If the radius of a wheel is 14 cm, the number of revolutions it makes in covering a distance of 1056 cm is?
Answer: D
Each revolution advances the wheel by one circumference.
\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \]
\[ n = \frac{1056}{88} = 12 \]
A radius that is a multiple of 7 makes the \( \tfrac{22}{7} \) cancel exactly, which is why the circumference lands on a whole 88 cm.
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The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
Answer: B
A square's diagonal and side are related by \( d = a\sqrt{2} \), so the area can be written directly in terms of the diagonal:
\[ A = a^{2} = \frac{d^{2}}{2} \]
The factor \( \tfrac{1}{2} \) is common to both squares and cancels:
\[ \frac{A_{1}}{A_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} = \frac{2^{2}}{5^{2}} = \frac{4}{25} \]
So the ratio is \( 4 : 25 \). Any two similar figures have areas in the square of their linear ratio, whichever length is used to compare them.
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The side and one of the diagonals of a rhombus are 13cm and 24cm respectively. The area of the rhombus (in sq.cm.) is
Answer: C
A rhombus's diagonals bisect each other at right angles, so half of each diagonal and a side form a right triangle.
\[ \text{half of the known diagonal} = \frac{24}{2} = 12 \text{ cm} \]
\[ \text{half of the other} = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = 5 \text{ cm} \]
\[ \text{other diagonal} = 10 \text{ cm} \]
\[ \text{area} = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 24 \times 10 = 120 \text{ cm}^{2} \]
5–12–13 is a Pythagorean triple, which keeps the arithmetic exact. Forgetting the \( \tfrac{1}{2} \) gives 240, the second option.
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The length of a rectangular plot is 20 metres more than its breadth. If fencing the plot at Rs. 26.50 per metre costs Rs. 5300, what is the length of the plot in metres?
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.
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The ratio of the area of a square to that of the square drawn on its diagonal is?
Answer: D
Let the original square have side \( a \), so its diagonal is \( a\sqrt{2} \).
The second square is built on that diagonal as its side:
\[ A_{1} = a^{2}, \qquad A_{2} = \left(a\sqrt{2}\right)^{2} = 2a^{2} \]
\[ \frac{A_{1}}{A_{2}} = \frac{a^{2}}{2a^{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]
The square on the diagonal is always exactly double - the diagonal is longer by \( \sqrt{2} \), and areas scale by the square of that.
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The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?
Answer: A
Multiply the two factors and compare with 1.
\[ \frac{A'}{A} = (1 + 0.25)(1 - 0.20) = 1.25 \times 0.80 = 1.00 \]
The area is therefore unchanged.
The percentages look unequal but exactly offset each other, because \( \tfrac{5}{4} \times \tfrac{4}{5} = 1 \). Adding them (\( +25 - 20 = +5 \)) would suggest a 5% rise, which is the distractor.
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The area of a rectangular park is 2400 sq. metres. If the length of the park is decreased by 5 metres, the area decreases by 200 sq. metres. The perimeter of the park is
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.
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The perimeter of a semi circle is 144 cm then the radius is?
Answer: B
\[ P = r(\pi + 2) = r \times \frac{36}{7} \]
\[ 144 = \frac{36r}{7} \quad\Longrightarrow\quad r = \frac{144 \times 7}{36} = 28 \text{ cm} \]
Using \( \pi r \) alone would give \( r = 45.8 \) cm - the diameter is part of a semicircle's boundary and must be included.
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The radius of a circle is increased by 1%. Find how much % does its area increases?
Answer: D
Area varies as the square of the radius.
\[ \frac{A'}{A} = (1.01)^{2} = 1.0201 \]
\[ \text{increase} = 0.0201 = 2.01\% \]
The exact rule is \( 2x + \tfrac{x^{2}}{100} = 2 + 0.01 \). For small changes the area rises by roughly twice the percentage change in the radius, with the tiny square term as the correction.
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How many minimum number of square tiles are needed to cover a rectangular floor of dimension 5.25m and 5.10m completely?
Answer: A
Square tiles must fit a whole number of times along both edges, so the tile's side must divide both dimensions. Fewest tiles means largest tiles, so take the highest common factor.
\[ 5.25 \text{ m} = 525 \text{ cm}, \qquad 5.10 \text{ m} = 510 \text{ cm} \]
\[ 525 = 3 \times 5^{2} \times 7, \qquad 510 = 2 \times 3 \times 5 \times 17 \]
\[ \text{HCF} = 3 \times 5 = 15 \text{ cm} \]
\[ n = \frac{525}{15} \times \frac{510}{15} = 35 \times 34 = 1190 \]
Converting to centimetres first is essential - the highest common factor of 5.25 and 5.10 is not a whole number of metres.
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