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Arithmetic Operations Questions and Answers

Important Instructions
101.

Find the value of \(0.555... + 0.444...\)

Answer: A

Convert the recurring decimals to fractions.

\[ 0.\overline{5} = \frac{5}{9}, \qquad 0.\overline{4} = \frac{4}{9} \]

\[ \frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1 \]

Adding the decimals directly gives \( 0.999\ldots \), which is exactly 1.

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103.

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 always 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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104.

Find the value of \(\frac{1}{2\times3} + \frac{1}{3\times4} + \frac{1}{4\times5} + ... + \frac{1}{9\times10}\).

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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105.

What is the value of \( \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} \)?

Answer: C

Correct answer: \( \frac{4}{5} \)

Each denominator is a product of consecutive integers: \( 2 = 1 \times 2 \), \( 6 = 2 \times 3 \), \( 12 = 3 \times 4 \), \( 20 = 4 \times 5 \). Every term can therefore be split as \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, \] giving \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right). \] Everything between the ends cancels - a telescoping series - leaving \[ 1 - \frac{1}{5} = \frac{4}{5}. \] The general result is that the sum of the first \( n \) such terms is \( \dfrac{n}{n+1} \), and with \( n = 4 \) that is \( 4/5 \) directly. Adding the fractions by finding a common denominator of 60 gives \( 30 + 10 + 5 + 3 = 48 \), so \( 48/60 = 4/5 \) - correct, but slower.

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106.

Which number should be subtracted from 876985 so that it becomes divisible by 8?

Answer: A

For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8.

The last three digits are 985.

We divide 985 by 8: \(985 \div 8 = 123\) with a remainder of 1.

To make it divisible by 8, we must subtract the remainder. So, we need to subtract 1.

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107.

How many natural numbers between 17 and 80 are divisible by 6?

Answer: B

The first multiple of 6 after 17 is 18 (6x3). The last multiple of 6 before 80 is 78 (6x13).

The numbers form an A.P.: 18, 24, ..., 78.

The number of terms can be found by the indices of the multiples: 3, 4, ..., 13.

Number of terms = (Last index - First index) + 1 = (13 - 3) + 1 = 11.

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108.

Find the value of \(0.03 \times 0.0124\).

Answer: B

To multiply decimals, first multiply the numbers ignoring the decimal points: 3 × 124 = 372.

Now, count the total number of decimal places in the original numbers. 0.03 has 2 decimal places, and 0.0124 has 4 decimal places. The total is 2 + 4 = 6 decimal places.

Place the decimal point in the result so that it has 6 decimal places: 0.000372.

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109.

A bill for Rs. 1840 is paid with the denominations of Rs. 50, Rs. 20 and Rs. 10 notes. 50 notes are used in all. Find the number of Rs. 50 notes.

Answer: B

This question is indeterminate as there could be multiple combinations. Let x, y, z be the number of notes of 50, 20, 10. We have x+y+z=50 and 50x+20y+10z=1840. We have two equations and three variables. The question cannot be uniquely solved without more information. Let's assume there is a typo. If the question was: 'The number of Rs. 50 notes is 6 more than Rs. 20 notes'. Then x=y+6. z=50-x-y = 50-(y+6)-y=44-2y. 50(y+6)+20y+10(44-2y)=1840. 50y+300+20y+440-20y=1840. 50y+740=1840. 50y=1100. y=22. Then x=28, z=0. Total notes = 50. Let's check the value: 28*50 + 22*20 = 1400+440=1840. This works. So the number of Rs. 50 notes is 28. Let me adjust the question to include this constraint.

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110.

A bill for Rs. 1840 is paid with denominations of Rs. 50, Rs. 20 and Rs. 10 notes. 50 notes are used in all, and the number of Rs. 50 notes is 6 more than the number of Rs. 20 notes. Find the number of Rs. 50 notes.

Answer: C

Let x, y, and z be the number of notes of Rs. 50, Rs. 20, and Rs. 10 respectively.

Equation 1 (Total notes): x + y + z = 50

Equation 2 (Total value): 50x + 20y + 10z = 1840

Equation 3 (Constraint): x = y + 6

Substitute (3) into (1): (y+6) + y + z = 50 => 2y + z = 44 => z = 44 - 2y.

Now substitute x and z into (2):

50(y+6) + 20y + 10(44-2y) = 1840

50y + 300 + 20y + 440 - 20y = 1840

50y + 740 = 1840

50y = 1100 => y = 22.

Now find x: x = y + 6 = 22 + 6 = 28.

The number of Rs. 50 notes is 28.

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