Arithmetic Operations Questions and Answers

Q: The sum of the place values of 7 in the number 470,752 is:

Correct answer: 70,700The place value of a digit is the digit multiplied by the value of the position it occupies. In 470,752 the digit 7 appears twice. The first is in the ten thousands place, giving \( 7 \times 10000 = 70000 \). The second is in the hundreds place, giving \( 7 \times 100 = 700 \). Their sum is \[ 70000 + 700 = 70700. \] The distinction to keep clear is between place value and face value. The face value of a digit is the digit itself, unchanged wherever it stands - so the sum of the face values of 7 here is simply \( 7 + 7 = 14 \). Reading the positions carefully is what the question tests, and the usual error is misplacing a digit by one position. Writing the number with its places labelled - 4 lakhs, 7 ten-thousands, 0 thousands, 7 hundreds, 5 tens, 2 units - removes the risk entirely.

Q: If \(x/y = 6/5\), the value of \(\frac{x^2+y^2}{x^2-y^2}\) is:

Take \( x = 6k \) and \( y = 5k \).\[ \frac{x^{2}+y^{2}}{x^{2}-y^{2}} = \frac{36k^{2}+25k^{2}}{36k^{2}-25k^{2}} \]\[ = \frac{61}{11} \]The \( k^{2} \) cancels, so only the ratio matters.

Q: The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

Let the numbers be 27a and 27b, where a and b are co-prime.Sum = 27a + 27b = 21627(a + b) = 216a + b = 216 / 27 = 8.We need to find pairs of co-prime numbers that sum to 8.The pairs are (1, 7) and (3, 5). The pair (2, 6) is not co-prime, and (4, 4) is not co-prime.There are 2 such pairs.

Q: Evaluate: \(1.07 \times 65 + 1.07 \times 26 + 1.07 \times 9\)

Take out the common factor.\[ 1.07(65 + 26 + 9) = 1.07 \times 100 \]\[ = 107 \]The three numbers were chosen to total exactly 100, which is what makes the factorisation worth spotting.

Q: What is the value of \(0.1 \times 0.01 \times 1000\)?

Count the decimal places as you go.\[ 0.1 \times 0.01 = 0.001 \]\[ 0.001 \times 1000 = 1 \]Equivalently \( 10^{-1} \times 10^{-2} \times 10^{3} = 10^{0} = 1 \).

Q: What is the result of dividing 0.025 by 100?

Dividing by 100 moves the decimal point two places left.\[ \frac{0.025}{100} = 0.00025 \]

Q: Half of 1 percent, written as a decimal, is:

\[ 1\% = 0.01 \]\[ \frac{1}{2} \times 0.01 = 0.005 \]Half of one per cent is 0.5%, not 0.5 - the per cent sign is worth a factor of 100.

Important Instructions

31.

The sum of the place values of 7 in the number 470,752 is:

A.70,700
B.70,007
C.7,700
D.77,000

Answer: A

Correct answer: 70,700

The place value of a digit is the digit multiplied by the value of the position it occupies. In 470,752 the digit 7 appears twice. The first is in the ten thousands place, giving \( 7 \times 10000 = 70000 \). The second is in the hundreds place, giving \( 7 \times 100 = 700 \). Their sum is \[ 70000 + 700 = 70700. \] The distinction to keep clear is between place value and face value. The face value of a digit is the digit itself, unchanged wherever it stands - so the sum of the face values of 7 here is simply \( 7 + 7 = 14 \). Reading the positions carefully is what the question tests, and the usual error is misplacing a digit by one position. Writing the number with its places labelled - 4 lakhs, 7 ten-thousands, 0 thousands, 7 hundreds, 5 tens, 2 units - removes the risk entirely.

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32.

If \(x/y = 6/5\), the value of \(\frac{x^2+y^2}{x^2-y^2}\) is:

A.11/61
B.61/11
C.31/1
D.1/31

Answer: B

Take \( x = 6k \) and \( y = 5k \).

\[ \frac{x^{2}+y^{2}}{x^{2}-y^{2}} = \frac{36k^{2}+25k^{2}}{36k^{2}-25k^{2}} \]

\[ = \frac{61}{11} \]

The \( k^{2} \) cancels, so only the ratio matters.

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33.

A number increased by 37.5% gives 99. The number is:

A.68
B.72
C.75
D.80

Answer: B

Let the number be x. An increase of 37.5% means the new number is 137.5% of the original.

Note that 37.5% = 3/8.

So, the new number is \(x + \frac{3}{8}x = \frac{11}{8}x\).

\(\frac{11}{8}x = 99\)

\(x = 99 \times \frac{8}{11} = 9 \times 8 = 72\).

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34.

The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

Answer: B

Let the numbers be 27a and 27b, where a and b are co-prime.

Sum = 27a + 27b = 216

27(a + b) = 216

a + b = 216 / 27 = 8.

We need to find pairs of co-prime numbers that sum to 8.

The pairs are (1, 7) and (3, 5). The pair (2, 6) is not co-prime, and (4, 4) is not co-prime.

There are 2 such pairs.

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35.

Evaluate: \(1.07 \times 65 + 1.07 \times 26 + 1.07 \times 9\)

A.10.7
B.107
C.1070
D.1.07

Answer: B

Take out the common factor.

\[ 1.07(65 + 26 + 9) = 1.07 \times 100 \]

\[ = 107 \]

The three numbers were chosen to total exactly 100, which is what makes the factorisation worth spotting.

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36.

What is the value of \(0.1 \times 0.01 \times 1000\)?

A.1
B.0.1
C.0.01
D.10

Answer: A

Count the decimal places as you go.

\[ 0.1 \times 0.01 = 0.001 \]

\[ 0.001 \times 1000 = 1 \]

Equivalently \( 10^{-1} \times 10^{-2} \times 10^{3} = 10^{0} = 1 \).

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37.

A student has to secure 40% marks to pass. He gets 178 marks and fails by 22 marks. The maximum marks are:

A.400
B.500
C.800
D.1000

Answer: B

The student gets 178 marks and fails by 22 marks. This means the passing marks are 178 + 22 = 200.

Let the maximum marks be M.

We are given that the passing mark is 40% of the maximum marks.

So, 40% of M = 200.

\(0.40 \times M = 200\).

\(M = \frac{200}{0.40} = \frac{2000}{4} = 500\).

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38.

What is the result of dividing 0.025 by 100?

A.2.5
B.0.25
C.0.0025
D.0.00025

Answer: D

Dividing by 100 moves the decimal point two places left.

\[ \frac{0.025}{100} = 0.00025 \]

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39.

A number is first increased by 20% and then decreased by 10%. Find the net percentage change.

A.10% increase
B.10% decrease
C.8% increase
D.8% decrease

Answer: C

We can use the successive percentage change formula: \(A + B + \frac{AB}{100}\).

Here, A = +20 (increase) and B = -10 (decrease).

Net change = \(20 + (-10) + \frac{20 \times (-10)}{100}\)

= \(10 - \frac{200}{100} = 10 - 2 = 8\)%.

Since the result is positive, it is an 8% increase.

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40.

Half of 1 percent, written as a decimal, is:

A.0.005
B.0.05
C.0.02
D.0.2

Answer: A

\[ 1\% = 0.01 \]

\[ \frac{1}{2} \times 0.01 = 0.005 \]

Half of one per cent is 0.5%, not 0.5 - the per cent sign is worth a factor of 100.

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Tags: Arithmetic Operations Questions and Answers || Arithmetic Operations MCQs || Basic Arithmetic Quantitative Aptitude || Arithmetic Problems GK || Quantitative Aptitude Arithmetic

Arithmetic Operations Questions and Answers

Important Instructions

Topics

Other Subjects

Tags: Arithmetic Operations Questions and Answers || Arithmetic Operations MCQs || Basic Arithmetic Quantitative Aptitude || Arithmetic Problems GK || Quantitative Aptitude Arithmetic

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