which is their midpoint).

Step 4 — Classify.
  Perpendicular diagonals that bisect each other ⇒ rhombus.
  Since the diagonals are UNEQUAL (6 ≠ 8), it is not a square.

Step 5 — Compute area.
  Area of a rhombus = (1/2) × d₁ × d₂ = (1/2) × 6 × 8 = 24 sq units.

✦ Answer: PQRS is a rhombus of area 24 sq units.

Question 3

20 students scored: 7, 8, 9, 7, 6, 8, 9, 5, 7, 8, 9, 7, 6, 5, 8, 9, 7, 7, 6, 8. Build a frequency table and find the mode.

Accepted Answer

Frequencies — 5: 2, 6: 3, 7: 6, 8: 5, 9: 4. Mode = 7 (highest frequency 6).

Question 4

State Euclid's Fifth Postulate and Playfair's Axiom. Show they are equivalent.

Accepted Answer

Step 1 — State Euclid's Fifth Postulate.
  If a straight line falling on two straight lines makes the interior angles on the same side together less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

Step 2 — State Playfair's Axiom.
  Through a given point not on a given line, exactly one line can be drawn parallel to the given line.

Step 3 — Outline of equivalence.
  Given P5, we can derive Playfair (suppose two parallels exist through the same external point; use P5 to show they must coincide).
  Given Playfair, we can derive P5 (suppose the two lines do NOT meet; show this contradicts the unique-parallel statement).

✦ Answer: Both stated above. They are logically equivalent — assuming one allows derivation of the other within Euclidean axioms.

Question 5

Find the remainder when p(x) = 3x³ − 2x² + 4x − 1 is divided by (3x − 1).

Accepted Answer

Step 1 — Find the root of the divisor.
  3x − 1 = 0 ⇒ x = 1/3.
  By the general Remainder Theorem, the remainder is p(1/3).

Step 2 — Compute p(1/3).
  p(1/3) = 3·(1/27) − 2·(1/9) + 4·(1/3) − 1
         = 1/9 − 2/9 + 4/3 − 1.

Step 3 — Combine over a common denominator of 9.
         = 1/9 − 2/9 + 12/9 − 9/9
         = (1 − 2 + 12 − 9)/9
         = 2/9.

✦ Answer: Remainder = 2/9.

Working with fractions feels slow but is much faster than long-dividing a cubic by (3x − 1).

Question 6

A rocket of mass 1000 kg is to be lifted vertically. To lift, the rocket must overcome gravity AND accelerate upward. If the upward acceleration is 5 m/s², find the thrust (force exerted by the engines). (g = 10 m/s²)

Accepted Answer

Step 1 — Forces on the rocket (vertical only).
  Weight (downward) = mg = 1000 × 10 = 10,000 N.
  Thrust (upward) = T.
  Net force = T − mg (upward positive).

Step 2 — Apply Newton's 2nd law.
  Net force = ma.
  T − mg = m × a.
  T = m(g + a) = 1000 × (10 + 5) = 15,000 N.

Step 3 — Interpret.
  Of the 15 kN thrust, 10 kN cancels gravity, and 5 kN provides the upward acceleration.

✦ Answer: Thrust required = 15,000 N (15 kN).

The rocket's thrust must always exceed its weight (mg) for any net upward motion. Just hovering needs T = mg; accelerating up needs T > mg.

Question 7

Find the kinetic energy of a 2 kg object moving at 10 m/s.

Accepted Answer

Step 1 — KE = ½mv².
  = ½ × 2 × 10² = ½ × 2 × 100 = 100 J.

✦ Answer: 100 J.

Question 8

If a + b + c = 0, find the value of a³ + b³ + c³ − 3abc.

Accepted Answer

Step 1 — Recall the three-variable identity.
  a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Step 2 — Substitute the given condition.
  a + b + c = 0 ⇒ the entire RHS is 0 × (anything) = 0.

Step 3 — Therefore a³ + b³ + c³ − 3abc = 0, i.e. a³ + b³ + c³ = 3abc.

✦ Answer: a³ + b³ + c³ − 3abc = 0 (equivalently, a³ + b³ + c³ = 3abc).

Follow-up: Find 5³ + (−2)³ + (−3)³. Since 5 + (−2) + (−3) = 0, the trick says 5³ + (−2)³ + (−3)³ = 3·5·(−2)·(−3) = 90. Verify: 125 − 8 − 27 = 90. ✓

Question 9

In a circle, AB is a diameter and P is a point on the circle. What is ∠APB?

Accepted Answer

Step 1 — Apply Thales' Theorem (Corollary 2).
  The angle in a semicircle (subtended by a diameter) is 90°.

✦ Answer: ∠APB = 90°.

This is one of the oldest theorems in geometry — attributed to Thales of Miletus circa 600 BCE.

Question 10

Explain why neon, argon and krypton are placed in the same group of the periodic table even though they have different atomic numbers.

Accepted Answer

Step 1 — Write the configurations.
  Ne (Z = 10): 2, 8.
  Ar (Z = 18): 2, 8, 8.
  Kr (Z = 36): 2, 8, 18, 8.

Step 2 — Identify the commonality.
  All three have a FILLED OUTERMOST SHELL with 8 electrons.

Step 3 — Consequence.
  Filled outermost shell → no tendency to gain or lose electrons → CHEMICALLY INERT.
  Same chemical behaviour → same group (Group 18, noble gases).

✦ Answer: All three have 8 electrons in their outermost shell (filled, stable configuration), giving them identical chemical inertness. Same group placement reflects same chemical behaviour, not same atomic number.

Today's daily practice

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