What is the sum of all prime numbers between 60 and 80?
Answer: A
We need to identify the prime numbers between 60 and 80.
Prime numbers are: 61, 67, 71, 73, 79.
Sum = \(61 + 67 + 71 + 73 + 79\).
Sum = \((60+1) + (60+7) + (70+1) + (70+3) + (70+9) = 120 + 210 + (1+7+1+3+9) = 330 + 21 = 351\).
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A number when divided by 296 gives a remainder of 75. When the same number is divided by 37, the remainder is:
Answer: A
Let the number be N. N = 296k + 75.
We need to find the remainder when N is divided by 37.
First, check if 296 is divisible by 37. \(296 = 37 \times 8\).
Since the first divisor is a multiple of the second divisor, we can simply divide the first remainder by the second divisor.
Remainder = \(75 \div 37\).
\(75 = 2 \times 37 + 1\).
The remainder is 1.
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A number whose fifth part increased by 4 is equal to its fourth part diminished by 10. The number is:
Answer: C
Let the number be x.
The equation is \(\frac{x}{5} + 4 = \frac{x}{4} - 10\).
\(4 + 10 = \frac{x}{4} - \frac{x}{5}\).
\(14 = \frac{5x - 4x}{20}\).
\(14 = \frac{x}{20}\).
\(x = 14 \times 20 = 280\).
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What is the value of \(\sqrt[3]{4 \frac{12}{125}}\)?
Answer: C
First, convert the mixed fraction to an improper fraction.
\(4 \frac{12}{125} = \frac{4 \times 125 + 12}{125} = \frac{500+12}{125} = \frac{512}{125}\).
Now find the cube root: \(\sqrt[3]{\frac{512}{125}} = \frac{\sqrt[3]{512}}{\sqrt[3]{125}}\).
We know \(8^3 = 512\) and \(5^3 = 125\).
So the value is \(\frac{8}{5} = 1.6\).
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Find the value of \(1/(2\times3) + 1/(3\times4) + 1/(4\times5) + ... + 1/(9\times10)\).
Answer: B
This is a telescoping series. Each term can be split using partial fractions: \(1/(n(n+1)) = 1/n - 1/(n+1)\).
The series becomes: \((\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + ... + (\frac{1}{9} - \frac{1}{10})\).
All the intermediate terms cancel out.
We are left with \(\frac{1}{2} - \frac{1}{10}\).
\(\frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}\).
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What will be the remainder when \(17^{200}\) is divided by 18?
Answer: A
We can use the remainder theorem.
\(17\) can be written as \((18 - 1)\).
So, we need to find the remainder of \((18-1)^{200} \div 18\).
Using the binomial expansion, every term except the last one will have 18 as a factor. The last term is \((-1)^{200}\).
\((-1)^{200} = 1\).
So the remainder is 1.
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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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What is the next number in the sequence: 1, 4, 9, 16, 25, ...?
Answer: B
The terms are the perfect squares.
\[ 1^{2},\ 2^{2},\ 3^{2},\ 4^{2},\ 5^{2} \]
\[ \text{next} = 6^{2} = 36 \]
The 49 on offer is \( 7^{2} \), one term too far.
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What is the equivalent of 0.75 as a fraction?
Answer: B
\(0.75\) can be written as \(\frac{75}{100}\).
To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 25.
\(75 \div 25 = 3\).
\(100 \div 25 = 4\).
So, the fraction is \(\frac{3}{4}\).
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Find the value of \((243)^{0.16} \times (243)^{0.04}\).
Answer: B
When multiplying powers with the same base, we add the exponents.
The expression becomes \((243)^{0.16 + 0.04} = (243)^{0.20}\).
\(0.20\) as a fraction is \(20/100 = 1/5\).
So we need to find \((243)^{1/5}\), which is the fifth root of 243.
We know that \(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243\).
Therefore, the value is 3.
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