1
A company had reserves of 68,000, 72,000, and 67,000 for three consecutive months. What was the average monthly reserve?
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Solution: Step 1: Calculate the sum of reserves for the three months: 68,000 + 72,000 + 67,000 = 207,000
Step 2: Divide the sum by the number of months to find the average: 207,000 / 3 = 69,000
The average reserve per month is 69,000.
5
A man's income for three months covers his expenses for four months. If his monthly income is Rs. 1,000, calculate his total annual savings.
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Solution: Step 1: Understand the spending pattern: The man spends 3 months' income over a period of 4 months.
Step 2: Calculate his income for 3 months: 3 * Rs. 1,000 = Rs. 3,000.
Step 3: This means his expenditure for 4 months is Rs. 3,000.
Step 4: Calculate his monthly expenditure: Rs. 3,000 / 4 = Rs. 750.
Step 5: Calculate his total annual income: 12 months * Rs. 1,000/month = Rs. 12,000.
Step 6: Calculate his total annual expenditure: 12 months * Rs. 750/month = Rs. 9,000.
Step 7: Calculate his annual savings: Annual Income - Annual Expenditure = Rs. 12,000 - Rs. 9,000 = Rs. 3,000.
7
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.
18
The average of 'n' numbers, x1, x2, ..., xn, is x_bar. What is the value of the summation from i=1 to n of (xi - x_bar)?
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Solution: Step 1: Understand the definition of the arithmetic mean.
The arithmetic mean (average) x_bar of 'n' numbers (x1, x2, ..., xn) is defined as:
x_bar = (x1 + x2 + ... + xn) / n
Step 2: Rewrite the sum of the numbers in terms of the average.
From the definition, Sum of numbers = x1 + x2 + ... + xn = n × x_bar.
Step 3: Expand the given summation expression.
Summation (xi - x_bar) from i=1 to n = (x1 - x_bar) + (x2 - x_bar) + ... + (xn - x_bar).
Step 4: Rearrange the terms in the expanded summation.
This can be rewritten as: (x1 + x2 + ... + xn) - (x_bar + x_bar + ... + x_bar) [n times]
Step 5: Substitute the sum of numbers and simplify.
Summation = (n × x_bar) - (n × x_bar) = 0.
Therefore, the sum of deviations of individual data points from their mean is always 0.