Circles – MBA CET questions with solutions

Approach: Distance per revolution = circumference = 2πr.

Solution:
Circumference = 2 × (22/7) × 0.25 = 11/7 m.
Distance = 11 km = 11,000 m.
Revolutions = 11,000 ÷ (11/7) = 11,000 × 7/11 = 7000.

Cetking shortcut: 11 km and 22/7 are built to cancel: 11000 × 7/11 = 7000.

Answer: C

Approach: Unoccupied region = square joining centres − four quarter circles.

Solution:
Let r = x + 1. Adjacent coins touch, so the side of the square = 2r.
Square area = 4r². Each corner holds a quarter coin; 4 quarters = one full circle = πr².
Unoccupied area = 4r² − πr² = r²(4 − π).
With π = 3.14 and r² = x² + 2x + 1: (x² + 2x + 1)(4 − 3.14).

Why the other options are wrong:
B adds instead of subtracting; E ignores the coins. C and D are algebraically the same as r²(4 − π), but A is the form that uses the given π = 3.14.

Cetking shortcut: Region between 4 touching equal circles = (4 − π)r² ≈ 0.86r² — memorise it.

Answer: A

Approach: Distance = number of revolutions × circumference.

Solution:
r = 1¾ m = 7/4 m.
Circumference = 2πr = 2 × 22/7 × 7/4 = 11 m.
11 km = 11,000 m.
Revolutions = 11,000/11 = 1000.

Why the other options are wrong:

• A (500): takes the diameter 3.5 m as the radius (circumference 22 m).
• C, D, E: wrong circumference or km→m conversion.

Cetking shortcut: r = 7/4 cancels with π = 22/7 → circumference 11 m, and 11 km ÷ 11 m = 1000.

Answer: B

Approach: Distance in one lap = circumference; then area = πr².

Solution:
Speed = 14.4 × 5/18 = 4 m/s
Time = 1 min 28 s = 88 s → circumference = 352 m
2 × 22/7 × r = 352 → r = 56 m
Area = 22/7 × 56 × 56 = 9856 m²

Cetking shortcut: 352 = 2πr with π = 22/7 → r = 352 × 7/44 = 56; area = 22 × 8 × 56 = 9856.

Answer: B

Approach: PQ is a diameter → ∠PSQ = 90°; then use 'angles on the same arc'.

Solution:
In the figure, PQ is a diameter, so ∠PSQ = 90° (angle in a semicircle)
In ΔPSQ: ∠SPQ = 180° − 90° − 35° = 55°
∠QRS and ∠QPS stand on the same arc QS → ∠QRS = ∠QPS = 55°

Why the other options are wrong:
A (35°): that is ∠PQS itself, which stands on arc PS, not QS.
B (135°): would be an angle in the opposite segment.
D, E: no theorem gives 60° or 65° here.

Cetking shortcut: Diameter + given angle 35° → the other acute angle of the right triangle is 55°, which carries over along the same arc.

Answer: C