Informative line

Operations On Rational Numbers

Addition of Rational Numbers

For the addition of the rational numbers, we follow the sign rule in which we consider different cases.

Case I: When two rational numbers are having the same sign.

Let's consider an example:

\(\left(-2\dfrac{1}{3}\right)+\left(-2\dfrac{7}{3}\right)\)

Step 1: Calculate their absolute values.

\(\left|-2\dfrac{1}{3}\right|=2\dfrac{1}{3}=\dfrac{7}{3}\)

\(\left|-2\dfrac{7}{3}\right|=2\dfrac{7}{3}=\dfrac{13}{3}\)

Step 2: Add the absolute values.

\(\dfrac{7}{3}+\dfrac{13}{3}=\dfrac{20}{3}\)

Step 3: Put the common sign to the resulting number.

\(\dfrac{-20}{3}\)

Thus, \(\left(-2\dfrac{1}{3}\right)+\left(-2\dfrac{7}{3}\right)=\dfrac{-20}{3}\)

Case II: When two rational numbers are having the opposite sign.

Let's consider an example:

\(\left(3\dfrac{2}{5}\right)+\left(-5\dfrac{3}{4}\right)\)

Step 1: We calculate the absolute values of both the rational numbers.

\(\left|3\dfrac{2}{5}\right|=3\dfrac{2}{5}=\dfrac{17}{5}\)

\(\left|-5\dfrac{3}{4}\right|=5\dfrac{3}{4}=\dfrac{23}{4}\)

We should have equivalent rational numbers since, denominators of both are different.

\(\dfrac{17}{5}×\dfrac{4}{4}=\dfrac{68}{20}\)

\(\dfrac{23}{4}×\dfrac{5}{5}=\dfrac{115}{20}\)

Step 2: Subtract the smaller absolute value from the higher absolute value.

\(\dfrac{115}{20}-\dfrac{68}{20}=\dfrac{47}{20}\)

Step 3: Put the sign of the higher rational number to the resulting value.

\(\dfrac{-47}{20}\)

Thus, \(\left(3\dfrac{2}{5}\right)+\left(-5\dfrac{3}{4}\right)=\dfrac{-47}{20}\)

Illustration Questions

Evaluate: \(\dfrac{29}{18}+\left(\dfrac{-1}{3}\right)\)

A \(1\dfrac{5}{18}\)

B \(2\dfrac{3}{18}\)

C \(\dfrac{39}{6}\)

D \(1\dfrac{4}{18}\)

×

Given: \(\dfrac{29}{18}+\left(\dfrac{-1}{3}\right)\)

Here, both the rational numbers have opposite signs.

\(\dfrac{29}{18}+\left(\dfrac{-1}{3}\right)\)

Calculating their absolute values,

\(\left|\dfrac{29}{18}\right|=\dfrac{29}{18}\)

\(\left|-\dfrac{1}{3}\right|=\dfrac{1}{3}\)

Converting the given numbers into equivalent rational numbers having the same denominator.

\(\dfrac{29}{18}×\dfrac{2}{2}=\dfrac{58}{36}\)

\(\dfrac{1}{3}×\dfrac{12}{12}=\dfrac{12}{36}\)

Now, subtracting the smaller absolute value from the higher absolute value.

\(\dfrac{58}{36}-\dfrac{12}{36}=\dfrac{46}{36}=\dfrac{23}{18}\)

Putting the sign of the higher rational number to the resulting value, i.e. positive sign.

Thus, \(\dfrac{29}{18}+\left(-\dfrac{1}{3}\right)=\dfrac{23}{18}\)

It can also be written as:

\(1\dfrac{5}{18}\)

Hence, option (A) is correct.

Evaluate: \(\dfrac{29}{18}+\left(\dfrac{-1}{3}\right)\)

A

\(1\dfrac{5}{18}\)

.

B

\(2\dfrac{3}{18}\)

C

\(\dfrac{39}{6}\)

D

\(1\dfrac{4}{18}\)

Option A is Correct

Subtraction of Rational Numbers

  • We can consider the subtraction problem as the addition problem after adding the additive inverse of that number which is to be subtracted from the first one.

For example: \(\dfrac{9}{4}-\dfrac{2}{3}\)

We can also write it as \(\dfrac{9}{4}+\left(\dfrac{-2}{3}\right)\)

Here, \(\dfrac{-2}{3}\) is the additive inverse of \(\dfrac{2}{3}.\)

Note: 

Additive inverse: It means adding the number with its opposite number which gives the result as zero.

For example:

\(\dfrac{2}{5}+\left(\dfrac{-2}{5}\right)=0\)

Thus, \(\dfrac{-2}{5}\) is the additive inverse of \(\dfrac{2}{5}.\)

To understand the subtraction of rational numbers, we consider an example: \(\dfrac{-23}{19}-\left(\dfrac{-12}{17}\right)\)

Step 1: We take the additive inverse of \(\left(\dfrac{-12}{17}\right)\), i.e. \(\dfrac{12}{17}\).

Now, we convert the subtraction into the addition by adding additive inverse.

Thus, we can write it as, \(\dfrac{-23}{19}+\left(\dfrac{12}{17}\right)\)

Step 2: Now, we apply the addition rule.

First we calculate the absolute values of both the numbers.

\(\left|\dfrac{-23}{19}\right|=\dfrac{23}{19}\)

\(\left|\dfrac{12}{17}\right|=\dfrac{12}{17}\)

Here, the denominators are not same. So, we convert the given numbers into equivalent rational numbers having the same denominator.

\(\dfrac{23}{19}×\dfrac{17}{17}=\dfrac{391}{323}\)

\(\dfrac{12}{17}×\dfrac{19}{19}=\dfrac{228}{323}\)

Step 3: Now, we apply the sign rule. According to this,

(1) If the rational numbers have the same sign, add their absolute values and put the common sign to the result.

(2) If the rational numbers have different signs, subtract their absolute values and put the sign of the number having higher absolute value to the result.

Since, both the rational numbers having different signs, therefore subtract their absolute values.

\(\dfrac{391}{323}-\dfrac{228}{323}=\dfrac{163}{323}\)

Step 4: Now, put the sign of the rational number having higher absolute value.

\(\dfrac{-163}{323}\)

Illustration Questions

Evaluate: \(\left(9\dfrac{3}{7}\right)-\left(\dfrac{23}{9}\right)\)

A \(\dfrac{44}{63}\)

B \(7\dfrac{7}{32}\)

C \(6\dfrac{55}{63}\)

D \(\dfrac{431}{63}\)

×

Given: \(\left(9\dfrac{3}{7}\right)-\left(\dfrac{23}{9}\right)\)

We can write it as, \(\left(\dfrac{66}{7}\right)-\left(\dfrac{23}{9}\right)\)

Subtraction of a number is same as adding the additive inverse of that number.

\(\because\;\dfrac{23}{9}+\left(\dfrac{-23}{9}\right)=0\)

Thus, \(\left(\dfrac{66}{7}\right)-\left(\dfrac{23}{9}\right)\) can be written as:

\(\left(\dfrac{66}{7}\right)+\left(\dfrac{-23}{9}\right)\)

Now apply the rules of addition.

Since, both the numbers are having different signs, thus subtract their absolute values.

Now, calculating their absolute values.

\(\left|\dfrac{66}{7}\right|=\dfrac{66}{7}\)

\(\left|\dfrac{-23}{9}\right|=\dfrac{23}{9}\)

Thus, \(\dfrac{66}{7}-\dfrac{23}{9}\)

Since, the denominators are not same. So, we should have their equivalent rational number having the same denominator.

\(\dfrac{66}{7}×\dfrac{9}{9}=\dfrac{594}{63}\)

\(\dfrac{23}{9}×\dfrac{7}{7}=\dfrac{161}{63}\)

Thus, \(\dfrac{594}{63}-\dfrac{161}{63}=\dfrac{433}{63}\)

Putting the sign of the rational number having higher absolute value, i.e. positive sign.

\(\dfrac{433}{63}=6\dfrac{55}{63}\)

Hence, option (C) is correct.

Evaluate: \(\left(9\dfrac{3}{7}\right)-\left(\dfrac{23}{9}\right)\)

A

\(\dfrac{44}{63}\)

.

B

\(7\dfrac{7}{32}\)

C

\(6\dfrac{55}{63}\)

D

\(\dfrac{431}{63}\)

Option C is Correct

Addition of Rational Numbers on the Number Line

  • Addition of rational numbers can be performed by using a horizontal arrow on the number line.
  • Let's consider an example.

Add \(\dfrac{1}{3}\) and \(\dfrac{-1}{5}\) i.e. \(\dfrac{1}{3}+\left(\dfrac{-1}{5}\right)\)

Here, we can not add them up on the number line because both the numbers have different denominators.

Thus, convert them into their equivalent rational numbers having the same denominator.

\(\dfrac{1}{3}=\dfrac{1×5}{3×5}=\dfrac{5}{15}\)

\(\dfrac{-1}{5}=\dfrac{-1×3}{5×3}=\dfrac{-3}{15}\)

Now, \(\dfrac{5}{15}+\left(\dfrac{-3}{15}\right)\)

Step 1: Draw a number line where each small segment represents \(\dfrac{1}{15}\).

Step 2:  \(\dfrac{5}{15}\) is a positive number, thus move the arrow \(5\) units forward from zero.

Step 3:  \(\dfrac{-3}{15}\) is a negative number, thus move the arrow \(3\) units backward from the point \(\dfrac{5}{15}\).

Step 4: Thus, the answer is \(\dfrac{2}{15}\).

Illustration Questions

Alex paints \(\dfrac{1}{4}\) part of the wall and Carl paints \(\dfrac{2}{3}\) part of the wall. Which one of the following number lines represents point \('P'\) as the total part of the wall painted?

A

B

C

D

×

To solve this problem, we will add \(\dfrac{1}{4}\) and \(\dfrac{2}{3}.\)

For addition, the denominators of both the numbers should be same.

So, converting them into the equivalent rational numbers having the same denominator.

\(\dfrac{1}{4}×\dfrac{3}{3}=\dfrac{3}{12}\)

\(\dfrac{2}{3}×\dfrac{4}{4}=\dfrac{8}{12}\),

\(\dfrac{3}{12}+\dfrac{8}{12}\)

Draw a number line where each small segment represents \(\dfrac{1}{12}.\)

image

\(\dfrac{3}{12}\) is a positive number, thus move the arrow \(3\) units forward from zero.

image

\(\dfrac{8}{12}\) is a positive number, thus move the arrow \(8\) units forward from the point \(\dfrac{3}{12}\).

image

Thus, \(\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{11}{12}\)

Hence, option (B) is correct.

Alex paints \(\dfrac{1}{4}\) part of the wall and Carl paints \(\dfrac{2}{3}\) part of the wall. Which one of the following number lines represents point \('P'\) as the total part of the wall painted?

A image
B image
C image
D image

Option B is Correct

Subtraction of Rational Numbers on the Number Line

  • Subtraction of rational numbers can be done easily on the number line with the help of the horizontal arrow.
  • We can done subtraction same as addition by taking the inverse of the second term.
  • Let's consider an example:

\(\dfrac{11}{5}-\left(\dfrac{7}{5}\right)\\\dfrac{11}{5}+\left(\dfrac{-7}{5}\right)\)

Step 1: Here, the denominators of both numbers are same so, we don't need to write their equivalent rational numbers.

Now, draw a number line where each small segment represents \(\dfrac{1}{5}\).

Step 2: Check whether the numbers are positive or negative. Now, start the arrow from zero and reach to the first number.

\(\dfrac{11}{5}\) is a positive number, thus move the arrow \(11\) units forward to the right.

Step 3: Now add \(\dfrac{-7}{5}\) to \(\dfrac{11}{5}.\)

\(\dfrac{-7}{5}\) is a negative number, thus move the arrow \(7\) units backward (to the left) from the point \(\dfrac{11}{5}\).

Step 4: Thus, \(\dfrac{11}{5}-\left(\dfrac{7}{5}\right)=\dfrac{4}{5}\)

So, the answer is \(\dfrac{4}{5}\).

Illustration Questions

To reach the school bus stop, Kevin walks \(\dfrac{3}{10}\) miles from his house. If the total distance between his school and house is \(2\dfrac{2}{5}\) miles, which one of the number lines represents the distance traveled by bus as point \(K\)?

A

B

C

D

×

Given: 

Distance covered by Kevin before taking the school bus \(=\dfrac{3}{10}\) miles

Total distance between Kevin's house and school \(=2\dfrac{2}{5}\) miles

Distance traveled by bus (as point \(K\)) = (Total distance between Kevin's house and school) – (Distance covered by Kevin before taking the school bus)

\(=\left(2\dfrac{2}{5}-\dfrac{3}{10}\right)\)

To solve \(2\dfrac{2}{5}-\dfrac{3}{10}\) on the number line both the numbers are to be changed into proper fractions.

\(2\dfrac{2}{5}=\dfrac{5×2+2}{5}=\dfrac{12}{5}\)

Now, the problem becomes \(\dfrac{12}{5}-\dfrac{3}{10}.\)

But we cannot subtract them on the number line because both the numbers have different denominators.

Thus, converting \(\dfrac{12}{5}\) into an equivalent rational number having the denominator \(10\).

\(\dfrac{12}{5}=\dfrac{12×2}{5×2}=\dfrac{24}{10}\)

Now, \(\dfrac{24}{10}-\dfrac{3}{10}\)

Draw a number line where each segment represents \(\dfrac{1}{10}.\)

image

Check whether the numbers are positive or negative.

Now, start the arrow from zero and reach to the first number.

\(\dfrac{24}{10}\) is a positive number, thus move the arrow \(24\) units forward to the right.

image

Now, add \(\dfrac{-3}{10}\) to \(\dfrac{24}{10}.\)

\(\dfrac{-3}{10}\) is a negative number, thus move the arrow \(3\) units backward (to the left) from the point \(\dfrac{24}{10}.\)

image

Thus, \(2\dfrac{2}{5}-\dfrac{3}{10}=\dfrac{21}{10}\)

Hence, option (A) is correct.

To reach the school bus stop, Kevin walks \(\dfrac{3}{10}\) miles from his house. If the total distance between his school and house is \(2\dfrac{2}{5}\) miles, which one of the number lines represents the distance traveled by bus as point \(K\)?

A image
B image
C image
D image

Option A is Correct

Multiplication of Rational Numbers

While multiplying rational numbers, we should follow the given rules:

Rule 1: If both the rational numbers are having the same sign then the result has a positive sign.

To understand it in a better way, we consider an example:

\(\left(\dfrac{-1}{3}\right)\;\left(\dfrac{-1}{4}\right)\)

Step-1: Calculate their absolute values.

\(\left|\dfrac{-1}{3}\right|=\dfrac{1}{3},\;\left|\dfrac{-1}{4}\right|=\dfrac{1}{4}\)

Step-2: Multiply their absolute values.

\(\dfrac{1}{3}×\dfrac{1}{4}=\dfrac{1}{12}\)

Step-3: Both the rational numbers have the same sign.

So, the result is positive.

Thus, \(\left(\dfrac{-1}{3}\right)×\left(\dfrac{-1}{4}\right)=\dfrac{1}{12}\)

Rule 2: If both the rational numbers are having different signs then the result has a negative sign.

To understand it in a better way, we consider an example:

\(\left(\dfrac{-1}{2}\right)×\left(\dfrac{3}{8}\right)\)

Step-1: Calculate their absolute values.

\(\left|\dfrac{-1}{2}\right|=\dfrac{1}{2}\)

\(\left|\dfrac{3}{8}\right|=\dfrac{3}{8}\)

Step-2: Multiply their absolute values.

\(\dfrac{1}{2}×\dfrac{3}{8}=\dfrac{3}{16}\)

Step-3: Both the rational numbers have different signs.

So, the result is negative.

Thus, \(\left(\dfrac{-1}{2}\right)×\left(\dfrac{3}{8}\right)=\dfrac{-3}{16}\)

Note: Multiplication is a repetitive addition.

Example:

\(\left(\dfrac{-3}{2}\right)×(2)=\left(\dfrac{-3}{2}\right)+\left(\dfrac{-3}{2}\right)\)

\(=\dfrac{-6}{2}\)

Illustration Questions

Evaluate: \(\left(\dfrac{-36}{20}\right)\;\left(\dfrac{4}{-9}\right)\)

A \(18\dfrac{3}{4}\)

B \(\dfrac{4}{5}\)

C \(4\dfrac{1}{5}\)

D \(\dfrac{44}{180}\)

×

Calculating the absolute values of the given rational numbers.

\(\left|\dfrac{-36}{20}\right|=\dfrac{36}{20}\\\left|\dfrac{4}{-9}\right|=\dfrac{4}{9}\)

Multiplying the absolute values.

\(\dfrac{36}{20}\times \dfrac{4}{9}=\dfrac{144}{180}=\dfrac{4}{5}\)

Both the rational numbers have the same signs, so the result is positive.

Thus,

\(\left(\dfrac{-36}{20}\right)\times\left(\dfrac{4}{-9}\right)=\dfrac{4}{5}\)

Hence, option (B) is correct.

Evaluate: \(\left(\dfrac{-36}{20}\right)\;\left(\dfrac{4}{-9}\right)\)

A

\(18\dfrac{3}{4}\)

.

B

\(\dfrac{4}{5}\)

C

\(4\dfrac{1}{5}\)

D

\(\dfrac{44}{180}\)

Option B is Correct

Division of Rational Numbers

  • Division is the inverse process of multiplication which means that dividing by a number is same as multiplying with its reciprocal.
  • Reciprocal of \(\dfrac{2}{3}\) is \(\dfrac{3}{2}.\)
  • It is also called multiplicative inverse.

To understand it in a better way, consider an example.

Example: Evaluate: \(-\dfrac{105}{148}\div\dfrac{7}{4}\)

Step 1: Calculate the absolute values.

\(\left|\dfrac{-105}{148}\right|=\dfrac{105}{148}\)

\(\left|\dfrac{7}{4}\right|=\dfrac{7}{4}\)

Step 2: Find the reciprocal of the absolute value of the divisor.

Reciprocal of \(\dfrac{7}{4}\) is \(\dfrac{4}{7}.\)

Step 3: Multiply the absolute value of dividend with the reciprocal of the absolute value of the divisor.

\(\dfrac{105}{148}×\dfrac{4}{7}=\dfrac{420}{1036}\)

\(=\dfrac{420}{1036}\)

Simplify: \(\dfrac{420\div28}{1036\div28}=\dfrac{15}{37}\)

Step 4: Apply the sign rule which is given below.

The sign rule is defined by \(2\) different cases.

Case-I: If both dividend and the divisor have the same sign then the quotient is positive.

Example:

\((-a)\div(-b)=\) Positive quotient

Case-II: If both dividend and the divisor have opposite signs then the quotient is negative.

Example:

\((a)\div(-b)=\) Negative quotient

Now, applying the sign rule,

\(\dfrac{-105}{148}\div\dfrac{7}{4}\)

Both dividend and the divisor are having opposite signs, so the quotient is negative.

\(\dfrac{-15}{37}\)

Thus, \(\left(\dfrac{-105}{148}\right)\div\left(\dfrac{7}{4}\right)=\dfrac{-15}{37}\)

Illustration Questions

Evaluate: \(15\dfrac{16}{17}\div\dfrac{19}{-51}\)

A \(-42\dfrac{15}{19}\)

B \(27\dfrac{8}{17}\)

C \(-271\)

D \(\dfrac{-271}{19}\)

×

Given: \(15\dfrac{16}{17}\div\dfrac{19}{-51}\)

Converting \(15\dfrac{16}{17}\) into an improper fraction,

\(15\dfrac{16}{17}=\dfrac{(15×17+16)}{17}\)

\(=\dfrac{255+16}{17}=\dfrac{271}{17}\)

Now, the problem can be written as \(\dfrac{271}{17}\div\dfrac{19}{-51}\)

Here, dividend \(=\dfrac{271}{17}\)

and divisor \(=\dfrac{19}{-51}\)

Calculate the absolute values.

\(\left|\dfrac{271}{17}\right|=\dfrac{271}{17},\;\left|\dfrac{19}{-51}\right|=\dfrac{19}{51}\)

Find the reciprocal of the absolute value of the divisor.

Reciprocal of \(\dfrac{19}{51}=\dfrac{51}{19}\)

Multiply the absolute value of dividend with the reciprocal of the absolute value of the divisor.

\(\dfrac{271}{17}×\dfrac{51}{19}\)

\(=\dfrac{271×3}{1×19}\)

\(=\dfrac{813}{19}\)

\(\dfrac{813}{19}\) can also be written as \(42\dfrac{15}{19}.\)

Apply the sign rule.

Both dividend and the divisor are having opposite signs, so the quotient is negative.

\(\therefore\) Quotient \(=-42\dfrac{15}{19}\)

Thus, \(15\dfrac{16}{17}\div\dfrac{19}{-51}=-42\dfrac{15}{19}\)

Hence, option (A) is correct.

Evaluate: \(15\dfrac{16}{17}\div\dfrac{19}{-51}\)

A

\(-42\dfrac{15}{19}\)

.

B

\(27\dfrac{8}{17}\)

C

\(-271\)

D

\(\dfrac{-271}{19}\)

Option A is Correct

Properties of Addition, Subtraction, Multiplication

  • We apply properties to simplify the expressions so that they become easier to solve.
  • Here, we are going to learn the properties of addition and multiplication.

Properties of Addition

(1) Associative property of addition: It states that changing the grouping of the addends (terms) will not change the sum.

\((a+b)+c=a+(b+c)\)

For example:

\(\left(\dfrac{1}{5}+\dfrac{2}{5}\right)+\dfrac{4}{5}=\dfrac{1}{5}+\left(\dfrac{2}{5}+\dfrac{4}{5}\right)=\dfrac{7}{5}\)

 

(2) Commutative property of addition: It states that if we change the position of the numbers being added, the result is same.

\(a+b=b+a\)

For example:

\(\dfrac{13}{15}+\dfrac{14}{15}=\dfrac{14}{15}+\dfrac{13}{15}=\dfrac{27}{15}\)

 

(3) Identity property of addition: It states that when zero is added to a number, the result is the number itself.

\(a+0=0+a=a\)

For example:

\(\dfrac{2}{5}+0=0+\dfrac{2}{5}=\dfrac{2}{5}\)

 

(4) Inverse property of addition: It states that when a number is added to its opposite, the sum is zero.

It is also called the property of additive inverse.

\(a+(-a)=(-a)+a=0\)

For example:

\(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)=\left(\dfrac{-4}{5}\right)+\dfrac{4}{5}=0\)

 

Properties of Multiplication

(1) Associative property of multiplication: It states that it does not matter how we group the numbers (i.e. where we put parentheses) but on multiplying them the product remains the same.

\((a.b).c=a.(b.c)\)

For example:

\(\left(\dfrac{1}{3}\cdot\dfrac{1}{4}\right)\cdot\dfrac{1}{5}=\dfrac{1}{3}\cdot\left(\dfrac{1}{4}\cdot\dfrac{1}{5}\right)=\dfrac{1}{60}\)

 

(2) Commutative property of multiplication: It states that while multiplying if we change the position of the numbers, the result remains the same.

\(a.b=b.a\)

For example:

\(\dfrac{1}{3}\cdot\dfrac{1}{7}=\dfrac{1}{7}\cdot\dfrac{1}{3}=\dfrac{1}{21}\)

 

(3) Identity property of multiplication:  It states that when a number is multiplied by \(1,\) the result is the number itself.

\(a\cdot1=1\cdot a=a\)

For example:

\(\dfrac{11}{12}\cdot1=1\cdot\dfrac{11}{12}=\dfrac{11}{12}\)

 

(4) Inverse property of multiplication: It states that when a number is multiplied by its reciprocal, the result will always be \(1.\)

\(b\cdot\dfrac{1}{b}=\dfrac{1}{b}\cdot b=1\)

where \(b\neq0\)

For example:

\(\dfrac{15}{2}\cdot\dfrac{2}{15}=\dfrac{2}{15}\cdot\dfrac{15}{2}=1\)

Distributive property

Distributive property of multiplication over addition

  • The property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the product together.

\(p\cdot(q+r)=p\cdot q+p\cdot r\)

For example:

\(\dfrac{5}{2}\left(\dfrac{2}{5}+\dfrac{4}{5}\right)=\dfrac{5}{2}\left(\dfrac{2}{5}\right)+\dfrac{5}{2}\left(\dfrac{4}{5}\right)\)

\(=1+2=3\)

Distributive property of multiplication over subtraction

  • The distributive property of multiplication over subtraction is like the distributive property of multiplication over addition.
  • In this case, either we find out the difference first and then multiply or we first multiply with each number and then subtract.

\(p\cdot(q-r)=p\cdot q-p\cdot r\)

For example:

\(\dfrac{3}{7}\left(\dfrac{1}{6}-\dfrac{1}{9}\right)=\dfrac{3}{7}\left(\dfrac{1}{6}\right)-\dfrac{3}{7}\left(\dfrac{1}{9}\right)\)

\(=\dfrac{1}{14}-\dfrac{1}{21}\)

\(=\dfrac{7}{294}=\dfrac{1}{42}\)

Note:

Opposite of a sum is the sum of its opposites.

It means that if we have a sum and want to take its opposite.

For example:

\(\left(\dfrac{-7}{11}+\dfrac{(-2)}{11}\right)=\dfrac{-9}{11}\)

Now, opposite of \(\dfrac{-9}{11}\) is \(\dfrac{9}{11}\)  ... (1)

So, the sum of the opposites of the numbers being added is,

\(\dfrac{7}{11}+\dfrac{2}{11}=\dfrac{9}{11}\)  ... (2)

Here, we can see that the results of both (1) and (2) are same.

Illustration Questions

\(18\dfrac{4}{5}-\left((-13.5)+\dfrac{4}{5}\right)+\dfrac{13}{10}\;\;\;\text{step I}\) \(18\dfrac{4}{5}+\left(13.5+\left(\dfrac{-4}{5}\right)\right)+\dfrac{13}{10}\;\;\;\text{step II}\) \(18+\dfrac{4}{5}+\left(13.5+\left(\dfrac{-4}{5}\right)\right)+\dfrac{13}{10}\;\;\;\text{step III}\) \(18+\left(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)\right)+13.5+\dfrac{13}{10}\;\;\;\text{step IV}\) \(18+0+13.5+\dfrac{13}{10}\;\;\;\text{step V}\) \(18+13.5+\dfrac{13}{10}\;\;\;\text{step VI}\) Which property is used in step IV?

A Commutative property of addition

B Property of additive inverse

C Associative property of addition

D Property of additive identity

×

In the question we are asked about which property is used in step IV i.e.

\(18+\left(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)\right)+13.5+\dfrac{13}{10}\)

Comparing one step below (step IV) i.e. with step V

image

Observing, that the numbers \(18,\;13.5\) and \(\dfrac{13}{10}\) remain same in both the steps and \(\left(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)\right)=0\)

Here, \(\dfrac{4}{5}\) is added to its opposite number \(\left(\dfrac{-4}{5}\right)\) and the result is zero.

Since, the property of additive inverse states that if a number is added to its opposite, the result is zero.

Hence, option (B) is correct.

\(18\dfrac{4}{5}-\left((-13.5)+\dfrac{4}{5}\right)+\dfrac{13}{10}\;\;\;\text{step I}\) \(18\dfrac{4}{5}+\left(13.5+\left(\dfrac{-4}{5}\right)\right)+\dfrac{13}{10}\;\;\;\text{step II}\) \(18+\dfrac{4}{5}+\left(13.5+\left(\dfrac{-4}{5}\right)\right)+\dfrac{13}{10}\;\;\;\text{step III}\) \(18+\left(\dfrac{4}{5}+\left(\dfrac{-4}{5}\right)\right)+13.5+\dfrac{13}{10}\;\;\;\text{step IV}\) \(18+0+13.5+\dfrac{13}{10}\;\;\;\text{step V}\) \(18+13.5+\dfrac{13}{10}\;\;\;\text{step VI}\) Which property is used in step IV?

A

Commutative property of addition

.

B

Property of additive inverse

C

Associative property of addition

D

Property of additive identity

Option B is Correct

Real World Problems

  • In real world problems, observations and analyses are as important as calculations.
  • To solve a real world problem, consider the following steps:

Step 1: First of all, observe the problem to pick out the given data and identify what is being asked to calculate.

Step 2: Identify the relevant keywords.

Here, we are mentioning some of the keywords, through which we can identify which operations are to be performed.

(1) For addition:

In all, increased by, altogether, sum, combined, total, added to, both, together etc.

(2) For subtraction:

Take away, left, change, difference, fewer, reduce, exceed by, remain, decreased by, deduct, etc.

(3) For multiplication:

Twice (double), thrice (triple), times, of, product etc.

(4) For division:

Separated, distribute, per, into, equally, split, each, shared, divided, every, fractions

(Example: half, forth) etc.

Note:

Emphasis on opposite words that represent the positive and negative numbers.

above - below

low - high

south - north

give - take

Step 4: Identify what is being asked and move ahead to compute it.

  • To understand these steps, consider an example given below:

For example:

Winter vacations are going on.

Jessica and Carl are making snowballs to build a fort. Together they can make \(28\) snowballs in an hour, but \(\dfrac{1}{7}\) of the total snowballs made in one hour get melted in every \(16\) minutes. If they need \(260\) snowballs to build the fort then how long will it take them to build that fort (calculate in hours)?

Observe the problem and pick out the given data.

Given:

Total snowballs made in one hour \(=28\)

Total snowballs needed to build the fort \(=260\)

Number of snowballs get melted in every  \(16\) minutes \(=\dfrac{1}{7}\) of the total snowballs made in one hour

Here, "of" means multiply.

So, number of snowballs get melted in every \(16\) minutes \(=\dfrac{1}{7}×28\) \(=4\)

Step 2: Identify what is being asked.

Since, we are asked "how long will it take to build that fort?", therefore, we need to compute the time required to make \(260\) snowballs (in hours).

Step 3: Convert minutes into hours.

\(1\) minute \(=\dfrac{1}{60}\) hours

\(16\) minutes \(=\dfrac{1}{60}×16\)

\(=\dfrac{4}{15}\) hours

\(\therefore\) Number of snowballs get melted in \(\dfrac{4}{15}\) hours \(=4\)

Let the number of snowballs get melted in one hour \(=x\)

Now, solving the proportion to calculate \(x.\)

\(4×1=x×\dfrac{4}{15}\)

\(x=15\)

Step 4: Number of snowballs Jessica and Carl have in one hour = Snowballs made in one hour – snowballs melted in one hour

\(=28-15\)

\(=13\)

Let it takes them \(y\) hours to build that fort.

\(13\) snowballs made in \(1\) hour and \(260\) snowballs made in \(y\) hours.

\(\therefore\) Solving the proportion to calculate \(y.\)

\(y×13=260×1\)

\(y=20\) hours

Therefore, they will take \(20\) hours to build that fort.

Illustration Questions

A monkey is hopping from one branch to another of a tree. Robin is observing the monkey's movements. According to Robin, at first the monkey was on a branch that was \(21\dfrac{1}{4}\) feet above the ground, then it climbed up \(9\) feet, again climbed up \(4\dfrac{1}{2}\) feet and then jumped down \(17\) feet and sat there. At this time, Robin stopped observing and wrote down his observations. At what distance above the ground is the monkey when Robin stopped his observation?

A \(17\dfrac{3}{4}\) feet

B \(118\dfrac{1}{2}\) feet

C \(\dfrac{66}{8}\) feet

D \(71\dfrac{1}{4}\) feet

×

Starting position of the monkey is \(21\dfrac{1}{4}\) feet above the ground and 'above' shows positive i.e.

\(21\dfrac{1}{4}=\dfrac{(21×4)+1}{4}\)

\(=\dfrac{84+1}{4}=\dfrac{85}{4}\)

First, the monkey climbs up \(9\) feet. Here, 'climb up' shows addition.

Current position of the monkey \(=\dfrac{85}{4}+9\)        \(\Big\{\dfrac{9}{1}=\dfrac{9×4}{1×4}=\dfrac{36}{4}\Big\}\)

\(=\dfrac{85}{4}+\dfrac{36}{4}\)

\(=\dfrac{85+36}{4}\)

\(=\dfrac{121}{4}\)

Second, it again climbs up \(4\dfrac{1}{2}\) feet and 'climb up' shows addition.

Second position of the monkey \(=\dfrac{121}{4}+4\dfrac{1}{2}\)

\(\left\{4\dfrac{1}{2}=\dfrac{(4×2)+1}{2}=\dfrac{9}{2}\right\}\)

\(\left\{\dfrac{9}{2}=\dfrac{9×2}{2×2}=\dfrac{18}{4}\right\}\)

\(=\dfrac{121}{4}+\dfrac{18}{4}\)

\(=\dfrac{139}{4}\)

Third, it jumps down \(17\) feet. Here, jumps 'down' shows subtraction.

Third position of the monkey \(=\dfrac{139}{4}-17=\dfrac{139}{4}-\dfrac{17}{1}\)

\(\left\{\dfrac{17}{1}=\dfrac{17×4}{1×4}=\dfrac{68}{4}\right\}\)

\(=\dfrac{139}{4}-\dfrac{68}{4}\)

\(=\dfrac{139-68}{4}\)

\(=\dfrac{71}{4}\)

\(=17\dfrac{3}{4}\) feet

image

Robin stopped observing after the third position.

So, the monkey is at \(17\dfrac{3}{4}\) feet above the ground at that time.

Hence, option (A) is correct.

A monkey is hopping from one branch to another of a tree. Robin is observing the monkey's movements. According to Robin, at first the monkey was on a branch that was \(21\dfrac{1}{4}\) feet above the ground, then it climbed up \(9\) feet, again climbed up \(4\dfrac{1}{2}\) feet and then jumped down \(17\) feet and sat there. At this time, Robin stopped observing and wrote down his observations. At what distance above the ground is the monkey when Robin stopped his observation?

A

\(17\dfrac{3}{4}\) feet

.

B

\(118\dfrac{1}{2}\) feet

C

\(\dfrac{66}{8}\) feet

D

\(71\dfrac{1}{4}\) feet

Option A is Correct

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