Practice ECAT questions with answers and explanations.
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The probability is 4/10 × 3/9.
This simplifies to 12/90 = 2/15.
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The intersection contains elements common to both sets.
Only 3 appears in both A and B.
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Three distinct objects can be arranged in 3! ways.
Thus, 3! = 3 × 2 × 1 = 6.
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The union contains every distinct element from either set.
Thus, A ∪ B = {1, 2, 3}.
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The number of arrangements of 5 distinct objects is 5!.
Thus, 5! = 120.
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Cube volume is side cubed.
Thus, 4³ = 64 cm³.
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A cube has six square faces, each of area 25 cm².
Thus, total area is 6 × 25 = 150 cm².
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Cuboid volume is length × width × height.
Thus, 8 × 5 × 3 = 120 cm³.
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Cylinder volume is πr²h.
Thus, π × 9 × 10 = 90π cm³.
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Curved surface area is 2πrh.
Thus, 2π × 4 × 7 = 56π cm².
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Rectangle perimeter is 2(length + width).
Thus, 2(8 + 5) = 26 cm.
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Cone volume is one-third πr²h.
Thus, 1/3 × π × 9 × 12 = 36π cm³.
