What is the value of the expression 101 + 102 + 103 + ... + 200?
Answer: B
The sum of an arithmetic progression is given by the formula \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term.
Number of terms (n) = (Last Term - First Term) + 1 = (200 - 101) + 1 = 100.
First term (\(a_1\)) = 101.
Last term (\(a_n\)) = 200.
Sum = \(\frac{100}{2} \times (101 + 200) = 50 \times 301 = 15050\).
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Evaluate \(123 \times 5 + 123 \times 15\).
Answer: A
Using the distributive property \(a \times b + a \times c = a \times (b+c)\).
Here, a = 123, b = 5, c = 15.
The expression is \(123 \times (5 + 15) = 123 \times 20\).
\(123 \times 20 = 2460\).
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The sum of two numbers is 25 and their product is 156. Find the smaller number.
Answer: A
Let the numbers be x and y. x + y = 25 and xy = 156.
We can form a quadratic equation: \(t^2 - (sum)t + (product) = 0\).
\(t^2 - 25t + 156 = 0\).
We need to find two numbers that multiply to 156 and add up to 25. These are 12 and 13.
\((t-12)(t-13) = 0\).
The numbers are 12 and 13. The smaller number is 12.
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What is the place value of 5 in the number 67.058?
Answer: B
The digit 5 is in the second position to the right of the decimal point. This is the hundredths place.
Its value is \(5 \times \frac{1}{100}\) or 0.05.
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The sum of three consecutive even numbers is 144. What is the largest of these numbers?
Answer: C
Let the three consecutive even numbers be \(x\), \(x+2\), and \(x+4\).
Their sum is \(x + (x+2) + (x+4) = 144\).
\(3x + 6 = 144\).
\(3x = 138\).
\(x = 46\).
The numbers are 46, 48, and 50. The largest number is 50.
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Find the value of: \(1 - 2 + 3 - 4 + 5 - 6 + ... + 99 - 100\).
Answer: A
We can group the terms in pairs:
(1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100).
Each pair sums to -1.
There are 100 terms in total, so there are 100/2 = 50 such pairs.
The total sum is 50 × (-1) = -50.
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The sum of the digits of a 2-digit number is 9. When 27 is added to the number, the digits get reversed. The number is:
Answer: B
Let the ten's digit be x and the unit's digit be y. Number = 10x + y.
Given: x + y = 9.
Given: (10x + y) + 27 = 10y + x (reversed number).
\(9x - 9y = -27\), which simplifies to \(x - y = -3\) or \(y - x = 3\).
We have two equations: x + y = 9 and y - x = 3.
Adding them: 2y = 12, so y = 6.
Substituting y=6 into x+y=9, we get x=3.
The number is 36.
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What is the result of \(3.5 \times (60 \div 2.5)\)?
Answer: A
Using BODMAS, we solve the bracket first.
\(60 \div 2.5 = 60 \div (\frac{5}{2}) = 60 \times \frac{2}{5} = 12 \times 2 = 24\).
Now, multiply by 3.5: \(3.5 \times 24 = (\frac{7}{2}) \times 24 = 7 \times 12 = 84\).
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The product of two numbers is 120 and the sum of their squares is 289. The sum of the two numbers is:
Answer: B
Let the numbers be x and y.
Given, xy = 120 and \(x^2 + y^2 = 289\).
We know that \((x+y)^2 = x^2 + y^2 + 2xy\).
Substitute the given values: \((x+y)^2 = 289 + 2(120) = 289 + 240 = 529\).
Therefore, x + y = \(\sqrt{529} = 23\).
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What is the average of the first 15 odd numbers?
Answer: A
The sum of the first n odd numbers is \(n^2\).
The average of the first n odd numbers is \(\frac{\text{Sum}}{n} = \frac{n^2}{n} = n\).
So, the average of the first 15 odd numbers is 15.
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