Find the value of \(\frac{0.5\times0.5+0.09-0.15}{0.125+0.027}\)
\( \frac{4}{5} \)
\( \frac{5}{4} \)
\( \frac{5}{6} \)
\( \frac{3}{4} \)
The problem requires us to evaluate a given expression involving decimal numbers. We will simplify the numerator and the denominator separately and then perform the division.
The numerator is \(0.5 \times 0.5 + 0.09 - 0.15\).
So, the numerator is \(0.19\).
The denominator is \(0.125 + 0.027\).
So, the denominator is \(0.152\).
Now, we need to calculate the value of the expression:
\[ \frac{0.19}{0.152} \]B) The simplified value of the expression is \(\frac{5}{4}\).
The question asks for the location of the headquarters of the Botanical Survey of India (BSI). This is a factual question requiring knowledge of the administrative structure of scientific organizations in India.
Correct Option: C) Kolkata is the correct answer. The headquarters of the Botanical Survey of India (BSI) is situated in Kolkata, West Bengal. It was established in 1890 and plays a crucial role in the exploration, inventorisation, and conservation of plant diversity in India.
To find the greatest possible length that can measure exactly the given lengths, we need to find the Highest Common Factor (HCF) of these lengths. The HCF is the largest number that divides each of the given numbers without leaving a remainder.
Correct Option: A) 20 cm is the HCF of 220, 260, and 300, meaning it can exactly measure all three lengths.
2 October 2019 marks the _____ birth anniversary of Mahatma Gandhi.
To determine the birth anniversary, we need to know Mahatma Gandhi's birth year and then calculate the number of years passed until 2019.
Correct Option: A) 150 th
The problem is a coding-decoding question. We need to identify the pattern used to transform "DOOR" into "FRSW" and then apply the same pattern to "FAN" to find its coded form.
Correct Option: A) HDR is the correct code for FAN, following the identified pattern of +2, +3, +4 shifts for each successive letter.
'Startup India' is a flagship initiative of the Government of India, intended to catalyse startup culture and build a strong and inclusive ecosystem for innovation and entrepreneurship in India. This flagship initiative was launched on:
26th January 2016
16th January 2015
26th January 2015
16th January 2016
The question asks for the launch date of the "Startup India" initiative. This is a factual question requiring knowledge of important government schemes and their launch dates. We need to identify the correct date from the given options.
Correct Option: D) 16 th January 2016
The "Startup India" initiative was indeed launched on January 16, 2016, by Prime Minister Narendra Modi. This initiative aims to build a strong ecosystem for nurturing innovation and startups in the country.
If a polynomial \( P(x) \) has a factor \( (x-k) \), then \( P(k) = 0 \) according to the Factor Theorem. Similarly, if \( (x^2 - 1) \) is a factor of a polynomial, then the roots of \( x^2 - 1 = 0 \) must also be roots of the given polynomial. The roots of \( x^2 - 1 = 0 \) are \( x = 1 \) and \( x = -1 \). We will substitute these values into the given polynomial and set the results to zero to find the relationship between the coefficients.
Let the given polynomial be \( P(x) = ax^{4} + bx^{3} + cx^{2} + dx + e \).
The factor is \( x^{2} - 1 \). We can write this as \( (x-1)(x+1) \). This means that \( x=1 \) and \( x=-1 \) are the roots of \( x^{2} - 1 = 0 \).
According to the Factor Theorem, if \( (x-1) \) is a factor of \( P(x) \), then \( P(1) = 0 \).
Substitute \( x=1 \) into \( P(x) \):
\[ P(1) = a(1)^{4} + b(1)^{3} + c(1)^{2} + d(1) + e = 0 \]\[ a + b + c + d + e = 0 \quad \text{(Equation 1)} \]
Similarly, if \( (x+1) \) is a factor of \( P(x) \), then \( P(-1) = 0 \).
Substitute \( x=-1 \) into \( P(x) \):
\[ P(-1) = a(-1)^{4} + b(-1)^{3} + c(-1)^{2} + d(-1) + e = 0 \]\[ a - b + c - d + e = 0 \quad \text{(Equation 2)} \]
Now, we have two equations:
(1) \( a + b + c + d + e = 0 \)
(2) \( a - b + c - d + e = 0 \)
Add Equation 1 and Equation 2:
\[ (a + b + c + d + e) + (a - b + c - d + e) = 0 + 0 \]\[ 2a + 2c + 2e = 0 \]\[ 2(a + c + e) = 0 \]\[ a + c + e = 0 \quad \text{(Equation 3)} \]
Subtract Equation 2 from Equation 1:
\[ (a + b + c + d + e) - (a - b + c - d + e) = 0 - 0 \]\[ a + b + c + d + e - a + b - c + d - e = 0 \]\[ 2b + 2d = 0 \]\[ 2(b + d) = 0 \]\[ b + d = 0 \quad \text{(Equation 4)} \]
From Equation 3, we have \( a + c + e = 0 \).
From Equation 4, we have \( b + d = 0 \). This implies \( b = -d \).
Since both \( a + c + e \) and \( b + d \) are equal to 0, we can equate them:
\[ a + c + e = b + d \]
A) \( a + c + e = b + d \). This matches the relation derived in step 9.
The question asks for the venue of the first India International Science Festival (IISF) held in December 2015. This is a factual recall question related to significant national events.
The first India International Science Festival (IISF) was organized by the Ministry of Science and Technology and Ministry of Earth Sciences, in association with Vijnana Bharati (VIBHA), and was held from December 4-8, 2015.
The primary venue for this inaugural festival was the Indian Institute of Technology (IIT) Delhi.
A) IIT Delhi is the correct answer. The first India International Science Festival (IISF) in December 2015 was hosted at IIT Delhi.
The question asks for the full form of 'PARIVESH', an environmental clearance system launched in August 2018. This requires knowledge of government initiatives and their acronyms.
Correct Option: B) Pro-Active and Responsive facilitation by Interactive, Virtuous and Environmental Single-window Hub
PARIVESH (Pro-Active and Responsive facilitation by Interactive, Virtuous and Environmental Single-window Hub) is a web-based, role-based workflow application that has been developed for online submission and monitoring of proposals submitted by proponents for seeking Environment, Forest, Wildlife and CRZ Clearances from Central, State and District Level authorities.
To find the distance between the college and the hostel, we can use the relationship between distance, speed, and time. Since the distance is the same for both going and coming, we can set up an equation based on the total time taken.
Let the distance between the college and the hostel be \(D\) km.
When the student goes to the college, his speed is \(S_1 = 4\) km/h.
The time taken to go to college is \[T_1 = \frac{D}{S_1} = \frac{D}{4}\] hours.
When the student returns from the college, his speed is \(S_2 = 3\) km/h.
The time taken to return from college is \[T_2 = \frac{D}{S_2} = \frac{D}{3}\] hours.
The total time taken for going and coming is given as 7 hours.
So, \(T_1 + T_2 = 7\).
Substitute the expressions for \(T_1\) and \(T_2\) into the total time equation:
\[ \frac{D}{4} + \frac{D}{3} = 7 \]To solve for \(D\), find a common denominator for the fractions, which is 12:
\[ \frac{3D}{12} + \frac{4D}{12} = 7 \]Combine the fractions:
\[ \frac{3D + 4D}{12} = 7 \]\[ \frac{7D}{12} = 7 \]Multiply both sides by 12:
\[ 7D = 7 \times 12 \]\[ 7D = 84 \]Divide by 7 to find \(D\):
\[ D = \frac{84}{7} \]\[ D = 12 \text{ km} \]A) 12 km is the correct distance between the college and the hostel.