To convert centimeters (cm) to kilometers (km), we need to use the standard conversion factors between these units of length. We know that 1 meter equals 100 centimeters and 1 kilometer equals 1000 meters. We will use these relationships to convert the given value step-by-step.
We know that \(1 \text{ m} = 100 \text{ cm}\). To convert 36 cm to meters, we divide by 100:
\[ 36 \text{ cm} = \frac{36}{100} \text{ m} = 0.36 \text{ m} \]We know that \(1 \text{ km} = 1000 \text{ m}\). To convert 0.36 m to kilometers, we divide by 1000:
\[ 0.36 \text{ m} = \frac{0.36}{1000} \text{ km} \]Dividing 0.36 by 1000 means moving the decimal point three places to the left:
\[ \frac{0.36}{1000} = 0.00036 \text{ km} \]C) 0.00036 km is the correct conversion of 36 cm to kilometers.
This question tests knowledge of Newlands' Law of Octaves, an early attempt at classifying elements. The key is to recall the specific elements that marked the beginning and end of this classification system as proposed by Newlands.
B) hydrogen, thorium. Newlands started his classification with hydrogen and ended it with thorium, which was the 56th element known at that time.
To find the sixth number, we will use the concept of averages. The sum of numbers can be calculated by multiplying the average by the count of numbers. We will find the total sum of all 11 numbers, the sum of the first five numbers, and the sum of the last five numbers. The sixth number will be the total sum minus the sum of the first five and the sum of the last five numbers.
A) 41 is the correct answer. The calculation shows that the sixth number is 41.
To find the ratio of Gold to Platinum in the third alloy Z, we need to calculate the total quantity of Gold and the total quantity of Platinum when equal quantities of alloys X and Y are melted together. We will first determine the proportion of Gold and Platinum in each alloy and then sum them up.
Alloy X: The ratio of Gold to Platinum is 5:2. The total parts are \(5+2=7\).
Alloy Y: The ratio of Gold to Platinum is 5:7. The total parts are \(5+7=12\).
Melting equal quantities: Let's assume we take 1 unit of Alloy X and 1 unit of Alloy Y. (The actual quantity doesn't matter as long as they are equal, the ratio will remain the same. We can also choose a common multiple of 7 and 12, like 84, to avoid fractions, but working with fractions is also fine).
Total Gold in Alloy Z: Sum the quantities of Gold from Alloy X and Alloy Y.
\[ \text{Total Gold} = \frac{5}{7} + \frac{5}{12} \]\[ \text{Total Gold} = \frac{5 \times 12 + 5 \times 7}{7 \times 12} \]\[ \text{Total Gold} = \frac{60 + 35}{84} \]\[ \text{Total Gold} = \frac{95}{84} \]Total Platinum in Alloy Z: Sum the quantities of Platinum from Alloy X and Alloy Y.
\[ \text{Total Platinum} = \frac{2}{7} + \frac{7}{12} \]\[ \text{Total Platinum} = \frac{2 \times 12 + 7 \times 7}{7 \times 12} \]\[ \text{Total Platinum} = \frac{24 + 49}{84} \]\[ \text{Total Platinum} = \frac{73}{84} \]Ratio of Gold to Platinum in Alloy Z:
\[ \text{Ratio} = \frac{\text{Total Gold}}{\text{Total Platinum}} = \frac{\frac{95}{84}}{\frac{73}{84}} \]\[ \text{Ratio} = \frac{95}{73} \]Thus, the ratio of Gold to Platinum in Alloy Z is 95:73.B) 95:73 is the correct ratio of Gold to Platinum in alloy Z, as calculated by summing the proportions of each metal from the two alloys.
The problem asks us to identify the number of symbols in the given series that satisfy two conditions simultaneously: they must be immediately preceded by a number and immediately followed by a letter. We need to scan the series and check each symbol against these conditions.
Therefore, the total number of such symbols is 1.
C) 1 — As identified in the step-by-step analysis, only the symbol '*' is immediately preceded by a number (9) and immediately followed by a letter (M).
The question asks to identify the chairman of the Planning Commission at the time of the First Five Year Plan's inception. This requires knowledge of the history of India's Five Year Plans and the structure of the Planning Commission.
Correct Option: C) Jawaharlal Nehru was the first Prime Minister of India and, by virtue of his office, the ex-officio Chairman of the Planning Commission from its inception in 1950 and throughout the First Five Year Plan (1951-1956).
To find the total number of students in a class when a person's rank from both the top and the bottom is given, we use a simple formula. The key is to understand that the person themselves is counted in both rankings, so we need to subtract one to avoid double-counting.
Correct Option: C) 62
The question describes the preparation and properties of a chemical compound X. It is prepared by heating gypsum, is a white powder, and is used as a fireproofing material. We need to identify compound X from the given options.
Correct Option: C) plaster of Paris
In the following decomposition reaction, identify the p, q, r and s values:
p FeSO4 (s) → q Fe2O3(s) + r SO2(g) + s SO3(g)
To identify the stoichiometric coefficients p, q, r, and s in the given decomposition reaction, we need to balance the chemical equation. This involves ensuring that the number of atoms of each element is the same on both the reactant and product sides of the equation.
Correct Option: D) 2, 1, 1, 1
To find the ratio of work efficiency between a man and a woman, we need to equate the total work done by women and men, as they complete the same work in the same number of days. The total work is calculated as the product of the number of workers, their individual efficiency, and the number of days.
Given that 20 women complete the work in 15 days.
\[ \text{Total Work} = \text{Number of Women} \times \text{Efficiency of one Woman} \times \text{Number of Days} \] \[ \text{Total Work} = 20 \times W_e \times 15 = 300 W_e \quad \text{(Equation 1)} \]Given that 16 men complete the same work in 15 days.
\[ \text{Total Work} = \text{Number of Men} \times \text{Efficiency of one Man} \times \text{Number of Days} \] \[ \text{Total Work} = 16 \times M_e \times 15 = 240 M_e \quad \text{(Equation 2)} \]Since the work is the same in both cases, we can equate Equation 1 and Equation 2:
\[ 300 W_e = 240 M_e \]We need to find the ratio \(M_e : W_e\), which is \(\frac{M_e}{W_e}\).
From the equation \(300 W_e = 240 M_e\), we can rearrange it to find the ratio:
\[ \frac{M_e}{W_e} = \frac{300}{240} \]Simplify the fraction:
\[ \frac{M_e}{W_e} = \frac{30}{24} = \frac{5 \times 6}{4 \times 6} = \frac{5}{4} \]So, the ratio of the work efficiency of a man to a woman is \(5:4\).