Question

Q1 Find the additive inverse of -6/-3

Q2 Subtract ⅔ from 5/4

Q3 What should be subtracted from 4/7 to get ¾?

Q4 Verify that-(-x)=x for i) x=11/15

Q5 Find the multiplicative inverse of i) -1×(-⅖)

Q6 Multiply 6/13 by the reciprocal of -7/16?

Q7 Fill in the blanks

i) Zero has _______ reciprocal .

ii) The number ____ and_____ are their own reciprocal.

Q8 Write five rational numbers which are smaller than 2?

Q9 Find -3/5×0

Q10 Verify ;i)3/4×(-⅓+⅚)={3/4×(-⅓)}+(3/4×5/6)

Q11 Verify a×(b+c)=(a×b)+(a×c) for:i)a=⅚, b=-⅘, c=-7/10

Q12 Simplify and express the result in the lowest form :i)5/8×-6/20×11/18×-4/33

Q13 By what number should 5/3 be multiplied to get the product as -5/11?

Q14 Evaluate (-7/11)÷(7/11)

Q15 Write the law of exponent of rational numbers (some important notes)

Answer 1

Solution : 1

$\sf The \: additive \: inverse \: of \: \dfrac{ - 6}{ - 3} \: is \: \dfrac{- 6}{3}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 2

$\sf \implies Subtract \: \dfrac{2}{3} \: from \: \dfrac{5}{4}$

$\sf \implies \dfrac{5}{4} - \dfrac{2}{3}$

$\sf \implies \dfrac{5 \times 3 = 15}{4 \times 3 = 12} - \dfrac{2 \times 4 = 8}{3 \times 4 = 12}$

$\sf \implies \dfrac{ 15}{ 12} - \dfrac{8}{12}$

$\sf \implies \dfrac{ 15 - 8}{ 12} = \dfrac{7}{12}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 3

$\sf \implies \dfrac{4}{7} - x = \dfrac{3}{4}$

$\sf \implies - x = \dfrac{3}{4} - \dfrac{4}{7}$

$\sf \implies - x = \dfrac{21 - 16}{28}$

$\sf \implies - x = \dfrac{5}{28}$

$\sf \implies x = \dfrac{ - 5}{28}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 4

$\sf Verify \: that, -(-x) = x$

$\sf Here, value \: of \: x = \dfrac{11}{15}$

$\sf So, - \bigg(-\dfrac{11}{15} \bigg) =\dfrac{11}{15}$

$\implies \bf [ \: Note : -,- \: is \: + \: ]$

$\sf Hence, \: \dfrac{11}{15} =\dfrac{11}{15}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 5

$\bf \underline{To \: find} : \sf Multiplicative \: inverse \: of \: - 1 \times \bigg( \dfrac{ - 2}{5} \bigg)$

$\implies - 1 \times \bigg( \dfrac{ - 2}{5} \bigg) = \dfrac{2}{5}$

$\sf Now, multiplicative \: inverse \: of \: \dfrac{2}{5} \: is \: \dfrac{5}{2}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 6

$\sf Multiply \: \dfrac{6}{13} \: by \: reciprocal \: of \: \dfrac{ - 7}{16}$

$\sf The \: reciprocal \: of \: \dfrac{ - 7}{16} \: is \: \dfrac{ - 16}{7}$

$\sf Now, \dfrac{6}{13} \times \dfrac{ - 16}{7} = \dfrac{ - 96}{91}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 7

$\bf \underline{ Fill \: in \: the \: blanks} :$

$\sf \: i) \: Zero \: has \: \underline{NO} \: reciprocal .$

$\sf ii) \: The \: number \: \underline{1 }\: and \: \underline{- 1 }\: are \: their \: own \: reciprocal.$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 8

Five rational numbers which are smaller than 2 are 1, 0, -1, -2, -3

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 9

$\sf Find \: \dfrac{ - 3}{5} \times 0$

$\sf \implies \dfrac{ - 3}{5} \times 0 = 0$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 10

$\bf \underline{ Verify \: that},$

$\sf \dfrac{3}{4} \times \bigg( \dfrac{ - 1}{3} + \dfrac{5}{6} \bigg) = \bigg[ \dfrac{3}{4} \times \bigg(\dfrac{ - 1}{3} \bigg) \bigg] + \bigg( \dfrac{3}{4} \times \dfrac{5}{6} \bigg)$

$\sf\implies \dfrac{3}{4} \times \bigg( \dfrac{ - 2 + 5}{6} \bigg) = \dfrac{3}{4} \times \dfrac{ - 1}{3} + \dfrac{3}{4} \times \dfrac{5}{6}$

$\implies \sf \dfrac{3}{4} \times \dfrac{3}{6} = \dfrac{ - 1}{4} + \dfrac{5}{8}$

$\implies\sf \dfrac{3}{8} = \dfrac{ - 2 + 5}{8}$

$\implies\sf \dfrac{3}{8} = \dfrac{3}{8}$

$\bf Hence \: proved \: !$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 11

$\sf Verify \: that, a \times (b + c) = (a \times b) + (a \times c)$

$\sf Here, a = \dfrac{5}{6 \: } \: , \: b = \dfrac{ - 4}{5} \: and \: c = \dfrac{7}{10}$

$\sf \implies \dfrac{5}{6 \: } \times \bigg(\dfrac{ - 4}{5} +\dfrac{7}{10} \bigg) = \bigg(\dfrac{5}{6 \: } \times \dfrac{ - 4}{5} \bigg) + \bigg(\dfrac{5}{6 \: } \times \dfrac{7}{10} \bigg)$

$\sf \implies \dfrac{5}{6 \: } \times \bigg(\dfrac{ - 4 + 7}{10} \bigg) = \dfrac{ - 4}{6} + \dfrac{7}{12}$

$\sf \implies \dfrac{5}{6 \: } \times \dfrac{ - 1}{10} = \dfrac{ - 2 \times 4 + 7}{12}$

$\sf \implies \dfrac{ - 1}{12} = \dfrac{ - 1}{12}$

$\bf Hence \: verified \: !$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 12

$\bf \underline{Given } \sf\implies \dfrac{5}{8} \times \dfrac{ - 6}{20} \times \dfrac{11}{18} \times \dfrac{ - 4}{33}$

$\sf\implies \dfrac{5 \times - 6 \times 11 \times - 4}{8 \times 20 \times 18 \times 33}$

$\sf\implies \dfrac{1320}{95040} = \dfrac{1}{72}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 13

$\sf \implies\dfrac{5}{3} \times x = \dfrac{ - 5}{11}$

$\implies \sf x = \dfrac{ - 5}{11} \times \dfrac{3}{5}$

$\implies \sf x = \dfrac{ - 3}{11}$

━━━━━━━━━━━━━━━━━━━━━━━━

Solution : 14

$\sf \implies \bigg(\dfrac{ - 7}{11} \bigg) \div \bigg( \dfrac{7}{11} \bigg)$

$\sf \implies \dfrac{ - 7}{11} \div \dfrac{7}{11}$

$\sf \implies \dfrac{ - 7}{11} \times \dfrac{11}{7} = 1$

━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Continuation of SachinGupta01's answer:

Solution : 15

$\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}$

$\sf 2^{nd} \: Law =$

$\sf \: Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}$

$\sf Case : (ii) \: if \: a<b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}$

$\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}$

$\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}$

$\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1$