Mensuration
- A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is
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Volume of bigger cube = 6 × 6 × 6 = 216 cu. cm.
Volume of unit cube = 1 × 1 × 1 = 1 cu. cm
Number of uncoloured cubes = 4 × 4 × 4 = 64
[because edge of uncoloured cube = 4 cm]Correct Option: B
Volume of bigger cube = 6 × 6 × 6 = 216 cu. cm.
Volume of unit cube = 1 × 1 × 1 = 1 cu. cm
Number of uncoloured cubes = 4 × 4 × 4 = 64
[because edge of uncoloured cube = 4 cm]
- A hall 25 metres long and 15 metres broad is surrounded by a verandah of uniform width of 3.5 metres. The cost of flooring the verandah, at ₹ 27.50 per square metre is
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Area of the verandah = (25 + 2 × 3.5) (15 + 2 × 3.5) – 25 × 15
= 32 × 22 – 25 × 15 = 704 – 375 = 329 sq.metre
∴ Cost of flooring = 329 × 27.5 = ₹ 9047.5Correct Option: C
Area of the verandah = (25 + 2 × 3.5) (15 + 2 × 3.5) – 25 × 15
= 32 × 22 – 25 × 15 = 704 – 375 = 329 sq.metre
∴ Cost of flooring = 329 × 27.5 = ₹ 9047.5
- The cost of carpeting a room is 120. If the width had been 4 metres less, the cost of the Carpet would have been 20 less. The width of the room is :
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Let the cost of carpeting per sq. metre be 1.
∴ Area of the room = 120 sq. metre
Let the breadth of the room be x metres. Then,Length of room = 120 metres x
New cost = 120 – 20 = ? 100
Breadth = (x – 4) metresThen, 120 × (x - 4) = 100 x ⇒ 6 (x - 4) = 5 x
⇒ 6x – 24 = 5x
⇒ x = 24
∴ Breadth of the room = 24 metresCorrect Option: A
Let the cost of carpeting per sq. metre be 1.
∴ Area of the room = 120 sq. metre
Let the breadth of the room be x metres. Then,Length of room = 120 metres x
New cost = 120 – 20 = ? 100
Breadth = (x – 4) metresThen, 120 × (x - 4) = 100 x ⇒ 6 (x - 4) = 5 x
⇒ 6x – 24 = 5x
⇒ x = 24
∴ Breadth of the room = 24 metres
- A soap cake is of size 8 cm × 5 cm × 4 cm. The number of such soap cakes that can be packed in a box measuring 56 cm × 35 cm × 28 cm is :
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Volume of the box = (56 × 35 × 28) cm³
Volume of a soap cake = (8 × 5 × 4) cm³∴ Number of soap cakes = 56 × 35× 28 = 343 8 × 5× 4 Correct Option: D
Volume of the box = (56 × 35 × 28) cm³
Volume of a soap cake = (8 × 5 × 4) cm³∴ Number of soap cakes = 56 × 35× 28 = 343 8 × 5× 4
- A cuboidal block of 6 cm × 9 cm × 12 cm is cut up into exact number of equal cubes. The least possible number of cubes will be
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The number of cubes will be least if each cube will be of maximum edge.
∴ Maximum possible length = HCF of 6, 9, 12 = 3
∴ Volume of cube = 3 × 3 × 3 cm³∴ Number of cubes = 6 × 9 × 12 cm³ = 24 3 × 3 × 3cm³ Correct Option: C
The number of cubes will be least if each cube will be of maximum edge.
∴ Maximum possible length = HCF of 6, 9, 12 = 3
∴ Volume of cube = 3 × 3 × 3 cm³∴ Number of cubes = 6 × 9 × 12 cm³ = 24 3 × 3 × 3cm³
