The question asks about the international collaboration behind India's first Pressurised Heavy Water Reactor (PHWR) in 1964. This requires knowledge of India's early nuclear program history and its international partners.
Correct Option: D) Canada. India's first PHWR, RAPS-1, was built with Canadian collaboration, utilizing their CANDU reactor technology.
This problem involves calculating the cost price of an article based on the difference in selling prices when sold at a gain versus a loss percentage. We can set up equations for the selling price in both scenarios and use the given difference to find the cost price.
Let the cost price of the article be \(CP\).
Scenario 1: Sold at a loss of 5%
If the article is sold at a loss of 5%, the selling price (\(SP_1\)) will be:
Scenario 2: Sold at a gain of 5%
If the article is sold at a gain of 5%, the selling price (\(SP_2\)) will be:
Given difference in selling prices
According to the problem, if the article is sold at a gain of 5% instead of a loss of 5%, the man gets ₹5 more. This means the difference between \(SP_2\) and \(SP_1\) is ₹5.
Substitute the expressions for \(SP_1\) and \(SP_2\) into the difference equation:
\[ \frac{105}{100} CP - \frac{95}{100} CP = 5 \]Simplify the equation:
\[ \frac{105 - 95}{100} CP = 5 \]\[ \frac{10}{100} CP = 5 \]\[ \frac{1}{10} CP = 5 \]Solve for \(CP\):
\[ CP = 5 \times 10 \]\[ CP = 50 \]D) ₹50. The cost price of the article is ₹50.
The problem asks for the value of a product of tangent functions. We can simplify this expression by using trigonometric identities, specifically the complementary angle identity \(\tan (90^\circ - \theta) = \cot \theta\) and the reciprocal identity \(\cot \theta = \frac{1}{\tan \theta}\).
\[ \text{Expression} = \tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ \]
Notice that \(5^\circ + 85^\circ = 90^\circ\) and \(25^\circ + 65^\circ = 90^\circ\).
We can rewrite the expression as:
\[ \text{Expression} = (\tan 5^\circ \tan 85^\circ) (\tan 25^\circ \tan 65^\circ) \tan 45^\circ \]
Recall that \[\tan (90^\circ - \theta) = \cot \theta = \frac{1}{\tan \theta}\].
\[ \text{Expression} = \left(\tan 5^\circ \cdot \frac{1}{\tan 5^\circ}\right) \left(\tan 25^\circ \cdot \frac{1}{\tan 25^\circ}\right) \tan 45^\circ \]
\[ \text{Expression} = (1) (1) \tan 45^\circ \]
We know that \(\tan 45^\circ = 1\).
\[ \text{Expression} = 1 \cdot 1 \cdot 1 = 1 \]
A) 1 — The value of the given expression simplifies to 1 using complementary angle identities and the known value of \(\tan 45^\circ\).
The problem presents a pattern-based question where a new operator 'A' is defined. We need to deduce the rule governing 'A' from the given examples and then apply it to find the result for the third expression.
Correct Option: D) 6429
This is a blood relation question. We need to trace the relationship step-by-step, starting from the last known relation and moving backward to establish how Radha is related to Abhinav.
Correct Option: C) Granddaughter. Based on the step-by-step analysis, Radha is the sister of Abhinav's grandson, making her Abhinav's granddaughter.
The question asks to identify the specific name of the special train service that connects prominent Buddhist pilgrimage sites: Lumbini, Bodhgaya, Sarnath, and Kushinagar. This requires knowledge of specific tourist train services offered in India.
C) Buddhist Circuit Tourist Train — This is the official and widely recognized name of the special train service operated by IRCTC (Indian Railway Catering and Tourism Corporation) that takes pilgrims to the major Buddhist sites, including Lumbini, Bodhgaya, Sarnath, and Kushinagar. The circuit covers places where Buddha was born, attained enlightenment, gave his first sermon, and passed away.
To determine the final profit compared to the initial profit, we can assume an initial profit value (e.g., 100) and then apply the given percentage changes sequentially. Alternatively, we can use a formula for successive percentage changes.
Let the profit of the book store at the beginning of April be \(P\). For simplicity, let's assume \(P = 100\).
In April, the profit increased by 25%. So, the profit at the end of April (or beginning of May) will be:
\[ \text{Profit}_{\text{April end}} = P + (0.25 \times P) = P(1 + 0.25) = 1.25P \]If \(P = 100\), then \(\text{Profit}_{\text{April end}} = 1.25 \times 100 = 125\).
In May, the profit decreased by 20%. This decrease is applied to the profit at the end of April (\(1.25P\)).
\[ \text{Profit}_{\text{May end}} = 1.25P - (0.20 \times 1.25P) \] \[ \text{Profit}_{\text{May end}} = 1.25P (1 - 0.20) \] \[ \text{Profit}_{\text{May end}} = 1.25P \times 0.80 \] \[ \text{Profit}_{\text{May end}} = 1.00P \]If we use the assumed value of 125:
\[ \text{Profit}_{\text{May end}} = 125 - (0.20 \times 125) \] \[ \text{Profit}_{\text{May end}} = 125 - 25 \] \[ \text{Profit}_{\text{May end}} = 100 \]The profit at the end of May is \(1.00P\) (or 100), which is the same as the initial profit \(P\) (or 100) at the beginning of April.
Correct Option: C) It was same.
This problem involves calculating work done by individuals and then by a team, followed by the remaining work done by one individual. We will find the per-day work rate of A and B, calculate the total work done when they work together, find the remaining work, and finally determine the time B takes to complete the remaining work.
Calculate the one-day work of A:
A can finish the work in 20 days. So, A's one day's work is:
\[ \text{A's 1 day work} = \frac{1}{20} \text{ of the total work} \]Calculate the one-day work of B:
B can finish the work in 24 days. So, B's one day's work is:
\[ \text{B's 1 day work} = \frac{1}{24} \text{ of the total work} \]Calculate their combined one-day work:
When A and B work together, their combined one day's work is:
\[ \text{(A + B)'s 1 day work} = \frac{1}{20} + \frac{1}{24} \]To add these fractions, find a common denominator, which is 120:
\[ \text{(A + B)'s 1 day work} = \frac{6}{120} + \frac{5}{120} = \frac{11}{120} \text{ of the total work} \]Calculate the work done by A and B in 10 days:
They work together for 10 days. So, the work completed in 10 days is:
\[ \text{Work done in 10 days} = 10 \times \frac{11}{120} = \frac{110}{120} = \frac{11}{12} \text{ of the total work} \]Calculate the remaining work:
The total work is considered as 1 unit. So, the remaining work is:
\[ \text{Remaining work} = 1 - \frac{11}{12} = \frac{12 - 11}{12} = \frac{1}{12} \text{ of the total work} \]Calculate the time B takes to finish the remaining work:
B's one day's work is \(\frac{1}{24}\). To find the number of days B will take to complete the remaining \(\frac{1}{12}\) of the work, we divide the remaining work by B's per-day work rate:
\[ \text{Days for B} = \frac{\text{Remaining work}}{\text{B's 1 day work}} = \frac{\frac{1}{12}}{\frac{1}{24}} \] \[ \text{Days for B} = \frac{1}{12} \times \frac{24}{1} = \frac{24}{12} = 2 \text{ days} \]Correct Option: D) 2 days
To calculate the compound interest, we first need to find the total amount after 2 years using the compound interest formula. Then, we subtract the principal amount from the total amount to get the compound interest.
The formula for the amount when interest is compounded annually is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]So, the total amount after 2 years is ₹28,090.
Compound Interest = Amount - Principal
\[ \text{CI} = A - P \] \[ \text{CI} = 28090 - 25000 \] \[ \text{CI} = 3090 \]The compound interest is ₹3,090.
Correct Option: B) ₹3,090
The question asks to identify the British chemist who proposed the atomic theory in 1808, incorporating the laws of conservation of mass and definite proportions. We need to recall the key figures in the history of atomic theory and their contributions.
A) John Dalton is the correct answer. He was a British chemist who presented his atomic theory in 1808, which explained the Law of Conservation of Mass and the Law of Definite Proportions. His theory was a significant milestone in the development of chemistry.