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Operations on Numbers Questions and Answers

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41.

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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42.

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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43.

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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44.

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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45.

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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46.

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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47.

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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48.

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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49.

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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50.

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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