This problem involves substituting given symbols with mathematical operators and then evaluating the resulting expression following the order of operations (BODMAS/PEMDAS).
The given expression is \(9\text{C}5\text{B}10\text{A}5\text{C}12\).
Replacing the symbols, we get:
\[ 9 \times 5 + 10 - 5 \times 12 \]First, perform multiplication, then addition and subtraction from left to right.
Step 1: Perform multiplications.
\[ 9 \times 5 = 45 \]\[ 5 \times 12 = 60 \]The expression becomes:
\[ 45 + 10 - 60 \]Step 1: Perform addition.
\[ 45 + 10 = 55 \]The expression becomes:
\[ 55 - 60 \]C) – 5 is the correct answer, as calculated by substituting the operators and following the order of operations.
If cot 35° = 2 - \(\sqrt{3}\) then the value of tan 35° cot 55° + tan 55° cot 35° is:
The problem asks us to find the value of an expression involving trigonometric ratios of 35° and 55°, given the value of \(\cot 35^\circ\). We will use trigonometric identities, specifically the complementary angle identity \(\cot \theta = \tan (90^\circ - \theta)\) and the reciprocal identity \(\tan \theta = 1/\cot \theta\), to simplify the expression and substitute the given value.
We are given \(\cot 35^\circ = 2 - \sqrt{3}\).
From the reciprocal identity, \(\tan 35^\circ = \frac{1}{\cot 35^\circ}\). So, \(\tan 35^\circ = \frac{1}{2 - \sqrt{3}}\).
Rationalize the denominator for \(\tan 35^\circ\):
\[ \tan 35^\circ = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 + \sqrt{3}}{4 - 3} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3} \]Now consider the terms involving 55°. We know that \(\cot 55^\circ = \cot (90^\circ - 35^\circ)\). Using the complementary angle identity, \(\cot (90^\circ - \theta) = \tan \theta\), so \(\cot 55^\circ = \tan 35^\circ\).
Similarly, \(\tan 55^\circ = \tan (90^\circ - 35^\circ)\). Using the complementary angle identity, \(\tan (90^\circ - \theta) = \cot \theta\), so \(\tan 55^\circ = \cot 35^\circ\).
Substitute these into the given expression: \(\tan 35^\circ \cot 55^\circ + \tan 55^\circ \cot 35^\circ\).
The expression becomes \(\tan 35^\circ (\tan 35^\circ) + (\cot 35^\circ) \cot 35^\circ = (\tan 35^\circ)^2 + (\cot 35^\circ)^2\).
Substitute the values we found for \(\tan 35^\circ\) and \(\cot 35^\circ\):
\[ (2 + \sqrt{3})^2 + (2 - \sqrt{3})^2 \]Expand the squares using the formula \((a+b)^2 = a^2 + 2ab + b^2\) and \((a-b)^2 = a^2 - 2ab + b^2\):
\[ (2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2) + (2^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2) \]\[ (4 + 4\sqrt{3} + 3) + (4 - 4\sqrt{3} + 3) \]\[ (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) \]Combine the terms:
\[ 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 7 + 7 = 14 \]D) The calculated value of the expression is 14.
The question asks to identify the first female judge appointed to the Supreme Court of India. This is a factual question related to Indian legal history and requires knowledge of prominent judicial figures.
Justice M. Fathima Beevi was the first woman judge to be appointed to the Supreme Court of India. She was appointed to the Supreme Court in 1989, making her the first female judge of the apex court and also the first Muslim woman to be appointed to any higher judiciary in India.
B) M Fathima Beevi was the first female judge of the Supreme Court of India, appointed in 1989.
The question asks about the primary cause of ozone layer depletion. We need to identify the substance or phenomenon most widely recognized for its destructive effect on stratospheric ozone.
The chemical reactions involved are:
This cycle continues, leading to significant ozone destruction.
A) Chlorofluorocarbons — CFCs are the primary man-made chemicals responsible for the depletion of the ozone layer. Their stability allows them to reach the stratosphere, where they release chlorine atoms that catalytically destroy ozone molecules.
The question asks to identify a film directed by Satyajit Ray from the given options. This requires knowledge of prominent Indian film directors and their filmographies.
Correct Option: C) Pather Panchali
Pather Panchali (1955) is a landmark film in Indian cinema, directed by Satyajit Ray. It is the first film in his Apu Trilogy and is widely regarded as one of the greatest films ever made. It introduced Ray's distinctive style to the world and brought Indian cinema to international attention.
The problem presents an analogy where a pattern needs to be identified between the first number and the second number in each pair. We need to apply this identified pattern to the third pair to find the missing number.
Let's try to find a relationship between 56 and 15. One common pattern in such analogies involves operations on the digits of the first number.
Apply the identified pattern to the numbers 76 and 21.
The pattern holds true for both given pairs.
Now, apply the established pattern to find the missing number for the pair 81 : ?
So, the missing number is 4.
Correct Option: A) 4 is the correct answer because applying the pattern (product of digits of the first number divided by 2) yields 4.
The question asks to identify the experimental satellite launched by ISRO in August 1979 from Satish Dhawan Space Centre. This requires knowledge of the history of ISRO's satellite launches.
The Rohini series of satellites were a series of experimental satellites launched by ISRO. The first of these, Rohini Technology Payload (RTP), was launched on August 10, 1979, by the Satellite Launch Vehicle-3 (SLV-3) from Satish Dhawan Space Centre, Sriharikota. This launch was a partial success as the satellite could not be placed into orbit due to a malfunction in the second stage of the launch vehicle. However, it was indeed the first experimental satellite launched by ISRO from Indian soil using an Indian-made launch vehicle.
B) Rohini — The Rohini Technology Payload (RTP) was the first experimental satellite launched by ISRO on August 10, 1979, from Satish Dhawan Space Centre, Sriharikota, using the SLV-3 rocket.
To find the gain percentage, we first need to calculate the gain amount by subtracting the cost price from the selling price. Then, we divide the gain by the cost price and multiply by 100 to express it as a percentage.
Gain is the difference between the Selling Price and the Cost Price.
\[ \text{Gain} = \text{SP} - \text{CP} \] \[ \text{Gain} = \text{₹}1,140 - \text{₹}950 \] \[ \text{Gain} = \text{₹}190 \]The gain percentage is calculated using the formula:
\[ \text{Gain Percentage} = \left( \frac{\text{Gain}}{\text{CP}} \right) \times 100 \]Substitute the values:
\[ \text{Gain Percentage} = \left( \frac{190}{950} \right) \times 100 \] \[ \text{Gain Percentage} = \left( \frac{19}{95} \right) \times 100 \]Since \(19 \times 5 = 95\), we can simplify the fraction:
\[ \text{Gain Percentage} = \left( \frac{1}{5} \right) \times 100 \] \[ \text{Gain Percentage} = 20\% \]Correct Option: C) 20%
To find the percentage reduction in the area of a circle when its radius is reduced by a certain percentage, we first calculate the initial area, then the new radius and new area after the reduction. Finally, we determine the percentage change in the area.
Define initial radius and area:
Let the original radius of the circle be \(R\).
The original area of the circle, \(A_1\), is given by the formula:
Calculate the new radius:
The radius is reduced by 25%. This means the new radius will be \(100\% - 25\% = 75\%\) of the original radius.
Calculate the new area:
The new area of the circle, \(A_2\), with the new radius \(R'\) is:
Calculate the reduction in area:
The reduction in area is the difference between the original area and the new area:
Calculate the percentage reduction in area:
The percentage reduction is the reduction in area divided by the original area, multiplied by 100:
Correct Option: A) 43.75%