Exercise 1 C
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
Q2 Ex-1C Class 8 Schand Composite mathematics Solution
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 :
$-6 \times 5 \frac{2}{3}$
$=-\frac{6}{1} \times \frac{17}{3}$
$=\frac{-2 \times 17}{1}$
$=\frac{-34}{1} or-34$
Q3 Ex-1C Class 8 Schand Composite mathematics Solution
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
(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 :
$\left(-\frac{4}{5} \times \frac{15}{8}\right)+\left(\frac{-1}{3} \times-\frac{9}{7}\right)-\left(\frac{2}{9} \times \frac{27}{14}\right)$
$=\left(\frac{4}{1} \times \frac{3}{8}\right)+\left(\frac{-1}{1} \times \frac{-3}{7}\right)-\left(\frac{2}{1} \times \frac{3}{14}\right)$
$=\left(-\frac{1}{1} \times \frac{3}{2}\right)+\left(\frac{-1}{1} \times \frac{-3}{7}\right)-\left(\frac{1}{1} \times \frac{3}{7}\right)$
$=-\frac{3}{2}+\frac{3}{7}-\frac{3}{7}$
$=-\frac{3}{2}$
Q4 Ex-1C Class 8 Schand Composite mathematics Solution
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) $\dfrac{2}{9}\times \left(\dfrac{-4}{9}\div\dfrac{6}{17}\right)=\dfrac{2}{9}\times \dfrac{-4}{9} \div \dfrac{2}{9} \times \dfrac{6}{17}$
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
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
(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
Q6 Ex-1C Class 8 Schand Composite mathematics Solution
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
(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
Q7 Ex-1C Class 8 Schand Composite mathematics Solution
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
$\frac{2}{3} \times\left(\frac{32+35}{40}\right)=\frac{(8 \times 8+14 \times 5)}{120}$
$\frac{2 \times 67}{120}=\frac{64+70}{120}$
$\frac{134}{120}=\frac{134}{120}$
(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
$-\frac{6}{15} \times \left(\frac{21-10}{24}\right)=\left(\frac{-42 \times 3+30 \times 2}{360}\right)$
$-\frac{6}{15} \times \frac{11}{24}=\frac{-126+60}{360}$
$-\frac{66}{360}=-\frac{66}{360}$
Q8 Ex-1C Class 8 Schand Composite mathematics Solution
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 :
$\frac{5}{8} \times\left(\frac{2}{3}+\frac{1}{2}\right)$
$=\left(\frac{5}{8} \times \frac{2}{3}\right)+\left(\frac{5}{8} \times \frac{1}{2}\right)$
$=\frac{10}{24}+\frac{5}{16}$
$\begin{array}{l|l}
2 &16,24 \\
\hline 2&8,12 \\
\hline 2&4,6 \\
\hline 2&2,3 \\
\hline 3&1,3\end{array}$
L.C.M of 24, 16 is 48
$=\frac{10 \times {2}+5 \times 3}{48}$
$=\frac{20+15}{48}$
$=\frac{35}{48}$
(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
$\begin{array}{l|l}2&21,42\\ \hline 3& 21,21 \\ \hline 7& 7,7 \\ \hline &1,1 \end{array}$
$=\frac{-8 \times {2}+22 \times {1}}{42}$
(iii) $\dfrac{-2}{9}\times \left(\dfrac{3}{4}- 36\right)$
Sol :
$=-\frac{2}{9} \times\left(\frac{3 \times 1-36 \times 4}{4}\right)$
$=-\frac{2}{9} \times\left(\frac{3-144}{4}\right)$
$=-\frac{2}{9} \times-\frac{141}{4}$
$=\frac{141}{18}=\frac{47}{6}=7 \frac{5}{6}$
Q9(i-iv) | Ex-1C | Class-8 | Composite | Schand Solution | Rational Numbers | Maths | Ch-1 | myhelper
Q9(v-viii) | Ex-1C | Class-8 | Composite | Schand Solution | Rational Numbers | Maths | Ch-1 | myhelper
Question 9
Find the multiplication inverse , i.e., the reciprocal of
(i) 8
Sol : $\dfrac{1}{8}$
(ii) -18
Sol : $\dfrac{-1}{18}$
(iii) $\dfrac{14}{23}$
Sol : $\dfrac{23}{14}$
(iv) $\dfrac{-29}{16}$
Sol : $\dfrac{16}{-29}$
(v) $\dfrac{14}{-9}$
Sol : $\dfrac{-9}{14}$
(vi) $\dfrac{-8}{-9}$
Sol : $\dfrac{-9}{-8}$ or $\dfrac{9}{8}$
(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 $=\frac{2}{3}$
(viii) $0\times \dfrac{2}{9}$
Sol : $=\frac{0}{9}$
⇒Multiplicative inverse $=\frac{9}{0}$
⇒∞ or do not exist
Q10 Ex-1C Class 8 Schand Composite mathematics Solution
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)
Q11 | Ex-1C | Class 8 | Schand Composite mathematics Solution | myhelper
Question 11
Divide:
(i) $\dfrac{6}{13}\text{ by }3$
Sol :
$=\frac{6}{13} \times \frac{1}{3}$
$=\frac{2}{13}$
(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 :
$=\frac{10}{33} \times \frac{-11}{2}$
$=\frac{5}{33}\times \frac{-11}{1}$
$=\frac{-5}{3} \text{ or}-1 \frac{2}{3}$
(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 :
$=\frac{9}{14} \times \frac{-26}{3}$
$=\frac{3}{14} \times \frac{-26}{1}$
$=\frac{3 \times-13}{7 \times 1}$
$=\frac{-39}{7}$
$=-5 \frac{4}{7}$
(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
Q12 | Ex-1C | Class 8 | Schand Composite mathematics Solution | myhelper
Question 12
Evaluate:
(i) $\left(\dfrac{5}{9}\div\dfrac{15}{36}\right)\div \left(-\dfrac{5}{6}\right)$
Sol :
$=\left(\frac{5}{9} \times \frac{36}{15}\right) \div\left(-\frac{5}{6}\right)$
$=\left(\frac{5}{1} \times \frac{4}{15}\right) \div\left(-\frac{5}{6}\right)$
$=\frac{4}{3} \div-\frac{5}{6}$
$=\frac{4}{3} -\frac{-6}{5}$
$=\frac{-8}{15}$
(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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