This problem involves the relationship between speed, distance, and time. The key is that the distance covered is the same in both scenarios. We can first calculate the total distance using the given speed and time, and then use this distance with the new time to find the required speed.
Calculate the total distance:
Given:
The formula for distance is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]So, the total distance (\(D\)) is:
\[ D = S_1 \times T_1 = 14 \text{ km/hr} \times 20 \text{ hours} = 280 \text{ km} \]Calculate the new speed:
We need to cover the same distance (\(D = 280 \text{ km}\)) in a new time (\(T_2\)) of \(5 \text{ hours}\).
The formula for speed is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]So, the new speed (\(S_2\)) should be:
\[ S_2 = \frac{D}{T_2} = \frac{280 \text{ km}}{5 \text{ hours}} = 56 \text{ km/hr} \]Correct Option: A) 56 km/hr
To find the average of the first 'n' odd natural numbers, we can use the property that the sum of the first 'n' odd natural numbers is \(n^2\). The average is then calculated by dividing this sum by 'n'.
Identify the sequence: The first odd natural numbers are 1, 3, 5, 7, ...
Determine the number of terms: We need to find the average of the first 149 odd natural numbers. So, \(n = 149\).
Recall the formula for the sum of the first 'n' odd natural numbers: The sum of the first 'n' odd natural numbers is given by \(S_n = n^2\).
For \(n = 149\), the sum of the first 149 odd natural numbers is:
\[ S_{149} = 149^2 \]Recall the formula for average: The average of a set of numbers is the sum of the numbers divided by the count of the numbers.
\[ \text{Average} = \frac{\text{Sum of numbers}}{\text{Count of numbers}} \]Calculate the average:
\[ \text{Average} = \frac{S_{149}}{149} = \frac{149^2}{149} = 149 \]C) 149. The average of the first 'n' odd natural numbers is simply 'n'. In this case, \(n = 149\), so the average is 149.
The question provides the principal amount, the final amount after 2 years and the compounding frequency. We need to calculate the annual rate of compound interest using the compound amount formula.
Identify the given information:
Apply the compound interest formula:
When interest is compounded annually, the amount is calculated using:
\[ A=P\left(1+\frac{R}{100}\right)^T \]Substituting the given values:
\[ 2809=2500\left(1+\frac{R}{100}\right)^2 \]Simplify the equation:
Divide both sides by 2500:
\[ \frac{2809}{2500}=\left(1+\frac{R}{100}\right)^2 \]Taking the positive square root of both sides:
\[ \frac{53}{50}=1+\frac{R}{100} \]Since \(53^2=2809\) and \(50^2=2500\).
Calculate the annual rate of interest:
\[ \frac{R}{100}=\frac{53}{50}-1 \] \[ =\frac{53-50}{50} \] \[ =\frac{3}{50} \]Multiplying both sides by 100:
\[ R=\frac{3}{50}\times100 \] \[ R=6\% \]Therefore, the annual rate of compound interest is 6%.
Correct Option: B) 6% — Using the compound amount formula \(A=P(1+\frac{R}{100})^T\), we get \(\frac{2809}{2500}=(1+\frac{R}{100})^2\). Taking the square root gives \(1+\frac{R}{100}=\frac{53}{50}\), resulting in \(R=6\%\). Hence, Option B is the correct answer.
A) 8%: This is incorrect because an annual interest rate of 8% would give an amount of \(2500(1.08)^2=2916\), which is greater than ₹2,809.
C) 7%: This is incorrect because an annual interest rate of 7% would give an amount of \(2500(1.07)^2=2862.25\), which does not match ₹2,809.
D) 3%: This is incorrect because an annual interest rate of 3% would give an amount of \(2500(1.03)^2=2652.25\), which is less than ₹2,809.
To determine divisibility, we will apply the divisibility rules for each given option to the number 877217. This method is generally faster than performing actual division for each option.
D) 11 is the correct option because, as per the divisibility rule for 11, the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is 0, which means the number 877217 is perfectly divisible by 11.
To find the difference, we need to calculate the final price after a single discount and the final price after two successive discounts on the original bill amount. Then, we find the difference between these two discounted amounts.
Original bill amount = ₹1,125
Single discount = 52%
Discount amount \(D_1\) = \(52\%\) of ₹1,125
\[ D_1 = \frac{52}{100} \times 1125 \] \[ D_1 = 0.52 \times 1125 \] \[ D_1 = \text{₹}585 \]Amount after single discount \(A_1\) = Original amount - \(D_1\)
\[ A_1 = 1125 - 585 \] \[ A_1 = \text{₹}540 \]First discount = 35%
Amount after first discount = \(1125 \times \left(1 - \frac{35}{100}\right)\)
\[ = 1125 \times \frac{65}{100} \] \[ = 1125 \times 0.65 \] \[ = \text{₹}731.25 \]Second discount = 4% (applied on the amount after the first discount)
Amount after second discount \(A_2\) = \(731.25 \times \left(1 - \frac{4}{100}\right)\)
\[ = 731.25 \times \frac{96}{100} \] \[ = 731.25 \times 0.96 \] \[ = \text{₹}702 \]Difference = Amount after two successive discounts - Amount after single discount
\[ \text{Difference} = A_2 - A_1 \] \[ \text{Difference} = 702 - 540 \] \[ \text{Difference} = \text{₹}162 \]A) ₹162 is the correct difference between the two scenarios.
To solve this problem, we will use the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
The given 6-digit number is N51M54.
Let's assign positions to the digits from right to left (units place as 1st, tens place as 2nd, and so on):
Sum of digits at odd places (1st, 3rd, 5th) = \(4 + M + 5 = 9 + M\).
Sum of digits at even places (2nd, 4th, 6th) = \(5 + 1 + N = 6 + N\).
According to the rule, the difference between these two sums must be 0 or a multiple of 11.
\[ (9 + M) - (6 + N) = 0 \text{ or } \pm 11, \pm 22, \dots \] \[ 9 + M - 6 - N = 0 \text{ or } \pm 11, \pm 22, \dots \] \[ 3 + M - N = 0 \text{ or } \pm 11, \pm 22, \dots \]We need to find a relation between M and N from the given options. M and N are digits, so they must be integers from 0 to 9. Also, N cannot be 0 if it is the leading digit of a 6-digit number, but in divisibility rules, sometimes N can be 0. Let's assume M and N are single digits (0-9).
Case 1: \(3 + M - N = 0\) \[ M - N = -3 \] This is a possible relation. For example, if \(M=0, N=3\); if \(M=1, N=4\); if \(M=6, N=9\).
Case 2: \(3 + M - N = 11\) \[ M - N = 8 \] This is also a possible relation. For example, if \(M=8, N=0\); if \(M=9, N=1\).
Case 3: \(3 + M - N = -11\) \[ M - N = -14 \] This is not possible because the minimum value for \(M-N\) (when \(M=0, N=9\)) is \(-9\), and the maximum value (when \(M=9, N=0\)) is \(9\).
Option A) \(M - N = 1\). This is not derived from our possible cases.
Option B) \(M - N = -3\). This matches Case 1.
Option C) \(M + N = 3\). This is not derived from our possible cases.
Option D) \(M = N\), which implies \(M - N = 0\). If \(M-N=0\), then \(3+0=0\), which means \(3=0\), which is false. So, \(M=N\) is not possible.
Correct Option: B) M - N = -3
This problem involves calculating the work rates of pipes. We'll use the concept that if a pipe can fill a cistern in \(T\) minutes, its filling rate is \(1/T\) of the cistern per minute. Similarly, if a pipe can empty a cistern in \(T\) minutes, its emptying rate is \(1/T\) of the cistern per minute. When multiple pipes operate simultaneously, their rates are added (for filling) or subtracted (for emptying) to find the net rate.
Pipe 1 fills the cistern in 93 min.
\[ \text{Rate of Pipe 1} = \frac{1}{93} \text{ cistern/min} \]Pipe 2 fills the cistern in 31 min.
\[ \text{Rate of Pipe 2} = \frac{1}{31} \text{ cistern/min} \]Let the emptying pipe (Pipe 3) take \(x\) minutes to empty the cistern. Its rate will be \(1/x\) cistern/min.
\[ \text{Rate of Pipe 3 (emptying)} = \frac{1}{x} \text{ cistern/min} \]When all three pipes are opened simultaneously, the cistern gets filled in 24 min. This means the net filling rate is \(1/24\) cistern/min.
The net rate is the sum of filling rates minus the emptying rate:
\[ \text{Net Rate} = \text{Rate of Pipe 1} + \text{Rate of Pipe 2} - \text{Rate of Pipe 3} \]\[ \frac{1}{24} = \frac{1}{93} + \frac{1}{31} - \frac{1}{x} \]First, combine the filling rates:
\[ \frac{1}{93} + \frac{1}{31} = \frac{1}{93} + \frac{3}{93} = \frac{4}{93} \]Now substitute this back into the net rate equation:
\[ \frac{1}{24} = \frac{4}{93} - \frac{1}{x} \]Rearrange to solve for \(\frac{1}{x}\):
\[ \frac{1}{x} = \frac{4}{93} - \frac{1}{24} \]Find a common denominator for 93 and 24. \(93 = 3 \times 31\), \(24 = 3 \times 8\). The LCM is \(3 \times 31 \times 8 = 744\).
\[ \frac{1}{x} = \frac{4 \times 8}{93 \times 8} - \frac{1 \times 31}{24 \times 31} \]\[ \frac{1}{x} = \frac{32}{744} - \frac{31}{744} \]\[ \frac{1}{x} = \frac{1}{744} \]Therefore,
\[ x = 744 \text{ min} \]Correct Option: B) 744 min
The height of a cylinder is 35 cm, and the radius of its circular base is 9 cm. Find its volume. (Take π = \( \frac{22}{7} \).)
To find the volume of a cylinder, we use the formula \(V = \pi r^2 h\), where \(r\) is the radius of the base and \(h\) is the height. We are given the height, the radius, and the value of \(\pi\).
Identify the given values:
Recall the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]Substitute the given values into the formula:
\[ V = \frac{22}{7} \times (9 \text{ cm})^2 \times 35 \text{ cm} \]Calculate the square of the radius:
\[ (9 \text{ cm})^2 = 81 \text{ cm}^2 \]Substitute this back into the volume formula:
\[ V = \frac{22}{7} \times 81 \text{ cm}^2 \times 35 \text{ cm} \]Simplify the expression. Notice that 35 is a multiple of 7:
\[ \frac{35}{7} = 5 \]Now, perform the multiplication:
\[ V = 22 \times 81 \times 5 \text{ cm}^3 \]Multiply 22 by 5:
\[ 22 \times 5 = 110 \]Finally, multiply 110 by 81:
\[ V = 110 \times 81 \text{ cm}^3 \]Calculation:
\[ 110 \times 81 = 8910 \]Therefore, the volume of the cylinder is \(8910 \text{ cm}^3\).
A) \(8910 \text{ cm}^3\) is the correct volume calculated using the given dimensions and the formula for the volume of a cylinder.
This problem involves relating distance, speed, and time. We are given two scenarios for Pranshu's travel and his time difference compared to Chandu. We can set up equations based on the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\) and solve for Pranshu's speed.
Let Pranshu's original speed be \(S_P\) km/hr and Chandu's speed be \(S_C\) km/hr.
The distance to be covered is 56 km.
Time taken by Pranshu: \(T_P = \frac{56}{S_P}\) hours.
Time taken by Chandu: \(T_C = \frac{56}{S_C}\) hours.
First condition: Pranshu takes 3 hours more than Chandu.
\[ T_P = T_C + 3 \]\[ \frac{56}{S_P} = \frac{56}{S_C} + 3 \quad \ldots(1) \]Second condition: If Pranshu doubles his speed (i.e., \(2S_P\)), he would take 4 hours less than Chandu.
New time for Pranshu: \(T_P' = \frac{56}{2S_P}\) hours.
\[ T_P' = T_C - 4 \]\[ \frac{56}{2S_P} = \frac{56}{S_C} - 4 \quad \ldots(2) \]We have a system of two equations. Let's subtract equation (2) from equation (1):
\[ \left(\frac{56}{S_P} - \frac{56}{2S_P}\right) = \left(\frac{56}{S_C} + 3\right) - \left(\frac{56}{S_C} - 4\right) \]\[ \frac{56}{S_P} - \frac{28}{S_P} = \frac{56}{S_C} + 3 - \frac{56}{S_C} + 4 \]\[ \frac{28}{S_P} = 7 \]Now, solve for \(S_P\):
\[ S_P = \frac{28}{7} \]\[ S_P = 4 \text{ km/hr} \]C) 4 km/hr — By setting up equations based on the given conditions and solving them simultaneously, Pranshu's speed is found to be 4 km/hr.
The mean proportional (or geometric mean) between two numbers, say \(a\) and \(b\), is given by the square root of their product, i.e., \(\sqrt{a \times b}\). We will apply this formula to the given numbers 25 and 196.
D) 70 is the correct answer, as calculated above.