The area of a triangle is with base 4m and height 5m?
Answer: B
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10 \text{ m}^{2} \]
The factor of one half is what distinguishes the triangle from the rectangle on the same base and height, whose area would be 20 m².
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Jay’s room has a floor of 8m by 4m. He decides to tile the floor with tiles of 25cm x 20 cm. How many tiles will he need?
Answer: B
Work in a single unit - centimetres - before dividing.
\[ \text{floor} = 800 \times 400 = 320{,}000 \text{ cm}^{2} \]
\[ \text{tile} = 25 \times 20 = 500 \text{ cm}^{2} \]
\[ n = \frac{320{,}000}{500} = 640 \]
Dividing the floor in square metres (32) by the tile in square centimetres (500) is the classic slip and produces the nonsensical 6.4 offered as the last option.
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What is the ratio between perimeters of two squares one having 3 times the diagonal then the other?
Answer: D
All lengths in a square scale together: if the diagonal is three times larger, so is the side, and so is the perimeter.
\[ d = a\sqrt{2} \quad\Longrightarrow\quad \frac{d_{1}}{d_{2}} = 3 \;\Longrightarrow\; \frac{a_{1}}{a_{2}} = 3 \]
\[ \frac{P_{1}}{P_{2}} = \frac{4a_{1}}{4a_{2}} = 3 \quad\Longrightarrow\quad 3 : 1 \]
Only the areas would be squared, in the ratio 9 : 1. Perimeter, side and diagonal are all first powers of length and share one ratio.
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The length of a rectangular plot is 60% more than its breadth. If the diffrent between the length and the breadth of that rectangle is 24 cm, what is the area of the rectangle?
Answer: C
Write the length in terms of the breadth and use the given difference.
\[ l = 1.6b \quad\Longrightarrow\quad l - b = 0.6b = 24 \]
\[ b = 40 \text{ cm}, \qquad l = 64 \text{ cm} \]
\[ A = 64 \times 40 = 2560 \text{ cm}^{2} \]
"60% more than the breadth" means \( 1.6b \), so the difference is 60% of the breadth, not of the length - taking 24 as 60% of the length would give \( l = 40 \) and a smaller rectangle.
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The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq metre.
Answer: D
\[ A = 5.5 \times 3.75 = 20.625 \text{ m}^{2} \]
\[ \text{cost} = 20.625 \times 800 \]
\[ = 20 \times 800 + 0.625 \times 800 = 16{,}000 + 500 = \text{Rs. } 16{,}500 \]
Splitting the multiplication into a whole part and a decimal part keeps the arithmetic manageable without a calculator.
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What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
Answer: C
Square tiles must divide both dimensions exactly, and the fewest tiles means the largest tile - the highest common factor.
\[ 15 \text{ m } 17 \text{ cm} = 1517 \text{ cm}, \qquad 9 \text{ m } 2 \text{ cm} = 902 \text{ cm} \]
\[ 1517 = 37 \times 41, \qquad 902 = 2 \times 11 \times 41 \]
\[ \text{HCF} = 41 \text{ cm} \]
\[ n = \frac{1517}{41} \times \frac{902}{41} = 37 \times 22 = 814 \]
The awkward 17 cm and 2 cm are deliberate: they force a common factor of 41 rather than a round number.
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The base of a right triangle is 8 and hypotenuse is 10. Its area is?
Answer: D
Find the missing perpendicular side by Pythagoras.
\[ \text{height} = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 \]
\[ A = \frac{1}{2} \times 8 \times 6 = 24 \]
The hypotenuse is never a side of the right angle, so it cannot be used as a height - using 8 and 10 as the two legs would give 40.
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What is the area of square field whose side of length 15 m?
Answer: A
\[ A = a^{2} = 15^{2} = 225 \text{ m}^{2} \]
The perimeter would be \( 4 \times 15 = 60 \) m - a different quantity with different units, and worth keeping distinct from the area.
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A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?
Answer: B
Poles stand along the boundary, so first find its length.
\[ \text{perimeter} = 2(90 + 50) = 280 \text{ m} \]
\[ n = \frac{280}{5} = 56 \]
On a closed loop the number of gaps equals the number of poles - the last pole is 5 m from the first - so no "+1" is needed here. That extra pole is required only for a fence along an open line.
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The side of a square is increased by 25% then how much % does its area increases?
Answer: B
a = 100 a² = 10000
a = 125 a² = 15625
----------------
10000 --------- 5625
100 -------? => 56.25 %
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