What is the unit digit in the product \((365 \times 459 \times 781)\)?
Answer: C
To find the unit digit of a product, we only need to multiply the unit digits of the numbers.
The unit digits are 5 (from 365), 9 (from 459), and 1 (from 781).
Unit digit of the product = Unit digit of \((5 \times 9 \times 1)\).
Unit digit of \((45 \times 1)\) = Unit digit of 45, which is 5.
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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 value of \(\frac{0.1 \times 0.1 \times 0.1 + 0.02 \times 0.02 \times 0.02}{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}\)?
Answer: A
Let a = 0.1 and b = 0.02. The numerator is \(a^3 + b^3\).
The denominator can be written in terms of a and b. \(0.2 = 2 \times 0.1 = 2a\) and \(0.04 = 2 \times 0.02 = 2b\).
The denominator is \((2a)^3 + (2b)^3 = 8a^3 + 8b^3 = 8(a^3+b^3)\).
The expression is \(\frac{a^3+b^3}{8(a^3+b^3)} = \frac{1}{8}\).
\(\frac{1}{8} = 0.125\).
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A number consists of two digits. If the digits interchange places and the new number is added to the original number, the resulting number will be divisible by:
Answer: B
Let the two-digit number be 10x + y, where x is the ten's digit and y is the unit's digit.
The reversed number is 10y + x.
Their sum is \((10x + y) + (10y + x) = 11x + 11y\).
Sum = 11(x + y).
Since the sum is a multiple of 11, it will always be divisible by 11.
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What is the sum of all positive integers from 1 to 100, inclusive, that are divisible by 3?
Answer: A
The numbers are 3, 6, 9, ..., 99. This is an AP.
First term a = 3, last term l = 99.
Number of terms n = \(99/3 = 33\).
Sum = \(\frac{n}{2}(a+l) = \frac{33}{2}(3+99) = \frac{33}{2}(102) = 33 \times 51 = 1683\).
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Simplify: \(1 + \frac{1}{1+\frac{1}{1+\frac{1}{2}}}\)
Answer: A
We solve this continued fraction from the bottom up.
Step 1: \(1+\frac{1}{2} = \frac{3}{2}\).
Step 2: The expression becomes \(1 + \frac{1}{1+\frac{1}{3/2}} = 1 + \frac{1}{1+\frac{2}{3}}\).
Step 3: \(1+\frac{2}{3} = \frac{5}{3}\).
Step 4: The expression becomes \(1 + \frac{1}{5/3} = 1 + \frac{3}{5}\).
Step 5: \(1 + \frac{3}{5} = \frac{5+3}{5} = \frac{8}{5}\).
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A student had to score 40% marks to pass. He gets 40 marks and fails by 40 marks. The maximum marks for the exam are:
Answer: C
The student gets 40 marks and fails by 40 marks. This means the passing marks are 40 + 40 = 80.
Let the maximum marks be M.
We are given that the passing mark is 40% of the maximum marks.
So, 40% of M = 80.
\(0.40 \times M = 80\).
\(M = \frac{80}{0.40} = \frac{8000}{40} = 200\).
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The product of two co-prime numbers is 117. Their LCM should be:
Answer: B
Co-prime numbers are numbers whose Highest Common Factor (HCF) is 1.
We know that for any two numbers a and b, Product of numbers = HCF(a,b) × LCM(a,b).
Given, Product = 117 and HCF = 1 (since they are co-prime).
117 = 1 × LCM.
Therefore, LCM = 117.
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What is the cube root of 0.000216?
Answer: B
We can write \(0.000216\) as \(\frac{216}{1000000}\).
The cube root is \(\sqrt[3]{\frac{216}{1000000}} = \frac{\sqrt[3]{216}}{\sqrt[3]{1000000}}\).
We know that \(6^3 = 216\) and \(100^3 = 1000000\).
So, \(\frac{6}{100} = 0.06\).
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How many times does the digit '3' appear in numbers from 1 to 100?
Answer: C
Let's count the occurrences of the digit '3'.
In the units place: 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 (10 times).
In the tens place: 30, 31, 32, 33, 34, 35, 36, 37, 38, 39 (10 times).
The number 33 is counted in both lists. So the total count is 10 + 10 = 20.
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