YRS : DAV | Class - 7 | Math | Chapter - 1 | Solutions - Rational Numbers

DAV Class 7 Maths – Chapter 1

Worksheet 2 – Complete Solutions

Question 1

In each of the following cases, show that the rational numbers are equivalent.

(i) $\frac{4}{9}$ and $\frac{44}{99}$

Solution:

$$\frac{4}{9}\times\frac{11}{11} = \frac{4\times11}{9\times11} = \frac{44}{99}$$

Therefore,

$$\boxed{\frac{4}{9}=\frac{44}{99}}$$

Hence, the rational numbers are equivalent.

(ii) $\frac{7}{-3}$ and $\frac{35}{-15}$

$$\frac{7}{-3}\times\frac{5}{5} = \frac{7\times5}{-3\times5} = \frac{35}{-15}$$

Therefore,

$$\boxed{\frac{7}{-3}=\frac{35}{-15}}$$

Hence, the rational numbers are equivalent.

(iii) $\frac{-3}{5}$ and $\frac{-12}{20}$

$$\frac{-3}{5}\times\frac{4}{4} = \frac{-3\times4}{5\times4} = \frac{-12}{20}$$

Therefore,

$$\boxed{\frac{-3}{5}=\frac{-12}{20}}$$

Hence, the rational numbers are equivalent.

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Question 2

In each of the following cases, show that the rational numbers are not equivalent.

(i) $\frac{4}{9}$ and $\frac{16}{27}$

For two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, they are equivalent if:

$$a\times d=b\times c$$

Here,

$$4\times27=108$$ $$9\times16=144$$

Since,

$$108\neq144$$

Therefore,

$$\boxed{\frac{4}{9}\neq\frac{16}{27}}$$

Hence, they are not equivalent rational numbers.

(ii) $\frac{-100}{3}$ and $\frac{300}{9}$

$$(-100)\times9=-900$$ $$3\times300=900$$

Since,

$$-900\neq900$$

Therefore,

$$\boxed{\frac{-100}{3}\neq\frac{300}{9}}$$

Hence, they are not equivalent rational numbers.

(iii) $\frac{3}{-17}$ and $\frac{8}{-51}$

$$3\times(-51)=-153$$ $$(-17)\times8=-136$$

Since,

$$-153\neq-136$$

Therefore,

$$\boxed{\frac{3}{-17}\neq\frac{8}{-51}}$$

Hence, they are not equivalent rational numbers.

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Question 3

Write three rational numbers equivalent to each of the following.

(i) $\frac{4}{7}$

Multiplying by 2:

$$\frac{4}{7} = \frac{4\times2}{7\times2} = \frac{8}{14}$$

Multiplying by 3:

$$\frac{4}{7} = \frac{4\times3}{7\times3} = \frac{12}{21}$$

Multiplying by 4:

$$\frac{4}{7} = \frac{4\times4}{7\times4} = \frac{16}{28}$$

Answer:

$$\boxed{\frac{8}{14},\frac{12}{21},\frac{16}{28}}$$

(ii) $\frac{36}{108}$

Dividing by 2:

$$\frac{36}{108} = \frac{36\div2}{108\div2} = \frac{18}{54}$$

Dividing by 3:

$$\frac{36}{108} = \frac{36\div3}{108\div3} = \frac{12}{36}$$

Dividing by 4:

$$\frac{36}{108} = \frac{36\div4}{108\div4} = \frac{9}{27}$$

Answer:

$$\boxed{\frac{18}{54},\frac{12}{36},\frac{9}{27}}$$

(iii) $\frac{-5}{-7}$

Multiplying by 2:

$$\frac{-5}{-7} = \frac{-10}{-14}$$

Multiplying by $-2$:

$$\frac{-5}{-7} = \frac{10}{14}$$

Multiplying by 3:

$$\frac{-5}{-7} = \frac{-15}{-21}$$

Answer:

$$\boxed{\frac{-10}{-14},\frac{10}{14},\frac{-15}{-21}}$$

(iv) $\frac{-72}{180}$

Dividing by 2:

$$\frac{-72}{180} = \frac{-36}{90}$$

Dividing by 3:

$$\frac{-72}{180} = \frac{-24}{60}$$

Dividing by 4:

$$\frac{-72}{180} = \frac{-18}{45}$$

Answer:

$$\boxed{\frac{-36}{90},\frac{-24}{60},\frac{-18}{45}}$$

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Question 4

Express $\frac{3}{5}$ as a rational number with numerator

(i) $-21$

We need to change 3 into $-21$.

$$3\times(-7)=-21$$

Therefore,

$$\frac{3}{5}\times\frac{-7}{-7} = \frac{-21}{-35}$$

Answer:

$$\boxed{\frac{-21}{-35}}$$

(ii) $150$

We need to change 3 into 150.

$$3\times50=150$$

Therefore,

$$\frac{3}{5}\times\frac{50}{50} = \frac{150}{250}$$

Answer:

$$\boxed{\frac{150}{250}}$$

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Question 5

Express $\frac{4}{-7}$ as a rational number with denominator

(i) $84$

We need to change $-7$ into 84.

$$-7\times(-12)=84$$

Therefore,

$$\frac{4}{-7}\times\frac{-12}{-12} = \frac{-48}{84}$$

Answer:

$$\boxed{\frac{-48}{84}}$$

(ii) $-28$

We need to change $-7$ into $-28$.

$$-7\times4=-28$$

Therefore,

$$\frac{4}{-7}\times\frac{4}{4} = \frac{16}{-28}$$

Answer:

$$\boxed{\frac{16}{-28}}$$

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Question 6

Express $\frac{90}{216}$ as a rational number with numerator 5.

We need to change 90 into 5.

$$90\div18=5$$

Divide both numerator and denominator by 18:

$$\frac{90}{216} = \frac{90\div18}{216\div18}$$ $$=\frac{5}{12}$$

Answer:

$$\boxed{\frac{5}{12}}$$

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Question 7

Express $\frac{-64}{256}$ as a rational number with denominator 8.

We need to change 256 into 8.

$$256\div32=8$$

Divide both numerator and denominator by 32:

$$\frac{-64}{256} = \frac{-64\div32}{256\div32}$$ $$=\frac{-2}{8}$$

Answer:

$$\boxed{\frac{-2}{8}}$$

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Question 8

Find the equivalent forms of the rational numbers having a common denominator in each of the following collections of rational numbers.

(i) $\frac{2}{5},\frac{6}{13}$

LCM of 5 and 13:

$$\text{LCM}=65$$

For $\frac{2}{5}$:

$$\frac{2}{5}\times\frac{13}{13} = \frac{26}{65}$$

For $\frac{6}{13}$:

$$\frac{6}{13}\times\frac{5}{5} = \frac{30}{65}$$

Answer:

$$\boxed{\frac{26}{65},\frac{30}{65}}$$

(ii) $\frac{1}{7},\frac{2}{8},\frac{3}{14}$

LCM of 7, 8 and 14:

$$\text{LCM}=56$$

For $\frac{1}{7}$:

$$\frac{1}{7}\times\frac{8}{8} = \frac{8}{56}$$

For $\frac{2}{8}$:

$$\frac{2}{8}\times\frac{7}{7} = \frac{14}{56}$$

For $\frac{3}{14}$:

$$\frac{3}{14}\times\frac{4}{4} = \frac{12}{56}$$

Answer:

$$\boxed{ \frac{8}{56}, \frac{14}{56}, \frac{12}{56} }$$

(iii) $\frac{5}{12},\frac{7}{4},\frac{9}{60},\frac{11}{3}$

LCM of 12, 4, 60 and 3:

$$\text{LCM}=60$$

For $\frac{5}{12}$:

$$\frac{5}{12}\times\frac{5}{5} = \frac{25}{60}$$

For $\frac{7}{4}$:

$$\frac{7}{4}\times\frac{15}{15} = \frac{105}{60}$$

For $\frac{9}{60}$:

$$\frac{9}{60}$$

For $\frac{11}{3}$:

$$\frac{11}{3}\times\frac{20}{20} = \frac{220}{60}$$

Answer:

$$\boxed{ \frac{25}{60}, \frac{105}{60}, \frac{9}{60}, \frac{220}{60} }$$