Area Questions and Answers

Answer: C

Divide the total area between the slabs, then take the square root.

\[ \text{area of one slab} = \frac{72}{50} = 1.44 \text{ m}^{2} \]

\[ \text{side} = \sqrt{1.44} = 1.2 \text{ m} = 120 \text{ cm} \]

The slabs are square, so the side is the square root of the area - and the answer is asked for in centimetres, hence the factor of 100.

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.

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.

Answer: C

The area is unchanged, so the new breadth follows from it directly.

\[ A = 18 \times 10 = 180 \text{ cm}^{2} \]

\[ b' = \frac{180}{25} = 7.2 \text{ cm} \]

Length and breadth are inversely proportional when the area is fixed: the length rose by a factor of \( \tfrac{25}{18} \), so the breadth falls by the same factor, \( 10 \times \tfrac{18}{25} = 7.2 \).

Answer: A

Cutting 5 m from the length removes a strip 5 m long and \( b \) wide.

\[ 5b = 200 \quad\Longrightarrow\quad b = 40 \text{ m} \]

\[ l = \frac{2400}{40} = 60 \text{ m} \]

\[ \text{perimeter} = 2(60 + 40) = 200 \text{ m} \]

The lost strip has the full breadth of the park, which is what makes the breadth recoverable from the area change alone.

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.

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.

Answer: C

Three sides are painted and one is not. The unpainted side is 9 ft, so the side opposite it is also 9 ft and is painted, along with the two equal widths.

\[ 9 + 2w = 37 \quad\Longrightarrow\quad 2w = 28 \quad\Longrightarrow\quad w = 14 \text{ ft} \]

\[ A = 9 \times 14 = 126 \text{ ft}^{2} \]

The 37 ft covers all three painted sides, not two - reading it as the two widths alone would give \( w = 18.5 \) and an area of 166.5 ft².

Answer: D

The perimeter gives the sum of the sides and the difference is given directly.

\[ 2(l + b) = 206 \quad\Longrightarrow\quad l + b = 103 \]

\[ l - b = 23 \]

Adding and subtracting the two:

\[ 2l = 126 \Rightarrow l = 63, \qquad 2b = 80 \Rightarrow b = 40 \]

\[ A = 63 \times 40 = 2520 \text{ m}^{2} \]

Halving the perimeter before pairing it with the difference is the step that turns two awkward numbers into a simple pair of simultaneous equations.

Answer: D

The fencing runs round the perimeter, so divide the cost by the rate.

\[ \text{perimeter} = \frac{5300}{26.50} = 200 \text{ m} \]

\[ 2\big((b + 20) + b\big) = 200 \quad\Longrightarrow\quad 2b + 20 = 100 \]

\[ b = 40 \text{ m}, \qquad l = 60 \text{ m} \]

The length is 60 m, which appears nowhere among the choices, so the answer is none of these. The 40 m on offer is the breadth - a genuine intermediate value, but not what was asked.