This problem can be solved by setting up a system of two linear equations with two variables, representing the ages of A and B. We will then solve this system using substitution or elimination to find the value of B's age.
Let A's age be \(A\) years.
Let B's age be \(B\) years.
The first statement says: "If 5 times A's age is added to B's age, the sum is 95 years."
\[ 5A + B = 95 \quad \text{(Equation 1)} \]The second statement says: "If 2.2 times B's age is added to A's age, the sum is 63 years."
\[ A + 2.2B = 63 \quad \text{(Equation 2)} \]From Equation 1, express \(B\) in terms of \(A\):
\[ B = 95 - 5A \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 2:
\[ A + 2.2(95 - 5A) = 63 \]Distribute 2.2:
\[ A + 2.2 \times 95 - 2.2 \times 5A = 63 \]\[ A + 209 - 11A = 63 \]Combine like terms:
\[ 209 - 10A = 63 \]Isolate the term with \(A\):
\[ -10A = 63 - 209 \]\[ -10A = -146 \]Solve for \(A\):
\[ A = \frac{-146}{-10} \]\[ A = 14.6 \]Now substitute the value of \(A\) back into Equation 3 to find \(B\):
\[ B = 95 - 5A \]\[ B = 95 - 5(14.6) \]\[ B = 95 - 73 \]\[ B = 22 \]So, B's age is 22 years.
A) 22 — As calculated, B's age is 22 years.
This problem involves the concept of work and time, specifically related to pipes and cisterns. We need to calculate the effective rate of filling when one tap fills and another empties the tank simultaneously. The net filling rate will be the difference between the filling rate and the emptying rate.
Correct Option: A) 63
The tank will be filled completely in 63 hours when both taps are open together.
A can complete a piece of work in 24 days and B can complete it in 28 days. If they work on alternate days, starting with B on the first day, then in how many days will the work be finished?
25 \( \frac{6}{7} \)
25 \( \frac{1}{6} \)
24 \( \frac{6}{7} \)
24 \( \frac{1}{6} \)
This problem involves calculating the time taken to complete a work when two individuals work on alternate days. We will first determine the individual work rates, then calculate the work done in one cycle (two days), and finally find out how many such cycles are needed to complete the work, accounting for any remaining work.
Calculate individual work rates:
A can complete the work in 24 days. So, A's one day's work is:
\[ \text{A's 1 day work} = \frac{1}{24} \]B can complete the work in 28 days. So, B's one day's work is:
\[ \text{B's 1 day work} = \frac{1}{28} \]Calculate work done in one cycle:
They work on alternate days, starting with B on the first day.
Day 1: B works.
Day 2: A works.
So, in 2 days (1 cycle), the total work done is:
\[ \text{Work in 1 cycle} = \text{B's 1 day work} + \text{A's 1 day work} \]\[ \text{Work in 1 cycle} = \frac{1}{28} + \frac{1}{24} \]To add these fractions, find a common denominator, which is the LCM of 28 and 24.
\(28 = 4 \times 7\)
\(24 = 4 \times 6\)
\(\text{LCM}(28, 24) = 4 \times 7 \times 6 = 168\)
\[ \text{Work in 1 cycle} = \frac{6}{168} + \frac{7}{168} = \frac{13}{168} \]Determine the number of cycles:
Let \(N\) be the number of cycles. We need to find the largest integer \(N\) such that \(N \times \frac{13}{168}\) is less than or equal to 1 (total work).
Multiply the total work (1) by the denominator of the work done in one cycle:
\(168 \div 13 \approx 12.92\)
So, 12 full cycles will be completed.
Work done in 12 cycles (12 \times 2 = 24 days):
\[ \text{Work in 12 cycles} = 12 \times \frac{13}{168} = \frac{156}{168} \]Remaining work after 24 days:
\[ \text{Remaining work} = 1 - \frac{156}{168} = \frac{168 - 156}{168} = \frac{12}{168} \]Calculate the time for the remaining work:
After 12 full cycles (24 days), it's the start of a new cycle, which means B will work on the 25th day.
B's 1 day work is \( \frac{1}{28} \).
The remaining work is \( \frac{12}{168} \).
Time taken by B to complete the remaining work is:
\[ \text{Time} = \frac{\text{Remaining Work}}{\text{B's 1 day work}} = \frac{12/168}{1/28} \]\[ \text{Time} = \frac{12}{168} \times 28 = \frac{12 \times 28}{168} \]Since \(168 = 6 \times 28\), we have:
\[ \text{Time} = \frac{12}{6} = 2 \text{ days} \]However, this calculation indicates B would take 2 days to complete the remaining work, which is more than B's one-day capacity. Let's recheck the calculation.
Remaining work: \( \frac{12}{168} \)
B's work rate: \( \frac{1}{28} \)
If B works on day 25, B will complete \( \frac{1}{28} \) of the work.
Work remaining after B works on day 25:
\[ \frac{12}{168} - \frac{1}{28} = \frac{12}{168} - \frac{6}{168} = \frac{6}{168} \]So, B works for 1 full day (day 25). The total days passed are \(24 + 1 = 25\) days.
Now, the remaining work is \( \frac{6}{168} \).
On day 26, A will work. A's 1 day work is \( \frac{1}{24} \).
Time taken by A to complete the remaining work \( \frac{6}{168} \) is:
\[ \text{Time} = \frac{\text{Remaining Work}}{\text{A's 1 day work}} = \frac{6/168}{1/24} \]\[ \text{Time} = \frac{6}{168} \times 24 = \frac{6 \times 24}{168} \]Since \(168 = 7 \times 24\), we have:
\[ \text{Time} = \frac{6}{7} \text{ days} \]Total time:
Total days = 24 days (from 12 cycles) + 1 day (B on day 25) + \( \frac{6}{7} \) day (A on day 26)
Total days = \[24 + 1 + \frac{6}{7} = 25 \frac{6}{7}\] days.
A) \( 25 \frac{6}{7} \)
The problem asks for the sum of squares of 100 terms, given their mean and standard deviation. We can use the formula that relates the standard deviation, mean, and sum of squares.
Recall the formula for standard deviation (\(\sigma\)) in terms of the sum of squares (\(\sum x_i^2\)), the sum of terms (\(\sum x_i\)), and the number of terms (\(n\)).
The formula for standard deviation is:
\[ \sigma = \sqrt{\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2} \]We know that the mean (\(\bar{x}\)) is given by \(\bar{x} = \frac{\sum x_i}{n}\). Substituting this into the standard deviation formula, we get:
\[ \sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2} \]Square both sides of the equation to eliminate the square root:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \]Rearrange the formula to solve for the sum of squares (\(\sum x_i^2\)):
\[ \frac{\sum x_i^2}{n} = \sigma^2 + (\bar{x})^2 \]\[ \sum x_i^2 = n(\sigma^2 + (\bar{x})^2) \]Substitute the given values into the formula:
Calculate the squares:
\[ 3^2 = 9 \]\[ 50^2 = 2500 \]Substitute these values back into the equation:
\[ \sum x_i^2 = 100(9 + 2500) \]\[ \sum x_i^2 = 100(2509) \]Perform the final multiplication:
\[ \sum x_i^2 = 250900 \]D) 2,50,900 is the correct sum of squares of the 100 terms.
This problem involves calculating simple interest and then finding the total amount received. We will use the formula for simple interest to find the interest earned and then add it to the principal amount to get the final amount.
The formula for simple interest is:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]The total amount (A) received will be the principal amount plus the simple interest.
\[ \text{Amount (A)} = \text{Principal (P)} + \text{Simple Interest (SI)} \]\[ \text{A} = 1380 + 124.2 \]\[ \text{A} = 1504.2 \]Therefore, Nirmal will get ₹1,504.2 after 3 years.
B) 1,504.2 — This is the correct amount obtained by adding the calculated simple interest (₹124.2) to the principal amount (₹1,380).
The problem asks us to identify a sequence of numbers that follows the same mathematical operations as two given sequences. We need to analyze the operations between consecutive numbers in the provided sets to find a consistent pattern.
Analyze the first given set: 5 - 6 - 10 - 20
The pattern for the first set is: +1, +4, \(\times 2\).
Analyze the second given set: 12 - 13 - 17 - 34
The pattern for the second set is also: +1, +4, \(\times 2\).
Conclusion on the pattern: Both given sets follow the same sequence of operations: add 1 to the first number, add 4 to the result, and multiply the new result by 2.
Apply the pattern to the options: We need to find an option that follows the +1, +4, \(\times 2\) pattern.
Correct Option: A) 17 - 18 - 22 - 44
This option perfectly matches the identified pattern.
A group of numbers and symbols is coded using letter codes as per the codes given below and the conditions that follow. Study the given codes and conditions and answer the question that follows.
Note: If none of the conditions apply, then codes for the respective number/symbol are to be followed directly as given in the table.
| Number/Symbol | * | 2 | 5 | 7 | @ | + | 8 | % | # | 9 | & | 6 | $ | 4 |
| Code | H | D | O | A | B | P | B | Q | Z | Y | R | B | U | F |
Conditions:
If the first element is a symbol and the last an odd number, the codes for these two (the first and the last elements) are to be interchanged. If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©.
If both second and third elements are perfect cubes, the third element is to be coded as the code for the second element.
What will be the code for the following group?
# 8 9 $
The problem requires us to code a given group of numbers and symbols based on a provided table and a set of conditions. We need to first identify the codes for each element from the table and then apply the conditions in the given order to see if any of them modify the initial coding. If no condition applies, the direct codes are used.
D) ZBYU is the correct code as derived from the direct mapping after verifying that none of the conditions apply.
The question asks to identify which option was NOT a significant cause of the Revolt of 1857. We need to evaluate each option based on its historical timing and its role in contributing to the revolt.
Based on this analysis, the Indian Councils Act, 1861, occurred after the revolt and thus cannot be its cause.
Correct Option: B) Introduction of the Indian Councils Act, 1861
The given series is an alternating series with two distinct patterns. We need to identify these patterns to find the missing term.
The series is 61, 54, 61, 47, 61, 40, 61, ?
We can observe that the number 61 appears at odd positions (1st, 3rd, 5th, 7th). This forms one sub-series.
The numbers at even positions (2nd, 4th, 6th, 8th) form another sub-series: 54, 47, 40, ?
The first sub-series consists of the number 61 repeated: 61, 61, 61, 61.
This pattern is constant.
The second sub-series is 54, 47, 40, ?
Let's find the difference between consecutive terms in this sub-series:
This indicates a consistent pattern where each term is obtained by subtracting 7 from the previous term.
Following the pattern of the second sub-series, the next term will be \(40 - 7\).
\[ 40 - 7 = 33 \]The missing term in the series is 33.
Correct Option: B) 33 — The series alternates between a constant number (61) and a decreasing arithmetic progression (54, 47, 40, ...). The decreasing series has a common difference of -7. Therefore, the next term in the decreasing series after 40 is \(40 - 7 = 33\).