The problem involves calculating the distance covered by a car and then determining the new speed required to cover the same distance in a shorter time. Finally, we need to find the increase in speed.
Given the initial speed and time, we can find the distance using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]Using the calculated distance and the new desired time, we can find the new speed required:
\[ \text{New Speed} = \frac{\text{Distance}}{\text{New Time}} \]The increase in speed is the difference between the new speed and the original speed.
\[ \text{Increase in Speed} = \text{New Speed} - \text{Original Speed} \]Original Speed \( = 50 \text{ kmph}\)
Original Time \( = 8 \text{ hours}\)
\[ \text{Distance} = 50 \text{ kmph} \times 8 \text{ hours} = 400 \text{ km} \]Desired Time \( = 5 \text{ hours}\)
Distance \( = 400 \text{ km}\)
\[ \text{New Speed} = \frac{400 \text{ km}}{5 \text{ hours}} = 80 \text{ kmph} \]New Speed \( = 80 \text{ kmph}\)
Original Speed \( = 50 \text{ kmph}\)
\[ \text{Increase in Speed} = 80 \text{ kmph} - 50 \text{ kmph} = 30 \text{ kmph} \]Correct Option: C) 30 kmph
This problem involves relative speed and distance. We need to determine the distance covered by bus X before bus Y starts, and then calculate the time it takes for bus Y to catch up to bus X. Once we have the time, we can find the distance from the starting point where they meet.
Calculate the distance covered by bus X before bus Y starts:
Bus X travels for 6 hours before bus Y leaves. Its speed is 20 kmph.
\[ \text{Distance covered by X} = \text{Speed of X} \times \text{Time} \\ \text{Distance covered by X} = 20 \text{ kmph} \times 6 \text{ hours} \\ \text{Distance covered by X} = 120 \text{ km} \]Determine the relative speed of bus Y with respect to bus X:
Since both buses are moving in the same direction, the relative speed is the difference between their speeds.
\[ \text{Relative Speed} = \text{Speed of Y} - \text{Speed of X} \\ \text{Relative Speed} = 50 \text{ kmph} - 20 \text{ kmph} \\ \text{Relative Speed} = 30 \text{ kmph} \]Calculate the time it takes for bus Y to catch up to bus X:
Bus Y needs to cover the 120 km lead that bus X has, at the relative speed.
\[ \text{Time to catch up} = \frac{\text{Distance to cover}}{\text{Relative Speed}} \\ \text{Time to catch up} = \frac{120 \text{ km}}{30 \text{ kmph}} \\ \text{Time to catch up} = 4 \text{ hours} \]Calculate the distance from the starting point where they meet:
This can be calculated using the speed of bus Y and the time it took to catch up.
\[ \text{Meeting Distance} = \text{Speed of Y} \times \text{Time to catch up} \\ \text{Meeting Distance} = 50 \text{ kmph} \times 4 \text{ hours} \\ \text{Meeting Distance} = 200 \text{ km} \]B) 200 km is the correct answer. Both buses will meet 200 km from the starting point.
To determine which article was the most profitable in percentage terms, we need to calculate the percentage profit for each article. The selling price for all articles is Rs. 2 above their respective purchase prices. The percentage profit is calculated as the ratio of profit to the cost price, multiplied by 100.
Correct Option: C) (iii) is the most profitable article in percentage terms because it yields a 25% profit, which is the highest among all options.
The question asks about the phenomenon related to the Tyndall effect. We need to identify which property of light is responsible for the Tyndall effect.
Correct Option: B) Scattering
The LCM of two numbers is 75. The numbers are in the ratio 1 : \( \frac{5}{3} \). The sum of the numbers is ______.
We are given the LCM of two numbers and their ratio. We need to find the sum of these numbers. We will first simplify the given ratio, then represent the numbers using a common factor. We will then use the property that for two numbers \(a\) and \(b\), \(\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b\). Alternatively, we can express the numbers in terms of their HCF and then use the definition of LCM.
The given ratio of the two numbers is \(1 : \frac{5}{3}\).
To simplify this ratio, multiply both sides by 3:
\[ 1 \times 3 : \frac{5}{3} \times 3 \]\[ 3 : 5 \]Let the two numbers be \(3x\) and \(5x\), where \(x\) is their Highest Common Factor (HCF).
The LCM of \(3x\) and \(5x\) is \(3 \times 5 \times x = 15x\).
We are given that the LCM of the two numbers is 75.
So, we have the equation:
\[ 15x = 75 \]Divide both sides by 15:
\[ x = \frac{75}{15} \]\[ x = 5 \]The first number is \(3x = 3 \times 5 = 15\).
The second number is \(5x = 5 \times 5 = 25\).
Sum \( = 15 + 25 = 40\).
B) The sum of the numbers is 40.
Answer the next 4 Questions.
Comprehension:
The following table represents the number of items sold by four divisions of a consumer goods dealer during 2010 to 2013.
| Year | W | X | Y | Z |
| 2010 | 100 | 96 | 110 | 136 |
| 2011 | 130 | 134 | 114 | 108 |
| 2012 | 140 | 144 | 164 | 96 |
| 2013 | 86 | 112 | 76 | 80 |
SubQuestion No: 1
The combined sales figures in 2011 and 2013 of which division was the lowest?
To find the division with the lowest combined sales in 2011 and 2013, we need to sum the sales figures for each division for these two years and then compare the totals.
C) Z is the correct answer because Division Z has the lowest combined sales of 188 in 2011 and 2013.
SubQuestion No: 2
During 2011, which division performed better comparatively?
The question asks to identify which division performed better comparatively in the year 2011. To determine this, we need to look at the bar chart provided and compare the performance (represented by the height of the bars) of each division (W, X, Y, Z) for the year 2011.
Therefore, Division X performed better comparatively in 2011.
A) X — Division X had the highest performance (35 units) in 2011 compared to divisions W (20 units), Y (30 units), and Z (25 units).
SubQuestion No: 3
What is the ratio of total number of items sold by Y in 2012 and 2013 to that of Z in the same period?
The problem requires us to calculate the ratio of the total number of items sold by company Y in 2012 and 2013 to the total number of items sold by company Z in the same period. We need to extract the relevant data from the provided (assumed) graph or table for companies Y and Z for the years 2012 and 2013, sum them up for each company, and then find the ratio.
A) 15:11 is the correct ratio calculated from the total items sold by Y and Z in the specified period.
SubQuestion No: 4
What is the average number of items sold by division W during the four years?
To find the average number of items sold by division W over four years, we need to sum the number of items sold by division W for each of the four years and then divide this sum by the total number of years, which is 4.
C) 114 — The average number of items sold by division W during the four years is 114.
This is a blood relation problem. We need to establish the relationships between the individuals mentioned and then deduce the relationship between Arun and Sonali. We can use a family tree diagram or list the relationships step-by-step.
B) Brother — Since Arun and Sonali share the same parents (Arjun and Sarda), Arun is the brother of Sonali.