What is the value of the expression \(101 + 102 + 103 + ... + 200\)?
Answer: B
This is an arithmetic progression. Number of terms (n) = (Last Term - First Term) + 1 = (200 - 101) + 1 = 100.
Sum = \(\frac{n}{2}(\text{first term} + \text{last term})\)
Sum = \(\frac{100}{2} \times (101 + 200) = 50 \times 301 = 15050\).
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If the sum of two numbers is 22 and the sum of their squares is 404, then the product of the numbers is:
Answer: A
Expand the square of the sum.
\[ (a+b)^{2} = a^{2} + b^{2} + 2ab \]
\[ 22^{2} = 404 + 2ab \quad\Longrightarrow\quad 484 - 404 = 2ab \]
\[ ab = 40 \]
The two numbers turn out to be 2 and 20.
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The number 311311311311 is divisible by:
Answer: D
The number is formed by repeating '311' four times.
Divisibility by 3: Sum of digits = (3+1+1) × 4 = 5 × 4 = 20. Since 20 is not divisible by 3, the number is not divisible by 3.
Divisibility by 11: Sum of alternate digits. (1+3+1+3+1+3) - (1+1+1+1) = 12 - 4 = 8. No. Let's do it right to left. (1+1+1+1+1+1) - (3+3+3+3) = 6-12=-6. No. Let's do odd/even positions. (3+1+1+1+1+1)-(1+3+1+3+1+3) = 8-12=-4. It is not divisible by 11.
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What is the value of \(0.796 \times 0.796 - 0.204 \times 0.204\) / \(0.796 - 0.204\)?
Answer: C
The numerator is in the form \(a^2 - b^2\), which factors to \((a-b)(a+b)\).
So, the expression is \(\frac{(0.796 - 0.204)(0.796 + 0.204)}{0.796 - 0.204}\).
We can cancel out the \((0.796 - 0.204)\) term.
We are left with \(0.796 + 0.204 = 1.000 = 1\).
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The sum of the digits of a number is subtracted from the number. The resulting number is always divisible by:
Answer: D
This is a property related to the divisibility rule of 9. Let's take an example. Number = 42. Sum of digits = 6. Result = 42-6=36. 36 is divisible by 9.
Let the number be 10x+y. Sum of digits is x+y. The difference is (10x+y)-(x+y) = 9x. This is always divisible by 9.
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A number when successively divided by 4, 5 and 6 leaves remainders 2, 3 and 4 respectively. The least such number is:
Answer: A
Correct answer: 94
In successive division each quotient becomes the next dividend, so work backwards from the last step.
Let the final quotient be \(c\). Then, reading the divisions in reverse:
\[b = 6c + 4, \qquad a = 5b + 3, \qquad N = 4a + 2\]
The least number comes from \(c = 0\), which gives \(b = 4\), then \(a = 5(4) + 3 = 23\), and
\[N = 4(23) + 2 = 94\]
Verifying the chain: \(94 \div 4\) gives quotient 23 remainder 2; \(23 \div 5\) gives quotient 4 remainder 3; \(4 \div 6\) gives quotient 0 remainder 4.
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What is the value of \((469 + 174)^2 - (469 - 174)^2\) / \((469 \times 174)\)?
Answer: B
Use the identity for the difference of two squares of a sum and a difference.
\[ (a+b)^{2} - (a-b)^{2} = 4ab \]
\[ \frac{4ab}{ab} = 4 \]
The actual values 469 and 174 never have to be multiplied out - they cancel entirely.
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Which of the following is a rational number?
Answer: C
A rational number is a number that can be expressed as a fraction p/q of two integers, where q is not zero. An irrational number cannot be expressed as such.
√2, √3, and √5 are irrational.
√4 = 2, which can be written as 2/1. Therefore, it is a rational number.
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What is the result of \(0.000064 \div 0.016\)?
Answer: C
Clear the decimals by scaling both numbers by the same power of ten.
\[ \frac{0.000064}{0.016} = \frac{64}{16000} \]
\[ = 0.004 \]
The numerator has six decimal places and the denominator three, so the quotient has three.
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Find the value of \(99 + 999 + 9999 + 99999\).
Answer: A
Write each term as a power of ten less one.
\[ (100-1) + (1000-1) + (10{,}000-1) + (100{,}000-1) \]
\[ = 111{,}100 - 4 = 111{,}096 \]
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