S.chand books class 8 maths solution chapter 1 Rational Numbers exercise 1 C

Exercise 1 C

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Q1 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 1

Answer True (T) or False (F)
(i) $\dfrac{8}{15}\div\dfrac{7}{8}=\dfrac{7}{8}\div\dfrac{8}{15}$

Sol : F

(ii) Addition distributes over multiplication in rational numbers .

Sol : F

(iii) The reciprocal of -8 is $\dfrac{1}{8}$

Sol : F

(iv) $\dfrac{3}{4}\div\dfrac{-8}{7}$ is a rational numbers .

Sol : T

(v) $0\times \dfrac{4}{5}=\dfrac{4}{5}\times 0$ implies that 0 is the multiplicative identity for rational numbers .

Sol : F

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Q2 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 2

(i) $\dfrac{1}{6}\text{ by }\dfrac{12}{17}$

Sol:

$=\frac{1}{6} \times \frac{12}{17}$

$=\frac{12}{102}=\frac{6}{51}=\frac{2}{17}$

(ii) $\dfrac{-5}{12}\text{ by }\dfrac{9}{10}$

Sol:

$=\frac{-5}{12} \times \frac{9}{10}$

$=\frac{-45}{120}$

$=-\frac{9}{24}=\frac{-3}{8}$

(iii) $\dfrac{-14}{33}\text{ by }\dfrac{-3}{28}$

Sol:

$\frac{-14}{33} \times \frac{-3}{28}$

$=\frac{1}{11 \times 2}=\frac{1}{22}$

(iv) $\dfrac{-80}{13}\text{ by }\dfrac{-65}{72}$

Sol:

$\frac{-80}{13} \times \frac{-65}{72}$

$=\frac{-80}{1} \times \frac{-5}{72}$

$=-\frac{10}{1} \times-\frac{5}{9}$

$=\frac{50}{9}$

(v) $5\dfrac{1}{2}\text{ by }-1\dfrac{5}{6}$

Sol:

$5 \frac{1}{2} x-1 \frac{5}{6}$

$=\frac{11}{2} \times-\frac{11}{6}$

$=-\frac{121}{12}$

$=-10 \frac{1}{12}$

(vi) -6 by $5\dfrac{2}{3}$

Sol :

$-6 \times 5 \frac{2}{3}$

$=-\frac{6}{1} \times \frac{17}{3}$

$=\frac{-2 \times 17}{1}$

$=\frac{-34}{1} or-34$

 

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Q3 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 3

Simplify :

(i) $\left(\dfrac{-3}{4}\times \dfrac{-24}{15}\right)+\left(\dfrac{-11}{13}\times \dfrac{78}{55}\right)$

Sol :

$\left(\frac{-3}{4} \times \frac{-24}{15}\right)$•$+\left(-\frac{11}{13} \times \frac{78}{55}\right)$

$\left(\frac{-1}{1} \times \frac{-6}{5}\right)+\left(-\frac{1}{1} \times \frac{6}{5}\right)$

$\left(\frac{6}{5}\right)+\left(-\frac{6}{5}\right)$

=0

 

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(ii) $\left(\dfrac{-4}{5}\times \dfrac{15}{8}\right)+\left(\dfrac{-1}{3}\times \dfrac{-9}{7}\right)-\left(\dfrac{2}{9}\times \dfrac{27}{14}\right)$

Sol :

$\left(-\frac{4}{5} \times \frac{15}{8}\right)+\left(\frac{-1}{3} \times-\frac{9}{7}\right)-\left(\frac{2}{9} \times \frac{27}{14}\right)$

$=\left(\frac{4}{1} \times \frac{3}{8}\right)+\left(\frac{-1}{1} \times \frac{-3}{7}\right)-\left(\frac{2}{1} \times \frac{3}{14}\right)$

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

$=-\frac{3}{2}+\frac{3}{7}-\frac{3}{7}$

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

 

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Q4 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 4

Fill in the blanks.
(i) $\dfrac{-19}{40}\times \dfrac{8}{11}=\dfrac{8}{11}\times \dfrac{-19}{40}$

(ii) $-25\times\dfrac{-7}{12}=\dfrac{-7}{12}\times (-25)$

(iii) $\left(\dfrac{6}{11}\times \dfrac{-20}{21}\right)\times \left(\dfrac{-7}{8}\right)=\dfrac{6}{11} \times\left(\dfrac{-20}{21}\times \dfrac{-7}{8}\right)$

(iv) $\dfrac{-21}{40}\times \left(\dfrac{3}{7}\times \dfrac{17}{-24}\right)=\left(\dfrac{-21}{40}\times \dfrac{17}{-24}\right)\times \dfrac{3}{7}$

(v) $\dfrac{2}{9}\times \left(\dfrac{-4}{9}\div\dfrac{6}{17}\right)=\dfrac{2}{9}\times \dfrac{-4}{9} \div \dfrac{2}{9} \times \dfrac{6}{17}$

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Verify the statements given in questions 5 to 7 . Also , name the properties of multiplication illustrated by these statements.

Q5 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 5

(i) $\dfrac{4}{5}\times \dfrac{7}{9}=\dfrac{7}{9}\times \dfrac{4}{5}$

Sol :

⇒$\dfrac{4\times 7}{5\times 9}=\dfrac{7\times 4}{9\times 5}$

⇒$\dfrac{28}{45}=\dfrac{28}{45}$

by commutative property

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(ii) $\dfrac{8}{7}\times \dfrac{9}{-10}=\dfrac{9}{-10}\times \dfrac{8}{7}$

Sol :

⇒$\dfrac{8\times 9}{7\times -10}=\dfrac{9\times 8}{-10\times 7}$

⇒$\dfrac{72}{-70}=\dfrac{72}{-70}$

by commutative property

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Q6 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 6

(i) $\left(\dfrac{3}{4}\times \dfrac{1}{2}\right)\times \dfrac{5}{7}=\dfrac{3}{4}\times \left(\dfrac{1}{2}\times \dfrac{5}{7}\right)$

Sol :

⇒$\left(\dfrac{3}{8}\right)\times \dfrac{5}{7}=\dfrac{3}{4}\times \left( \dfrac{5}{14}\right)$

⇒$\dfrac{15}{56}=\dfrac{15}{56}$

By associative property

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(ii) $\left(-\dfrac{7}{6}\times \dfrac{-2}{5}\right)\times \dfrac{3}{8}=\dfrac{-7}{6}\times \left(\dfrac{-2}{5}\times \dfrac{3}{8}\right)$

Sol :

⇒$\left(\dfrac{14}{30}\right)\times \dfrac{3}{8}=\dfrac{-7}{6}\times \left(\dfrac{-6}{40}\right)$

⇒$\dfrac{42}{240}=\dfrac{42}{240}$

By associative property

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Q7 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 7

(i) $\dfrac{2}{3}\times \left(\dfrac{4}{5}+ \dfrac{7}{8}\right)=\left(\dfrac{2}{3}\times \dfrac{4}{5}\right)+ \left(\dfrac{2}{3}\times \dfrac{7}{8}\right)$

Sol :

$\frac{2}{3} \times\left(\frac{4}{5}+\frac{7}{8}\right)=\left(\frac{2}{3} \times \frac{4}{5}\right)+\left(\frac{2}{3} \times \frac{7}{8}\right)$

LCM OF 5,8 is 40

$\frac{2}{3} \times\left(\frac{4 \times 8+7 \times 5}{40}\right)=\left(\frac{8}{15}+\frac{14}{24}\right)$

LCM of 15 , 24  is 120

$\frac{2}{3} \times\left(\frac{32+35}{40}\right)=\frac{(8 \times 8+14 \times 5)}{120}$

$\frac{2 \times 67}{120}=\frac{64+70}{120}$

$\frac{134}{120}=\frac{134}{120}$

 

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(ii) $\dfrac{-6}{15}\times \left(\dfrac{7}{8}+ \dfrac{-5}{12}\right)=\left(\dfrac{-6}{15}\times\dfrac{7}{8}\right)+ \left(\dfrac{-6}{15}\times \dfrac{-5}{12}\right)$

Sol :

$-\frac{6}{15} \times\left(\frac{7}{8}+\frac{-5}{12}\right)=\left(\frac{-6}{15} \times \frac{7}{8}\right)+\left(\frac{-6}{15} \times \frac{-5}{12}\right)$

L.C.M of 8,12 is 24

$\frac{-6}{15} \times\left(\frac{7 \times 3+(-5)(2)}{24}\right)=\frac{-42}{120}+\frac{30}{180}$

L.C.M of 120,180 is 360

$-\frac{6}{15} \times \left(\frac{21-10}{24}\right)=\left(\frac{-42 \times 3+30 \times 2}{360}\right)$

 $-\frac{6}{15} \times \frac{11}{24}=\frac{-126+60}{360}$

$-\frac{66}{360}=-\frac{66}{360}$

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Q8 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 8

Use the distributivity of multiplication of rational numbers over addition to simplify:

(i) $\dfrac{5}{8}\times \left(\dfrac{2}{3}+ \dfrac{1}{2}\right)$

Sol :

$\frac{5}{8} \times\left(\frac{2}{3}+\frac{1}{2}\right)$

$=\left(\frac{5}{8} \times \frac{2}{3}\right)+\left(\frac{5}{8} \times \frac{1}{2}\right)$

$=\frac{10}{24}+\frac{5}{16}$

$\begin{array}{l|l} 2 &16,24 \\ \hline 2&8,12 \\ \hline 2&4,6 \\ \hline 2&2,3 \\ \hline 3&1,3\end{array}$

=2×2×2×2×3

=48

L.C.M of 24, 16 is 48

$=\frac{10 \times {2}+5 \times 3}{48}$

$=\frac{20+15}{48}$

$=\frac{35}{48}$

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(ii) $\dfrac{-2}{3}\times \left(\dfrac{4}{7}+ \dfrac{-11}{14}\right)$

Sol :

$=\left(-\frac{2}{3} \times \frac{4}{7}\right) +\left(-\frac{2}{3} \times \frac{-11}{14}\right)$

$=\frac{-8}{21}+\frac{22}{42}$

L.C.M of 21,42 is 42

$\begin{array}{l|l}2&21,42\\ \hline 3& 21,21 \\ \hline 7& 7,7 \\ \hline &1,1 \end{array}$

$=\frac{-8 \times {2}+22 \times {1}}{42}$

$=\frac{-16+22}{42}$

$=\frac{6}{42}$ or $\frac{1}{7}$

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(iii) $\dfrac{-2}{9}\times \left(\dfrac{3}{4}- 36\right)$

Sol :

$=-\frac{2}{9} \times\left(\frac{3 \times 1-36 \times 4}{4}\right)$

$=-\frac{2}{9} \times\left(\frac{3-144}{4}\right)$

$=-\frac{2}{9} \times-\frac{141}{4}$

$=\frac{141}{18}=\frac{47}{6}=7 \frac{5}{6}$

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Q9(i-iv) | Ex-1C | Class-8 | Composite | Schand Solution | Rational Numbers | Maths | Ch-1 | myhelper

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Q9(v-viii) | Ex-1C | Class-8 | Composite | Schand Solution | Rational Numbers | Maths | Ch-1 | myhelper

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Question 9

Find the multiplication inverse , i.e., the reciprocal of

(i) 8

Sol : $\dfrac{1}{8}$

 

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(ii) -18

Sol : $\dfrac{-1}{18}$

 

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(iii) $\dfrac{14}{23}$

Sol : $\dfrac{23}{14}$

 

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(iv) $\dfrac{-29}{16}$

Sol : $\dfrac{16}{-29}$

 

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(v) $\dfrac{14}{-9}$

Sol : $\dfrac{-9}{14}$

 

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(vi) $\dfrac{-8}{-9}$

Sol : $\dfrac{-9}{-8}$ or $\dfrac{9}{8}$

 

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(vii) $\dfrac{4}{5}\times \dfrac{15}{8}$

Sol :

⇒$\dfrac{\not{4}}{5} \times \dfrac{15}{\not{8}}$

⇒$\dfrac{1}{\not{5}} \times \dfrac{\not{15}}{2}$

⇒$\dfrac{1}{1} \times \dfrac{3}{2}$

⇒$\dfrac{3}{2}$

⇒Multiplicative inverse $=\frac{2}{3}$

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(viii) $0\times \dfrac{2}{9}$

Sol : $=\frac{0}{9}$

 ⇒Multiplicative inverse $=\frac{9}{0}$ 

 ⇒∞ or do not exist

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Q10 Ex-1C Class 8 Schand Composite mathematics Solution

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Question 10

Match the mathematical sentence in Column A with the property illustrated by the statement in Column B

Column A	Column B
(i) $\dfrac{1}{3}\times 0=0$	(a) Property of 1
(ii) $\dfrac{-2}{5}\times 1=\dfrac{-2}{5}$	(b) Closure Property
(iii) $\dfrac{1}{4}\times 4=1$	(c) Property of 0
(iv) $\dfrac{-6}{7}\times \dfrac{2}{3}=-\dfrac{4}{5}$	(d) Commutative Property
(v) $0\times -\dfrac{7}{8}=\dfrac{-7}{8}\times 0$	(e) Multiplicative Inverse

Sol :

(i)→ (c)

(ii)→ (a)

(iii)→ (e)

(iv)→ (b)

(v)→ (d)

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Q11 | Ex-1C | Class 8 | Schand Composite mathematics Solution | myhelper

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Question 11

Divide:

(i) $\dfrac{6}{13}\text{ by }3$

Sol :

$=\frac{6}{13} \times \frac{1}{3}$

$=\frac{2}{13}$

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(ii) $\dfrac{5}{6}\text{ by }\dfrac{-10}{21}$

Sol :

$=\frac{5}{6} \times \frac{-21}{10}$

$=\frac{1}{6} \times \frac{-21}{2}$

$=-\frac{7}{4}\text{ or }-1\frac{3}{4}$

(iii) $\dfrac{10}{33}\text{ by }\dfrac{2}{-11}$

Sol :

$=\frac{10}{33} \times \frac{-11}{2}$

$=\frac{5}{33}\times \frac{-11}{1}$

$=\frac{-5}{3} \text{ or}-1 \frac{2}{3}$

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(iv) $-\dfrac{21}{22}\text{ by }\dfrac{-7}{11}$

Sol :

$=-\frac{21}{22} \times \frac{-11}{7}$

$=-\frac{21}{2} \times \frac{-1}{7}$

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

(v) $\dfrac{9}{14}\text{ by }\dfrac{-3}{28}$

Sol :

$=\frac{9}{14} \times \frac{-26}{3}$

$=\frac{3}{14} \times \frac{-26}{1}$

$=\frac{3 \times-13}{7 \times 1}$

$=\frac{-39}{7}$

$=-5 \frac{4}{7}$

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(vi) $-15\dfrac{3}{4}\text{ by }-2\dfrac{5}{8}$

Sol :

$=-\left(\frac{15 \times 4+3}{4}\right) \div-\left(\frac{2 \times 8+5}{8}\right)$

$=-\frac{63}{4} \div-\frac{21}{8}$

$=-\frac{63}{4} \times \frac{-8}{21}$

$=\frac{63}{1} \times \frac{2}{2 1}$

=3×2

=6

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Q12 | Ex-1C | Class 8 | Schand Composite mathematics Solution | myhelper

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Question 12

Evaluate:

(i) $\left(\dfrac{5}{9}\div\dfrac{15}{36}\right)\div \left(-\dfrac{5}{6}\right)$

Sol :

$=\left(\frac{5}{9} \times \frac{36}{15}\right) \div\left(-\frac{5}{6}\right)$

$=\left(\frac{5}{1} \times \frac{4}{15}\right) \div\left(-\frac{5}{6}\right)$

$=\frac{4}{3} \div-\frac{5}{6}$

$=\frac{4}{3} -\frac{-6}{5}$

$=\frac{-8}{15}$

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(ii) $\left(\dfrac{-3}{29}\div\dfrac{9}{87}\right)\div\dfrac{-1}{\phantom{-}7}$

Sol :

$=\left(\frac{-3}{29} \times \frac{87}{9}\right) \div\left(-\frac{1}{7}\right)$

$=\left(\frac{-1}{29} \times \frac{87}{3}\right) \div\left(\frac{-1}{7}\right)$

$=\left(\frac{-1}{29} \times \frac{29}{1}\right) \div\left(-\frac{1}{7}\right)$

$=-1 \div\left(\frac{-1}{7}\right)$

=-1×-7

=7

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