Find the missing number in the series: 4, 10, ?, 82, 244, 730.
Answer: B
The pattern in the series is: multiply by 3 and subtract 2.
\(4 \times 3 - 2 = 12 - 2 = 10\).
\(10 \times 3 - 2 = 30 - 2 = 28\).
\(28 \times 3 - 2 = 84 - 2 = 82\).
\(82 \times 3 - 2 = 246 - 2 = 244\).
\(244 \times 3 - 2 = 732 - 2 = 730\).
The missing number is 28.
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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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What is the value of \((32 \times 12) + (17^2) - 95\)?
Answer: A
Follow the BODMAS rule. First, perform multiplication and find the square, then do addition and subtraction.
Step 1: \(32 \times 12 = 384\).
Step 2: \(17^2 = 289\).
Step 3: The expression becomes \(384 + 289 - 95\).
Step 4: \(384 + 289 = 673\).
Step 5: \(673 - 95 = 578\).
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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 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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If the product of three consecutive integers is 120, then the sum of the integers is:
Answer: C
Let the consecutive integers be \(n-1, n, n+1\).
We need to find three consecutive numbers whose product is 120. We can estimate the cube root of 120. \(5^3=125\), so the numbers should be around 5.
Let's try 4, 5, and 6.
\(4 \times 5 \times 6 = 20 \times 6 = 120\). This is correct.
The integers are 4, 5, and 6.
Their sum is \(4 + 5 + 6 = 15\).
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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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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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If \(x\) and \(y\) are the two digits of the number \(653xy\) such that this number is divisible by 80, then what is the value of \(x+y\)?
Answer: D
A number is divisible by 80 if it is divisible by both 10 and 8.
For the number to be divisible by 10, its last digit must be 0. So, \(y=0\).
The number now is \(653x0\).
For the number to be divisible by 8, the number formed by its last three digits must be divisible by 8. So, \(3x0\) must be divisible by 8.
Let's check values for x:
If x=2, the number is 65320. \(65320 \div 80 = 816.5\) (Not divisible).
If x=6, the number is 65360. \(65360 \div 80 = 817\) (Divisible).
So the number is 65360, which means \(x=6\) and \(y=0\).
The value of \(x+y = 6+0 = 6\).
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