Unit: 2: Fractions and Decimals

In questions 1 to 20, out of four options, only one is correct. Write the correct answer.

$\frac{2}{5}\times 5\frac{1}{5}$ is equal to:

a.       $\frac{26}{25}$

b.       $\frac{52}{25}$

c.       $\frac{2}{5}$

d.       $6$

(b)

$5\frac{1}{5}=\frac{\left(5\times 5\right)+1}{5}$

      $=\frac{25+1}{5}=\frac{26}{5}$
Now,

      $\frac{2}{5}\times 5\frac{1}{5}$$=\frac{2}{5}\times 5\frac{1}{5}=\frac{2}{5}\times \frac{26}{5}=\frac{52}{25}$

$3\frac{3}{4}÷\frac{3}{4}$ is equal to:

a.  $3$

b.  $4$

c.  $5$

d.  $\frac{45}{16}$

(c)

Given expression = $3\frac{3}{4}÷\frac{3}{4}$

$\begin{array}{c}3\frac{3}{4}=\frac{\left(3\times 4\right)+3}{4}\\ =\frac{12+3}{4}\\ =\frac{15}{4}\end{array}$
Now,

$\begin{array}{l}3\frac{3}{4}÷\frac{3}{4}\\ =\frac{15}{4}\times \frac{4}{3}\\ =5\end{array}$

A ribbon of length $5\frac{1}{4}\text{ m}$ is cut into small pieces each of length $\frac{3}{4}\text{ m}\text{.}$

Number of pieces will be:

a.  $5$

b.  $6$

c.  $7$

d.  $8$

(c)

Number of pieces $=$ (Total length of ribbon) ÷   
                                (Length of $1$ piece)

                                   $\begin{array}{l}=\frac{\left(5\frac{1}{4}\right)}{\frac{3}{4}}=\frac{\frac{\left(5\times 4\right)+1}{4}}{\frac{3}{4}}\\ =\frac{\left(\frac{21}{4}\right)}{\frac{3}{4}}\\ =\frac{21}{4}\times \frac{4}{3}=7\end{array}$

The ascending arrangement of $\frac{2}{3},\frac{6}{7},\frac{13}{21}$ is

a.    $\frac{6}{7},\frac{2}{3},\frac{13}{21}$

b.   $\frac{13}{21},\frac{2}{3},\frac{6}{7}$

c.    $\frac{6}{7},\frac{13}{21},\frac{2}{3}$

d.   $\frac{2}{3},\frac{6}{7},\frac{13}{21}$

(b)

Given fractions are: $\frac{2}{3},\frac{6}{7},\frac{13}{21}$

LCM of denominators $3,7\text{ and }21=21$

$\begin{array}{l}\frac{2}{3}=\frac{2}{3}\times \frac{7}{7}=\frac{14}{21}\\ \frac{6}{7}=\frac{6}{7}\times \frac{3}{3}=\frac{18}{21}\end{array}$

And $\frac{13}{21}=\frac{13}{21}$

Now, compare $\frac{14}{21},\frac{18}{21}$ and $\frac{13}{21}$.

So, $\frac{13}{21}<\frac{14}{21}<\frac{18}{21}$

Hence, $\frac{13}{21}<\frac{2}{3}<\frac{6}{7}$

Reciprocal of the fraction $\frac{2}{3}$ is:

a.    $2$

b.   $3$

c.    $\frac{2}{3}$

d.   $\frac{3}{2}$

(d)

The reciprocal of a non-zero fraction is obtained by interchanging its numerator and denominator.

Hence, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$

The product of $\frac{11}{13}$ and $4$ is

a.    $3\frac{5}{13}$

b.   $5\frac{3}{13}$

c.    $13\frac{3}{5}$

d.   $13\frac{5}{3}$

(a)

We have, $\frac{11}{13}\times 4$

$\frac{11}{13}\times 4=\frac{44}{13}=3\frac{5}{13}$

Hence, the product of $\frac{11}{13}$ and $4$ is $3\frac{5}{13}$.

The product of $3$ and $4\frac{2}{5}$ is

a.    $17\frac{2}{5}$

b.   $\frac{24}{5}$

c.    $13\frac{1}{5}$

d.   $5\frac{1}{3}$

(c)

Given: $3\times 4\frac{2}{5}$

$\begin{array}{l}4\frac{2}{5}=\frac{\left(4\times 5\right)+2}{5}=\frac{22}{5}\\ 3\times 4\frac{2}{5}=3\times \frac{22}{5}=\frac{66}{5}=13\frac{1}{5}\end{array}$

Hence, the product of $3$ and $4\frac{2}{5}$ is $13\frac{1}{5}$.

Pictorial representation of $3\times \frac{2}{3}$ is

a.   

 

b.  

 

 

c.   

d.  

 

 

 

(b)

$3\times \frac{2}{3}$ means $3$ times the two third part of anything.

$\frac{1}{5}÷\frac{4}{5}$ is equal to:

a.    $\frac{4}{5}$

b.   $\frac{1}{5}$

c.    $\frac{5}{4}$

d.   $\frac{1}{4}$

(d)

$\frac{1}{5}÷\frac{4}{5}=\frac{1}{5}\times \frac{5}{4}=\frac{1}{4}$

The product of $0.03\times 0.9$ is:

a.    $2.7$

b.   $0.27$

c.    $0.027$

d.   $0.0027$

(c)

Given, $0.03\times 0.9$

Here, $3\times 9=27$

Sum of the decimal places to the right of the decimal points in $0.03$ and $0.9$ is $3.$

So, $0.03\times 0.9=0.027$

$\frac{5}{7}÷6$ is equal to:

a.    $\frac{30}{7}$

b.   $\frac{5}{42}$

c.    $\frac{30}{42}$

d.   $\frac{6}{7}$

(b)

$\frac{5}{7}÷6=\frac{5}{7}\times \frac{1}{6}$

$=\frac{5}{42}$

$5\frac{1}{6}÷\frac{9}{2}$ is equal to:

a.    $\frac{31}{6}$

b.   $\frac{1}{27}$

c.    $5\frac{1}{27}$

d.   $\frac{31}{27}$

(d)

$\begin{array}{l}5\frac{1}{6}÷\frac{9}{2}=\frac{\left(5\times 6\right)+1}{6}÷\frac{9}{2}\\ =\frac{30+1}{6}÷\frac{9}{2}\\ =\frac{31}{6}÷\frac{9}{2}\\ 5\frac{1}{6}÷\frac{9}{2}=\frac{31}{6}\times \frac{2}{9}=\frac{31}{27}\end{array}$

Which of the following represents $\frac{1}{3}$ of $\frac{1}{6}$

a.    $\frac{1}{3}+\frac{1}{6}$

b.   $\frac{1}{3}-\frac{1}{6}$

c.    $\frac{1}{3}\times \frac{1}{6}$

d.   $\frac{1}{3}÷\frac{1}{6}$

(c) $\frac{1}{3}$ of $\frac{1}{6}=\frac{1}{3}\times \frac{1}{6}$

$\frac{3}{7}$ of $\frac{2}{5}$ is equal to:

a.    $\frac{5}{12}$

b.   $\frac{5}{35}$

c.    $\frac{1}{35}$

d.   $\frac{6}{35}$

(d)

$\frac{3}{7}$ of $\frac{2}{5}=\frac{3}{7}\times \frac{2}{5}=\frac{6}{35}$

One packet of biscuits requires $2\frac{1}{2}$ cups of flour and $1\frac{2}{3}$ cups of sugar. Estimated total quantity of both ingredients used in $10$ such packets of biscuits will be

a.    less than $30$ cups

b.   between $30$ cups and $40$ cups

c.    between $40$ cups and $50$ cups

d.   above $50$ cups

(c)

Total quantity of both ingredients in $1$ packet of biscuits
$=$ quantity of flour $+$ quantity of sugar $=2\frac{1}{2}\text{ cups}+1\frac{2}{3}\text{ cups}$

$\begin{array}{l}=\frac{\left(2\times 2\right)+1}{2}+\frac{\left(1\times 3\right)+2}{3}\\ =\frac{4+1}{2}+\frac{3+2}{3}\\ =\frac{5}{2}+\frac{5}{3}\end{array}$

$\begin{array}{l}=\frac{5\times 3+2\times 5}{6}\\ =\frac{15+10}{6}\\ =\frac{25}{6}\end{array}$

Total quantity of both ingredients used in $10$ packets

$=10\times$ (Total quantity of ingredients in one packet)

$=10\times \frac{25}{6}=\frac{250}{6}$

$\frac{250}{6}$ lies between $40$ and $50.$

The product of $7$ and $6\frac{3}{4}$ is

a.           $42\frac{1}{4}$

b.          $47\frac{1}{4}$

c.           $42\frac{3}{4}$

d.          $47\frac{3}{4}$

(b)

Given expression: $7\times 6\frac{3}{4}$

$\begin{array}{l}6\frac{3}{4}=\frac{\left(6\times 4\right)+3}{4}=\frac{24+3}{4}=\frac{27}{4}\\ 7\times 6\frac{3}{4}=7\times \frac{27}{4}=\frac{189}{4}=47\frac{1}{4}\end{array}$

Hence, the product of $7$ $6\frac{3}{4}$ is $47\frac{1}{4}$.

On dividing $7$ by $\frac{2}{5}$, the result is

a.    $\frac{14}{2}$

b.   $\frac{35}{4}$

c.    $\frac{14}{5}$

d.   $\frac{35}{2}$

(d)

Given, $7÷\frac{2}{5}=7\times \frac{5}{2}$

                    $=\frac{35}{2}$

Hence, on dividing $7$ by $\frac{2}{5}\text{,}$ we get $\frac{35}{2}$.

$2\frac{2}{3}÷5$ is equal to:

a.    $\frac{8}{15}$

b.   $\frac{40}{30}$

c.    $\frac{40}{5}$

d.   $\frac{8}{3}$

(a)

Given, $2\frac{2}{3}÷5=\frac{\left(2\times 3\right)+2}{3}÷5=\frac{6+2}{3}÷5$

                       $=\frac{8}{3}\times \frac{1}{5}=\frac{8}{15}$

Hence, $2\frac{2}{3}÷5$ is equal to $\frac{8}{15}$.

$\frac{4}{5}$ of $5$ kg apples were used on Monday. The next day $\frac{1}{3}$ of what was left was used. Weight (in kg) of apples left now is

a.    $\frac{2}{7}$

b.   $\frac{1}{14}$

c.    $\frac{2}{3}$

d.   $\frac{14}{21}$

(c)

Apples used on Monday $=\frac{4}{5}$ of $5=\frac{4}{5}\times 5$

                                        $=4$ kg

Remaining apples (in kg) $=5-4=1$ kg

Apples used next day  $=\frac{1}{3}$ of remaining apples

                                        $=\frac{1}{3}\times 1$ kg $=\frac{1}{3}$ kg

So, weight of apples left now

$=$ Total weight of apples
$=$ Apples used on Monday
$=$ Apples used on next day

$=\left(5-4-\frac{1}{3}\right)$ kg

$=\frac{15-12-1}{3}=\frac{2}{3}$ kg

The picture

 

interprets

a.    $\frac{1}{4}÷3$

b.   $3\times \frac{1}{4}$

c.    $\frac{3}{4}\times 3$

d.   $3÷\frac{1}{4}$

(b)

$\frac{1}{4}\text{ th}$ part of circle:

 

 

 $=3\times \frac{1}{4}$

 

 

Also, picture represents $3\times \frac{1}{4}$ i.e., $\frac{3}{4}th$ part of the circle.

 

 

 

 

In Questions 21 to 44, fill in the blanks to make the statements true.

Rani ate $\frac{2}{7}$ part of a cake while her brother Ravi ate $\frac{4}{5}$ of the remaining. Part of the cake left is _______

 Rani ate $\frac{2}{7}$ part of a cake, then

remaining part $=1-\frac{2}{7}$

                          $\begin{array}{l}=\frac{7-2}{7}\\ =\frac{5}{7}\end{array}$

 Part of cake ate by Ravi
= $\frac{4}{5}$ of $\frac{5}{7}=\frac{4}{5}\times \frac{5}{7}$$=\frac{4}{7}$

So, remaining part of the cake $=\frac{5}{7}-\frac{4}{7}$

                                                  $\begin{array}{l}=\frac{5-4}{7}\\ =\frac{1}{7}\end{array}$

Hence, part of the cake left is $\frac{1}{7}\text{.}$

The reciprocal of $\frac{3}{7}$ is _______.

The reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.

$\frac{2}{3}$ of $27$ is _______

$\frac{2}{3}$ of $27$

$\begin{array}{l}=\frac{2}{3}\times 27\\ =18\end{array}$

Hence, $\frac{2}{3}$ of $27$ is $18.$

$\frac{4}{5}$ of $45$ is _______

$\frac{4}{5}$ of $45=\frac{4}{5}\times 45=36$

Hence, $\frac{4}{5}$ of $45$ is $36.$

$4\times 6\frac{1}{3}$ is equal to _______

$4\times 6\frac{1}{3}$

$\begin{array}{l}=4\times \frac{\left(6\times 3\right)+1}{3}\\ =4\times \frac{19}{3}\\ =\frac{76}{3}\\ =25\frac{1}{3}\end{array}$

Hence, $4\times 6\frac{1}{3}=25\frac{1}{3}$

$\frac{1}{2}$ of $4\frac{2}{7}$ is _______

$\frac{1}{2}$ of $4\frac{2}{7}$

$=\frac{1}{2}\times \frac{\left(4\times 7\right)+2}{7}$

$\begin{array}{l}=\frac{1}{2}\times \frac{30}{7}\\ =\frac{30}{14}\\ =\frac{15}{7}\end{array}$

Hence, $\frac{1}{2}$ of $4\frac{2}{7}$ is $\frac{15}{7}$.

$\frac{1}{9}$ of $\frac{6}{5}$ is _______.

$\frac{1}{9}$ of $\frac{6}{5}$

$\begin{array}{l}=\frac{1}{9}\times \frac{6}{5}\\ =\frac{2}{15}\end{array}$

Hence, $\frac{1}{9}$ of $\frac{6}{5}$ is $\frac{2}{15}$.

The lowest form of the product $2\frac{3}{7}\times \frac{7}{9}$ is _______

$2\frac{3}{7}\times \frac{7}{9}$

$\begin{array}{l}=\frac{\left(2\times 7\right)+3}{7}\times \frac{7}{9}\\ =\frac{17}{7}\times \frac{7}{9}\\ =\frac{17}{9}\\ =1\frac{8}{9}\end{array}$

Hence, the lowest form of the product $2\frac{3}{7}\times \frac{7}{9}$ is $1\frac{8}{9}$.

$\frac{4}{5}÷4$ is equal to _______

$\frac{4}{5}÷4=\frac{4}{5}\times \frac{1}{4}=\frac{1}{5}$ 

Hence, $\frac{4}{5}÷4$ is equal to $\frac{1}{5}$.

$\frac{2}{5}$ of $25$ is _______

$\frac{2}{5}$ of $25=\frac{2}{5}\times 25=10$

Hence, $\frac{2}{5}$ of $25$ is $10$.

$\frac{1}{5}÷\frac{5}{6}=\frac{1}{5}$ _______ $\frac{6}{5}$

$\frac{1}{5}÷\frac{6}{5}$

$\frac{1}{5}÷\frac{5}{6}=\frac{1}{5}\times \frac{6}{5}$

$3.2\times 10$ is _______

$3.2\times 10=\frac{32}{10}\times 10=32$

Hence, $3.2\times 10=32$

$25.4\times 1000=$ _______

$25.4\times 1000=\frac{254}{10}\times 1000=25400$

Hence, $25.4\times 1000=25400$

$93.5\times 100=$ _______

$93.5\times 100=\frac{935}{10}\times 100=9350$

Hence, $93.5\times 100=9350$

$4.7÷10=$ _______

$4.7÷10=\frac{47}{10}\times \frac{1}{10}$

                         $\begin{array}{l}=\frac{47}{100}\\ =0.47\end{array}$

Hence, $4.7÷10=0.47$

$4.7÷100=$ _______

Given, $4.7÷100=\frac{47}{10}\times \frac{1}{100}$

                           $\begin{array}{l}=\frac{47}{1000}\\ =0.047\end{array}$

Hence, $4.7÷100=0.047$

$4.7÷1000=$ _______

Given, $4.7÷1000=\frac{47}{10}\times \frac{1}{1000}$

                             $\begin{array}{l}=\frac{47}{10000}\\ =0.0047\end{array}$

Hence, $4.7÷1000=0.0047$

The product of two proper fractions is _______ than each of the fractions that are multiplied.

The product of two proper fractions is less than each of the fractions that are multiplied.
For example, $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

$\frac{1}{6}<\frac{1}{2}$ and $\frac{1}{6}<\frac{1}{3}$

While dividing a fraction by another fraction, we _________ the first fraction by the _______ of the other fraction.

While dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the other fraction.

$8.4$ _______ $=2.1$

Let $x$ be the missing number, then

$\begin{array}{l}8.4÷x=2.1\\ 8.4\times \frac{1}{x}=2.1\\ 8.4=2.1x\\ x=\frac{8.4}{2.1}\\ x=\frac{84}{21}\times \frac{10}{10}=4\\ x=4\end{array}$

Hence, $8.4÷4=2.1$

$52.7÷$ _______ $=0.527$

Let $x$ be the missing number, then $52.7÷x=0.527$

$\begin{array}{l}\frac{527}{10}\times \frac{1}{x}=\frac{527}{1000}\\ \frac{527}{10}\times \frac{1000}{527}=x\\ x=100\end{array}$

Hence, $52.7÷100=0.527$

$0.5$ _______ $0.7=0.35$

$0.5=\frac{5}{10}$ and $0.7=\frac{7}{10}$

$0.5\times 0.7=\frac{5}{10}\times \frac{7}{10}=\frac{35}{100}=0.35$

Hence, $0.5\times 0.7=0.35$

$2$ _______ $\frac{5}{3}=\frac{10}{3}$

On multiplying $2$ with $\frac{5}{3}$, we get $\frac{10}{3}$.

Hence, $2\times \frac{5}{3}=\frac{10}{3}$

$2.001÷0.003=$ _______

$2.001÷0.003$

$2.001=\frac{2001}{1000}$

And $0.003=\frac{3}{1000}$

$\begin{array}{c}2.001÷0.003=\frac{2001}{1000}÷\frac{3}{1000}\\ =\frac{2001}{1000}\times \frac{1000}{3}=667\end{array}$

Hence, $2.001÷0.003=667$

 

In each of the Questions 45 to 54, state whether the statement is True or False.

The reciprocal of a proper fraction is a proper fraction.

False

The reciprocal of a proper fraction is always an improper fraction, e.g. $\frac{5}{6}$ is a proper fraction, but its reciprocal is $\frac{6}{5}\text{,}$ which is. an improper fraction.

The reciprocal of an improper fraction is an improper fraction.

False

The reciprocal of an improper fraction is a proper fraction. E.g. $\frac{6}{5}\text{,}$ is an improper fraction, but its reciprocal is $\frac{5}{6}$ which is a proper fraction.

Product of two fractions $=\frac{\text{Product of their denominators}}{\text{Product of their numerators}}$

False

Two fractions are multiplied by multiplying their numerators and denominators separately and writing the product as,

Product of two fractions $=\frac{\text{Product of their numerators}}{\text{Product of their denominators}}$

The product of two improper fractions is less than both the fractions.

False

The product of two improper fractions are greater than both the fractions.

For example: $\frac{3}{2}\times \frac{7}{4}=\frac{21}{8}$

Hence, $\frac{21}{8}$ is greater than both $\frac{3}{2}$ and $\frac{7}{4}$.

A reciprocal of a fraction is obtained by inverting it upside down.

True

Let $\frac{a}{b}$ be the fraction. Then, for obtaining its reciprocal, numerator and denominator are interchanged.

$\therefore$ Reciprocal of $\frac{a}{b}=\frac{b}{a}$

To multiply a decimal number by $1000\text{,}$ we move the decimal point in the number to the right by three places.

True

$2.732\times 1000=2732$ (moving the decimal to the right by three places)

To divide a decimal number by $100\text{,}$ we move the decimal point in the number to the left by two places.

True

Example: $\frac{273.2}{100}=2.732$ (moving decimal to the left by two places.)

$1$ is the only number which is its own reciprocal.

True

For obtaining the reciprocal of a number, we simply interchange the numerator and denominator. Hence, reciprocal of $1$ will be $\frac{1}{1}\text{,}$ i.e. $1$.

$\frac{2}{3}$ of $8$ is same as $\frac{2}{3}÷8.$

False

$\frac{2}{3}$ of $8=\frac{2}{3}\times 8=\frac{16}{3}$

$\frac{2}{3}÷8=\frac{2}{3}\times \frac{1}{8}=\frac{2}{24}=\frac{1}{12}$

Therefore, $\frac{2}{3}$ of $8\ne \frac{2}{3}÷8$

Hence, $\frac{2}{3}$ of $8$ is not same as $\frac{2}{3}÷8.$

The reciprocal of $\frac{4}{7}$ is $\frac{4}{7}\text{.}$

False

Reciprocal of $\frac{4}{7}$ is $\frac{7}{4}$.

If $5$ is added to both the numerator and the denominator of the fraction $\frac{5}{9}\text{,}$ will the value of the fraction be changed? If so, will the value increase or decrease?

Given fraction $=\frac{5}{9}$

Add $5$ to the numerator and denominator. $\frac{5+5}{9+5}=\frac{10}{14}=\frac{5}{7}$

$\frac{5}{7}>\frac{5}{9}$

So, the value will increase.

What happens to the value of a fraction if the denominator of the fraction is decreased while numerator is kept unchanged?

The value of fraction would increase if the denominator of the fraction is decreased while numerator is kept unchanged.

For example, consider a fraction $\frac{2}{3}$.

$\frac{2}{2}>\frac{2}{3}$

Which letter comes $\frac{2}{5}$ of the way among A and J?

The total number of letters from A to j is $10$.

Determine the letter that would fall $\frac{2}{5}$ of the way from A to J:

$\frac{2}{5}$ of $10=\frac{2}{5}\times 10$

$=\frac{20}{5}=4$ 

Two-fifth along the way is $4$ th letter which is ‘d’.

If $\frac{2}{3}$ of a number is $10\text{,}$ then what is $1.75$ times of that number?

Let the number be $x$.

According to the question, $\frac{2}{3}$ of $x=10$

                                             $\frac{2}{3}\times x=10$

On multiplying both sides by $\frac{3}{2}\text{,}$ we get

$\begin{array}{l}\frac{2}{3}\times x\times \frac{3}{2}=10\times \frac{3}{2}\\ x=5\times 3=15\end{array}$

$1.75$ times of $15=1.75\times 15=\frac{175}{100}\times 15=\frac{2625}{100}=26.25$

In a class of 40 students, $\frac{1}{5}$ of the total number of students like to eat rice only, $\frac{2}{5}$ of the total number of students like to eat chapatti only and the remaining students like to eat both. What fraction of the total number of students like to eat both?

Total number of students $=40$

Students who eat rice only $=\frac{1}{5}$ of total students $\frac{1}{5}\times 40=8$

Students who eat chapatti only $=\frac{2}{5}$ of total students

                                                    $=\frac{2}{5}\times 40=16$

Students who eat both chapatti and rice $=$ total no. of students $-$ (students who eat rice only $+$ students who eat chapatti only)

$\begin{array}{l}=40-\left(8+16\right)\\ =40-24=16\end{array}$

Fraction of students who eat both chapatti & rice

$=\frac{\text{No}\text{. of students who eat both chapatti \& rice}}{\text{Total no}\text{. of students}}=\frac{16}{40}=\frac{2}{5}$

Renu completed $\frac{2}{3}$ part of her home work in $2$ hours. How much part of her home work had she completed in $1\frac{1}{4}$ hours?

The part of the work finished by Renu in $2\text{ hr}=\frac{2}{3}$

So, the part of the work finished by Renu in $1\text{ hr}=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$

The part of the work finished by Renu in $1\frac{1}{4}\text{ hr}=\frac{1}{3}\times 1\frac{1}{4}$

         $\begin{array}{l}=\frac{1}{3}\times \frac{\left(1\times 4\right)+1}{4}\\ =\frac{1}{3}\times \frac{5}{4}=\frac{5}{12}\text{ part}\end{array}$

Hence, $\frac{5}{12}\text{ part}$ of Renu’s home-work is completed by her in $1\frac{1}{4}\text{ hr}\text{.}$

Reemu read $\frac{1}{5}\text{th}$ pages of a book. If she reads further $40$ pages, she would have read $\frac{7}{10}\text{th}$ pages of the book. How many pages are left to be read?

Let total pages of the book be $x$.

According to question, $\frac{1}{5}x+40=\frac{7}{10}x$

$\begin{array}{l}40=\frac{7}{10}x-\frac{1}{5}x=\frac{7x-2x}{10}\\ 40=\frac{5x}{10}\\ x=\frac{400}{5}=80\end{array}$

Total pages of book $=80$

Hence,

pages left to be read
$=$ total pages of book $-\left(\frac{7}{10}\right)x$

$\begin{array}{l}=80-\frac{7}{10}\times 80\\ =80-56=24\text{ pages}\end{array}$

Write the number in the box $\boxed{}$ such that

$\frac{3}{7}\times \boxed{}=\frac{15}{98}$

Let the missing number be $x$.

Then, $\frac{3}{7}\times x=\frac{15}{98}$

$\begin{array}{l}x=\frac{15}{98}÷\frac{3}{7}=\frac{15}{98}\times \frac{7}{3}\\ x=\frac{5}{14}\end{array}$

Hence, $\frac{3}{7}\times \frac{5}{14}=\frac{15}{98}$

Will the quotient $7\frac{1}{6}÷3\frac{2}{3}$ be a fraction greater than $1.5$ or less than $1.5?$ Explain.

$7\frac{1}{6}÷3\frac{2}{3}=\frac{\left(7\times 6\right)+1}{6}÷\frac{\left(3\times 3\right)+2}{3}$

$\begin{array}{l}=\frac{42+1}{6}÷\frac{9+2}{3}\\ =\frac{43}{6}÷\frac{11}{3}\\ =\frac{43}{6}\times \frac{3}{11}=\frac{43}{22}=1.95\\ 1.95>1.5\end{array}$

Hence, $7\frac{1}{6}÷3\frac{2}{3}>1.5$

Describe two methods to compare $\frac{13}{17}$ and $\mathrm{0.82.}$ Which do you think is easier and why?

Method $1$

Convert both numbers into decimal form.

$\begin{array}{l}\frac{13}{17}=0.76\\ 0.76<0.82\\ \frac{13}{17}<0.76\end{array}$

Method $2$

Convert both numbers into fraction form.

$0.82=\frac{82}{100}=\frac{41}{50}$

Now, compare $\frac{13}{17}$ and $\frac{41}{50}$

To compare these fractions, we have to make the denominators same.

$\begin{array}{l}\frac{13}{17}=\frac{13}{17}\times \frac{50}{50}=\frac{650}{850}\\ \frac{41}{50}=\frac{41}{50}\times \frac{17}{17}=\frac{697}{850}\\ \frac{697}{850}>\frac{650}{850}\end{array}$

Conclusion, method 1 is easier in this case.

Health: The directions for a pain reliever recommend that an adult of $60\text{ kg}$ and over, take $4$ tablets every $4$ hours as needed, and an adult who weighs between $40$ and $50$ kg take only $2\frac{1}{2}$ tablets every $4$ hours as needed. Each tablet weighs $\frac{4}{25}$ gram.

a.    If a $72\text{ kg}$ adult takes $4$ tablets, how many grams of pain reliever is he or she receiving?

b.   How many grams of pain reliever is the recommended dose for an adult weighing $46\text{ kg?}$

a.    Given, $4$ tablets for a $72\text{ kg}$ adult & each tablet weighs $\frac{4}{25}\text{ g}$
Total weight of pain reliever, he/she is receiving $=4\times \frac{4}{25}\text{ g}=\frac{16}{25}\text{ g}$

b.   Given, adult weighing $46\text{ kg}$ takes $2\frac{1}{2}$ tablets & each tablet weighs $\frac{4}{25}\text{ g}$.
Total weight of pain reliever, he/she is receiving

$\begin{array}{l}=\left(\frac{4}{25}\times 2\frac{1}{2}\right)\text{ g}=\left[\frac{4}{25}\times \frac{\left(2\times 2\right)+1}{2}\right]\text{ g}\\ =\left(\frac{4}{25}\times \frac{5}{2}\right)\text{ g}\\ =\frac{2}{5}\text{ g}\end{array}$

Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals?

a.    a $18\text{ kg}$ adult dog

b.   a $6\text{ kg}$ cat

c.    a $18\text{ kg}$ pregnant dog

 

 

 

 

 

 

a.    Dosage prescribed for an adult dog is $\frac{1}{2}\text{ tsp}$ per $9\text{ kg}$ body weight.

For a $18\text{ kg}$ adult dog,
dosage $=\frac{\left(\frac{1}{2}\right)}{9}\times 18$

$\begin{array}{l}=\frac{1}{2\times 9}\times 18\\ =\frac{1}{18}\times 18\\ =1\text{ tsp}\end{array}$

b.   For a $6\text{ kg}$ cat,
dosage $=\frac{\left(\frac{1}{4}\right)}{9}\times 6$

$\begin{array}{l}=\frac{1}{4}\times 6\text{ tsp}\\ =\frac{6}{4}=\frac{3}{2}\text{ tsp}\\ =1\frac{1}{2}\text{ tsp}\end{array}$

c.    For a $18\text{ kg}$ pregnant dog,
dosage $=\frac{\left(\frac{1}{2}\right)}{4.5}\times 18$

$\begin{array}{l}=\frac{1}{2\times 4.5}\times 18\\ =\frac{1}{9}\times 18\\ =\frac{18}{9}\text{ tsp}\\ =2\text{ tsp}\end{array}$

How many $\frac{1}{16}\text{ kg}$ boxes of chocolates can be made with $1\frac{1}{2}\text{ kg}$ chocolates?

Total chocolates $=1\frac{1}{2}\text{ kg}$

$\begin{array}{l}=\frac{\left(1\times 2\right)+1}{2}\\ =\frac{3}{2}\text{ kg}\end{array}$

Number of boxes of chocolates $=\frac{\text{Total weight of chocolates}}{\text{Weight of }1\text{ box}}=\left(\frac{3}{2}÷\frac{1}{16}\right)$

$=\frac{3}{2}\times 16=3\times 8=24$

Anvi is making bookmarker like the one shown in $\text{Fig}\text{. }\mathrm{2.6.}$ How many bookmarker can she make from a $15\text{ m}$ long ribbon?

 

 

 

 

Length of $1$ bookmarker $=10\frac{1}{2}\text{ cm}$

                                        $\begin{array}{l}=\frac{10\times 2+1}{2}\\ =\frac{21}{2}\text{ cm}\end{array}$

Length of ribbon $=15\text{ m}=1500\text{ cm}$

Number of bookmarkers $=$ Length of ribbon/Length of $1$ bookmarker $=\frac{1500}{\frac{21}{2}}$

                             $\begin{array}{l}=\frac{1500}{21}\times 2\\ =142.85\\ =142\end{array}$

Hence, $142$ bookmarkers can be made from a $15\text{ m}$ long ribbon.

A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by $\mathrm{1.414.}$ Find the length of the diagonal when:

a.    The length of a side of the square is $8.3\text{ cm}\text{.}$

b.   The length of a side of the square is exactly $7.875\text{ cm}\text{.}$

a.    Side of a square $=8.3\text{ cm}$

Length of diagonal
$=$ Length of side of the square $\times 1.414$

$\begin{array}{l}=8.3\times 1.414\\ =11.7362=11.74\text{ cm (approx}\text{.)}\end{array}$

b.   Side of a square $=7.875\text{ cm}$

Length of diagonal

$=$ Length of side of the square $\times 1.414$

$\begin{array}{l}=7.875\times 1.414\\ =11.13525=11.14\text{ (approx}\text{.)}\end{array}$

The largest square that can be drawn in a circle has a side whose length is $0.707$ times the diameter of the circle. By this rule, find the length of the side of such a square when the diameter of the circle is

a.    $14.35\text{ cm}$

b.   $8.63\text{ cm}$

Given,

Side of square $=0.707\times \text{diameter of circle}$

a.    Diameter of circle $=14.35\text{ cm}$

Side of square $=0.707\times 14.35$

                        $=10.15\text{ cm}$

b.   Diameter of circle $=8.63\text{ cm}$

Side of square $=0.707\times 8.63$

                        $=6.10\text{ cm}$

To find the distance around a circular disc, multiply the diameter of the disc by $\mathrm{3.14.}$ What is the distance around the disc when :

a.    the diameter is $18.7\text{ cm?}$

b.   the radius is $6.45\text{ cm?}$

Given,

a.    Diameter of  a circular disc $=18.7\text{ cm}$

Distance around the circular disc $=18.7\times 3.14$

                                                       $=58.718\text{ cm}$

b.   Radius of disc $=6.45\text{ cm}$

Diameter of disc $=2\times \text{Radius of disc}$

                            $\begin{array}{l}=2\times 6.45\\ =12.9\text{ cm}\end{array}$

Distance around the circular disc $=12.9\times 3.14$

                                                      $=40.506\text{ cm}$

What is the cost of $27.5\text{ m}$ of cloth at $`53.50$ per metre?

Cost of $1\text{ m}$ cloth $=`53.50$

Cost of $27.5\text{ m}$ cloth $=`\left(53.50\times 27.5\right)$

                                   $\begin{array}{l}=`\left(\frac{5350}{100}\times \frac{275}{10}\right)\\ =`\left(\frac{1471250}{1000}\right)\\ =`1471.25\end{array}$

In a hurdle race, Nidhi is over hurdle B and $\frac{2}{6}$ of the way through the race, as shown in $\text{Fig}\text{. }\mathrm{2.7.}$

Then, answer the following:

a.    Where will Nidhi be, when she is $\frac{4}{6}$ of the way through the race?

b.   Where will Nidhi be when she is $\frac{5}{6}$ of the way through the race?

c.    Give two fractions to tell what part of the race Nidhi has finished when she is over hurdle C.

As per the given information and figure, 5 hurdles are placed at equal distances between starting and finishing point.

These 5 hurdles divide the whole distance in 6 equal parts.

So, hurdle A is placed after $\frac{1}{6}$ of the way from starting point.

So, if Nidhi is at A, she will cover $\frac{1}{6}$ way.

If Nidhi is at B, then $\frac{2}{6}$ of way is completed.

                                            

a.    When she is $\frac{4}{6}$ of the way, she will be at $\frac{\frac{4}{6}}{\frac{1}{6}}$ position

$=\left(\frac{4}{6}\times \frac{6}{1}\right)=4^{\text{th}}\text{ position}=\text{D}$

b.   When she is $\frac{5}{6}$ of the way, she will be at $\frac{\frac{5}{6}}{\frac{1}{6}}$ position

$=\frac{5}{6}\times \frac{6}{1}=5^{\text{th}}\text{ position}=\text{E}$

c.    When she is over hurdle C, she has completed half race.

Hence, she will be at $\frac{3}{6}\text{way}=\frac{3}{6}=\frac{1}{2}\text{way}$

Diameter of Earth is $12756000\text{ m}\text{.}$ In $1996\text{,}$ a new planet was discovered whose diameter is $\frac{5}{86}$ of the diameter of Earth. Find the diameter of this planet in km.

Give, diameter of Earth $=12756000\text{ m}$

                                       $\begin{array}{l}=\frac{12756000}{1000}\\ =12756\text{ km}\end{array}$

According to question,

Diameter of new planet $=\frac{5}{86}$ of diameter of earth

                                       $\begin{array}{l}=\frac{5}{86}\times 12756\\ =\frac{63780}{86}\\ =741.62\text{ km}\end{array}$

What is the product of $\frac{5}{129}$ and its reciprocal?

Reciprocal of $\frac{5}{129}=\frac{129}{5}$

Product of $\frac{5}{129}$ and its reciprocal

$=\frac{5}{129}\times \frac{129}{5}=1$

Simplify: $\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}÷\frac{1}{5}}$

Given, $\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}÷\frac{1}{5}}=\frac{\frac{\left(2\times 2\right)+1}{2}+\frac{1}{5}}{\frac{\left(2\times 2\right)+1}{2}÷\frac{1}{5}}$

                        $\begin{array}{l}=\frac{\frac{5}{2}+\frac{1}{5}}{\frac{5}{2}÷\frac{1}{5}}=\frac{\frac{25+2}{10}}{\frac{5}{2}\times 5}\\ =\frac{\frac{27}{10}}{\frac{25}{2}}=\frac{27}{10}\times \frac{2}{25}\\ =\frac{27}{125}\end{array}$

Simplify: $\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}\times \frac{3}{5}}$

Given, $\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}\times \frac{3}{5}}=\frac{\frac{5+4}{20}}{1-\frac{9}{40}}$

                          $\begin{array}{l}=\frac{\frac{9}{20}}{\frac{31}{40}}=\frac{9}{20}\times \frac{40}{31}\\ =\frac{18}{31}\end{array}$

Divide $\frac{3}{10}$ by $\left(\frac{1}{4}\text{ of }\frac{3}{5}\right)$

 $\frac{3}{10}÷\left(\frac{1}{4}\text{ of }\frac{3}{5}\right)=\frac{3}{10}÷\left(\frac{1}{4}\times \frac{3}{5}\right)$

                     $\begin{array}{l}=\frac{3}{10}÷\left(\frac{3}{20}\right)\\ =\frac{3}{10}\times \frac{20}{3}=2\end{array}$

$\frac{1}{8}$ of a number equals $\frac{2}{5}÷\frac{1}{20}\text{.}$ What is the number?

Let the number be $x$

Then, $\frac{1}{8}$ of a number $=\frac{2}{5}÷\frac{1}{20}$

$\begin{array}{l}\frac{1}{8}\times x=\frac{2}{5}\times 20\\ \frac{x}{8}=8\\ x=8\times 8=64\end{array}$

Heena’s father paid an electric bill of $\text{Rs}385.70$ out of a $500$ rupee note. How much change should he have received?

Given, total rupees $=\text{Rs}500$ and money paid $=\text{Rs}385.70$

$\therefore$ Change he received $=\text{Rs}\left(500-385.70\right)=\text{Rs}114.30$

The normal body temperature is $98.6°\text{F}\text{.}$ When Savitri was ill her temperature rose to $103.1°\text{F}\text{.}$ How many degrees above normal was that?

Given, normal body temperature $=98.6°\text{F}$ and temperature rise to $=103.1°\text{F}$

$\therefore$ Rise in temperature $=\left(103.1-98.6\right)°\text{F}=4.5°\text{F}$

Meteorology: One measure of average global temperature shows how each year varies from a base measure. The table shows results for several years.

Year	$1958$	$1964$	$1965$	$1978$	$2002$
Difference from Base	$0.10°\text{C}$	$-0.17°\text{C}$	$-0.10°\text{C}$	$\left(\frac{1}{50}\right)°\text{C}$	$0.54°\text{C}$

See the table and answer the following:

a.    Order the five years from coldest to warmest.

b.   In $1946\text{,}$ the average temperature varied by $-0.03°\text{C}$ from the base measure. Between which two years should $1946$ fall when the years are ordered from coldest to warmest?

In year $1978\text{,}$ temperature is $\left(\frac{1}{50}\right)°\text{C}=0.02°\text{C}$

(a) By observing the given temperatures, coldest to warmest in ascending order is:

$-0.17°\text{C}<-0.10°\text{C}<0.02°\text{C}<0.10°\text{C}<0.54°\text{C}$

Order of years is:

$1964<1965<1978<1958<2002$

(b) In $1946\text{,}$ the average temperature varied by $-0.03°\text{C}$ from the base measure. So, temperature from coldest to warmest in ascending order is:

$-0.17°\text{C}<-0.10°\text{C}<-0.03°\text{C}<0.02°\text{C}<0.10°\text{C}<0.54°\text{C}$

Therefore, order of years is:

$1964<1965<1946<1978<1958<2002$

 

Thus, year $1946$ fall between years 1965 and 1978 when the years are ordered from coldest to warmest.

 

In her science class, Jyoti learned that the atomic weight of Helium is $4.0030\text{;}$ of Hydrogen is $1.0080\text{;}$ and of Oxygen is $\mathrm{16.0000.}$ Find the difference between the atomic weights of:

a.    Oxygen and Hydrogen

b.   Oxygen and Helium

c.    Helium and Hydrogen

Given, atomic weights of Helium $=4.0030\text{,}$ Hydrogen $=1.0080$ and Oxygen $=16.0000$

a.    Difference between atomic weights of Oxygen & Hydrogen $=16.0000-1.0080=14.9920$

b.   Difference between atomic weights of Oxygen & Helium $=16.0000-04.0030=11.9970$

c.    Difference between atomic weights of Helium & Hydrogen $=4.0030-1.0080=2.9950$

Measurement made in science lab must be as accurate as possible. Ravi measured the length of an iron rod and said it was $19.34\text{ cm}$ long; Kamal said $19.25\text{ cm;}$ and Tabish said $19.27\text{ cm}\text{.}$ The correct length was $19.33\text{ cm}\text{.}$ How much of error was made by each of the boys?

The actual length of an iron rod $=19.33\text{ cm}$

Length measured by Ravi $=19.34\text{ cm}$

Error $=$ Measured value $-$ Actual value $=\left(19.34-19.33\right)\text{ cm}=0.01\text{ cm}$

Length measured by Kamal $=19.25\text{ cm}$

Error $=\left(19.25-19.33\right)\text{ cm}=-0.08\text{ cm}$

Length measured by Tabish $=19.27\text{ cm}$

Error $=\left(19.27-19.33\right)\text{ cm}=-0.06\text{ cm}$

When $0.02964$ is divided by $0.004\text{,}$ what will be the quotient?

Given, $0.02964÷0.004=\frac{2964}{100000}÷\frac{4}{1000}$

                                      $\begin{array}{l}=\frac{2964}{100000}\times \frac{1000}{4}\\ =\frac{741}{100}\\ =7.41\end{array}$

What number divided by $520$ gives the same quotient as $85$ divided by $0.625\text{?}$

Let the number be $x$.

According to the question,

$\begin{array}{l}\frac{x}{520}=\frac{85}{0.625}\\ x=\frac{85\times 520\times 1000}{625}\\ x=\frac{44200000}{625}\\ x=70720\end{array}$

Hence, the number is $70720.$

A floor is $4.5\text{ m}$ long and $3.6\text{ m}$ wide. A $6\text{ cm}$ square tile costs $\text{Rs}\mathrm{23.25.}$ What will be the cost to cover the floor with these tiles?

Area of floor $=4.5\times 3.6=16.2{\text{ m}}^2$

Convert $16.2{\text{ m}}^2{\text{ to cm}}^2\left(\because 1{\text{ m}}^2=10000{\text{ cm}}^2\right)$

$\begin{array}{l}=16.2\times 10000\\ =162000{\text{ cm}}^2\end{array}$

Cost of $6{\text{ cm}}^2$ tile is $\text{Rs}23.25$.

Area of each square tile $=6\times 6{\text{cm}}^2$

So, the cost for $162000{\text{ cm}}^2$ floor

$=\frac{\left(162000\times 23.25\right)}{6\times 6}=\text{Rs}104625$

Sunita and Rehana want to make dresses for their dolls. Sunita has $\frac{3}{4}\text{ m}$ of cloth, and she gave $\frac{1}{3}$ of it to Rehana. How much did Rehana have?

Given that, Sunita has $\frac{3}{4}\text{ m}$ of cloth.

She gave one-third of cloth to Rehana.

$\frac{1}{3}\text{ of }\frac{3}{4}$

$\begin{array}{l}=\frac{1}{3}\times \frac{3}{4}\\ =\frac{1}{4}\text{ m}\end{array}$

Hence, Rehana has $\frac{1}{4}\text{ m}$ of cloth.

A flower garden is $22.50\text{ m}$ long. Sheela wants to make a border along one side using bricks that are $0.25\text{ m}$ long. How many bricks will be needed?

Length of flower garden $=22.50\text{ m}$

Length of $1$ brick $=0.25\text{ m}$

Number of bricks used in $1$ side
$=$ length of flower garden/length of $1$ brick

$=\frac{22.50}{0.25}=\frac{2250}{25}=90$

Hence, $90$ bricks will be needed.

How much cloth will be used in making $6$ shirts, if each required $2\frac{1}{4}\text{ m}$ of cloth, allowing $\frac{1}{8}\text{ m}$ for waste in cutting and finishing in each shirt?

Cloth required in making $1$ shirt $=\left(2\frac{1}{4}+\frac{1}{8}\right)=\frac{\left(2\times 4\right)+1}{4}+\frac{1}{8}$

$\begin{array}{l}=\frac{9}{4}+\frac{1}{8}=\frac{18+1}{8}\\ =\frac{19}{8}\text{ m}\end{array}$

Total cloth required in making such $6$ shirts $=6\times \frac{19}{8}$

$\begin{array}{l}=\frac{114}{8}\\ =14\frac{1}{4}\text{ m}\end{array}$

Hence, $14\frac{1}{4}\text{ m}$ cloth will be used in making $6$ shirts.

A picture hall has seats for $820$ persons. At a recent film show, one usher guessed it was $\frac{3}{4}$ full, another that it was $\frac{2}{3}$ full. The ticket office reported $648$ sales. Which usher (first or second) made the better guess?

Given, picture hall has seats $=820$

$1$ usher guessed, picture hall was $\frac{3}{4}$ full.

$\frac{3}{4}\text{ of }820=\frac{3}{4}\times 820=\frac{2460}{4}=615$

Another usher guessed, picture hall was $\frac{2}{3}$ full.

$\frac{2}{3}\text{ of }820=\frac{2}{3}\times 820=\frac{1640}{3}=546.66$

Since, $648$ tickets are sold that is near to $615.$

So, first usher made  better guess.

For the celebrating children’s students of Class VII bought sweets for $\text{Rs}740.25$ and cold drink for $\text{Rs}70.$ If $35$ students contributed equally what amount was contributed by each student?

Cost of sweets $=\text{Rs}740.25$

Cost of cold drink $=\text{Rs}70$

Total cost $=\text{Rs}\left(740.25+70\right)$

                  $=\text{Rs}810.25$

Given, $35$ students are contributing equally.

$\begin{array}{l}=\text{Rs}\frac{810.25}{35}\\ =\text{Rs}\frac{81025}{35\times 100}\\ =\text{Rs}\frac{2315}{100}\\ =\text{Rs}23.15\end{array}$

The time taken by Rohan in five different races to run a distance of $500\text{ m}$ was $3.20$ minutes, $3.37$ minutes, $3.29$ minutes, $3.17$ minutes and $3.32$ minutes. Find the average time taken by him in the races.

Total time taken by Rohan $5$ races $=\left(3.20+3.37+3.29+3.17+3.32\right)$

$=16.35\text{ min}$

Average time taken by Rohan $=$ total time taken/total number of observation

$=$ total time taken/ $5$

$\begin{array}{l}=\frac{16.35}{5}\\ =\frac{1635}{5\times 100}\\ =\frac{327}{100}\\ =3.27\text{ min}\end{array}$

A public sewer line is being installed along $80\frac{1}{4}\text{ m}$ of road. The supervisor says that the labourers will be able to complete $7.5\text{ m}$ in one day. How long will the project take to complete?

 

 

Total sewer line to be installed $=80\frac{1}{4}\text{ m}$

                                                   $\begin{array}{l}=\frac{\left(80\times 4\right)+1}{4}\\ =\frac{321}{4}\text{ m}\end{array}$

In one day, labourers can complete $=7.5\text{ m}$

Number of days to complete the project $=$ total sewer line to be installed/ $1$ day work

$\begin{array}{l}=\frac{321}{4\times 7.5}\\ =\frac{312}{30}\\ =10.4\text{ days}\\ =11\text{ days}\end{array}$

Hence, the number of days to complete the project will be $11\text{ days}\text{.}$

The weight of an object on moon is $\frac{1}{6}$ its weight on Earth. If an object weighs $5\frac{3}{5}\text{ kg}$ on Earth, how much would it weigh on the moon?

Weight of an object on the moon is $\frac{1}{6}$ of its weight on Earth.

Object weights on Earth $=5\frac{3}{5}\text{ kg}$

                                         $\begin{array}{l}=\frac{\left(5\times 5\right)+3}{5}\\ =\frac{28}{5}\text{ kg}\end{array}$

Weight of the object on Moon $=\frac{1}{6}\text{ of }\frac{28}{5}\text{ kg}$

                             $\begin{array}{l}=\frac{1}{6}\times \frac{28}{5}\\ =\frac{28}{30}\\ =\frac{14}{15}\\ =0.93\text{ kg}\end{array}$

Hence, weight of the object on the Moon is $0.93\text{ kg}$.

In a survey, $200$ students were asked what influenced them most to buy their latest CD. The results are shown in the circle graph.

a.    How many students said radio influenced them most?

b.   How many more students were influenced by radio than by a music video channel?

c.    How many said a friend or relative influenced them or they heard the CD in a shop?

 

 

(a)       Fraction of radio (in figure) $=\frac{9}{20}$

Total number of students $=200$

Number of students influenced by radio the most $\frac{9}{20}\text{ of }200$

$\begin{array}{l}=\frac{9}{20}\times 200\\ =90\end{array}$

(b)       We have to find,

(Students influenced by radio) $-$ (Students influenced by music video channel) $=$ (fraction of radio $\times$ total number of students) $-$ (fraction of music video channel $\times$ total number of students)

$=\frac{9}{20}\text{ of }200-\frac{2}{25}\text{ of }200$

$\begin{array}{l}=\frac{9}{20}\times 200-\frac{2}{25}\times 200\\ =90-16\\ =74\end{array}$

 

(c)
Number of students influenced by friend or relative or CD in a shop
= Number of students influenced by friend or relative + Number of students who heard a CD in a shop

$\begin{array}{l}=\frac{3}{20}\times 200+\frac{1}{10}\times 200\\ =3\times 10+20\\ =30+20=50\end{array}$

In the morning, a milkman filled $5\frac{1}{2}\text{ L}$ of milk in his can. He sold to Renu, Kamla and Renuka $\frac{3}{4}\text{ L}$ each; to Shadma he sold $\frac{7}{8}\text{ L;}$ and to Jassi he gave $1\frac{1}{2}\text{ L}\text{.}$ How much milk is left in the can?

Given, milk in can $=5\frac{1}{2}\text{ L}$

$\begin{array}{l}=\frac{\left(5\times 2\right)+1}{2}\\ =\frac{10+1}{2}\\ =\frac{11}{2}\text{ L}\end{array}$

If $\frac{3}{4}\text{ L}$ sold to Renu, Kamla and Renuka

Then, total milk sold $=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$

                                   $\begin{array}{l}=\frac{3+3+3}{4}\\ =\frac{9}{4}\text{ L}\end{array}$

Milk sold to Shadma $=\frac{7}{8}\text{ L}$

Milk sold to Jassi $=1\frac{1}{2}\text{ L}$

                       $\begin{array}{l}=\frac{\left(1\times 2\right)+1}{2}\\ =\frac{3}{2}\text{ L}\end{array}$

Total milk sold $=\frac{9}{4}+\frac{7}{8}+\frac{3}{2}$

                          $\begin{array}{l}=\frac{18+7+12}{8}\\ =\frac{37}{2}\text{ L}\end{array}$

Total mink left in can $=\frac{11}{2}-\left(\frac{37}{8}\right)$

$\begin{array}{l}=\frac{44-37}{8}\\ =\frac{7}{8}\text{ L}\end{array}$

Hence, $\frac{7}{8}\text{ L}$ milk is left in the can.

Anuradha can do a piece of work in $6$ hours. What part of the work can she do in $1$ hour, in $5$ hours, in $6$ hours?

It is given that, Anuradha can do a piece of work in $6\text{ h}\text{.}$

In $6\text{ hr,}$ Anuradha can do $=$ $1$ complete  work

In $1\text{ hr,}$ Anuradha can do $=\frac{1}{6}\text{ part of work}$

In $5\text{ hr,}$ Anuradha can do $=\frac{1}{6}\times 5=\frac{5}{6}\text{ part of work}$

What portion of a ‘saree’ can Rehana paint in $1$ hour if it requires $5$ hours to paint the whole saree? In $4\frac{3}{5}$ hours? In $3\frac{1}{2}$ hours?

In $5\text{ hr,}$ Rehana paints $=$ 1 whole saree

In $1\text{ hr,}$ she paints $=\frac{1}{5}\text{ part of saree}$

In $4\frac{3}{5}\text{ hr,}$ she paints $=\frac{1}{5}\times 4\frac{3}{5}$

                                 $\begin{array}{l}=\frac{1}{5}\times \frac{\left(5\times 4\right)+3}{5}\\ =\frac{1}{5}\times \frac{23}{5}\end{array}$

                                 $=\frac{23}{25}\text{ part of saree}$

In $3\frac{1}{2}\text{ hr,}$ she paints $=\frac{1}{5}\times 3\frac{1}{2}$

                                 $\begin{array}{l}=\frac{1}{5}\times \frac{\left(3\times 2\right)+1}{2}\\ =\frac{1}{5}\times \frac{7}{2}\end{array}$

                                 $=\frac{7}{10}\text{ part of saree}$

Rama has $6\frac{1}{4}\text{ kg}$ of cotton wool for making pillows. If one pillow takes $1\frac{1}{4}\text{ kg,}$ how many pillows can she make?

Give, Rama has $6\frac{1}{4}\text{ kg}$ of cotton for making pillows

i.e., $6\frac{1}{4}\text{ kg}=\frac{\left(6\times 4\right)+1}{4}$

 

                  $\begin{array}{l}=\frac{24+1}{4}\\ =\frac{25}{4}\text{ kg}\end{array}$

Where, $1$ pillow can be made from $1\frac{1}{4}\text{ kg}$

i.e., $1\frac{1}{4}\text{ kg}=\frac{\left(1\times 4\right)+1}{4}$

 

                  $\begin{array}{l}=\frac{4+1}{4}\\ =\frac{5}{4}\text{ kg}\end{array}$

Number of pillows $=\frac{\text{total quantity of cotton available}}{\text{cotton used in one pillow}}$ 

$\begin{array}{l}=\frac{\left(\frac{25}{4}\right)}{\frac{5}{4}}\\ =\frac{25}{4}\times \frac{4}{5}\\ =\frac{25}{5}\\ =5\end{array}$

It takes $2\frac{1}{3}\text{ m}$ of cloth to make a shirt. How many shirts can Radhika make from a piece of cloth $9\frac{1}{3}\text{ m}$ long?

Radhika takes $2\frac{1}{3}\text{ m}$ cloth to make a shirt

i.e., $2\frac{1}{3}\text{ m}=\frac{\left(2\times 3\right)+1}{3}$

 

                 $\begin{array}{l}=\frac{6+1}{3}\\ =\frac{7}{3}\text{ m}\end{array}$

If Radhika has $9\frac{1}{3}\text{ m}$ long cloth

i.e., $9\frac{1}{3}\text{ m}=\frac{\left(9\times 3\right)+1}{3}$

                 $\begin{array}{l}=\frac{27+1}{3}\\ =\frac{28}{3}\text{ m}\end{array}$

Then, number of shirts that can be made $=\frac{\text{available cloth}}{\text{required cloth to make }1\text{ shirt}}$

$\begin{array}{l}=\frac{\frac{28}{3}}{\frac{7}{3}}\\ =\frac{28}{3}\times \frac{3}{7}\\ =\frac{28}{7}\\ =4\end{array}$

Ravi can walk $3\frac{1}{3}\text{ km}$ in one hour. How long will it take him to walk to his office which is $10\text{ km}$ from his home?

Given, Ravi can walk $3\frac{1}{3}\text{ km}$ in $1\text{ hr}\text{.}$

Ravi’s speed $=3\frac{1}{3}\text{ km/hr}$

$\begin{array}{l}=\frac{\left(3\times 3\right)+1}{3}\\ =\frac{9+1}{3}\\ =\frac{10}{3}\text{ km/hr}\end{array}$

Distance between home & office $=10\text{ km}$

Time $=\frac{\text{Distance between home \& office}}{\text{Ravi’s speed in km/hr}}$ 

$\begin{array}{l}=\frac{10}{\frac{10}{3}}\\ =\frac{10}{1}\times \frac{3}{10}\\ =\frac{30}{10}\\ =3\text{ hr}\end{array}$

Raj travels $360\text{ km}$ on three fifths of his petrol tank. How far would he travel at the same rate with a full tank of petrol?

Raj travels $360\text{ km}$ on $\frac{3}{5}$ of his petrol tank.

Total distance travelled with full tank of petrol
$=$ $\left(\text{Reciprocal of}\frac{3}{5}\right)\times 360\text{ km}$

$=\frac{5}{3}\times 360=5\times 120=600\text{ km}$

Hence, total distance travelled by Raj with the available petrol tank is $600\text{ km}\text{.}$

Kajol has $\text{Rs}75.$ This is $\frac{3}{8}$ of the amount she earned. How much did she earn?

Kajol has $\text{Rs}75$.

According to question, $75=\frac{3}{8}\text{ of amount earned}$

$75=\frac{3}{8}\times \text{amount earned}$

Amount earned $=\frac{75}{3}\times 8=\text{Rs}200$

It takes $17$ full specific type of trees to make one tonne of paper. If there are $221$ such trees in a forest, then

      i.      What fraction of forest will be used to make;

a.    $5$ tonnes of paper.

b.   $10$ tonnes of paper.

   ii.            To save $\frac{7}{13}$ part of the forest how much of paper we have to save.

(i)         

a.    $1$ tonne of paper require $=17$ trees

$5$ tonne of paper require $=17\times 5$ trees

                                         $=85$ trees

Now, there are $221$ trees on the forest.

So, $85$ trees covers $=\frac{85}{221}$ fraction of forest

                                $=\frac{5}{13}$ fraction of forest

b.   Similarly

$10$ tonne of paper require $=17\times 10$ trees

                                           $=170$ trees

So, $170$ trees covers $=\frac{170}{221}$ fraction of forest

                                  $=\frac{10}{13}$ fraction of forest

(ii)          $\frac{7}{13}$ part of forest $=\frac{7}{13}\times 221\text{ trees}$

                         $=119\text{ trees}$

Number of tonnes of paper which can be made by $119\text{ trees}=\frac{119}{17}=7$

Simplify and write the result in decimal form :

$\left(1÷\frac{2}{9}\right)+\left(1÷3\frac{1}{5}\right)+\left(1÷2\frac{2}{3}\right)$

$\left(1÷\frac{2}{9}\right)+\left(1÷3\frac{1}{5}\right)+\left(1÷2\frac{2}{3}\right)$

$\begin{array}{l}=\left(1÷\frac{2}{9}\right)+\left(1÷\frac{\left(5\times 3\right)+1}{5}\right)+\left(1÷\frac{\left(2\times 3\right)+2}{3}\right)\\ =\left(1\times \frac{9}{2}\right)+\left(1÷\frac{16}{5}\right)+\left(1÷\frac{8}{3}\right)\\ =\left(1\times \frac{9}{2}\right)+\left(1\times \frac{5}{16}\right)+\left(1\times \frac{3}{8}\right)\\ =\frac{9}{2}+\frac{5}{16}+\frac{3}{8}\\ =\frac{72+5+6}{16}\\ =\frac{83}{16}\\ =5.1875\end{array}$

Some pictures (a) to (f) are given below. Tell which of them show:

1.   $2\times \frac{1}{4}$

2.   $2\times \frac{3}{7}$

3.   $2\times \frac{1}{3}$

4.   $\frac{1}{4}\times 4$

5.   $3\times \frac{2}{9}$

6.   $\frac{1}{4}\times 3$

a.   

 

b.  

 

 

c.   

 

d.  

 

 

e.   

 

f.    

 

 

 

1.   d.

 

$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$

 

 

2.   f.

 

$\frac{3}{7}+\frac{3}{7}=\frac{6}{7}$

 

 

3.   c.

$\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

 

 

4.   b.

 

$\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{4}{4}$

 

 

5.   a.

 

$\frac{2}{9}+\frac{2}{9}+\frac{2}{9}=\frac{6}{9}$

 

 

6.   e.

 

$\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$

 

 

 

 

Evaluate: $\left(0.3\right)\times \left(0.3\right)-\left(0.2\right)\times \left(0.2\right)$

Given, $\left(0.3\right)\times \left(0.3\right)-\left(0.2\right)\times \left(0.2\right)$

$\begin{array}{l}0.3=\frac{3}{10}\text{ and }0.2=\frac{2}{10}\\ \left(\frac{3}{10}\times \frac{3}{10}\right)-\left(\frac{2}{10}\times \frac{2}{10}\right)\\ =\frac{9}{100}-\frac{4}{100}\\ =\frac{9-4}{100}\\ =\frac{5}{100}\\ =0.05\end{array}$

Evaluate: $\frac{0.6}{0.3}+\frac{0.16}{0.4}$

Given, $\frac{0.6}{0.3}+\frac{0.16}{0.4}$

$\begin{array}{l}0.6=\frac{6}{10}\text{ and }0.3=\frac{3}{10}\\ 0.16=\frac{16}{100}\text{ and }0.4=\frac{4}{10}\end{array}$

$\begin{array}{l}\frac{0.6}{0.3}+\frac{0.16}{0.4}\\ =\frac{\frac{6}{10}}{\frac{3}{10}}+\frac{\frac{16}{100}}{\frac{4}{10}}\\ \left(\frac{6}{10}\times \frac{10}{3}\right)+\left(\frac{16}{100}\times \frac{10}{4}\right)\\ =\frac{60}{30}+\frac{160}{400}\\ =\frac{6}{3}+\frac{16}{40}\\ =\frac{2}{1}+\frac{4}{10}\\ =\frac{20+4}{10}\\ =\frac{24}{10}\\ =\frac{12}{5}\\ =2.4\end{array}$

Find the value: $\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}$

Given, $\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}$

$\begin{array}{l}0.2=\frac{2}{10}\text{, }0.14=\frac{14}{100}\text{ and }0.5=\frac{5}{10}\\ 0.91=\frac{91}{100}\text{, }0.1=\frac{1}{10}\text{ and }0.2=\frac{2}{10}\\ \frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}\\ =\frac{\left(\frac{2}{10}\times \frac{14}{100}\right)+\left(\frac{5}{10}\times \frac{91}{100}\right)}{\left(\frac{1}{10}\times \frac{2}{10}\right)}\\ =\frac{\frac{2\times 14}{1000}+\frac{5\times 91}{1000}}{\frac{1\times 2}{100}}\\ =\left(\frac{\frac{28}{1000}+\frac{455}{1000}}{\frac{2}{100}}\right)\\ =\frac{\frac{28+455}{1000}}{\frac{2}{100}}\end{array}$

$\begin{array}{l}=\frac{\frac{483}{1000}}{\frac{2}{100}}\\ =\frac{483}{1000}\times \frac{100}{2}\\ =\frac{483}{10\times 2}\\ =\frac{241.5}{10}\\ =24.15\end{array}$

A square and an equilateral triangle have a side in common. If side of triangle is $\frac{4}{3}\text{ cm}$ long, find the perimeter of figure formed (Fig. 2.8).

Fig. 2.8

The given square & equilateral triangle both have a common side  BC.

So, all the side of square & triangle will be equal and of measure $\frac{4}{3}\text{ cm}$.

Perimeter of the figure $=\text{AB}+\text{BD}+\text{DE}+\text{EC}+\text{AC}$

                                $\begin{array}{l}=5\times \text{AB}\\ =5\times \frac{4}{3}\\ =\frac{20}{3}\text{ cm}\end{array}$

Rita has bought a carpet of size $4\text{ m}\times 6\frac{2}{3}\text{ m}\text{.}$ But her room size is $3\frac{1}{3}\text{ m}\times 5\frac{1}{3}\text{ m}\text{.}$ What fraction of area should be cut off to fit wall to wall carpet into the room?

Carpet size $=4\text{ m}\times 6\frac{2}{3}\text{ m}$

                   $\begin{array}{l}=4\times \frac{\left(6\times 3\right)+2}{3}\\ =4\times \frac{18+2}{3}\\ =4\times \frac{20}{3}\end{array}$

                   $\begin{array}{l}=\frac{4}{1}\times \frac{20}{3}\\ =\frac{80}{3}\\ =\frac{80}{3}{\text{ m}}^2\end{array}$

Room size $=3\frac{1}{3}\text{ m}\times 5\frac{1}{3}\text{ m}$

$\begin{array}{l}=\frac{\left(3\times 3\right)+1}{3}\times \frac{\left(5\times 3\right)+1}{3}\\ =\frac{9+1}{3}\times \frac{15+1}{3}\\ =\frac{10}{3}\times \frac{16}{3}\\ =\frac{160}{9}{\text{ m}}^2\end{array}$

Difference between the area of carpet & room

$\begin{array}{l}=\frac{80}{3}-\frac{160}{9}\\ =\frac{240-160}{9}\\ =\frac{80}{9}{\text{ m}}^2\end{array}$

In fraction,

Area that will be cut off $=\frac{\text{Difference between the area of carpet \& room}}{\text{Original}\text{area}}=\frac{\frac{80}{9}}{\frac{80}{3}}$

                                                                                $\begin{array}{l}=\frac{80}{9}\times \frac{3}{80}\\ =\frac{1}{3}\end{array}$

Hence, $\frac{1}{3}$ of area should be cut off.

Family photograph has length $14\frac{2}{5}\text{ cm}$ and breadth $10\frac{2}{5}\text{ cm}\text{.}$ It has border of uniform width $2\frac{3}{5}\text{ cm}\text{.}$ Find the area of framed photograph.

Length of family photograph $=14\frac{2}{5}\text{ cm (given)}$

                                                $\begin{array}{l}=\frac{\left(14\times 5\right)+2}{5}\\ =\frac{72}{5}\text{ cm}\end{array}$

Breadth of family photograph $=10\frac{2}{5}\text{ cm}$

                                                  $\begin{array}{l}=\frac{\left(10\times 5\right)+2}{5}\\ =\frac{52}{5}\text{ cm}\end{array}$

New length including border (from both sides)

$\begin{array}{l}=\frac{72}{5}+\left(2\frac{3}{5}\times 2\right)\\ =\frac{72}{5}+\left(\frac{13}{5}\times 2\right)\\ =\frac{72}{5}+\frac{26}{5}\\ =\frac{72+26}{5}\\ =\frac{98}{5}\text{ cm}\end{array}$

New width including border (from both sides)

$\begin{array}{l}=\frac{52}{5}+\left(2\frac{3}{5}\times 2\right)\\ =\frac{52}{5}+\frac{26}{5}\\ =\frac{52+26}{5}\\ =\frac{78}{5}\text{ cm}\end{array}$

Area of framed photograph $=$ length $\times$ breadth

$=\frac{98}{5}\times \frac{78}{5}$

$\begin{array}{l}=\frac{7644}{25}\\ =305\frac{19}{25}{\text{ cm}}^2\end{array}$

Hence, the area of framed photograph is $305\frac{19}{25}{\text{ cm}}^2$

Cost of a burger is $\text{Rs}20\frac{3}{4}$ and of Macpuff is $\text{Rs}15\frac{1}{2}\text{.}$ Find the cost of $4$ burgers and $14$ macpuffs.

Cost of $1$ burger $=\text{Rs}20\frac{3}{4}$

                            $\begin{array}{l}=\text{Rs}\frac{\left(20\times 4\right)+3}{4}\\ =\text{Rs}\frac{83}{4}\end{array}$

Cost of $4$ burger $=\text{Rs}4\times \frac{83}{4}$

                            $=\text{Rs}83$

Cost of $1$ macpuff $=\text{Rs}15\frac{1}{2}$

                               $=\text{Rs}\frac{31}{2}$

Cost of $14$ macpuff $=\text{Rs}14\times \frac{31}{2}$

                                 $=\text{Rs}217$

Total cost of $4$ burgers & $14$ macpuffs $=\text{Rs}\left(83+217\right)$

$=\text{Rs}300$

A hill, $101\frac{1}{3}\text{ m}$ in height, has $\frac{1}{4}\text{th}$ of its height under water. What is the height of the hill visible above the water?

Given, height of the hill $=101\frac{1}{3}\text{ m}$

                                       $\begin{array}{l}=\frac{\left(101\times 3\right)+1}{3}\\ =\frac{303+1}{3}\\ =\frac{304}{3}\text{ m}\end{array}$

Height of the hill under water $=\frac{1}{4}$ of the height of the hill

                                                $\begin{array}{l}=\frac{1}{4}\times \frac{304}{3}\\ =\frac{76}{3}\text{ m}\end{array}$

Height of the hill above the water
$=$ height of the hill – height of the hill under water

$\begin{array}{l}=\frac{304}{3}-\frac{76}{3}\\ =\frac{228}{3}\\ =76\text{ m}\end{array}$

Sports: Reaction time measures how quickly a runner reacts to the starter pistol. In the $100\text{ m}$ dash at the $2004$ Olympic Games, Lauryn Williams had a reaction time of $0.214$ second. Her total race time, including reaction time, was $11.03$ seconds. How long did it take her to run the actual distance?

Time taken to run the actual distance $=$ Total race time $-$ reaction time

$\begin{array}{l}=\left(11.03-0.214\right)\text{s}\\ =10.816\text{ s}\end{array}$

State whether the answer is greater than $1$ or less than $1.$ Put a ‘√’ mark in appropriate box.

      i.       $\frac{2}{3}÷\frac{1}{2}=\frac{2}{3}\times \frac{2}{1}=\frac{4}{3}=1.33\left(>1\right)$

   ii.       $\frac{2}{3}÷\frac{2}{1}=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}=0.33\left(<1\right)$

iii.       $6÷\frac{1}{4}=6\times \frac{4}{1}=24\left(>1\right)$

iv.       $\frac{1}{5}÷\frac{1}{2}=\frac{1}{5}\times \frac{2}{1}=\frac{2}{5}=0.4\left(<1\right)$

   v.       $4\frac{1}{3}÷3\frac{1}{2}=\frac{13}{3}÷\frac{7}{2}=\frac{13}{3}\times \frac{2}{7}=\frac{26}{21}=1.24\left(>1\right)$

vi.       $\frac{2}{3}\times 8\frac{1}{2}=\frac{2}{3}\times \frac{17}{2}=\frac{17}{3}=5.67\left(>1\right)$

There are four containers that are arranged in the ascending order of their heights. If the height of the smallest container given in the figure is expressed as $\frac{7}{25}x=10.5\text{ cm}\text{.}$ Find the height of the largest container.

It is given that, the smallest cylinder is $10.5\text{ cm}$ high.

Height of the smallest cylinder in terms of $x$ is $\frac{7}{25}x\text{,}$ where $x$ is height of largest cylinder.

Then, $\frac{7}{25}x=10.5$

$\begin{array}{l}x=\frac{10.5}{1}\times \frac{25}{7}\\ =\frac{10.5\times 25}{7}\\ =\frac{262.5}{7}\\ =37.5\text{ cm}\end{array}$

Hence, height of the largest container is $37.5\text{ cm}$

 

In Questions 119 to 122, replace ‘?’ with appropriate fraction.

 

 

Given sequence is $\frac{7}{8}\text{, }\frac{7}{24}\text{, }\frac{7}{72}\text{, }\frac{7}{216}\text{,?}$

We observe that each fraction is divided by $3$ to get next fraction

So, $\text{?}=\frac{7}{216}÷3$

         $\begin{array}{l}=\frac{7}{216}\times \frac{1}{3}\\ =\frac{7}{648}\end{array}$

Question: 120

 

 

Given sequence is $\frac{3}{32}\text{, }\frac{3}{16}\text{, }\frac{3}{8}\text{, }\frac{3}{4}\text{,?}$

Each fraction is multiplied by $2$ to get the next fraction

So, $\text{?}=\frac{3}{4}\times 2$

      $=\frac{3}{2}$

 

 

Given sequence is $0.05\text{, }0.5\text{, }5\text{, }50\text{,?}$

We observe that, each fraction is multiplied by $10$ to get the next number.

$\text{?}=50\times 10$

  $=500$

 

 

Given sequence is $0.1\text{, }0.01\text{, }0.001\text{, }0.0001\text{,?}$

We observe that, each fraction is divided by $10$ to get the next number

$\text{?}=\frac{0.0001}{10}$

  $=0.00001$

 

What is the Error in each of question 123 to 125?

A student compared $-\frac{1}{4}$ and $-\mathrm{0.3.}$ He changed $-\frac{1}{4}$ to the decimal $-0.25$ and wrote, “Since $0.3$ is greater than $0.25\text{,}-0.3$ is greater than $-0.25\text{"}\text{.}$ What was the student’s error?

If the numbers are negative, then the numbers whose absolute value is greater, will be smaller.

Hence, $-0.25$ is greater than $-0.3$

So, the student made the error that $\left(-0.3\right)>-\left(0.25\right)$.

A student multiplied two mixed fractions in the following manner: $2\frac{4}{7}\times 3\frac{1}{4}=6\frac{1}{7}\text{.}$ What error the student has done?

To multiply two mixed fractions, first convert them into improper fraction.

So, $2\frac{4}{7}\times 3\frac{1}{4}=\frac{2\times 7+4}{7}\times \frac{3\times 4+1}{4}$

                     $\begin{array}{l}=\frac{18}{7}\times \frac{13}{4}\\ =\frac{9\times 2}{7}\times \frac{13}{2\times 2}\\ =\frac{117}{14}\\ =\frac{14\times 8+5}{14}\\ =\frac{14\times 8}{14}+\frac{5}{14}\\ =8\frac{5}{14}\end{array}$

In the pattern $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots \mathrm{..}$ which fraction makes the sum greater than $1$ (first time)? Explain.

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\\ =\frac{20+15+12}{60}\\ =\frac{47}{60}<1\end{array}$

According to the pattern, next number will be $\frac{1}{6}$

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\\ =\frac{40+30+24+20}{120}\\ =\frac{114}{120}<1\end{array}$

Now, according to the pattern, next number after $\frac{1}{6}\text{ is }\frac{1}{7}$

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\\ =\frac{280+210+168+140+120}{840}\\ =\frac{918}{840}>1\end{array}$

Hence, first time, $\frac{1}{7}$ makes the sum greater than $1$ .