To find the number of moles of magnesium in a given mass of magnesium carbonate (MgCO₃), we first need to calculate the molar mass of MgCO₃. Then, we can determine the number of moles of MgCO₃. Since there is one magnesium atom per molecule of MgCO₃, the number of moles of magnesium will be equal to the number of moles of MgCO₃.
Correct Option: C) 0.25
To find the cost price, we can use the formula relating selling price, cost price, and profit percentage. If the profit is 6%, it means the selling price is 106% of the cost price. We can set up an equation to solve for the cost price.
The selling price is the cost price plus the profit. If the profit is a percentage of the cost price, the formula is:
\[ \text{SP} = \text{CP} \times \left(1 + \frac{\text{P%}}{100}\right) \]To find the cost price, divide the selling price by 1.06:
\[ \text{CP} = \frac{689}{1.06} \]\[ \text{CP} = 650 \]The cost price of the washing machine was Rs 650.
C) 650 is the correct cost price, calculated by dividing the selling price (Rs 689) by (1 + 0.06).
The question asks to identify where the reproductive parts of angiosperms are located. Angiosperms are flowering plants, and their reproduction involves specific structures.
Correct Option: B) Flowers
The reproductive parts of angiosperms, namely the stamens (male) and carpels/pistils (female), are located within the flower. The flower is the primary reproductive organ of angiosperms, responsible for producing seeds through sexual reproduction.
The question asks to identify the word that does not belong to the group formed by the other words. This is a classification type of verbal reasoning question. We need to find a common characteristic shared by most of the words and then identify the outlier.
Correct Option: A) D is the correct option because "Think" refers to a mental activity, whereas "Ride," "Crawl," and "Swim" are all verbs describing physical forms of movement or locomotion.
The question asks to identify the specific mobile application launched by the Government of India for road safety and highway users. This requires knowledge of government initiatives related to road transport and digital services.
Correct Option: D) Sukhad Yatra
Find the odd one out.
| A | B | C | D |
| 6F | P14 | U21 | W23 |
The mirror formula is ______.
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{f}+\frac{1}{v}=\frac{1}{u}\)
\(\frac{1}{f}+\frac{1}{u}=\frac{1}{v}\)
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
The question asks for the correct mirror formula. The mirror formula is a fundamental equation in optics that relates the object distance, image distance, and focal length of a spherical mirror (concave or convex).
A) \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\]
This is the universally accepted mirror formula that correctly relates the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) of a spherical mirror.
To find the average speed of an object, we need to calculate the total distance traveled and divide it by the total time taken. The problem provides two segments of travel with their respective distances and times.
The object travels 15 m in the first segment and 16 m in the second segment.
\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 \]\[ \text{Total Distance} = 15 \text{ m} + 16 \text{ m} = 31 \text{ m} \]The object takes 4 s for the first segment and 2 s for the second segment.
\[ \text{Total Time} = \text{Time}_1 + \text{Time}_2 \]\[ \text{Total Time} = 4 \text{ s} + 2 \text{ s} = 6 \text{ s} \]Average speed is defined as the total distance divided by the total time.
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]\[ \text{Average Speed} = \frac{31 \text{ m}}{6 \text{ s}} \]\[ \text{Average Speed} \approx 5.1666... \text{ m/s} \]Rounding to two decimal places, the average speed is approximately 5.17 m/s. The unit for speed is meters per second, which can also be written as \( \text{ms}^{-1} \).
Correct Option: B) 5.17 ms⁻¹
The width of each of the continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. There are 7 classes in all. What is the upper-class limit of the highest class?
To find the upper-class limit of the highest class, we need to determine all the class intervals. We are given the width of each class, the lower limit of the lowest class, and the total number of classes. We can construct the class intervals sequentially.
Identify given information:
Determine the first class interval:
Since the lower class limit of the lowest class is 10 and the class width is 5, the first class interval is \(10 - (10+5)\), which is \(10 - 15\).
List all class intervals:
We have 7 classes. We can list them by adding the class width to the previous upper limit to get the next lower limit, and then adding the class width to that lower limit to get the new upper limit.
Identify the upper-class limit of the highest class:
From the list above, the highest class is Class 7, which is \(40 - 45\). The upper-class limit of this class is 45.
Lower limit of the highest class:
The lower limit of the highest (7th) class can be found by adding the class width \((N-1)\) times to the lower limit of the first class.
\[ \text{Lower limit of highest class} = \text{Lower limit of 1st class} + (\text{Number of classes} - 1) \times \text{Class width} \]\[ = 10 + (7 - 1) \times 5 \]\[ = 10 + 6 \times 5 \]\[ = 10 + 30 \]\[ = 40 \]Upper limit of the highest class:
The upper limit of the highest class is its lower limit plus the class width.
\[ \text{Upper limit of highest class} = \text{Lower limit of highest class} + \text{Class width} \]\[ = 40 + 5 \]\[ = 45 \]Correct Option: A) 45
What day would it be on 19th April 2020?
To determine the day of the week for a given date, we use the concept of odd days. Odd days are the number of days remaining after dividing the total number of days by 7. We calculate the total odd days from a reference point (e.g., year 0 or a known date) up to the day before the target date, and then use a day code to find the day of the week.
Reference Date: We know that 1st January 0001 was a Monday. However, it's more common and easier to use the fact that 1st January 1 AD was a Monday. For competitive exams, often a common reference like 1st January 1 AD being Monday or Sunday is used. Let's use the standard method of calculating odd days from the year 0, where 0000 was a Sunday (or 0 odd days).
Odd days in centuries:
Odd days up to 2000 years: \[ \text{Odd days in 2000 years} = \text{Odd days in } (4 \times 500) \text{ years} = 0 \text{ odd days} \]
Odd days in 2019 years (from 2000 to 2019): We need to calculate odd days for the years 2001 to 2019. This is 19 years. Number of leap years in this period (2004, 2008, 2012, 2016) = 4. Number of ordinary years = \(19 - 4 = 15\). \[ \text{Odd days in 19 years} = (4 \times 2) + (15 \times 1) = 8 + 15 = 23 \text{ odd days} \] \[ 23 \div 7 = 3 \text{ weeks and } 2 \text{ odd days} \] So, up to the end of 2019, we have \(0 + 2 = 2\) odd days.
Odd days in 2020 until 19th April: 2020 is a leap year (divisible by 4).
Total Odd Days: Total odd days up to 19th April 2020 = (Odd days up to 2019) + (Odd days in 2020 until April 19th) \[ \text{Total odd days} = 2 + 5 = 7 \text{ odd days} \] \[ 7 \div 7 = 1 \text{ week and } 0 \text{ odd days} \]
Day of the week: We use the standard day code where: 0 = Sunday 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday Since we have 0 odd days, the day is Sunday.
Correct Option: B) Sunday