To multiply decimals, we can first multiply the numbers as if they were whole numbers, ignoring the decimal points. Then, we count the total number of decimal places in the original numbers and place the decimal point in the product accordingly.
We can write \(6.5\) as \(\frac{65}{10}\) and \(0.0004\) as \(\frac{4}{10000}\).
\[ \frac{65}{10} \times \frac{4}{10000} = \frac{65 \times 4}{10 \times 10000} = \frac{260}{100000} \]
\[ \frac{260}{100000} = \frac{26}{10000} \]
To convert \(\frac{26}{10000}\) to a decimal, we place the decimal point such that there are 4 digits after it (because there are 4 zeros in the denominator).
So, \(\frac{26}{10000} = 0.0026\).
Multiply \(65\) by \(4\):
\[ 65 \times 4 = 260 \]
Count the decimal places in the original numbers:
Total number of decimal places in the product will be \(1 + 4 = 5\).
Starting from the right of \(260\), move the decimal point 5 places to the left. We need to add leading zeros:
\[ 260 \rightarrow 0.00260 \]
The trailing zero can be removed, so the result is \(0.0026\).
Correct Option: C) 0.0026
This problem involves work and time. We are given the combined time taken by Milena and Mashuka to complete a work, and also a scenario where they work separately for parts of the work, leading to a total time. We need to find the time Milena takes to complete the work alone, given that Mashuka is more efficient.
Let the time taken by Milena to complete the work alone be \(M\) days and by Mashuka alone be \(H\) days.
The work rate of Milena is \(\frac{1}{M}\) and Mashuka is \(\frac{1}{H}\).
When they work together, they complete the work in 36 days. So, their combined work rate is \(\frac{1}{36}\).
\[ \frac{1}{M} + \frac{1}{H} = \frac{1}{36} \quad \text{(Equation 1)} \]
In the second scenario, Mashuka completes \(\frac{2}{5}\) of the work. Let \(t_H\) be the time Mashuka worked.
\[ \frac{t_H}{H} = \frac{2}{5} \implies t_H = \frac{2H}{5} \]
Milena completes the remaining work, which is \[1 - \frac{2}{5} = \frac{3}{5}\] of the work. Let \(t_M\) be the time Milena worked.
\[ \frac{t_M}{M} = \frac{3}{5} \implies t_M = \frac{3M}{5} \]
The total time taken in this scenario is 78 days.
\[ t_H + t_M = 78 \]
\[ \frac{2H}{5} + \frac{3M}{5} = 78 \]
\[ 2H + 3M = 390 \quad \text{(Equation 2)} \]
From Equation 1, we can express \(H\) in terms of \(M\):
\[ \frac{1}{H} = \frac{1}{36} - \frac{1}{M} = \frac{M - 36}{36M} \]
\[ H = \frac{36M}{M - 36} \]
Substitute this expression for \(H\) into Equation 2:
\[ 2\left(\frac{36M}{M - 36}\right) + 3M = 390 \]
\[ \frac{72M}{M - 36} + 3M = 390 \]
Divide the entire equation by 3 to simplify:
\[ \frac{24M}{M - 36} + M = 130 \]
\[ 24M + M(M - 36) = 130(M - 36) \]
\[ 24M + M^2 - 36M = 130M - 4680 \]
\[ M^2 - 12M = 130M - 4680 \]
\[ M^2 - 142M + 4680 = 0 \]
Solve this quadratic equation for \(M\). We can use the quadratic formula or factorization.
We look for two numbers that multiply to 4680 and add up to 142. Let's try factors of 4680. Notice that \(102 \times 40 = 4080\) and \(102 \times 42 = 4284\). Let's try \(90 \times 52 = 4680\). And \(90 + 52 = 142\). So, the factors are 90 and 52.
\[ (M - 90)(M - 52) = 0 \]
So, \(M = 90\) or \(M = 52\).
Now we need to check the condition that Mashuka is more efficient than Milena. This means Mashuka takes less time than Milena to complete the work alone, i.e., \(H < M\).
Case 1: If \(M = 90\) days
\[ H = \frac{36 \times 90}{90 - 36} = \frac{36 \times 90}{54} = \frac{2 \times 90}{3} = 2 \times 30 = 60 \text{ days} \]
In this case, \(H = 60\) days and \(M = 90\) days. Since \(60 < 90\), Mashuka is more efficient than Milena. This solution is consistent with the given condition.
Case 2: If \(M = 52\) days
\[ H = \frac{36 \times 52}{52 - 36} = \frac{36 \times 52}{16} = \frac{9 \times 52}{4} = 9 \times 13 = 117 \text{ days} \]
In this case, \(H = 117\) days and \(M = 52\) days. Since \(117 > 52\), Milena is more efficient than Mashuka. This contradicts the given condition that Mashuka is more efficient than Milena.
Therefore, the only valid solution is \(M = 90\) days.
Correct Option: C) 90
To find the number of twin prime pairs between 100 and 200, we first need to understand what twin primes are. A twin prime pair consists of two prime numbers that differ by 2. We will list all prime numbers between 100 and 200 and then identify the pairs that satisfy the twin prime condition.
C) There are 7 twin prime pairs between 100 and 200: (101, 103), (107, 109), (137, 139), (149, 151), (179, 181), (191, 193), and (197, 199).
The question asks about Saikhom Mirabai Chanu's achievement at a specific World Weightlifting Championships. To answer this, one needs to recall or know the historical performance of the athlete in that particular event.
Correct Option: D) Gold Medal
The question asks for the year in which the Right to Education (RTE) Act came into force. This is a factual question related to Indian legislation and its implementation date.
Correct Option: B) 2010. The Right to Education Act, 2009, came into force on 1st April 2010. This Act mandates free and compulsory education for all children between the ages of 6 and 14 years in India.
The question asks to identify the type of protein responsible for muscle movement through compression and relaxation. This points directly to the fundamental mechanism of muscle contraction.
Correct Option: A) Contractile protein is compressed and relaxed, causing the muscles to move. Contractile proteins, primarily actin and myosin, are responsible for the mechanical work of muscle contraction and relaxation.
The mass of the earth is _____.
6 × 10-24 kg
6 × 10-23 kg
6 × 1023 kg
6 × 1024 kg
This question asks for the approximate mass of the Earth. This is a fundamental constant in physics and astronomy that candidates for competitive exams are expected to know or be able to recognize from given options.
Correct Option: D) The mass of the Earth is approximately \(6 \times 10^{24} \text{ kg}\). This value is a standard astronomical constant.
This question tests knowledge about the chemical composition of cell walls in different biological organisms. Specifically, it asks about the primary component of fungal cell walls.
Correct Option: B) Chitin
The cell walls of fungi are primarily composed of chitin. Chitin is a strong, nitrogen-containing polysaccharide, similar in structure to cellulose, but with acetylglucosamine units instead of glucose units. This provides structural rigidity and protection to the fungal cell.
The question asks to identify the Chief Minister of Goa as of September 2018. This requires knowledge of political officeholders during that specific period.
Correct Option: C) Manohar Parrikar
To determine if a number is divisible by 6, we need to check two conditions: divisibility by 2 and divisibility by 3. A number is divisible by 6 if and only if it is divisible by both 2 and 3.
Only 12384 is divisible by both 2 and 3.
A) 12384 is divisible by 2 (ends in 4) and the sum of its digits (\(1+2+3+8+4=18\)) is divisible by 3. Therefore, 12384 is divisible by 6.