The question asks to identify the political party of Imran Khan, who was the Prime Minister of Pakistan as of October 2020. This requires knowledge of contemporary political leaders and their affiliations.
Imran Khan founded the Pakistan Tehreek-e-Insaf (PTI) party in 1996. He became the Prime Minister of Pakistan in August 2018 after his party won the general elections. He remained Prime Minister until April 2022. Therefore, as of October 2020, he was indeed the Prime Minister and belonged to the Pakistan Tehreek-e-Insaf party.
A) Pakistan Tehreek-e-Insaf
Imran Khan is the founder and chairman of the Pakistan Tehreek-e-Insaf (PTI) party, under whose banner he served as the Prime Minister of Pakistan from August 2018 to April 2022.
The question asks to identify the odd word-pair out of four given options. We need to find a common relationship among three pairs and identify the one that does not fit this pattern.
Correct Option: B) Humayun : Mughal — This pair represents a ruler and the dynasty he belonged to, which is different from the 'Religion : Founder' relationship seen in the other options.
The question asks to identify the Indian fort where the Musamman Burj is located. This requires knowledge of prominent historical structures within famous Indian forts.
Correct Option: C) Red Fort of Agra
The Musamman Burj (also known as Saman Burj or Shah Burj) is a white marble octagonal tower located in the Agra Fort. It was built by Shah Jahan for his beloved wife Mumtaz Mahal. It is particularly famous as the place where Shah Jahan spent his last years as a prisoner, gazing at the Taj Mahal, which he had built for Mumtaz Mahal.
The question asks to identify which of the given options does NOT grow under the soil. This requires knowledge of common vegetables and the part of the plant that is consumed.
Correct Option: A) Tomato. Tomatoes are fruits that grow on plants above the ground, not under the soil.
An observer at the top of a tower observes that two cars are running towards the foot of the tower at a distance of 120 m from each other making angles of depression α and β such that α > β and tan α = \(\sqrt{3}\) and tan β = \(\frac{1}{\sqrt{3}}\). Find the height of the tower.
120\(\sqrt{3}\) m
40\(\sqrt{3}\) m
80\(\sqrt{3}\) m
60\(\sqrt{3}\)m
We need to find the height of the tower using the given angles of depression and the distance between the two cars. We will use trigonometric ratios (tangent) in the right-angled triangles formed by the tower and the cars.
Let the height of the tower be \(h\). Let the observer be at point A at the top of the tower, and the foot of the tower be at point B. Let the two cars be at points C and D, such that C is closer to the tower than D.
The angles of depression are \(\alpha\) and \(\beta\). Since \(\alpha > \beta\), the car making angle \(\alpha\) is closer to the tower. So, \(\angle ACB = \alpha\) and \(\angle ADB = \beta\).
We are given \(\tan \alpha = \sqrt{3}\). This implies \(\alpha = 60^\circ\).
We are given \(\tan \beta = \frac{1}{\sqrt{3}}\). This implies \(\beta = 30^\circ\).
In right-angled triangle \(\triangle ABC\):
\[ \tan \alpha = \frac{AB}{BC} \]\[ \sqrt{3} = \frac{h}{BC} \]\[ BC = \frac{h}{\sqrt{3}} \quad \ldots(1) \]In right-angled triangle \(\triangle ABD\):
\[ \tan \beta = \frac{AB}{BD} \]\[ \frac{1}{\sqrt{3}} = \frac{h}{BD} \]\[ BD = h\sqrt{3} \quad \ldots(2) \]The distance between the two cars is 120 m. So, \(CD = 120\) m. From the diagram, \(CD = BD - BC\).
\[ 120 = h\sqrt{3} - \frac{h}{\sqrt{3}} \]Now, solve for \(h\):
\[ 120 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) \]\[ 120 = h \left( \frac{3 - 1}{\sqrt{3}} \right) \]\[ 120 = h \left( \frac{2}{\sqrt{3}} \right) \]\[ h = \frac{120 \times \sqrt{3}}{2} \]\[ h = 60\sqrt{3} \text{ m} \]D) 60\(\sqrt{3}\) m — The height of the tower is calculated to be \(60\sqrt{3}\) m using the trigonometric relations and the given information.
The question asks to identify the Indian airline that ceased operations or was grounded in April 2019. This requires knowledge of significant events in the Indian aviation sector.
Correct Option: C) Jet Airways — Jet Airways, once India's second-largest airline, suspended all its flight operations indefinitely on April 17, 2019, due to severe financial crisis and inability to secure emergency funding. This event was widely reported and is a significant point in Indian aviation history.
Find the area of a trapezium whose parallel sides are 10 cm and 20 cm and non-parallel sides are equal to 10 cm.
70\(\sqrt{3}\) cm²
75\(\sqrt{3}\) cm²
60\(\sqrt{3}\) cm²
65\(\sqrt{3}\) cm²
To find the area of a trapezium, we need the lengths of its parallel sides and its height. We are given the lengths of the parallel sides and the non-parallel sides. Since the non-parallel sides are equal, this is an isosceles trapezium. We can use the Pythagorean theorem to find the height of the trapezium.
Let the parallel sides be \(a = 20 \text{ cm}\) and \(b = 10 \text{ cm}\). Let the non-parallel sides be \(c = 10 \text{ cm}\).
Draw perpendiculars from the ends of the shorter parallel side to the longer parallel side. This will form two right-angled triangles and a rectangle in the middle.
The length of the base of each right-angled triangle can be found by subtracting the shorter parallel side from the longer parallel side and dividing by 2 (since it's an isosceles trapezium).
\[ \text{Base of triangle} = \frac{a - b}{2} = \frac{20 - 10}{2} = \frac{10}{2} = 5 \text{ cm} \]
Now, consider one of the right-angled triangles. The hypotenuse is the non-parallel side, which is \(10 \text{ cm}\). The base of the triangle is \(5 \text{ cm}\). Let \(h\) be the height of the trapezium (which is also the height of the triangle).
Apply the Pythagorean theorem: \((\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2\).
\[ 10^2 = 5^2 + h^2 \]
\[ 100 = 25 + h^2 \]
\[ h^2 = 100 - 25 \]
\[ h^2 = 75 \]
\[ h = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \text{ cm} \]
The formula for the area of a trapezium is:
\[ \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \]
\[ \text{Area} = \frac{1}{2} \times (a + b) \times h \]
\[ \text{Area} = \frac{1}{2} \times (20 + 10) \times 5\sqrt{3} \]
\[ \text{Area} = \frac{1}{2} \times 30 \times 5\sqrt{3} \]
\[ \text{Area} = 15 \times 5\sqrt{3} \]
\[ \text{Area} = 75\sqrt{3} \text{ cm}^2 \]
Correct Option: B) \(75\sqrt{3}\) cm²
Using BODMAS, simplify the following.
\( \frac{7}{9} \times \frac{21}{5} \times 25 \left(65^{2} - 55^{2}\right) \)
To simplify the given expression, we will follow the BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction) rule. First, we will simplify the terms inside the bracket, then perform multiplication.
The expression inside the bracket is \( \left(65^{2} - 55^{2}\right) \). We can use the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \).
\[ 65^2 - 55^2 = (65 - 55)(65 + 55) \]\[ = (10)(120) \]\[ = 1200 \]The original expression is \[\frac{7}{9} \times \frac{21}{5} \times 25 \left(65^{2} - 55^{2}\right)\].
Substitute \( \left(65^{2} - 55^{2}\right) = 1200 \):
\[ \frac{7}{9} \times \frac{21}{5} \times 25 \times 1200 \]We can simplify the terms by canceling common factors.
\[ \frac{7}{\cancel{9}_{\text{1}}} \times \frac{\cancel{21}^{\text{7}}}{\cancel{5}_{\text{1}}} \times \cancel{25}^{\text{5}} \times 1200 \]\[ = 7 \times \frac{7}{1} \times 5 \times 1200 \]\[ = 7 \times 7 \times 5 \times 1200 \]\[ = 49 \times 5 \times 1200 \]\[ = 245 \times 1200 \]Wait, there was a mistake in the cancellation in step 3. Let's re-evaluate step 3 carefully.
The expression is \[\frac{7}{9} \times \frac{21}{5} \times 25 \times 1200\].
First, simplify the fractions:
\[ \frac{7}{9} \times \frac{21}{5} = \frac{7 \times 21}{9 \times 5} = \frac{7 \times \cancel{3}^{\text{7}}}{\cancel{9}_{\text{3}} \times 5} = \frac{7 \times 7}{3 \times 5} = \frac{49}{15} \]Now, multiply this by \( 25 \times 1200 \):\[ \frac{49}{15} \times 25 \times 1200 \]\[ = \frac{49}{\cancel{15}_{\text{3}}} \times \cancel{25}^{\text{5}} \times 1200 \]\[ = \frac{49 \times 5 \times 1200}{3} \]\[ = \frac{245 \times 1200}{3} \]\[ = 245 \times \frac{1200}{3} \]\[ = 245 \times 400 \]\[ = 98000 \]A) The simplified value of the expression is \(98000\).
This is a problem involving inverse proportionality. The total amount of provisions is constant. If fewer men consume the provisions, they will last for a longer duration. We can calculate the total 'man-days' of provisions available and then divide by the actual number of men to find out how many days the provisions will last.
Initially, there were provisions for 425 men for 30 days.
\[ \text{Total Provisions} = \text{Number of Men} \times \text{Number of Days} \]\[ \text{Total Provisions} = 425 \text{ men} \times 30 \text{ days} = 12750 \text{ man-days} \]The problem states that 375 men attended the camp.
Now, divide the total provisions (in man-days) by the actual number of men to find the number of days the provisions will last.
\[ \text{Days Provisions Last} = \frac{\text{Total Provisions}}{\text{Actual Number of Men}} \]\[ \text{Days Provisions Last} = \frac{12750 \text{ man-days}}{375 \text{ men}} = 34 \text{ days} \]C) 34 days. The calculation shows that the provisions will last for 34 days when consumed by 375 men.
The question asks to identify the largest delta in the world. This requires knowledge of major river deltas and their relative sizes.
Correct Option: B) Ganges-Brahmaputra