Formula used to measure circumference of circle is
Answer: A
\[ C = 2\pi r \]
A circumference is a length, so it must be proportional to \( r \) rather than \( r^{2} \) - which rules out the area formula \( \pi r^{2} \) on dimensional grounds alone.
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A circle is inscribed in a square. If the difference between area of the square and circle is 262.5 cm², then find the area of the rectangle whose perimeter is same as that of circle while length of rectangle is 20% more than the breadth of rectangle (in cm²).
Answer: D
Let the side of square be 2a.
So, the diameter of circle = 2a
Radius of circle = a
WKT, Area of square = 4a
Area of circle = π r²
Perimeter of circle = 2π r
Area of rectangle = l x b
Perimeter of rectangle = 2(l + b)
Given that difference between area of the square and circle is 262.5 cm².
Area of square - Area of circle = 262.5
4a² - (22/7)a² = 262.5
28a² - 22a² = 1837.5
6a² = 1837.5
a² = 306.25
a = 17.5 cm
Perimeter of circle = 2πa = 2 x (22/7) x 17.5 = 110 cm
Also given length of rectangle is 20% more than the breadth of rectangle.
Let the breadth of rectangle be X.
So, length of rectangle = (120/100)X = 1.2X
As per the question,
Perimeter of rectangle = Perimeter of circle
2(1.2X + X) = 110
2.4X + 2X = 110
4.4X = 110
X = 25 cm
Breadth of rectangle = 25 cm
Length of rectangle = 1.2(25) = 30 cm
Area of rectangle = 25 x 30 = 750 cm².
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What is the curved surface area of a cone whose radius is 7 cm and slant height 10 cm ? (Take \(\pi = 22/7\))
Answer: A
Correct answer: 220 sq. cm
\[\text{Curved surface area} = \pi r l = \frac{22}{7} \times 7 \times 10 = 220 \text{ sq. cm}\]
Adding the base, \(\pi r^2 = 154\) sq. cm, would give a total surface area of 374 sq. cm, so read carefully which is asked.
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A solid rubber sphere weighs 6kg when its diameter is 6cm. Using the same material, a hollow sphere is made with outer diameter 18cm and inner diameter 12 cm. What is its weight?
Answer: A
Same material means the same density, so weight is proportional to volume.
\[ V_{\text{solid}} = \frac{4}{3}\pi (3)^{3} = 36\pi \text{ cm}^{3} \;\longrightarrow\; 6 \text{ kg} \]
The hollow sphere has outer radius 9 cm and inner radius 6 cm:
\[ V_{\text{hollow}} = \frac{4}{3}\pi\left(9^{3} - 6^{3}\right) = \frac{4}{3}\pi (729 - 216) = \frac{4}{3}\pi (513) = 684\pi \text{ cm}^{3} \]
\[ \text{weight} = 6 \times \frac{684\pi}{36\pi} = 6 \times 19 = 114 \text{ kg} \]
The cavity must be subtracted; treating the object as a solid 18 cm sphere would give 162 kg.
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The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.
Answer: D
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.
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Formula used to measure area of circle is
Answer: B
\[ A = \pi r^{2} \]
An area must vary as the square of a length, which alone rules out \( \pi r^{2} + 2r \) as dimensionally inconsistent. The other two are real formulas from elsewhere: \( 4\pi r^{2} \) is a sphere's surface area and \( 2\pi r^{2} \) a hemisphere's curved surface.
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If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?
Answer: A
Check the sides first: \( 10^{2} + 24^{2} = 100 + 576 = 676 = 26^{2} \), so this is a right-angled triangle with legs 10 cm and 24 cm.
\[ A = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^{2} \]
Heron's formula agrees: with \( s = 30 \), \( \sqrt{30 \times 4 \times 6 \times 20} = \sqrt{14400} = 120 \). Spotting the 5–12–13 triple doubled saves that work.
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Find the number of bricks, each measuring 24 cm×12 cm × 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick if 10% of the wall is filled with mortar?
Answer: C
Work in centimetres throughout.
\[ V_{\text{wall}} = 2400 \times 800 \times 60 = 115{,}200{,}000 \text{ cm}^{3} \]
Mortar fills 10%, so bricks occupy 90%:
\[ 0.9 \times 115{,}200{,}000 = 103{,}680{,}000 \text{ cm}^{3} \]
\[ V_{\text{brick}} = 24 \times 12 \times 8 = 2304 \text{ cm}^{3} \]
\[ n = \frac{103{,}680{,}000}{2304} = 45{,}000 \]
Ignoring the mortar would give 50,000 - a tenth too many.
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Which of the following is equal to 1 kilolitre?
Answer: D
Work through the metric chain of volume units.
\[ 1 \text{ litre} = 1 \text{ dm}^{3} \]
\[ 1 \text{ kilolitre} = 1000 \text{ litres} = 1000 \text{ dm}^{3} \]
That is also 1 cubic metre, since \( 1 \text{ m}^{3} = 1000 \text{ dm}^{3} \). The 1000 millilitres on offer is only one litre - a thousandth of the answer.
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If the parallel sides of a parallelogram are 2 cm apart and their sum is 10 cm then its area is:
Answer: C
The two parallel sides of a parallelogram are equal, so a sum of 10 cm makes each one 5 cm, and the 2 cm between them is the height.
\[ A = \text{base} \times \text{height} = 5 \times 2 = 10 \text{ cm}^{2} \]
The trapezium formula gives the same figure here, since
\[ A = \frac{1}{2}(a + b)h = \frac{1}{2} \times 10 \times 2 = 10 \text{ cm}^{2} \]
and for a parallelogram the two parallel sides are equal.
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