Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 2 Fractions & Decimals are provided here with simple step-by-step explanations. These solutions for Fractions & Decimals are extremely popular among class 7 students for Maths Fractions & Decimals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 7 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 38:

#### Question 1:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $\frac{26}{25}$

(b) $\frac{52}{25}$

(c) $\frac{2}{5}$

(d) 6

#### Answer:

$\begin{array}{rcl}\frac{2}{5}\times 5\frac{1}{5}& =& \frac{2}{5}\times \frac{26}{5}\\ & =& \frac{2\times 26}{5\times 5}\\ & =& \frac{52}{25}\end{array}$

Hence, the correct answer is option (b).

#### Page No 38:

#### Question 2:

**In the given question out of four options, only one is correct. Write the correct answer.**

$3\frac{3}{4}\xf7\frac{3}{4}$ is equal to:

(a) 3

(b) 4

(c) 5

(d) $\frac{45}{16}$

#### Answer:

$3\frac{3}{4}\xf7\frac{3}{4}=\frac{15}{4}\times \frac{4}{3}\phantom{\rule{0ex}{0ex}}=5$

Hence, the correct answer is option (c).

#### Page No 38:

#### Question 3:

A ribbon of length $5\frac{1}{4}$m is cut into small pieces each of length $\frac{3}{4}\mathrm{m}$. Number of pieces will be:

(a) 5

(b) 6

(c) 7

(d) 8

#### Answer:

Total length of ribbon = $5\frac{1}{4}$m

Length of each small piece = $\frac{3}{4}\mathrm{m}$

∴ Number of pieces $=5\frac{1}{4}\xf7\frac{3}{4}$

$=\frac{21}{4}\times \frac{4}{3}\phantom{\rule{0ex}{0ex}}=7$

Thus, number of pieces will be 7.

Hence, the correct answer is option (c).

#### Page No 39:

#### Question 4:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(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) $\frac{2}{3},\frac{6}{7},\frac{13}{21}$

#### Answer:

LCM of 3, 7 and 21 is 21.

$\frac{2}{3}=\frac{2\times 7}{3\times 7}=\frac{14}{21},\phantom{\rule{0ex}{0ex}}\frac{6}{7}=\frac{6\times 3}{7\times 3}=\frac{18}{21},\phantom{\rule{0ex}{0ex}}\frac{13}{21}$

Now,

$\because \frac{13}{21}<\frac{14}{21}<\frac{18}{21}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{13}{21}\frac{2}{3}\frac{6}{7}\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (b).

#### Page No 39:

#### Question 5:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) 2

(b) 3

(c) $\frac{2}{3}$

(d) $\frac{3}{2}$

#### Answer:

$\begin{array}{rcl}\mathrm{Reciprocal}\mathrm{of}\frac{2}{3}& =& \frac{1}{\frac{2}{3}}\\ & =& \frac{3}{2}\end{array}$

Hence, the correct answer is option (d).

#### Page No 39:

#### Question 6:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $3\frac{5}{13}$

(b) $5\frac{3}{13}$

(c) $13\frac{3}{5}$

(d) $13\frac{5}{3}$

#### Answer:

$\begin{array}{rcl}\frac{11}{13}\times 4& =& \frac{11}{13}\times \frac{4}{1}\\ & =& \frac{44}{13}\\ & =& 3\frac{5}{13}\end{array}$

Hence, the correct answer is option (d).

#### Page No 39:

#### Question 7:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $17\frac{2}{5}$

(b) $\frac{24}{5}$

(c) $13\frac{1}{5}$

(d) $5\frac{1}{13}$

#### Answer:

$\begin{array}{rcl}3\times 4\frac{2}{5}& =& \frac{3}{1}\times \frac{22}{5}\\ & =& \frac{66}{5}\\ & =& 13\frac{1}{5}\end{array}$

Hence, the correct answer is option (c).

#### Page No 39:

#### Question 8:

**In the given question out of four options, only one is correct. Write the correct answer.**

Pictorial representation of 3×$\frac{2}{3}\mathrm{is}$

#### Answer:

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

Thus, the Pictorial representation of 3×$\frac{2}{3}\mathrm{is}$

Hence, the correct answer is option (b).

#### Page No 39:

#### Question 9:

**In the given question out of four options, only one is correct. Write the correct answer.**

$\frac{1}{5}\xf7\frac{4}{5}$equal to:

(a) $\frac{4}{5}$

(b) $\frac{1}{5}$

(c) $\frac{5}{4}$

(d) $\frac{1}{4}$

#### Answer:

$\begin{array}{rcl}\frac{1}{5}\xf7\frac{4}{5}& =& \frac{1}{5}\times \frac{5}{4}\\ & =& \frac{1}{4}\end{array}$

Hence, the correct answer is option (d).

#### Page No 39:

#### Question 10:

**In the given question out of four options, only one is correct. Write the correct answer.**

The product of 0.03 × 0.9 is:

(a) 2.7

(b) 0.27

(c) 0.027

(d) 0.0027

#### Answer:

$\begin{array}{rcl}0.03\times 0.9& =& \frac{3}{100}\times \frac{9}{10}\\ & =& \frac{27}{1000}\\ & =& 0.027\end{array}$

Hence, the correct answer is option (c).

#### Page No 39:

#### Question 11:

**In the given question out of four options, only one is correct. Write the correct answer.**

$\frac{5}{7}\xf76$ is equal to:

(a) $\frac{30}{7}$

(b) $\frac{5}{42}$

(c) $\frac{30}{42}$

(d) $\frac{6}{7}$

#### Answer:

$\begin{array}{rcl}\frac{5}{7}\xf76& =& \frac{5}{7}\times \frac{1}{6}\\ & =& \frac{5}{42}\end{array}$

Hence, the correct answer is option (b).

#### Page No 40:

#### Question 12:

**In the given question out of four options, only one is correct. Write the correct answer.**

$5\frac{1}{6}\xf7\frac{9}{2}$is equal to

(a) $\frac{31}{6}$

(b) $\frac{1}{27}$

(c) $5\frac{1}{27}$

(d) $\frac{31}{27}$

#### Answer:

$\begin{array}{rcl}5\frac{1}{6}\xf7\frac{9}{2}& =& 5\frac{1}{6}\times \frac{2}{9}\\ & =& \frac{31}{6}\times \frac{2}{9}\\ & =& \frac{31}{27}\end{array}$

Hence, the correct answer is option (d).

#### Page No 40:

#### Question 13:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $\frac{1}{3}+\frac{1}{6}$

(b) $\frac{1}{3}-\frac{1}{6}$

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

(d) $\frac{1}{3}\xf7\frac{1}{6}$

#### Answer:

$\frac{1}{3}\mathrm{of}\frac{1}{6}=\frac{1}{3}\times \frac{1}{6}$

Hence, the correct answer is option (c).

#### Page No 40:

#### Question 14:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $\frac{5}{12}$

(b) $\frac{5}{35}$

(c) $\frac{1}{35}$

(d) $\frac{6}{35}$

#### Answer:

Hence, the correct answer is option (d).

#### Page No 40:

#### Question 15:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

#### Answer:

Flour required in 1 packet = $2\frac{1}{2}$ cups

Sugar required in 1 packet = â€‹$1\frac{2}{3}$ cups

Total quantity of both ingredients used in 1 packet = quantity of flour + quantity of sugar

$=2\frac{1}{2}+1\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}+\frac{5}{3}\phantom{\rule{0ex}{0ex}}=\frac{15+10}{6}\phantom{\rule{0ex}{0ex}}=\frac{25}{6}\phantom{\rule{0ex}{0ex}}=4\frac{1}{6}\mathrm{cups}$

Total quantity of both ingredients used in 10 packet = 10 × Total quantity of both ingredients used in 1 packet

$=10\times 4\frac{1}{6}\mathrm{cups}\phantom{\rule{0ex}{0ex}}=10\times \frac{25}{6}\mathrm{cups}\phantom{\rule{0ex}{0ex}}=\frac{125}{3}\mathrm{cups}\phantom{\rule{0ex}{0ex}}=41\frac{2}{3}\mathrm{cups}$

âˆµ 40 < $41\frac{2}{3}$< 50

Thus, the estimated total quantity of both ingredients used in 10 such packets of biscuits will be between 40 cups and 50 cups.

Hence, the correct answer is option (c).

#### Page No 41:

#### Question 16:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $42\frac{1}{2}$

(b) $47\frac{1}{4}$

(c) $42\frac{3}{4}$

(d) $47\frac{3}{4}$

#### Answer:

$\begin{array}{rcl}7\times 6\frac{3}{4}& =& \frac{7}{1}\times \frac{27}{4}\\ & =& \frac{189}{4}\\ & =& 47\frac{1}{4}\end{array}$

Hence, the correct answer is option (b).

#### Page No 41:

#### Question 17:

**In the given question out of four options, only one is correct. Write the correct answer.**

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

(a) $\frac{14}{2}$

(b)$\frac{35}{4}$

(c) $\frac{14}{5}$

(d) $\frac{35}{2}$

#### Answer:

$\begin{array}{rcl}7\xf7\frac{2}{5}& =& 7\times \frac{5}{2}\\ & =& \frac{7}{1}\times \frac{5}{2}\\ & =& \frac{35}{2}\\ & & \end{array}$

Hence, the correct answer is option (d).

#### Page No 41:

#### Question 18:

**In the given question out of four options, only one is correct. Write the correct answer.**

$2\frac{2}{3}\xf75$ is equal to

(a) $\frac{8}{15}$

(b) $\frac{40}{3}$

(c) $\frac{40}{5}$

(d) $\frac{8}{3}$

#### Answer:

$\begin{array}{rcl}2\frac{2}{3}\xf75& =& \frac{8}{3}\xf75\\ & =& \frac{8}{3}\times \frac{1}{5}\\ & =& \frac{8}{15}\end{array}$

Hence, the correct answer is option (a).

#### Page No 41:

#### Question 19:

**In the given question out of four options, only one is correct. Write the correct answer.**

$\frac{4}{5}$ 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) $\frac{4}{21}$

#### Answer:

The quantity of apples used on Monday = $\frac{4}{5}$ of 5 kg

$=\frac{4}{5}\times 5\mathrm{kg}\phantom{\rule{0ex}{0ex}}=4\mathrm{kg}$

The quantity of apples left after Monday = 5 kg − 4 kg = 1 kg

The quantity of apples used on the next day, i.e., Tuesday = $\frac{1}{3}$ of what was left was used

$\begin{array}{rcl}& =& \frac{1}{3}\times 1\mathrm{kg}\\ & =& \frac{1}{3}\mathrm{kg}\\ & & \end{array}$

The quantity of apples left after Tuesday =$1\mathrm{kg}-\frac{1}{3}\mathrm{kg}$

$=\frac{3-1}{3}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\mathrm{kg}$

Thus, weight (in kg) of apples left now is $\frac{2}{3}\mathrm{kg}$.

Hence, the correct answer is option (c).

#### Page No 41:

#### Question 20:

**In the given question out of four options, only one is correct. Write the correct answer.**

(a) $\frac{1}{4}\xf73$

(b) $3\times \frac{1}{4}$

(c) $\frac{3}{4}\times 3$

(d) $3\xf7\frac{1}{4}$

#### Answer:

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

Hence, the correct answer is option (b).

#### Page No 41:

#### Question 21:

**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 __________

#### Answer:

Let the whole part of the cake is 1.

Part of the cake Rani ate

∴ Part of the cake left

Now, part of the cake Ravi ate

Thus, part of the cake left is $\frac{4}{7}$.

#### Page No 41:

#### Question 22:

**Fill in the blanks to make the statements true.**

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

#### Answer:

Reciprocal of $\frac{3}{7}=\frac{1}{\frac{3}{7}}=\frac{7}{3}$

#### Page No 41:

#### Question 23:

**Fill in the blanks to make the statements true.**

$\frac{2}{3}\mathrm{of}27\mathrm{is\_\_\_\_\_\_\_\_\_\_\_\_}$

#### Answer:

$\begin{array}{rcl}\frac{2}{3}\mathrm{of}27& =& \frac{2}{3}\times 27\\ & =& 2\times 9\\ & =& 18\end{array}$

Thus, $\frac{2}{3}\mathrm{of}27\mathrm{is}\overline{)\mathbf{}\mathbf{18}\mathbf{}}$.

#### Page No 42:

#### Question 24:

**Fill in the blanks to make the statements true.**

$\frac{4}{5}\mathrm{of}45\mathrm{is}\mathrm{\_\_\_\_\_\_\_\_\_\_\_}$

#### Answer:

$\begin{array}{rcl}\frac{4}{5}\mathrm{of}45& =& \frac{4}{5}\times 45\\ & =& 4\times 9\\ & =& 36\end{array}$

Thus, $\frac{4}{5}\mathrm{of}45\mathrm{is}\overline{)36}.$

#### Page No 42:

#### Question 25:

**Fill in the blanks to make the statements true.**

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

#### Answer:

$\begin{array}{rcl}4\times 6\frac{1}{3}& =& 4\times \frac{19}{3}\\ & =& \frac{76}{3}\\ & =& 25\frac{1}{3}\end{array}$

Thus, $4\times 6\frac{1}{3}\mathrm{is}\mathrm{equal}\mathrm{to}\overline{)25\frac{1}{3}}.$

#### Page No 42:

#### Question 26:

**Fill in the blanks to make the statements true.**

$\frac{1}{2}\mathrm{of}4\frac{2}{7}\mathrm{is\_\_\_\_\_\_\_\_\_\_}$

#### Answer:

$\begin{array}{rcl}\frac{1}{2}\mathrm{of}4\frac{2}{7}& =& \frac{1}{2}\times \frac{30}{7}\\ & =& \frac{15}{7}\\ & =& 2\frac{1}{7}\end{array}$

Thus, $\frac{1}{2}\mathrm{of}4\frac{2}{7}\mathrm{is}\overline{)2\frac{1}{7}}.$

#### Page No 42:

#### Question 27:

**Fill in the blanks to make the statements true.**

$\frac{1}{9}\mathrm{of}\frac{6}{5}\mathrm{is\_\_\_\_\_\_\_\_\_}$

#### Answer:

$\begin{array}{rcl}\frac{1}{9}\mathrm{of}\frac{6}{5}& =& \frac{1}{9}\times \frac{6}{5}\\ & =& \frac{2}{15}\end{array}$

Thus, $\frac{1}{9}\mathrm{of}\frac{6}{5}\mathrm{is}\overline{)\frac{2}{15}}.$

#### Page No 42:

#### Question 28:

**Fill in the blanks to make the statements true.**

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

#### Answer:

$\begin{array}{rcl}2\frac{3}{7}\times \frac{7}{9}& =& \frac{17}{7}\times \frac{7}{9}\\ & =& \frac{17}{9}\\ & =& 1\frac{8}{9}\end{array}$

Thus, $2\frac{3}{7}\times \frac{7}{9}\mathrm{is}\overline{)1\frac{8}{9}}.$

#### Page No 42:

#### Question 29:

**Fill in the blanks to make the statements true.**

$\frac{4}{5}\xf74$ is equal to _______

#### Answer:

$\begin{array}{rcl}\frac{4}{5}\xf74& =& \frac{4}{5}\times \frac{1}{4}\\ & =& \frac{1}{5}\end{array}$

Thus, $\frac{4}{5}\xf74\mathrm{is}\mathrm{equal}\mathrm{to}\overline{)\frac{1}{5}}.$

#### Page No 42:

#### Question 30:

**Fill in the blanks to make the statements true.**

$\frac{2}{5}$ of 25 is ________

#### Answer:

$\begin{array}{rcl}\frac{2}{5}\mathrm{of}25& =& \frac{2}{5}\times 25\\ & =& 10\end{array}$

Thus, $\frac{2}{5}\mathrm{of}25\mathrm{is}\mathrm{equals}\mathrm{to}\overline{)10}.$

#### Page No 42:

#### Question 31:

**Fill in the blanks to make the statements true.**

$\frac{1}{5}\xf7\frac{5}{6}=\frac{1}{5}\_\_\_\_\_\frac{6}{5}$

#### Answer:

$\frac{1}{5}\xf7\frac{5}{6}=\frac{1}{5}\overline{)\mathbf{}\mathbf{\times}\mathbf{}}\frac{6}{5}$

#### Page No 42:

#### Question 32:

**Fill in the blanks to make the statements true.**

3.2 × 10 = _______

#### Answer:

$3.2\times 10=\frac{32}{10}\times 10=\overline{)\mathbf{}\mathbf{32}\mathbf{}}$

#### Page No 42:

#### Question 33:

**Fill in the blanks to make the statements true.**

25.4 × 1000 = _______

#### Answer:

$\begin{array}{rcl}25.4\times 1000& =& \frac{254}{10}\times 1000\\ & =& 25400\end{array}$

Thus, 25.4 × 1000 = 25400.

#### Page No 42:

#### Question 34:

**Fill in the blanks to make the statements true.**

93.5 × 100 = _______

#### Answer:

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

Thus, 93.5 × 100 = 9350

#### Page No 42:

#### Question 35:

**Fill in the blanks to make the statements true.**

4.7 ÷ 10 = ______

#### Answer:

$\begin{array}{rcl}4.7\xf710& =& \frac{47}{10}\times \frac{1}{10}\\ & =& \frac{47}{100}\\ & =& 0.47\end{array}$

Thus, 4.7 ÷ 10 = ** 0.47 **** .**

#### Page No 42:

#### Question 36:

**Fill in the blanks to make the statements true.**

4.7 ÷ 100 = _____

#### Answer:

$\begin{array}{rcl}4.7\xf7100& =& \frac{47}{10}\times \frac{1}{100}\\ & =& \frac{47}{1000}\\ & =& 0.047\end{array}$

Thus, 4.7 ÷ 100 =** 0.047 **.

#### Page No 42:

#### Question 37:

**Fill in the blanks to make the statements true.**

4.7 ÷ 1000 = ______

#### Answer:

$\begin{array}{rcl}4.7\xf71000& =& \frac{47}{10}\times \frac{1}{1000}\\ & =& \frac{47}{10000}\\ & =& 0.0047\end{array}$

Thus, 4.7 ÷ 100 =** 0.0047 **.

#### Page No 42:

#### Question 38:

**Fill in the blanks to make the statements true.**

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

#### Answer:

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

#### Page No 42:

#### Question 39:

**Fill in the blanks to make the statements true.**

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

#### Answer:

While dividing a fraction by another fraction, we ** multiply** the first fraction by the

**of the other fraction.**

__reciprocal__#### Page No 42:

#### Question 40:

**Fill in the blanks to make the statements true.**

8.4 ÷ ______ = 2.1

#### Answer:

8.4 ÷ ______ = 2.1

Let *x* be the required number.

$8.4\xf7x=2.1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{84}{10}\times \frac{1}{x}=\frac{21}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{84}{10}\times \frac{10}{21}=x\phantom{\rule{0ex}{0ex}}\Rightarrow 4=x\phantom{\rule{0ex}{0ex}}$

Thus, 8.4 ÷ ** 4** = 2.1

#### Page No 43:

#### Question 41:

**Fill in the blanks to make the statements true.**

52.7 ÷ _______ = 0.527

#### Answer:

52.7 ÷ _______ = 0.527

Let *x* be the required number.

$\Rightarrow 52.7\xf7x=0.527\phantom{\rule{0ex}{0ex}}\Rightarrow 52.7\times \frac{1}{x}=0.527\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{527}{10}\xf7\frac{527}{1000}=x\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{527}{10}\times \frac{1000}{527}=x\phantom{\rule{0ex}{0ex}}\Rightarrow 100=x$

Thus, 52.7 ÷ __ 100 __= 0.527

#### Page No 43:

#### Question 42:

**Fill in the blanks to make the statements true.**

0.5 _____ 0.7 = 0.35

#### Answer:

$\begin{array}{rcl}0.5\times 0.7& =& \frac{5}{10}\times \frac{7}{10}\\ & =& \frac{35}{100}\\ & =& 0.35\end{array}$

Thus, 0.5 __×__ 0.7 = 0.35

#### Page No 43:

#### Question 43:

**Fill in the blanks to make the statements true.**

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

#### Answer:

$\begin{array}{rcl}2\times \frac{5}{3}& =& \frac{2}{1}\times \frac{5}{3}\\ & =& \frac{10}{3}\end{array}$

Thus, 2 __ ×__ $\frac{5}{3}=\frac{10}{3}$

#### Page No 43:

#### Question 44:

**Fill in the blanks to make the statements true.**

2.001 ÷ 0.003 = __________

#### Answer:

$\begin{array}{rcl}2.001\xf70.003& =& \frac{2001}{1000}\xf7\frac{3}{1000}\\ & =& \frac{2001}{1000}\times \frac{1000}{3}\\ & =& \frac{2001}{3}\\ & =& 667\end{array}$

Thus, 2.001 ÷ 0.003 = **667**

#### Page No 43:

#### Question 45:

**State whether the statement is True or False.**

The reciprocal of a proper fraction is a proper fraction.

#### Answer:

Let $\frac{a}{b}\mathrm{be}\mathrm{a}\mathrm{proper}\mathrm{fraction}\mathrm{such}\mathrm{that}ab.$

$\begin{array}{rcl}\mathrm{Reciprocal}\mathrm{of}\frac{a}{b}& =& \frac{1}{\frac{a}{b}}\\ & =& \frac{b}{a}\end{array}$

Since,

$b>a\Rightarrow \frac{b}{a}>1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{b}{a}\mathrm{is}\mathrm{an}\mathrm{improper}\mathrm{fraction}$

Thus, the reciprocal of a proper fraction is a improper fraction.

So, the statement is False.

#### Page No 43:

#### Question 46:

**State whether the statement is True or False.**

The reciprocal of an improper fraction is an improper fraction.

#### Answer:

Let $\frac{a}{b}\mathrm{be}\mathrm{an}\mathrm{improper}\mathrm{fraction}\mathrm{such}\mathrm{that}ab.$

$\begin{array}{rcl}\mathrm{Reciprocal}\mathrm{of}\frac{a}{b}& =& \frac{1}{\frac{a}{b}}\\ & =& \frac{b}{a}\end{array}$

Since,

$b<a\Rightarrow \frac{b}{a}<1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{b}{a}\mathrm{is}a\mathrm{proper}\mathrm{fraction}$

Thus, the reciprocal of an improper fraction is a proper fraction.

So, the statement is False.

#### Page No 43:

#### Question 47:

**State whether the statement is True or False.**

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

#### Answer:

Let the two fractions be $\frac{a}{b}\mathrm{and}\frac{c}{d}$.

So, $\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}$

Thus, Product of two fractions = $\frac{\mathrm{Product}\mathrm{of}\mathrm{their}\mathrm{numerators}}{\mathrm{Product}\mathrm{of}\mathrm{their}\mathrm{denominators}}$

So, the statement is False.

#### Page No 43:

#### Question 48:

**State whether the statement is True or False.**

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

#### Answer:

Let the two improper fractions be $\frac{a}{b}\mathrm{and}\frac{c}{d}$ where *a *< *b* and *c* < *d*.

Since, *a *< *b* and *c* < *d*

⇒ *a ×* *c* < *b** ×* *d*

$\Rightarrow \frac{ac}{bd}<1$

As improper fractions are greater than 1.

Therefore, $\frac{ac}{bd}<1<\frac{a}{b}\mathrm{and}\frac{ac}{bd}1\frac{c}{d}$.

Thus, the product of two improper fractions is less than both fractions.

So, the statement is False.

#### Page No 43:

#### Question 49:

**State whether the statement is True or False.**

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

#### Answer:

Let a fraction be $\frac{a}{b}$.

The reciprocal of the fraction $\begin{array}{rcl}\frac{a}{b}& =& \frac{1}{\frac{a}{b}}=\frac{b}{a}\end{array}$.

Thus, the reciprocal of a fraction is obtained by inverting it upside down.

So, the statement is True.

#### Page No 43:

#### Question 50:

**State whether the statement is True or False.**

To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.

#### Answer:

For example, consider the decimal number 8.9765.

On multiplying 8.9765 by 1000

8.9765 × 1000 = 8976.5

Thus, to multiply a decimal number by 1000, we have moved the decimal point in the number to the right by three places.

So, the statement is True.

#### Page No 43:

#### Question 51:

**State whether the statement is True or False.**

To divide a decimal number by 100, we move the decimal point in the number to the left by two places.

#### Answer:

For example, consider the decimal number 456.7.

On dividing 456.7 by 100

456.7 ÷ 100 = 4.567

Thus, to divide a decimal number by 100, we move the decimal point in the number to the left by two places.

So, the statement is True.

#### Page No 43:

#### Question 52:

**State whether the statement is True or False.**

1 is the only number which is it's own reciprocal.

#### Answer:

True, 1 is the only number which is it's own reciprocal.

#### Page No 43:

#### Question 53:

**State whether the statement is True or False.**

$\frac{2}{3}$ of 8 is same as $\frac{2}{3}\xf78.$

#### Answer:

$\begin{array}{rcl}\frac{2}{3}\mathrm{of}8& =& \frac{2}{3}\times \frac{8}{1}\\ & =& \frac{16}{3}\end{array}$

$\begin{array}{rcl}\frac{2}{3}\xf78& =& \frac{2}{3}\times \frac{1}{8}\\ & =& \frac{1}{4}\end{array}$

$\frac{16}{3}\ne \frac{1}{4}$

∴ $\frac{2}{3}$ of 8 is not same as $\frac{2}{3}\xf78.$

So, the statement is False.

#### Page No 43:

#### Question 54:

**State whether the statement is True or False.**

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

#### Answer:

The reciprocal of $\frac{4}{7}=\frac{1}{{\displaystyle \frac{4}{7}}}=\frac{7}{4}.$

Thus, $\frac{4}{7}$ is not reciprocal of $\frac{4}{7}$.

#### Page No 43:

#### Question 55:

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

#### Answer:

Adding 5 to the numerator and denominator of the fraction $\frac{5}{9}$.

$\frac{5+5}{9+5}=\frac{10}{14}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}$

Yes, the value of the fraction changes.

$\frac{5}{9}<\frac{5}{7}\phantom{\rule{0ex}{0ex}}$

Thus, the value increases.

#### Page No 43:

#### Question 56:

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

#### Answer:

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

#### Page No 43:

#### Question 57:

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

#### Answer:

Letter among A and J are A, B, C, D, E, F, G, H, I, J.

Ttotal letters among A and J =10

$\frac{2}{5}\times 10=4$

∴ 4th letter among A and J = D

#### Page No 44:

#### Question 58:

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

#### Answer:

Let the number be *x*.

$\frac{2}{3}\mathrm{of}x=10\phantom{\rule{0ex}{0ex}}\Rightarrow x=10\times \frac{3}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=15\phantom{\rule{0ex}{0ex}}$

1.75 times of the number 15 =1.75 × 15

= 26.25

Thus, 1.75 times of that number is 26.25.

#### Page No 44:

#### Question 59:

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 chapati only and the remaining students like to eat both. What fraction of the total number of students like to eat both?

#### Answer:

Total number of students in the class = 40

Fraction of students who like to eat rice only = $\frac{1}{5}$

Number of students who like to eat rice only = $\frac{1}{5}\times 40=8$

Fraction of students who like to eat chapati only = $\frac{2}{5}$

Number of students who like to eat chapati only = $\frac{2}{5}\times 40=16$

The total number of students like to eat both = 40 − (8 + 16)

= 40 − 24

= 16

Thus, the total number of students who like to eat both are 16.

#### Page No 44:

#### Question 60:

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?

#### Answer:

Part of her homework in 2 hours = $\frac{2}{3}$

Part of her homework in 1 hours = $\frac{2}{3}\xf72=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$

Part of her homework in $1\frac{1}{4}$ hours = $\frac{1}{3}\times 1\frac{1}{4}=\frac{1}{3}\times \frac{5}{4}=\frac{5}{12}$

Thus, the part of her homework she had completed in $1\frac{1}{4}$ hours is $\frac{5}{12}$.

#### Page No 44:

#### Question 61:

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

#### Answer:

Let total number of pages be x.

Number of pages read by Reemu

According to question,

$\frac{1}{5}x+40=\frac{7}{10}x\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7}{10}x-\frac{1}{5}x=40\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{7-2}{10}\right)x=40\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{5}{10}\right)x=40\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}x=40\phantom{\rule{0ex}{0ex}}\Rightarrow x=80$

Number of pages read by Reemu = $\frac{7}{10}\times 80=56$

Number of pages left to be read by Reemu = $80-56=24$

#### Page No 44:

#### Question 62:

Write the number in the box â˜ such that

$\frac{3}{7}\times \overline{)}=\frac{15}{98}$

#### Answer:

Let the number in the box â˜ be *x*.

$\frac{3}{7}\times x=\frac{15}{98}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{15}{98}\times \frac{7}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{14}$

#### Page No 44:

#### Question 63:

Will the quotient $7\frac{1}{6}\xf73\frac{2}{3}$ be a fraction greater than 1.5 or less than 1.5? Explain.

#### Answer:

$7\frac{1}{6}\xf73\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{43}{6}\xf7\frac{11}{3}\phantom{\rule{0ex}{0ex}}=\frac{43}{6}\times \frac{3}{11}\phantom{\rule{0ex}{0ex}}=\frac{43}{22}\phantom{\rule{0ex}{0ex}}=1.95$

1.95 > 1.5

Thus, the quotient $7\frac{1}{6}\xf73\frac{2}{3}$ of the given fractions is greater than 1.5.

#### Page No 44:

#### Question 64:

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

#### Answer:

Method I: Conversion of both numbers $\frac{13}{17}$ and 0.82 into decimals.

$\frac{13}{17}$ = 0.76 and 0.82 into decimals.

Method II: Conversion of both numbers $\frac{13}{17}$ and 0.82 into fractions.

∴ Method of converting numbers into decimals is easier because it can easily be compared in this method but in other method it is also required to convert the fractions into like fractions to compare the numbers.

#### Page No 45:

#### Question 65:

**Health:** The directions for a pain reliever recommend that an adult of 60 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 kg adult takes 4 tablets, how many grams of pain reliever is he or she receivings?

(b) How many grams of pain reliever is the recommended dose for an adult weighing 46 kg?

#### Answer:

(a) Each tablet weighs

∴Weight of 4 tablets

(b) An adult weighing 46 kg takes only

∴ Weight of $2\frac{1}{2}$ tablets = $\frac{4}{25}\times \frac{5}{2}=\frac{2}{5}\mathrm{grams}$

#### Page No 45:

#### Question 66:

**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 kg adult dog

(b) a 6 kg cat

(c) a 18 kg pregnant dog

graphic missing

#### Answer:

(a) Total dosage given to a 18 kg adult dog

(b) Total dosage given to a 6 kg cat$=6\times \frac{1}{4}=\frac{3}{2}=1\frac{1}{2}$ tsp

(c) Total dosage given to a 18 kg pregnant dog $=\frac{1}{2}\times \frac{1}{4.5}\times 18=\frac{1}{9}\times 18=2$ tsp

#### Page No 45:

#### Question 67:

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

#### Answer:

∴ 1 kg chocolates to be filled in $\frac{1}{16}=1\xf7\frac{1}{16}=1\times \frac{16}{1}=16\mathrm{boxes}$

Now, $1\frac{1}{2}$ kg chocolates to be filled in $16\times 1\frac{1}{2}=16\times \frac{3}{2}=24$ boxes

#### Page No 45:

#### Question 68:

Anvi is making bookmarker like the one shown in the given figure. How many bookmarker can she make from a 15 m long ribbon?

#### Answer:

Length of one bookmarker = $10\frac{1}{2}\mathrm{cm}=\frac{21}{2}\mathrm{cm}$

∴ Number of bookmarkers Anvi can make from 15 m ix., 15 × 100 cm long ribbon

$=\left(\frac{1}{{\displaystyle \frac{21}{2}}}\times 15\times 100\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{2}{21}\times 15\times 100\right)\phantom{\rule{0ex}{0ex}}=\frac{1000}{7}\phantom{\rule{0ex}{0ex}}=142.85$

=142

Thus, Anvi can make 142 bookmaker from a 15 m long ribbon

#### Page No 46:

#### Question 69:

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

(a) The length of a side of the square is 8.3 cm.

(b) The length of a side of the square is exactly 7.875 cm.

#### Answer:

(a) Length of a side of the square = 83 cm

∴ Length of the diagonal of the square = 1.414 × 8.3 cm

= 11.7362 cm

≈ 11.74 cm

(b) Length of a side of the square = 7.875 cm

Length of the diagonal of the square = 1.414 × 7.875 cm

= 11.13525 cm

≈ 11.14 cm

#### Page No 46:

#### Question 70:

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 cm (b) 8.63 cm

#### Answer:

(a) Diameter of the circle = 14.35 cm

∴ Length of the side of the square = 0.707 × 14.35 cm

= 10.14545

= 10.15 cm

(b) Diameter of the circle = 8.63 cm

∴ Length of the side of the square = 0.707 × 8.63 cm

= 6.10141 cm

= 6.10 cm

#### Page No 46:

#### Question 71:

To find the distance around a circular disc, multiply the diameter of the disc by 3.14. What is the distance around the disc when :

(a) the diameter is 18.7 cm?

(b) the radius is 6.45 cm?

#### Answer:

(a) Diameter of the disc = 18.7 cm

∴ The distance around the disc = 3.14 × 18.7 cm

= 58.718 cm

(b) Radius of the disc = 6.45 cm

∴ Diameter of the disc = 2 × 6.45 cm = 12.9 cm

∴ The distance around the disc = 3.14 × 12.9 cm

= 40.506 cm

#### Page No 46:

#### Question 72:

What is the cost of 27.5 m of cloth at â‚¹53.50 per metre?

#### Answer:

Cost of 1 metre of cloth = â‚¹53.50

∴ Cost of 27.5 metres of cloth = â‚¹ (53.50 × 27.5) = â‚¹1471.25

#### Page No 46:

#### Question 73:

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

#### Answer:

(a) When Nidhi is

(b) When Nidhi is

(c) When Nidhi is over hurdle C, she would finish the $\frac{3}{6}=\frac{1}{2}$ or middle part of the race.

#### Page No 46:

#### Question 74:

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

#### Answer:

Diameter of Earth = 12756000 m

Diameter of a new planet

= 741627.90 m

= 741.6 km

Thus, the diameter of this planet is 741.6 km.

#### Page No 46:

#### Question 75:

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

#### Answer:

Reciprocal of $\frac{5}{129}=\frac{1}{\frac{5}{129}}=\frac{129}{5}$

The product of $\frac{5}{129}$ and its reciprocal = $\frac{5}{129}\times \frac{129}{5}=1$

Thus, the product of $\frac{5}{129}$ and its reciprocal is 1.

#### Page No 47:

#### Question 76:

Simplify: $\frac{2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}}{2\frac{1}{2}\xf7\frac{1}{5}}$

#### Answer:

$\begin{array}{rcl}\frac{2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}}{2\frac{1}{2}\xf7\frac{1}{5}}& =& \frac{{\displaystyle \frac{5}{2}}+{\displaystyle \frac{1}{5}}}{\frac{5}{2}\xf7\frac{1}{5}}\\ & =& \frac{{\displaystyle \frac{25+2}{10}}}{\frac{5}{2}\times \frac{5}{1}}\\ & =& \frac{{\displaystyle \frac{27}{10}}}{\frac{25}{2}}\\ & =& \frac{27}{10}\times \frac{2}{25}\\ & =& \frac{27}{125}\end{array}$

Thus, $\frac{2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}}{2\frac{1}{2}\xf7\frac{1}{5}}=\begin{array}{rcl}& & \frac{27}{125}\end{array}$

#### Page No 47:

#### Question 77:

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

#### Answer:

$\begin{array}{rcl}\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{3}{8}}\times {\displaystyle \frac{3}{5}}}& =& \frac{{\displaystyle \frac{5}{20}+\frac{4}{20}}}{1-{\displaystyle \frac{9}{40}}}\\ & =& \frac{{\displaystyle \frac{9}{20}}}{\frac{40-9}{40}}\\ & =& \frac{{\displaystyle \frac{9}{20}}}{\frac{31}{40}}\\ & =& \frac{9}{20}\times \frac{40}{31}\\ & =& \frac{18}{31}\end{array}$

Thus, $\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{3}{8}}\times {\displaystyle \frac{3}{5}}}=\frac{18}{31}$

#### Page No 47:

#### Question 78:

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

#### Answer:

$\begin{array}{rcl}\frac{3}{10}\mathrm{by}\left(\frac{1}{4}\mathrm{of}\frac{3}{5}\right)& =& \frac{3}{10}\xf7\left(\frac{1}{4}\times \frac{3}{5}\right)\\ & =& \frac{3}{10}\xf7\left(\frac{3}{20}\right)\\ & =& \frac{3}{10}\times \frac{20}{3}\\ & =& 2\end{array}$

Thus, $\frac{3}{10}\mathrm{by}\left(\frac{1}{4}\mathrm{of}\frac{3}{5}\right)=2$

#### Page No 47:

#### Question 79:

$\frac{1}{8}$ of a number equals $\frac{2}{5}\xf7\frac{1}{20}$. What is the number?

#### Answer:

Let the number be *x*.

According to the question,

$\frac{1}{8}\text{o}\mathrm{f}x=\frac{2}{5}\xf7\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}\times x=\frac{2}{5}\times \frac{20}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2}{5}\times \frac{20}{1}\times \frac{8}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\times 4\times 8\phantom{\rule{0ex}{0ex}}x=64\phantom{\rule{0ex}{0ex}}$

Thus, the number is x.

#### Page No 47:

#### Question 80:

Heena’s father paid an electric bill of â‚¹ 385.70 out of a 500 rupee note. How much change should he have received?

#### Answer:

The amount of money Heena’s father has = â‚¹500

He paid an electric bill of â‚¹385.70

∴ The amount of money he should have received = â‚¹(500 − 385.70) = â‚¹114.30

Thus, the change he should have received is â‚¹114.30.

#### Page No 47:

#### Question 81:

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

#### Answer:

The normal body temperature = 98.6°F

Savitri’s temperature, when she was ill = 103.1°F

∴ Savitri’s temperature above normal = 103.1°F – 98.6°F

= 4.5°F

#### Page No 47:

#### Question 82:

**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°C | –0.17°C | –0.10°C | ${\left(\frac{1}{50}\right)}^{{0}^{}}\mathrm{C}$ | 0.54°C |

(b) In 1946, the average temperature varied by –0.03°C from the base measure. Between which two years should 1946 fall when the years are ordered from coldest to warmest?

#### Answer:

(a) –0.17°C < –0.10°C < 0.02°C < 0.10°C < 0.54°C

–0.17°C < –0.10°C < ${\left(\frac{1}{50}\right)}^{{0}^{}}\mathrm{C}$ < 0.10°C < 0.54°C

The order of five years from coldest to warmest is 1964, 1965, 1978, 1958, 2002.

(b) In 1946, the average temperature varied by –0.03°C.

Since, –0.10°C< –0.03°C < ${\left(\frac{1}{50}\right)}^{{0}^{}}\mathrm{C}$

∴ 1946 should fall between 1965 and 1978.

#### Page No 48:

#### Question 83:

**Science Application**

In her science class, Jyoti learned that the atomic weight of Helium is 4.0030; of Hydrogen is 1.0080; and of Oxygen is 16.0000. Find the

difference between the atomic weights of:

(a) Oxygen and Hydrogen

(b) Oxygen and Helium

(c) Helium and Hydrogen

#### Answer:

The atomic weight of Helium = 4.0030

The atomic weight of Hydrogen = 1.0080

The atomic weight of Oxygen – 16.0000

(a) Difference between the atomic weights of Oxygen and Hydrogen = 16.0000 – 1.0080

= 14.9920

(b) Difference between atomic weights of Oxygen and Helium = 16.0000 – 4.0030 = 11.9970

(c) Difference between atomic weights of Helium and Hydrogen = 4.0030 – 1.0080 = 2.9950

#### Page No 48:

#### Question 84:

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 cm long; Kamal said 19.25 cm; and Tabish said 19.27 cm. The correct length was 19.33 cm. How much of error was made by each of the boys?

#### Answer:

The correct length of the iron rod = 19.33 cm

The length of the rod measured by Ravi = 19.34 cm

∴ The error made by Ravi = (19.34 − 19.33) cm = +0.01 cm

The length of the rod measured by Kamal = 19.25 cm

∴ The error made by Kamal = (19.25 − 19.33) cm = −0.08 cm

The length of the rod measured by Tabish = 19.27 cm

∴ The error made by Tabish = (19.27 – 19.33) cm = −0.06 cm

#### Page No 48:

#### Question 85:

When 0.02964 is divided by 0.004, what will be the quotient?

#### Answer:

$\begin{array}{rcl}0.02964\xf70.004& =& \frac{2964}{100000}\xf7\frac{4}{1000}\\ & =& \frac{2964}{100000}\times \frac{1000}{4}\\ & =& \frac{741}{100}\\ & =& 7.41\end{array}$

Thus, when 0.02964 is divided by 0.004, the quotient will be 7.41.

#### Page No 48:

#### Question 86:

What number divided by 520 gives the same quotient as 85 divided by 0.625?

#### Answer:

Let the number be *x*.*x* divided by 520 = 85 divided by 0.625

⇒ *x *÷ 520 = 85 ÷ 0.625

$\Rightarrow \frac{x}{520}=\frac{85}{0.625}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{520}=136\phantom{\rule{0ex}{0ex}}\Rightarrow x=136\times 520\phantom{\rule{0ex}{0ex}}\Rightarrow x=70720$

Thus, the required number is 70720.

#### Page No 48:

#### Question 87:

A floor is 4.5 m long and 3.6 m wide. A 6 cm square tile costs â‚¹ 23.25. What will be the cost to cover the floor with these tiles?

#### Answer:

Length of the floor = 4.5 m

= (4.5 × 100) cm

= 450 cm

Width of the floor = 3.6 m = (3.6 × 100) cm = 360 cm

∴ Area of the floor = (450 × 360) cm^{2 }= 162000 cm^{2}

Side of a square tile = 6 cm

∴ Area of one square tile = (6 × 6) cm^{2 }= 36 cm^{2}

Number of tiles required to cover the floor ^{2 }÷ 36 cm^{2}

= 4500

Cost of 1 tile = â‚¹23.25

∴Cost of 4500 tiles = â‚¹(23.25 × 4500)

= â‚¹104625

#### Page No 48:

#### Question 88:

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

#### Answer:

Length of cloth Sunita has

∴ Length of cloth Rehana has $\frac{1}{3}\mathrm{of}\frac{3}{4}$ m

$=\frac{1}{3}\times \frac{3}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{m}$

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

#### Page No 48:

#### Question 89:

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

#### Answer:

Length of the garden = 22.50 m

Length of one brick = 0.25 m

∴ Number of bricks required to make the border

$=\frac{22.50\mathrm{m}}{0.25\mathrm{m}}\phantom{\rule{0ex}{0ex}}=\frac{2250}{25}\phantom{\rule{0ex}{0ex}}=90$

Thus, 90 bricks will be needed.

#### Page No 48:

#### Question 90:

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

#### Answer:

Length of cloth required for one shirt = $2\frac{1}{4}$m + $\frac{1}{8}$m

$=\frac{9}{4}+\frac{1}{8}\phantom{\rule{0ex}{0ex}}=\frac{18}{8}+\frac{1}{8}\phantom{\rule{0ex}{0ex}}=\frac{19}{8}\phantom{\rule{0ex}{0ex}}=2\frac{3}{8}\mathrm{m}$

∴ Length of cloth required for 6 shirts$=6\times 2\frac{3}{8}\mathrm{m}$

$=\frac{6}{1}\times \frac{19}{8}\phantom{\rule{0ex}{0ex}}=\frac{57}{4}\phantom{\rule{0ex}{0ex}}=14\frac{1}{4}\mathrm{m}$

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

#### Page No 48:

#### Question 91:

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 full $\frac{2}{3}$. The ticket office reported 648 sales. Which usher (first or second) made the better guess?

#### Answer:

Total number of seats = 820

Sale of total tickets = 648

Number of sold tickets guessed by the first usher = $\frac{3}{4}$ of 820

$=\frac{3}{4}\times 820\phantom{\rule{0ex}{0ex}}=615$

Number of sold tickets guessed by the second usher = $\frac{2}{3}$ of 820

$=\frac{2}{3}\times 820\phantom{\rule{0ex}{0ex}}=546.67\phantom{\rule{0ex}{0ex}}\approx 547$

Since, 615 is more close to 648 than 547.

Thus, the first usher made the better guess.

#### Page No 48:

#### Question 92:

For the celebrating children’s students of Class VII bought sweets for â‚¹740.25 and cold drink for â‚¹ 70. If 35 students contributed equally what amount was contributed by each student?

#### Answer:

Total spending amount = â‚¹740.25 + â‚¹70

= â‚¹810.25

Total number of students who contributed money = â‚¹35

Contribution of each student = $=\frac{\u20b9810.25}{35}$

= â‚¹23.14

Thus, the equal amount contributed by each student is â‚¹23.14.

#### Page No 48:

#### Question 93:

The time taken by Rohan in five different races to run a distance of 500 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.

#### Answer:

Total time taken by Rohan in five races = (3.20 + 3.37 + 3.29 + 3.17 + 3.32) minutes

= 16.35 minutes

Total number of races = 5

∴ The average time taken by Rohan

#### Page No 49:

#### Question 94:

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

#### Answer:

Total length of a sewer line = $80\frac{1}{4}$m

7.5 m long sewer line completed in 1 day.

$80\frac{1}{4}$m long sewer line completed in

$=\frac{10}{75}\times \frac{321}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\times \frac{107}{2}\phantom{\rule{0ex}{0ex}}=\frac{107}{10}\phantom{\rule{0ex}{0ex}}=10.7\phantom{\rule{0ex}{0ex}}\approx 11\mathrm{days}$

Thus, the project will get complete in 3 days.

#### Page No 49:

#### Question 95:

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

figure

#### Answer:

Weight of the object on Earth = $5\frac{3}{5}$ kg

$=\frac{28}{5}\mathrm{kg}$

Weight of an object on moon = $\frac{1}{6}$ × weight on Earth

$=\frac{1}{6}\times \frac{28}{5}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\frac{14}{15}\mathrm{kg}$

#### Page No 49:

#### Question 96:

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?

#### Answer:

Total number of students = 200

(a) Number of students who were influenced by radio$=\frac{9}{20}\times 200=90$

(b) Number of students who were influenced by music video channel$=\frac{2}{25}\times 200=16$

Number of students who were influenced by radio = 90

∴ Number of students that were influenced more by radio than by a music video channel = 90 − 16 = 74

(c) Numberofstudents who were influenced by friend or relative $=\frac{3}{20}\times 200=30$

Number of students who were influenced by hearing or seeing the CD in a shop$=\frac{1}{10}\times 200=20$

∴ Total number of students who were influenced by both friend or relative and hearing the CD in a shop = 30 + 20 = 50

#### Page No 50:

#### Question 97:

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

#### Answer:

The can of milk filled by the milkman in the morning = $5\frac{1}{2}\mathrm{L}=\frac{11}{2}\mathrm{L}$

Milk sold by him to Renu = $\frac{3}{4}$ L

Milk sold by him to Kamla = $\frac{3}{4}$ L

Milk sold by him to Renuka = $\frac{3}{4}$ L

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

Milk sold by him to Jassi = $1\frac{1}{2}\mathrm{L}=\frac{3}{2}\mathrm{L}$

Total milk sold by him = $\left(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{7}{8}+\frac{3}{2}\right)\mathrm{L}$

$=\left(\frac{9}{4}+\frac{7}{8}+\frac{3}{2}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\left(\frac{18}{8}+\frac{7}{8}+\frac{12}{8}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\left(\frac{18+7+12}{8}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\frac{37}{8}\mathrm{L}$

Milk left in the can = The can of milk filled by the milkman in the morning − Total milk sold by him

$=\left(\frac{11}{2}-\frac{37}{8}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\left(\frac{44}{8}-\frac{37}{8}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\left(\frac{44-37}{8}\right)\mathrm{L}\phantom{\rule{0ex}{0ex}}=\left(\frac{7}{8}\right)\mathrm{L}$

Thus, milk left in the can is $\frac{7}{8}\mathrm{L}$

#### Page No 50:

#### Question 98:

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?

#### Answer:

Let the whole work be represented by *x*.

The part of work done by Anuradha in 6 hours = *x*

∴ The part of work done by her in 1 hour

The part of work done by her in 5 hours *$\frac{5x}{6}$*

The part of work done by her in 6 hours = *$\frac{x}{6}\times 6=x$*

Thus, the work can she do ${\frac{1}{6}}^{\mathrm{th}}$ part of the work in 1 hour, ${\frac{5}{6}}^{\mathrm{th}}$ part of the work in 5 hours and complete work in 6 hours.

#### Page No 50:

#### Question 99:

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?

#### Answer:

Let the total portion of the saree be represented by 1.

The portion of saree painted by Rehana in 5 hours = 1

The portion of saree painted by her in 1 hour

Since, $4\frac{3}{5}\mathrm{hours}=\frac{23}{5}\mathrm{hours}$

The portion of saree painted by her in $4\frac{3}{5}$ hours$=\frac{1}{5}\times 4\frac{3}{5}$

$=\frac{1}{5}\times \frac{23}{5}\phantom{\rule{0ex}{0ex}}=\frac{23}{25}$

Also, $3\frac{1}{2}\mathrm{hours}=\frac{7}{2}\mathrm{hours}$

The portion of saree painted by her in $3\frac{1}{2}$ hours$=\frac{1}{5}\times 3\frac{1}{2}$

$=\frac{1}{5}\times \frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{7}{10}$

#### Page No 50:

#### Question 100:

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

#### Answer:

Quantity of cotton wool Rama has = $6\frac{1}{4}\mathrm{kg}=\frac{25}{4}\mathrm{kg}$

Quantity of cotton wool one pillow takes = $1\frac{1}{4}\mathrm{kg}=\frac{5}{4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}$

Number of pillows can she make $=6\frac{1}{4}\mathrm{kg}\xf71\frac{1}{4}\mathrm{kg}$

$=\frac{25}{4}\mathrm{kg}\xf7\frac{5}{4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\frac{25}{4}\times \frac{4}{5}\phantom{\rule{0ex}{0ex}}=5$

Thus, she cam make 5 such pillows.

#### Page No 50:

#### Question 101:

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

#### Answer:

Length of cloth required to make a shirt = $2\frac{1}{3}\mathrm{m}=\frac{7}{3}\mathrm{m}$

Length of piece of cloth Radhika has = $9\frac{1}{3}\mathrm{m}=\frac{28}{3}\mathrm{m}$

Number of shirts she can make$=9\frac{1}{3}\mathrm{m}\xf72\frac{1}{3}\mathrm{m}$

$=\left(\frac{28}{3}\xf7\frac{7}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{28}{3}\times \frac{3}{7}\right)\phantom{\rule{0ex}{0ex}}=4$

Thus, she can make 4 such shirts.

#### Page No 50:

#### Question 102:

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

#### Answer:

Time taken by Ravi to walk $3\frac{1}{3}$km = 1 hour

Time taken by Ravi to walk 1 km = $1\xf73\frac{1}{3}$

$=1\xf7\frac{10}{3}\phantom{\rule{0ex}{0ex}}=1\times \frac{3}{10}\phantom{\rule{0ex}{0ex}}=\frac{3}{10}\mathrm{hours}$

Time taken by Ravi to walk 10 km = $\frac{3}{10}\times 10\mathrm{hours}$

= 3 hours

Thus, time taken by him to walk to his office which is 10 km from his home is 3 hours.

#### Page No 50:

#### Question 103:

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

#### Answer:

Distance travelled by Raj with ${\frac{3}{5}}^{\mathrm{th}}$petrol tank = 360 km

Distance travelled by him with a full petrol tank = 360 km ÷ $\frac{3}{5}$

$=360\times \frac{5}{3}\phantom{\rule{0ex}{0ex}}=120\times 5\phantom{\rule{0ex}{0ex}}=600\mathrm{km}$

Thus, at the same rate with a full tank of petrol he can travel 600 km.

#### Page No 50:

#### Question 104:

Kajol has â‚¹ 75. This is$\frac{3}{8}$ of the amount she earned. How much did she earn?

#### Answer:

Let the amount earned by Kajol be â‚¹*x*.

Amount she has = â‚¹75

According to the question

$\frac{3}{8}\mathrm{of}x=75\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{8}\times x=75\phantom{\rule{0ex}{0ex}}\Rightarrow x=75\times \frac{8}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=25\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow x=200$

Thus, she earns â‚¹200.

#### Page No 51:

#### Question 105:

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.

#### Answer:

(i) Number of trees required to make one tonne of paper = 17

Total number of trees = 221

∴ Fraction of forest will be used to make one tonne of paper

(a) Fraction of forest will be used to make 5 tonnes of paper

(b) Fraction of forest will be used to make 10 tonnes of paper

(ii) Let we have to save *x *tonnes of paper to save $\frac{7}{13}$ part of the forest

Fraction of forest will be used to make *x* tonnes of paper

Here, $\frac{x}{13}=\frac{7}{13}\phantom{\rule{0ex}{0ex}}\Rightarrow x=7$

#### Page No 51:

#### Question 106:

Simplify and write the result in decimal form :

$\left(1\xf7\frac{2}{9}\right)+\left(1\xf73\frac{1}{5}\right)+\left(1\xf72\frac{2}{3}\right)$

#### Answer:

$\left(1\xf7\frac{2}{9}\right)+\left(1\xf73\frac{1}{5}\right)+\left(1\xf72\frac{2}{3}\right)=\left(1\times \frac{9}{2}\right)+\left(1\xf7\frac{16}{5}\right)+\left(1\xf7\frac{8}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(1\times \frac{9}{2}\right)+\left(1\times \frac{5}{16}\right)+\left(1\times \frac{3}{8}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{9}{2}\right)+\left(\frac{5}{16}\right)+\left(\frac{3}{8}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{72}{16}\right)+\left(\frac{5}{16}\right)+\left(\frac{6}{16}\right)\phantom{\rule{0ex}{0ex}}=\frac{72+5+6}{16}\phantom{\rule{0ex}{0ex}}=\frac{83}{16}\phantom{\rule{0ex}{0ex}}=5.1875\phantom{\rule{0ex}{0ex}}$

#### Page No 51:

#### Question 107:

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

$\left(1\right)2\times \frac{1}{4}\left(2\right)2\times \frac{3}{7}\left(3\right)2\times \frac{1}{3}\phantom{\rule{0ex}{0ex}}\left(4\right)\frac{1}{4}\times 4\left(5\right)3\times \frac{2}{9}\left(6\right)\frac{1}{4}\times 3$

#### Answer:

$\left(1\right)2\times \frac{1}{4}$ represents the addition of 2 figures, each representing 1 shaded part out of 4 equal parts. Thus, (1) is represented by (d)

$\left(2\right)2\times \frac{3}{7}$represents the addition of 2 figures, each representing 3 shaded parts out of 7 equal parts. Thus, (2) is represented by (f).

$\left(4\right)\frac{1}{4}\times 4$represents the addition of 4 figures, each representing 1 shaded part out of 4 equal parts. Thus, (

$\left(5\right)3\times \frac{2}{9}$ represents the addition of 3 figures, each representing 2 shaded parts out of 9 equal parts. Thus, (5) is represented by (a).

$\left(5\right)\frac{1}{4}\times 3$represents the addition of 3 figures, each representing 1 shaded part out of 3 equal parts. Thus, (6) is represented by (e).

#### Page No 52:

#### Question 108:

Evaluate : (0.3) × (0.3) – (0.2) × (0.2)

#### Answer:

(0.3) × (0.3) – (0.2) × (0.2)

$=\frac{3}{10}\times \frac{3}{10}-\frac{2}{10}\times \frac{2}{10}\phantom{\rule{0ex}{0ex}}=\frac{9}{100}-\frac{4}{100}\phantom{\rule{0ex}{0ex}}=\frac{5}{100}\phantom{\rule{0ex}{0ex}}=0.05$

Thus, (0.3) × (0.3) – (0.2) × (0.2) = 0.05

#### Page No 52:

#### Question 109:

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

#### Answer:

$\frac{0.6}{0.3}+\frac{0.16}{0.4}=\frac{6}{3}+\frac{16}{40}\phantom{\rule{0ex}{0ex}}=\frac{2}{1}+\frac{2}{5}\phantom{\rule{0ex}{0ex}}=\frac{10+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{12}{5}\phantom{\rule{0ex}{0ex}}=2.4$

Thus, $\frac{0.6}{0.3}+\frac{0.16}{0.4}$ = 2.4

#### Page No 52:

#### Question 110:

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

#### Answer:

$\begin{array}{rcl}\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}& =& \frac{\left({\displaystyle \frac{2}{10}}\times {\displaystyle \frac{14}{100}}\right)+\left({\displaystyle \frac{5}{10}}+{\displaystyle \frac{91}{100}}\right)}{\left(\frac{1}{10}\times \frac{2}{10}\right)}\\ & =& \frac{\left(\frac{28}{1000}\right)+\left({\displaystyle \frac{455}{1000}}\right)}{{\displaystyle \frac{2}{100}}}\\ & =& \frac{\frac{483}{1000}}{\frac{2}{100}}\\ & =& \frac{483}{1000}\times \frac{100}{2}\\ & =& \frac{241.5}{10}\\ & =& 24.15\\ & & \end{array}$

Thus, $\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}$= 24.15

#### Page No 52:

#### Question 111:

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

#### Answer:

Side of the equilateral triangle = $\frac{4}{3}$cm

∴ Side of the square = $\frac{4}{3}$cm

∴ Perimeter of the figure formed = $\left(\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}\right)$ cm

$=\left(\frac{20}{3}\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=6\frac{2}{3}\mathrm{cm}$

Thus, the perimeter of figure formed is $6\frac{2}{3}\mathrm{cm}$.

#### Page No 52:

#### Question 112:

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

#### Answer:

Area of the carpet bought by Rita = 4 m × $6\frac{2}{3}$m

$=\frac{4}{1}\times \frac{20}{3}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{80}{3}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

Area of the room = $3\frac{1}{3}$m × $5\frac{1}{3}$m

$=\frac{10}{3}\times \frac{16}{3}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{160}{9}{\mathrm{m}}^{2}$

Area to be cut off to fit wall to wall carpet into the room = $\left(\frac{80}{3}-\frac{160}{9}\right){\mathrm{m}}^{2}$

$=\left(\frac{240}{9}-\frac{160}{9}\right){\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{240-160}{9}\right){\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{80}{9}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=8\frac{8}{9}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

Thus, area to be cut off to fit wall to wall carpet into the room is $8\frac{8}{9}{\mathrm{m}}^{2}$.

#### Page No 52:

#### Question 113:

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

#### Answer:

Length of the photograph = $14\frac{2}{5}$cm $=\frac{72}{5}\mathrm{cm}$

Breadth of the photograph = $10\frac{2}{5}$cm$=\frac{52}{5}\mathrm{cm}$

Width of the boarded = $2\frac{3}{5}$ cm = $\frac{13}{5}\mathrm{cm}$

Length of the framed photograph $=\left(\frac{72}{5}+\frac{13}{5}+\frac{13}{5}\right)\mathrm{cm}=\left(\frac{98}{5}\right)\mathrm{cm}$

Breadth of the framed photograph $=\left(\frac{52}{5}+\frac{13}{5}+\frac{13}{5}\right)\mathrm{cm}=\left(\frac{78}{5}\right)\mathrm{cm}$

Area of the framed photograph $=\frac{98}{5}\times \frac{78}{5}{\mathrm{cm}}^{2}$

$=\frac{7644}{25}{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=305\frac{19}{25}{\mathrm{cm}}^{2}$

#### Page No 52:

#### Question 114:

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

#### Answer:

Cost of 1 burger = â‚¹$20\frac{3}{4}$

Cost of 4 burger = â‚¹$20\frac{3}{4}$ × 4

$=\u20b9\frac{83}{4}\times 4\phantom{\rule{0ex}{0ex}}=\u20b983$

Cost of 1 Macpuff = â‚¹ $15\frac{1}{2}$

Cost of 14 Macpuff = â‚¹ $15\frac{1}{2}$ × 14

$=\u20b9\frac{31}{2}\times 14\phantom{\rule{0ex}{0ex}}=\u20b9217$

So, the cost of 4 burgers and 14 macpuffs = â‚¹83 + â‚¹217

= â‚¹300

Thus, the cost of 4 burgers and 14 macpuffs is â‚¹300.

#### Page No 52:

#### Question 115:

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

#### Answer:

Total height of hill = $101\frac{1}{3}$ m

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

$=\frac{1}{4}\times \frac{304}{3}\mathrm{m}\phantom{\rule{0ex}{0ex}}=\frac{76}{3}\mathrm{m}$

Height of hill above water$=\left(101\frac{1}{3}-\frac{76}{3}\right)\mathrm{m}$

$=\left(\frac{304}{3}-\frac{76}{3}\right)\mathrm{m}\phantom{\rule{0ex}{0ex}}=\left(\frac{228}{3}\right)\mathrm{m}\phantom{\rule{0ex}{0ex}}=76\mathrm{m}$

Thus, the height of the hill visible above the water is 76 m.

#### Page No 52:

#### Question 116:

*Sports*: Reaction time measures how quickly a runner reacts to the starter pistol. In the 100 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?

#### Answer:

Total race time of Lauryn Williams = 11.03 seconds

Her reaction time = 0.214 second

Time taken by her to run the actual distance = (11.03 – 0.214) seconds

= 10.816 seconds

#### Page No 53:

#### Question 117:

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

Questions |
Greater than 1 |
Less than 1 |

$\frac{2}{3}\xf7\frac{1}{2}$ | ||

$\frac{2}{3}\xf7\frac{2}{1}$ | ||

$6\xf7\frac{1}{4}$ | ||

$\frac{1}{5}\xf7\frac{1}{2}$ | ||

$4\frac{1}{3}\xf73\frac{1}{2}$ | ||

$\frac{2}{3}\times 8\frac{1}{2}$ |

#### Answer:

(i)

$\begin{array}{rcl}\frac{2}{3}\xf7\frac{1}{2}& =& \frac{2}{3}\times \frac{2}{1}\\ & =& \frac{4}{3}\\ & >& 1\end{array}$

(ii)

$\begin{array}{rcl}\frac{2}{3}\xf7\frac{2}{1}& =& \frac{2}{3}\times \frac{1}{2}\\ & =& \frac{1}{3}\\ & <& 1\end{array}$

(iii)

$\begin{array}{rcl}6\xf7\frac{1}{4}& =& 6\times \frac{4}{1}\\ & =& 24\\ & >& 1\end{array}$

(iv)

$\begin{array}{rcl}\frac{1}{5}\xf7\frac{1}{2}& =& \frac{1}{5}\times \frac{2}{1}\\ & =& \frac{2}{5}\\ & <& 1\end{array}$

(v)

$\begin{array}{rcl}4\frac{1}{3}\xf73\frac{1}{2}& =& \frac{13}{3}\xf7\frac{7}{2}\\ & =& \frac{13}{3}\times \frac{2}{7}\\ & =& \frac{26}{21}\\ & >& 1\end{array}$

(vi)

$\begin{array}{rcl}\frac{2}{3}\times 8\frac{1}{2}& =& \frac{2}{3}\times \frac{17}{2}\\ & =& \frac{17}{3}\\ & >& 1\end{array}$

#### Page No 53:

#### Question 118:

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* = $10.5$ cm. Find the height of the largest container.

#### Answer:

Height of the smallest container = 10.5 cm

We have given that

$\frac{7}{25}x=10.5\phantom{\rule{0ex}{0ex}}\Rightarrow x=10.5\times \frac{25}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{105}{10}\times \frac{25}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{15}{2}\times \frac{5}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow x=37.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

∴ Height of the largest container is 37.5 cm.

#### Page No 53:

#### Question 119:

In the given question replace ‘?’ with appropriate fraction.

#### Answer:

In the given figure a pattern is formed

$\frac{7}{8}\phantom{\rule{0ex}{0ex}}\frac{7}{24}=\frac{7}{8\times 3}\phantom{\rule{0ex}{0ex}}\frac{7}{72}=\frac{7}{24\times 3}\phantom{\rule{0ex}{0ex}}\frac{7}{216}=\frac{7}{72\times 3}\phantom{\rule{0ex}{0ex}}\Rightarrow ?=\frac{7}{216\times 3}=\frac{7}{648}$

#### Page No 53:

#### Question 120:

In the given question replace ‘?’ with appropriate fraction.

#### Answer:

In the given figure a pattern is formed

$\frac{3}{32}\phantom{\rule{0ex}{0ex}}\frac{3}{16}=\frac{3}{32\xf72}\phantom{\rule{0ex}{0ex}}\frac{3}{8}=\frac{3}{16\xf72}\phantom{\rule{0ex}{0ex}}\frac{3}{4}=\frac{3}{8\xf72}\phantom{\rule{0ex}{0ex}}\Rightarrow ?=\frac{3}{4\xf72}=\frac{3}{2}$

#### Page No 54:

#### Question 121:

In the given question replace ‘?’ with appropriate fraction.

#### Answer:

In the given figure a pattern is formed

$0.05\phantom{\rule{0ex}{0ex}}0.5=0.05\times 10\phantom{\rule{0ex}{0ex}}5=0.5\times 10\phantom{\rule{0ex}{0ex}}50=5\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow ?=50\times 10=500$

#### Page No 54:

#### Question 122:

In the given question replace ‘?’ with appropriate fraction.

#### Answer:

In the given figure a pattern is formed

$0.1\phantom{\rule{0ex}{0ex}}0.01=\frac{0.1}{10}\phantom{\rule{0ex}{0ex}}0.001=\frac{0.01}{10}\phantom{\rule{0ex}{0ex}}0.0001=\frac{0.001}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow ?=\frac{0.0001}{10}=0.00001$

#### Page No 54:

#### Question 123:

**What is the Error in the given question.**

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

#### Answer:

Since 0.3 is greater than 0.25.

∴ −0.3 is less than −0.25

Thus, the error is −0.30 > −0.25

#### Page No 54:

#### Question 124:

**What is the Error in the given question.**

A student multiplied two mixed fractions in the following manner:

$2\frac{4}{7}\times 3\frac{1}{4}=6\frac{1}{7}$ ,What error the student has done?

#### Answer:

$\begin{array}{rcl}2\frac{4}{7}\times 3\frac{1}{4}& =& \frac{18}{7}\times \frac{13}{4}\\ & =& \frac{9}{7}\times \frac{13}{2}\\ & =& \frac{117}{14}\\ & =& 8\frac{5}{14}\end{array}$

But student has multiplied two mixed fractions $2\frac{4}{7}\mathrm{and}3\frac{1}{4}\mathrm{to}\mathrm{get}6\frac{1}{7}$.

So, the student has done error that mixed fractions are not converted into improper fractions.

#### Page No 54:

#### Question 125:

**What is the Error in the given question.**

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

#### Answer:

The given pattern is $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........$**Case I:**

$\begin{array}{rcl}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}& =& \frac{20}{60}+\frac{15}{60}+\frac{12}{60}\\ & =& \frac{20+15+12}{60}\\ & =& \frac{47}{60}\\ & <& 1\end{array}$**Case II:**

$\begin{array}{rcl}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}& =& \frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{10}{60}\\ & =& \frac{20+15+12+10}{60}\\ & =& \frac{57}{60}\\ & <& 1\end{array}$**Case III: **

$\begin{array}{rcl}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}& =& \frac{140}{420}+\frac{105}{420}+\frac{84}{420}+\frac{70}{420}+\frac{60}{420}\\ & =& \frac{140+105+84+70+60}{420}\\ & =& \frac{459}{420}\\ & >& 1\end{array}$

Thus, in the pattern $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........$thr fraction that makes the sum greater than 1 (first time) is $\frac{1}{7}$.

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