\( \frac{11}{13} \) + \( \frac{13}{11} \) =?
\( \frac{290}{143} \)
\( \frac{24}{143} \)
2
\( \frac{245}{24} \)
To add two fractions with different denominators, we need to find a common denominator. The easiest way to do this is to find the least common multiple (LCM) of the denominators. In this case, since the denominators are prime numbers relative to each other (11 and 13), their LCM is simply their product.
Identify the fractions to be added:
\[ \frac{11}{13} + \frac{13}{11} \]Find the common denominator. The denominators are 13 and 11. Since 11 and 13 are prime numbers, their least common multiple (LCM) is their product:
\[ \text{LCM}(13, 11) = 13 \times 11 = 143 \]Convert each fraction to an equivalent fraction with the common denominator (143).
For the first fraction, \( \frac{11}{13} \), multiply the numerator and denominator by 11:
For the second fraction, \( \frac{13}{11} \), multiply the numerator and denominator by 13:
\[ \frac{13}{11} = \frac{13 \times 13}{11 \times 13} = \frac{169}{143} \]Add the two new fractions:
\[ \frac{121}{143} + \frac{169}{143} = \frac{121 + 169}{143} \]Perform the addition in the numerator:
\[ 121 + 169 = 290 \]The final sum is:
\[ \frac{290}{143} \]A) \( \frac{290}{143} \) is the correct result obtained from the addition of the two fractions.
The question asks for the city where City Montessori School, known for its record-breaking student enrollment, is located. This is a factual recall question related to general knowledge about famous institutions in India.
C) Lucknow is the correct answer. City Montessori School, a prominent educational institution recognized by the Guinness Book of World Records for its large student body, is headquartered in Lucknow, Uttar Pradesh.
This is a direction-based reasoning question. We need to track the cow's initial direction, its turn, and then determine its final facing direction. Visualizing the directions or drawing a simple diagram can be helpful.
Correct Option: A) North
The problem involves calculating the amount of zinc to be added to a German silver alloy to change the ratio of its components. We will first determine the initial quantities of copper, zinc, and nickel in the 54 kg alloy. Then, we will set up equations based on the new ratio after adding zinc, keeping in mind that the amounts of copper and nickel remain unchanged.
Calculate initial quantities of copper, zinc, and nickel:
The initial ratio of copper : zinc : nickel is 4 : 3 : 2.
Total parts in the initial ratio \( = 4 + 3 + 2 = 9 \).
Total mass of the alloy \( = 54 \text{ kg} \).
Mass of copper \[= \frac{4}{9} \times 54 \text{ kg} = 4 \times 6 \text{ kg} = 24 \text{ kg}\].
Mass of zinc \[= \frac{3}{9} \times 54 \text{ kg} = 3 \times 6 \text{ kg} = 18 \text{ kg}\].
Mass of nickel \[= \frac{2}{9} \times 54 \text{ kg} = 2 \times 6 \text{ kg} = 12 \text{ kg}\].
Set up the new ratio:
Let \(x\) be the amount of zinc added in kg.
The new mass of zinc \( = (18 + x) \text{ kg} \).
The masses of copper and nickel remain unchanged: Copper \( = 24 \text{ kg} \), Nickel \( = 12 \text{ kg} \).
The new ratio of copper : zinc : nickel is 2 : 5 : 1.
We can use the ratio of copper to nickel to find a common multiplier for the new ratio:
\[ \frac{\text{Copper}}{\text{Nickel}} = \frac{24 \text{ kg}}{12 \text{ kg}} = \frac{2}{1} \]Calculate the new amount of zinc:
From the new ratio 2 : 5 : 1, if copper is 2 parts and nickel is 1 part, then zinc is 5 parts.
Since the actual amount of copper is 24 kg (which is \(2 \times 12\)), and nickel is 12 kg (which is \(1 \times 12\)), the multiplier for the new ratio is 12.
Therefore, the new amount of zinc should be \(5 \times 12 \text{ kg} = 60 \text{ kg} \).
Determine the amount of zinc added:
The new mass of zinc \( = 60 \text{ kg} \).
The initial mass of zinc \( = 18 \text{ kg} \).
Amount of zinc added \( = \text{New mass of zinc} - \text{Initial mass of zinc} \)
Amount of zinc added \( = 60 \text{ kg} - 18 \text{ kg} = 42 \text{ kg} \).
D) 42
The calculations show that 42 kg of zinc needs to be added to achieve the desired new ratio.
A) 48 — This value would result in a different final ratio. If 48 kg of zinc were added, the new zinc amount would be \(18 + 48 = 66 \text{ kg}\). The ratio would be \(24 : 66 : 12\), which simplifies to \(4 : 11 : 2\), not \(2 : 5 : 1\).
B) 50 — If 50 kg of zinc were added, the new zinc amount would be \(18 + 50 = 68 \text{ kg}\). The ratio would be \(24 : 68 : 12\), which simplifies to \(6 : 17 : 3\), not \(2 : 5 : 1\).
C) 36 — If 36 kg of zinc were added, the new zinc amount would be \(18 + 36 = 54 \text{ kg}\). The ratio would be \(24 : 54 : 12\), which simplifies to \(4 : 9 : 2\), not \(2 : 5 : 1\).
This problem involves calculating the combined work rate of two pipes, one filling and one emptying, to determine the time taken to fill a specific portion of a tank. We will first find the individual rates of filling and emptying, then their net combined rate, and finally use this net rate to calculate the time required to fill half the tank.
Pipe M fills the tank in 45 hours. So, in 1 hour, Pipe M fills:
\[ \text{Rate}_M = \frac{1}{45} \text{ of the tank per hour} \]Pipe N empties the tank in 90 hours. So, in 1 hour, Pipe N empties:
\[ \text{Rate}_N = \frac{1}{90} \text{ of the tank per hour} \]Since Pipe M fills and Pipe N empties, their combined effect is the difference between their rates:
\[ \text{Net Rate} = \text{Rate}_M - \text{Rate}_N = \frac{1}{45} - \frac{1}{90} \]To subtract, find a common denominator, which is 90:
\[ \text{Net Rate} = \frac{2}{90} - \frac{1}{90} = \frac{2-1}{90} = \frac{1}{90} \text{ of the tank per hour} \]This means that when both pipes are open, the tank fills at a rate of 1/90 of the tank per hour.
If the net rate is \(1/90\) of the tank per hour, then the time taken to fill the entire tank (1 unit) is the reciprocal of the net rate:
\[ \text{Time to fill full tank} = \frac{1}{\text{Net Rate}} = \frac{1}{1/90} = 90 \text{ hours} \]The question asks for the time to fill a half-empty tank completely. This means we need to fill \(1/2\) of the tank.
\[ \text{Time to fill half tank} = \frac{1}{2} \times \text{Time to fill full tank} \] \[ \text{Time to fill half tank} = \frac{1}{2} \times 90 \text{ hours} = 45 \text{ hours} \]Correct Option: C) 45 hours
This is a logical deduction problem. We need to match the four couples and their clothing colors based on the given statements. We will create a table or list to keep track of the information as we deduce it.
| Man | Shirt Color | Woman | Sari Color |
|---|---|---|---|
| Ramesh | Red | Rama | ? |
| Anup | Green | Shreya | Green |
| Sandesh | ? | Mrunmayi | Not Yellow |
| Viraj | Yellow | Purva | Blue |
A) Anup is Shreya's husband, as deduced from Statement 3 where they are both wearing green clothes, implying they are a couple.
To separate a mixture of two solids, we need to identify a property that is significantly different between them. In the case of ammonium chloride and sodium chloride, ammonium chloride has a unique property called sublimation, which can be used for separation.
Correct Option: B) Sublimation is the process where a solid changes directly into a gas without passing through the liquid state. Ammonium chloride is a sublimable substance, while sodium chloride is not. Therefore, heating a mixture of ammonium chloride and sodium chloride will cause the ammonium chloride to sublime, allowing it to be separated from the sodium chloride.
What is the molecular formula of isobutane?
The question asks for the molecular formula of isobutane. Isobutane is an isomer of butane, meaning it has the same molecular formula but a different structural arrangement. Both butane and isobutane are alkanes, which follow the general formula \(C_nH_{2n+2}\).
B) C 4 H 10
Isobutane is an isomer of butane. Alkanes follow the general formula \(C_nH_{2n+2}\). For butane, \(n=4\), so the molecular formula is \(C_4H_{2(4)+2} = C_4H_{10}\).
163 ÷ 83 ÷ 22 =?
The problem involves simplifying an expression with exponents and division. We will use the properties of exponents, specifically the rule for division of powers with the same base: \(a^m \div a^n = a^{m-n}\), and the rule for power of a power: \((a^m)^n = a^{mn}\). We will convert all numbers to a common base, which is 2 in this case, and then apply the exponent rules.
B) \(2^1\) is the correct answer, as derived from the step-by-step simplification.
The question asks to identify the block of elements to which Lanthanides and Actinides belong in the periodic table. This requires knowledge of the periodic table's structure and the classification of elements based on their electronic configurations.
B) f - Block element
Lanthanides and Actinides are characterized by the filling of their f-orbitals (4f for lanthanides and 5f for actinides). Therefore, they are classified as f-block elements.