11
Among three numbers, the second number is double the first and also triple the third. If the average of these three numbers is 44, determine the largest number.
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Solution: Step 1: Let the three numbers be N1, N2, and N3.
Step 2: According to the problem, N2 = 2 × N1 and N2 = 3 × N3.
Step 3: Express all numbers in terms of a common variable. Let N1 = 3x. Then N2 = 2 × (3x) = 6x.
Step 4: Since N2 = 3 × N3, 6x = 3 × N3, which implies N3 = 2x.
Step 5: The three numbers are 3x, 6x, and 2x.
Step 6: The average of the three numbers is 44. So, (N1 + N2 + N3) / 3 = 44.
Step 7: Substitute the expressions: (3x + 6x + 2x) / 3 = 44.
Step 8: Simplify: 11x / 3 = 44.
Step 9: Solve for x: 11x = 132, so x = 12.
Step 10: Find the largest number. The numbers are 3x=36, 6x=72, and 2x=24.
Step 11: The largest number is 6x = 6 × 12 = 72.
14
For five consecutive years, the revenues of a company are $21,37,100, $19,28,700, $26,34,500, $22,85,400, and $24,14,300. Find by what percentage the highest revenue exceeds the average revenue.
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Solution: Step 1: Calculate the total revenue = $21,37,100 + $19,28,700 + $26,34,500 + $22,85,400 + $24,14,300 = $114,00,000
Step 2: Calculate the average revenue = Total revenue / Number of years = $114,00,000 / 5 = $22,80,000
Step 3: Identify the highest revenue = $26,34,500
Step 4: Calculate the difference between the highest and average revenue = $26,34,500 - $22,80,000 = $3,54,500
Step 5: Calculate the percentage by which the highest revenue exceeds the average revenue = ($3,54,500 / $22,80,000) * 100 = 15.54%
15
What is the arithmetic mean of the first 11 natural numbers?
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Solution: Step 1: Identify the first 11 natural numbers.
These are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
Step 2: Calculate the sum of these 11 numbers using the sum of an arithmetic progression formula.
Sum of first 'n' natural numbers = n × (n + 1) / 2
Here, n = 11.
Sum = 11 × (11 + 1) / 2 = 11 × 12 / 2 = 11 × 6 = 66.
Step 3: Divide the sum by the count of numbers (11) to find the mean.
Mean = Sum / Number of terms = 66 / 11 = 6.0.
16
The average of ten positive numbers is 'x'. If each of these numbers is increased by 10%, how is 'x' affected?
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Solution: Step 1: Let the ten positive numbers be n1, n2, ..., n10.
Step 2: The original average 'x' is given by: x = (n1 + n2 + ... + n10) / 10.
Step 3: If each number is increased by 10%, each new number becomes 1.1 times its original value (e.g., 1.1 * n1, 1.1 * n2, etc.).
Step 4: The new average, let's call it x', will be: x' = (1.1*n1 + 1.1*n2 + ... + 1.1*n10) / 10.
Step 5: Factor out 1.1 from the numerator: x' = 1.1 * (n1 + n2 + ... + n10) / 10.
Step 6: Recognize that (n1 + n2 + ... + n10) / 10 is the original average 'x'.
Step 7: Substitute 'x' back into the equation: x' = 1.1 * x.
Step 8: This shows that the new average is 1.1 times the original average, which means the average 'x' is increased by 10%.
18
Five distinct positive numbers have an average of 25. If the smallest number among them is replaced by 0, the average decreases by 'x'. What statement is true about 'x'?
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Solution: Step 1: Let the five distinct positive numbers be n1, n2, n3, n4, n5, where n1 is the smallest. Since they are distinct positive numbers and their average is 25, the smallest number (n1) must be positive and less than 25.
* 0 < n1 < 25
Step 2: The average of the five numbers is 25. So, their sum is:
* Sum (S) = n1 + n2 + n3 + n4 + n5 = 5 * 25 = 125.
Step 3: When the smallest number (n1) is replaced by 0, the new sum (S') becomes:
* S' = 0 + n2 + n3 + n4 + n5 = (n1 + n2 + n3 + n4 + n5) - n1 = S - n1 = 125 - n1.
Step 4: The new average (Avg') is the new sum divided by 5.
* Avg' = S' / 5 = (125 - n1) / 5 = 25 - (n1 / 5).
Step 5: The decrease in average, 'x', is the original average minus the new average.
* x = 25 - Avg' = 25 - (25 - n1/5) = n1/5.
Step 6: From Step 1, we established that 0 < n1 < 25. Divide this inequality by 5:
* 0/5 < n1/5 < 25/5
* 0 < n1/5 < 5.
Step 7: Since x = n1/5, it follows that 0 < x < 5. Therefore, 'x is less than 5'.