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(i) Given (-7/5)

Numerator of (-7/5) is -7

(ii) Given (14/-4)

Numerator of (14/-4) is 1

(iii) Given (-17/-21)

Numerator of (-17/-21) is -17

(iv) Given (8/9)

Numerator of (8/9) is 8

(v) Given 5

Numerator of 5 is 5

(i) (-5/26)

Explanation:

Consider (-4/13) – (-3/26)

(-4/13) = (-4/13) × (2/2) = (-8/26)

(-4/13) – (-3/26) = (-8/26) – (-3/26)

= (-5/26)

(ii) (-5/14)

Explanation:

Given (-9/14) + ….. = -1

(-9/14) + 1 = ….

(-9/14) + (14/14) = (5/14)

(-9/14) + (-5/14) = -1

(iii) (34/9)

Explanation:

Given (-7/9) + ….. = 3

(-7/9) + x = 3

x = 3 + (7/9)

(3/1) = (3/1) × (9/9) = (27/9)

x = (27/9) + (7/9) = (34/9)

(iv) (77/23)

Explanation:

Given ….. + (15/23) = 4

x + (15/23) = 4

x = 4 – (15/23)

(4/1) = (4/1) × (23/23) = (92/23)

x = (92/23) – (15/23)

= (77/23)

(i)

  \((\frac{5}{9}\div\frac{1}{3})\) \(\div\frac{5}{2}= \frac{5}{9}\div(\frac{1}{3}\div\frac{5}{2})\)

LHS = \((\frac{5}{9}\div\frac{1}{3})\div\frac{5}{2}\)

= \((\frac{5}{9}\times\frac{3}{1}) \div\frac{5}{2}\)

= \((\frac{5\times3}{9\times1})\div\frac{5}{2}\)

= \(\frac{15}{9}\div\frac{5}{2}\)

= \(\frac{15}{9}\times\frac{2}{5}\)

= \(\frac{15\times2}{9\times5}\)

= \(\frac{30}{45}=\frac{30\div15}{45\div15}= \frac{2}{3}\)

RHS = \(\frac{5}{9}\div(\frac{1}{3}\div\frac{5}{2})\)

= \(\frac{5}{9}\div(\frac{1}{3}\times\frac{2}{5})\)

= \(\frac{5}{9}\div\frac{2}{15}\)

= \(\frac{5\times15}{9\times2}\)

= \(\frac{75}{18}= \frac{75\div3}{18\div3}=\frac{25}{6}\)

RHS ≠ LHS

Hence, False

(ii)

\(\{(-16)\div\frac{6}{5}\}\div\frac{-9}{10}= (-16)\div\{\frac{6}{5}\div\frac{-9}{10}\}\)

LHS = \((-16\div\frac{6}{5})\div\frac{-9}{10}\)

= \((\frac{-16}{1}\div\frac{6}{5})\div\frac{-9}{10}\)

= \((\frac{-16}{1}\times\frac{5}{6})\div\frac{-9}{10}\)

= \((\frac{-16\times5}{1\times6})\div\frac{-9}{10}\)

= \(\frac{-80}{6}\div\frac{-9}{10}\)

= \(\frac{-80}{6}\times\frac{-9}{10}\)

= \(\frac{-80\times10}{6\times-9}\)

= \(\frac{-800}{-54}=\frac{800}{54}= \frac{800\div2}{54\div2}=\frac{400}{27}\)

RHS = \(-16\div(\frac{6}{5}\div\frac{-9}{10})\)

= \(\frac{-16}{1}\div(\frac{6}{5}\times\frac{10}{-9})\)

= \(\frac{-16}{1}\div\frac{60}{-45}\)

= \(\frac{-16}{1}\times\frac{-45}{60}\)

= \(\frac{-16\times-45}{1\times60}\)

= \(\frac{720}{60}=\frac{720\div60}{60\div60}=12\)

RHS ≠ LHS 

Hence, False

(iii)

\((\frac{-3}{5}\div\frac{-12}{35})\div\frac{1}{14}= \frac{-3}{5}\div(\frac{-12}{35}\div\frac{1}{14})\)

LHS = \((\frac{-3}{5}\div\frac{-12}{35})\div\frac{1}{14}\)

= \((\frac{-3}{5}\times\frac{35}{-12})\div\frac{1}{14}\)

= \((\frac{-3\times35}{5\times-12}) \div\frac{1}{14}\)

= \(\frac{-105}{-60}\div\frac{1}{14}\)

= \(\frac{-105}{-60}\times\frac{14}{1}\)

= \(\frac{-105\times14}{-60\times1}\)

= \(\frac{-1470}{-60}=\frac{-1470}{-60}=\frac{1470\div30}{60\div30}\) = \(\frac{49}{2}\)

RHS = \(\frac{-3}{5}\div(\frac{-12}{35}\div\frac{1}{14})\)

= \(\frac{-3}{5}\div(\frac{-12}{35}\times\frac{14}{1})\)

= \(\frac{-3}{5}\div(\frac{-12\times1+14\times35}{35})\)

= \(\frac{-3}{5}\div\frac{-12+490}{35}\)

= \(\frac{-3}{5}\div\frac{478}{35}\)

= \(\frac{-3}{5}\times\frac{35}{478}\)

= \(\frac{-3\times35}{5\times478}\)

= \(\frac{-105}{2390}=\frac{-105\div5}{2390\div5}=\frac{-21}{478}\)

RHS ≠ LHS 

Hence, False

Let there be a rational number \(\frac{a}{b}\) then \((\frac{a}{b})^{-1}\) = \(\frac{b}{a}\)

Therefore,

(i)

\((\frac{5}{8})^{-1}=\frac{8}{5}\)

(ii)

\((\frac{-4}{9})^{-1}=\frac{9}{-4} =\frac{9\times-1}{-4\times-1}=\frac{-9}{4}\)

(iii)

\((-7)^{-1}=(\frac{-7}{1})^{-1}=\frac{1}{-7}= \frac{1\times-1}{-7\times-1}=\frac{-1}{7}\)

(iv)

\((\frac{1}{-3})^{-1}=\frac{-3}{1}=-3\)

(i) Commutative law i.e., a b = b a

(ii) Associative law i.e., a(bc) = (ab)c

(iii) Distributive law i.e., a(b + c) = ab + ac

(iv) Property of multiplicative identity i.e., a× 1=1× a

(v) Property of multiplicative inverse i.e., \(\frac{a}{b}\times\frac{b}{a}\) =1

(vi) Multiplicative property of 0 i.e., a× 0=0

(i)

\(\frac{13}{5} \div\frac{26}{10}= \frac{26}{10}\div\frac{13}{5}\)

LHS = \(\frac{13}{5}\div\frac{26}{10}\)

= \(\frac{13}{5}\times\frac{10}{26}\)

= \(\frac{13\times10}{5\times26}\)

= \(\frac{130}{130}= 1\)

RHS = \(\frac{26}{10}\div\frac{13}{5}\)

= \(\frac{26}{10}\times\frac{5}{13}\)

= \(\frac{26\times5}{10\times13}\)

= \(\frac{130}{130}=1\)

Since, RHS = LHS 

Therefore, True

(ii)

\(-9\div\frac{3}{4}=\frac{3}{4}(-9)\)

LHS = \(-9\div\frac{4}{3}\)

= \(-9\times\frac{4}{3}\)

= \(\frac{-9\times4}{3}\)

= \(\frac{-36}{3}=-12\)

RHS = \(\frac{3}{4}\div(-9)\)

= \(\frac{3}{4}\times\frac{1}{-9}\)

= \(\frac{3\times1}{4\times-9}\)

= \(\frac{3}{-36}=\frac{-1}{12}\)

Since, RHS ≠ LHS

Therefore, False

(iii)

\(\frac{-8}{9}\div\frac{-4}{3}= \frac{-4}{3}\div\frac{-8}{9}\)

LHS = \(\frac{-8}{9}\div\frac{-4}{3}\)

= \(\frac{-8}{9}\times\frac{3}{-4}\)

= \(\frac{-8\times3}{9\times-4}\)

= \(\frac{-24}{-36} =\frac{2}{3}\)

RHS = \(\frac{-4}{3}\div\frac{-8}{9}\)

= \(\frac{-4}{9}\times\frac{9}{-8}\)

= \(\frac{-4\times9}{3\times-8}\)

= \(\frac{-36}{-24}=\frac{3}{2}\)

Since, RHS ≠ LHS

(iv)

\(\frac{-7}{24}\div\frac{3}{-16}=\frac{3}{-16}\div\frac{-7}{24}\)

LHS = \(\frac{-7}{24}\div\frac{3}{-16}\)

= \(\frac{-7}{24}\times\frac{-16}{3}\)

= \(\frac{-7\times-16}{24\times3}\)

= \(\frac{112}{72}=\frac{14}{9}\)

RHS = \(\frac{3}{-16}\div\frac{-7}{24}\)

= \(\frac{3}{-16}\times\frac{24}{-7}\)

= \(\frac{3\times24}{-16\times-7}\)

= \(\frac{72}{112}=\frac{9}{14}\)

Since, RHS ≠ LHS

Therefore, False

(i)

\(\frac{3}{7}\times(\frac{5}{6}+\frac{12}{13})=(\frac{3}{7}\times\frac{5}{6})+(\frac{3}{7}\times\frac{12}{13})\)

LHS = \(\frac{3}{7}\times(\frac{5}{6}+\frac{12}{13})\)

\(=\frac{3}{7}\times(\frac{5\times13+12\times6}{78})\)

= \(\frac{3}{7}\times(\frac{65+72}{78})\)

= \(\frac{3}{7}\times(\frac{137}{78})\)

= \(\frac{3\times137}{7\times78}\)

= \(\frac{411}{546}\)

In lowest terms,

\(\frac{411}{546}= \frac{411\div3}{546\div3}=\frac{137}{182}\)

RHS = \((\frac{3}{7}\times\frac{5}{6})+(\frac{3}{7}\times\frac{12}{13})\)

= \((\frac{3\times5}{7\times6})+(\frac{3\times12}{7\times13})\)

= \(\frac{15}{42}+(\frac{36}{91})\)

= \(\frac{15\times13+36\times6}{546}\)

= \(\frac{195+216}{546}\)

= \(\frac{411}{546}\) 

In lowest terms,

\(\frac{411}{546}= \frac{411\div3}{546\div3}=\frac{137}{182}\)

LHS=RHS

(ii)

\(\frac{-15}{4}\times(\frac{3}{7}+\frac{-12}{5}) =(\frac{-15}{4}\times\frac{3}{7})+(\frac{-15}{4}\times\frac{-12}{5})\)

LHS = \(\frac{-15}{4}\times(\frac{3}{7}+\frac{-12}{5})\)

= \(\frac{-15}{4}\) \(\times(\frac{3\times5+(-12)\times7}{35})\)

= \(\frac{-15}{4}\times(\frac{15-84}{35})\)

= \(\frac{-15}{4}\times(\frac{-69}{35})\)

= \((\frac{-15\times-69}{4\times35})\)

= \(\frac{1035}{140}\)

In lowest terms,

\(\frac{1035}{140}=\frac{1035\div5}{140\div5} =\frac{207}{28}\)

RHS = \((\frac{-15}{4}\times\frac{3}{7})+(\frac{-15}{4}\times\frac{-12}{5})\)

= \((\frac{-15\times3}{4\times7})+(\frac{-1500}{4}\times\frac{-12}{5})\)

= \(\frac{-45}{28}+(\frac{180}{20})\)

= \(\frac{-45\times5+180\times7}{140}\)

= \(\frac{-225+1260}{140}\)

= \(\frac{1035}{140}\)

In lowest terms,

\(\frac{1035}{140}=\frac{1035\div5}{140\div5} =\frac{207}{28}\)

LHS=RHS

(iii)

\((\frac{-8}{3}+\frac{-13}{12})\times\frac{5}{6} =(\frac{-8}{3}\times\frac{5}{6})+(\frac{-13}{12}\times\frac{5}{6})\)

LHS = \((\frac{-8}{3}+\frac{-13}{12})\times\frac{5}{6}\)

= \((\frac{-8\times4+(-13)\times1}{12})\times\frac{5}{6}\)

= \((\frac{-32-13}{12})\times(\frac{5}{6})\)

= \(\frac{-45}{12}\times\frac{5}{6}\)

= \(\frac{-45\times5}{12\times6}\)

= \(\frac{-225}{72}\)

In lowest terms,

\(\frac{-225}{72}= \frac{-225\div9}{72\div9}=\frac{-25}{8}\)

RHS = \((\frac{-8}{3}\times\frac{5}{6})+(\frac{-13}{12}\times\frac{5}{6})\)

= \((\frac{-8\times5}{3\times6})+(\frac{-13\times5}{12\times6})\)

= \(\frac{-40}{18}+(\frac{-65}{72})\)

= \(\frac{-40\times4+(-65)\times1}{72}\)

= \(\frac{-160-65}{72}\)

= \(\frac{-225}{72}\)

In lowest terms,

\(\frac{-225}{72}= \frac{-225\div9}{72\div9}=\frac{-25}{8}\)

LHS=RHS

(iv)

\(\frac{-16}{7}\times(\frac{-8}{9}+\frac{-7}{6}) =(\frac{-16}{7}\times\frac{-7}{6})\)

LHS = \(\frac{-16}{7}\times(\frac{-8}{9}+\frac{-7}{6})\)

= \(\frac{-16}{7}\times(\frac{-8\times2+(-7)\times3}{18})\)

= \(\frac{-16}{7}\times(\frac{-16-21}{18})\)

= \(\frac{-16}{7}(\frac{-37}{18})\)

= \(\frac{-16\times-37}{7\times18}\)

= \(\frac{592}{126}\)

In lowest terms,

\(\frac{592}{126}=\frac{592\div2}{126\div2}=\frac{296}{63}\)

RHS = \((\frac{-16}{7}\times\frac{-8}{9})+(\frac{-16}{7}\times\frac{-7}{6})\)

= \((\frac{-16\times-18}{7\times9})+ (\frac{-16\times-7}{7\times6})\)

= \(\frac{128}{63}+(\frac{112}{42})\)

= \(\frac{128\times2+112\times3}{126}\)

= \(\frac{592}{126}\)

In lowest terms,

\(\frac{592}{126}=\frac{592\div2}{126\div2}=\frac{296}{63}\)

LHS=RHS

Let the numbers be 2x, 3x and 4x

(2x)³ + (3x)³ + (4x)³ = 33957

8x³ + 27x³ + 64x³ = 33957

99x³ = 33957

x³ = 33957/99

x³ = 343

x³ = 7 × 7 × 7

x³ = 7³

x = 7

∴ The numbers are 2x = 2 × 7 = 14

3x = 3 × 7 = 21

4x = 4 × 7 = 28

(i) 17

17 ends with 7, so its cube ends with 3. i.e, ones digit in 17³ is 3.

(ii) 12

12 ends with 2, so its cube ends with 8. i.e, ones digit in 12³ is 8.

(iii) 38

38 ends with 8, so its cube ends with 2. i.e, ones digit in 38³ is 2.

(iv) 53

53 ends with 3, so its cube ends with 7. i.e, ones digit in 53³ is 7.

(v) 71

71 ends with 1, so its cube ends with 1. i.e, ones digit in 71³ is 1

(vi) 84

84 ends with 4, so its cube ends with 4. i.e, ones digit in 84³ is 4.

(i) 7

(ii) 6

(iii) 30

(iv) 0.017

(v) 42

(i) √1.2321 = √12321/10000 =111/100 = 1.11

(ii) √11.9025 = √119025/√10000 = 345/100 = 3.45

(i) Clearly,

\(\frac{6}{-13}=\frac{6}{-13}\)

(ii)

\(\frac{5}{-13}=\frac{5\times-1}{-13\times-1}=\frac{-5}{13}\)

\(\frac{-5}{13}\) and \(\frac{-35}{91}\)have different denominators.

Therefore, we take LCM of 13 and 91 that is 91. 

Now,

\(\frac{-5}{13}=\frac{-5\times7}{13\times7}=\frac{-35}{91}\)

And,

\(\frac{-35}{91}=\frac{-35\times1}{91\times1}=\frac{-35}{91}\)
Clearly,

\(\frac{-35}{91}=\frac{-35}{91}\)

Hence,

\(\frac{5}{-13}=\frac{-35}{91}\)

(iii) We can write \(-2=\frac{-2}{1}\)

\(\frac{-2}{1}\) and \(\frac{-13}{5}\) have different denominators.

Now,

\(\frac{-2}{1}=\frac{-2\times5}{1\times5}=\frac{-10}{5}\)
And,

\(\frac{-13}{5}=\frac{-13\times1}{5\times1}=\frac{-13}{5}\)

Since, -10 > -13

Therefore, \( \frac{-10}{5}>\frac{-13}{5}\)

Hence, \( -2>\frac{-13}{5}\)

(iv)  \(\frac{5}{-8}=\frac{5\times-1}{-8\times-1}=\frac{-5}{8}\)

 \(\frac{-2}{3}\) and \(\frac{-5}{8}\) have different denominators.

Therefore, we take LCM of 3 and 8 that is 24. 

Now,

\(\frac{2}{-3}=\frac{-2\times8}{3\times8}=\frac{-16}{24}\)
And,

\(\frac{-5}{8}=\frac{-5\times3}{8\times3}=\frac{-15}{24}\)
Since, -16 < -15

Therefore, \( \frac{-16}{24}<\frac{-15}{24}\)

Hence, \( \frac{-2}{3}<\frac{-5}{8}\)

(v)

\(\frac{-3}{-5}=\frac{-3\times-1}{-5\times-1}=\frac{3}{5}\)

\(\frac{3}{5}\)is a positive number and all positive numbers are greater than 0.

Therefore, \( 0<\frac{3}{5}\)

Hence, \( 0<\frac{-3}{-5}\)

(vi) 

 \(\frac{-8}{9}\) and \(\frac{-9}{10}\) have different denominators.

Therefore, we take LCM of 9 and 10 that is 90.

 Now,

\(\frac{-8}{9}=\frac{-8\times10}{9\times10}=\frac{-80}{90}\)

And,

\(\frac{-9}{10}=\frac{-9\times9}{10\times9}=\frac{-81}{90}\)

Since, -80 > -81

 Therefore, \( \frac{-80}{90}<\frac{-81}{90}\)

Hence, \( \frac{-8}{9}<\frac{-9}{10}\)

(…… ) ÷ (-3) = (-4/15)

Let the required number be (a/b). Then,

(a/b) ÷ (-3/1) = (-4/15)

⇒ (a/b) = (-4/15) × (-3/1)

⇒ (a/b) = (-4 × -3) / (15 × 1)

⇒ (a/b) = (-4× -1) / (5×1)

⇒ (a/b) = (4/5)

(-12/7) ÷ (-18)

We have,

= (-12/7) × (1/-18) … [∵ reciprocal of (-18) is (1/-18)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-12×1) / (7×-18)

On simplifying,

= (-4×1) / (7×-6)

= (-2×1) / (7×-3)

= (-2/-21)

= (2/21)

(…… ) ÷ (-7/5) = (10/19)

Let the required number be (a/b). Then,

(a/b) ÷ (-7/5) = (10/19)

⇒ (a/b) = (10/19) × (-7/5)

⇒ (a/b) = (10 × -7) / (19 × 5)

⇒ (a/b) = (2× -7) / (19×1)

⇒ (a/b) = (-14/19)

(i) \(\frac{25\times 2}{8\times 5}-\frac{3\times -10}{5\times 9}\)

= \(\frac{50}{40}+\frac{30}{45}\)

= \(\frac{5}{4}+\frac{2}{3}\)

= \(\frac{15+8}{12}\)

= \(\frac{23}{12}\)

 (ii) \(\frac{1\times 1}{2\times 4}+\frac{1\times 6}{2\times 1}\)

= \(\frac{1}{8}+\frac{6}{2}\)

= \(\frac{1}{8}+3\)

= \(\frac{1+24}{8}\)

= \(\frac{25}{8}\)

 (iii) \(\frac{-5\times 2}{1\times 15}+\frac{6\times 2}{1\times 9}\)

= \(\frac{-10}{15}+\frac{12}{9}\)

= \(\frac{-2}{3}+\frac{4}{3}\)

= \(\frac{-2+4}{3}\)

= \(\frac{2}{3}\)

 (iv) \(\frac{-9\times 5}{4\times 3}+\frac{13\times 5}{2\times 6}\)

= \(\frac{-45}{12}+\frac{65}{12}\)

= \(\frac{-45+65}{12}\)

= \(\frac{20}{12}\)

= \(\frac{5}{3}\)

(v) \(\frac{-4\times 12}{3\times -5}+\frac{3\times 21}{7\times 15}\)

= \(\frac{4\times 4}{1\times 5}+\frac{1\times 3}{1\times 5}\)

= \(\frac{16}{5}+\frac{3}{5}\)

= \(\frac{19}{5}\)

(vi) \(\frac{13\times 8}{5\times 3}+\frac{5\times 11}{2\times 3}\)

= \(\frac{104}{15}+\frac{15}{6}\)

 = \(\frac{104\times 2+55\times 6}{30}\)

= \(\frac{483}{30}\)

(vii) \(\frac{13\times 11}{7\times 26}+\frac{4\times 5}{3\times 6}\)

= \(\frac{1\times 11}{7\times 2}+\frac{4\times 5}{3\times 6}\)

 = \(\frac{11}{14}+\frac{20}{18}\)

 = \(\frac{11\times 9+20\times 7}{126}\)

= \(\frac{99+140}{126}\)

= \(\frac{239}{126}\)

(viii) \(\frac{8\times -3}{5\times 2}-\frac{3\times 11}{10\times 16}\)

= \(\frac{-24}{10}-\frac{33}{160}\)

= \(\frac{-24\times 16-33\times 1}{160}\)

= \(\frac{-384-33}{160}\)

= \(\frac{-417}{160}\)

(i) \((\frac{3}{2}\times \frac{1}{6})+(\frac{5}{3}\times \frac{7}{2})-(\frac{13}{8}\times \frac{4}{3})\)

= \(\frac{3}{12}+\frac{35}{6}-\frac{52}{24}\)

= \(\frac{3\times 2+35\times 4-52\times 1}{24}\)

= \(\frac{6+140-52}{24}\)

= \(\frac{94}{24}\)

= \(\frac{47}{12}\)

 (ii) \((\frac{1}{4}\times \frac{2}{7})-(\frac{5}{14}\times \frac{-2}{3})+(\frac{3}{7}\times \frac{9}{2})\)

= \(\frac{2}{28}+\frac{10}{42}+\frac{27}{14}\)

= \(\frac{3\times 2+10\times 2+27\times 6}{84}\)

= \(\frac{6+20+162}{84}\)

= \(\frac{188}{84}\)

= \(\frac{47}{21}\)

 (iii) \((\frac{13}{9}\times \frac{-15}{2})+(\frac{7}{3}\times \frac{8}{5})+(\frac{3}{5}\times \frac{1}{2})\)

= \(\frac{13\times -15}{9\times 2}+\frac{7\times 8}{3\times 5}+\frac{3\times 1}{5\times 2}\)

= \(\frac{-65}{6}+\frac{56}{15}+\frac{3}{10}\)

= \(\frac{-65\times 5+56\times 2-3\times 3}{30}\)

= \(\frac{-204}{30}\)

(iv) \((\frac{3}{11}\times \frac{5}{6})-(\frac{9}{12}\times \frac{4}{3})+(\frac{5}{13}\times \frac{6}{15})\)

= \(\frac{3\times 5}{11\times 6}-\frac{36}{36}+\frac{30}{13\times 15}\)

= \(\frac{15}{66}-1+\frac{1}{13}\)

= \(\frac{15\times 13-858+66}{858}\)

= \(\frac{-597}{858}\)

(-16/35) ÷ (-15/14)

We have,

= (-16/35) × (14/-15) … [∵ reciprocal of (-15/14) is (14/-15)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-16×14) / (35×-15)

On simplifying,

= (-16×2) / (5×-15)

= (-32/-75)

= (32/75)

(-9/16) × (-64/27)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-9×-64)/ (16×27)

On simplifying,

= (-1×-4)/ (1×3)

= 4/3

(i) \((\frac{3}{2}\times \frac{1}{6})+(\frac{5}{3}\times \frac{7}{2})-(\frac{13}{8}\times \frac{4}{3})\)

= \(\frac{3}{12}+\frac{35}{6}-\frac{52}{24}\)

= \(\frac{3\times 2+35\times 4-52\times 1}{24}\)

= \(\frac{6+140-52}{24}\)

= \(\frac{94}{24}\)

 = \(\frac{47}{12}\) 

(ii) \((\frac{1}{4}\times \frac{2}{7})-(\frac{5}{14}\times \frac{-2}{3})+(\frac{3}{7}\times \frac{9}{2})\)

= \(\frac{2}{28}+\frac{10}{42}+\frac{27}{14}\)

= \(\frac{3\times 2+10\times 2+27\times 6}{84}\)

= \(\frac{6+20+162}{84}\)

= \(\frac{188}{84}\)

 = \(\frac{47}{21}\) 

(iii) \((\frac{13}{9}\times \frac{-15}{2})+(\frac{7}{3}\times \frac{8}{5})+(\frac{3}{5}\times \frac{1}{2})\)

= \(\frac{13\times -15}{9\times 2}+\frac{7\times 8}{3\times 5}+\frac{3\times 1}{5\times 2}\)

= \(\frac{-65}{6}+\frac{56}{15}+\frac{3}{10}\)

= \(\frac{-65\times 5+56\times 2+3\times 3}{30}\)

= \(\frac{-204}{30}\) 

(iv) \((\frac{3}{11}\times \frac{5}{6})-(\frac{9}{12}\times \frac{4}{3})+(\frac{5}{13}\times \frac{6}{15})\)

= \(\frac{3\times 5}{11\times 6}-\frac{36}{36}+\frac{30}{13\times 15}\)

= \(\frac{15}{66}\) -1+\(\frac{1}{13}\)

= \(\frac{15\times 3-858+66}{858}\)

= \(\frac{-597}{858}\)

(-65/14) ÷ (13/-7)

First we write (13/-7) in standard form

= (13×-1)/ (-7×-1)

= (-13/7)

We have,

= (-65/14) × (7/-13) … [∵ reciprocal of (-13/7) is (7/-13)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-65×7) / (14×-13)

On simplifying,

= (-5×1) / (2×-1)

= (-5) / (-2)

= (5/2)

(-13/5) × (-10)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-13×5)/ (10×1)

On simplifying,

= (-13×10)/ (5×1)

= (-13×2)/ (1×1)

= -26

(i) Given (-5 × (2/15)) – (-6 × (2/9))

(-5 × (2/15)) – (-6 × (2/9)) 

= (-10/15) – (-12/9)

= (-2/3) + (12/9)

= (-6/9) + (12/9)

= (6/9)

= (2/3)

(ii) Given ((-9/4) × (5/3)) + ((13/2) × (5/6))

((-9/4) × (5/3)) + ((13/2) × (5/6)) 

= ((-3/4) × 5) + ((13/2) × (5/6))

= (-15/4) + (65/12)

= (-15/4) × (3/3) + (65/12)

= (-45/12) + (65/12)

= (65 – 45)/12

= (20/12)

= (5/3)

(-13) × (17/26)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-13×1)/ (17×26)

On simplifying,

= (-13×17)/ (1×26)

= (-1×17)/ (1×2)

= -17/2

For a fraction, \(\frac{a}{b}\)

\(\frac{a}{b} = \) \(\frac{a\times{n}}{b\times{n}}\)

Where, n ≠ 0

(i) We have to express \(\frac{-3}{5}\) as a rational number with denominator 20.

In order to make the denominator 20, multiply 5 by 4.

Therefore,

\(\frac{-3}{5}\) = \(\frac{-3\times4}{5\times4}\)

\(\Rightarrow\) \(\frac{-3}{5}\) = \(\frac{-12}{20}\)

(ii) We have to express \(\frac{-3}{5}\) as a rational number with denominator -30.

In order to make the denominator -30, multiply 5 by -6.

\(\frac{-3}{5}\) = \(\frac{-3\times4}{5\times-6}\)

\(\Rightarrow\) \(\frac{-3}{5}\) = \(\frac{18}{-30}\)

(iii) We have to express \(\frac{-3}{5}\) as a rational number with denominator 35.

In order to make the denominator 35, multiply 5 by 7. 

Therefore,

\(\frac{-3}{5}\) = \(\frac{-3\times7}{5\times7}\)

\(\Rightarrow\) \(\frac{-3}{5}\) = \(\frac{-21}{35}\)

(iv) We have to express \(\frac{-3}{5}\) as a rational number with denominator -40.

In order to make the denominator 20, multiply 5 by -8.

Therefore,

\(\frac{-3}{5}\) = \(\frac{-3\times8}{5\times-8}\)

\(\Rightarrow\) \(\frac{-3}{5}\) = \(\frac{24}{-40}\)

For a fraction, \(\frac{a}{b}\)

\(\frac{a}{b}=\frac{a \div n}{b \div n}\)

Where, n ≠ 0 and n divides both a and b 

(i) We have to express \(\frac{-42}{98}\) as a rational number with denominator 7. 

In order to make the denominator 7, divide 98 by 14. Therefore,

\(\frac{-42}{98}=\frac{-42\div14}{98\div14}\)

\(\Rightarrow\)\(\frac{-42}{98}=\frac{-3}{7}\)

(i) Given (5/7)

To get denominator -14 we have to multiply both numerator and denominator by -2

Then we get, (5/7) × (-2/-2)

= (-10/-14)

Therefore (5/7) as a rational number with denominator -14 is (-10/-14)

(ii) Given (5/7)

To get denominator 70 we have to multiply both numerator and denominator by -2

Then we get, (5/7) × (10/10)

= (50/70)

Therefore (5/7) as a rational number with denominator 70 is (50/70)

(iii) Given (5/7)

To get denominator -28 we have to multiply both numerator and denominator by -4

Then we get, (5/7) × (-4/-4)

= (-20/-28)

Therefore (5/7) as a rational number with denominator -28 is (-20/-28)

(iv) Given (5/7)

To get denominator -84 we have to multiply both numerator and denominator by -12

Then we get, (5/7) × (-12/-12)

= (-60/-84)

Therefore (5/7) as a rational number with denominator -84 is (-60/-84)

5/7 x (reciprocal of -7/15)

= 5/7 x 15/-7

= 75/-49

= -75/49

(a) To make the denominator 36, we have to multiply numerator and denominator by 9.

(3 x 9)/(4 x 9) = 27/36

(b) To make the denominator -80, we have to multiply numerator and denominator by -20.

3 x (-20)/4 x (-20) = -60/-80

(i) Given ((13/9) × (-15/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2))

((13/9) × (-15/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2)) = (-195/18) + (56/15) + (3/10)

= (-65/6) + (56/15) + (3/10)

= (-65/6) × (5/5) + (56/15) × (2/2) + (3/10) × (3/3).

= (-325/30) + (112/30) + (9/30)

= (-325 + 112 + 9)/30

= (-204/30)

= (-34/5)

(ii) Given ((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15))

((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15)) = (15/66) – (36/36) + (30/195)

= (5/22) – (12/12) + (1/11)

= (5/22) – 1 + (2/13)

= (5/22) × (13/13) + (1/1) × (286/286) + (2/13) × (22/22)

= (65/286) – (286/286) + (44/286)

= (-177/286)

(i) Given (2/5) ÷ (26/15)

(2/5) ÷ (26/15) 

= (2/5) × (15/26)

= (3/13)

(ii) Given (10/3) ÷ (-35/12)

(10/3) ÷ (-35/12) 

= (10/3) × (12/-35)

= (-40/35)

= (- 8/7)

(iii) Given -6 ÷ (-8/17)

-6 ÷ (-8/17) 

= -6 × (17/-8)

= (102/8)

= (51/4)

(iv) Given (40/98) ÷ -20

(40/98) ÷ -20 

= (40/98) × (1/-20)

= (-2/98)

= (-1/49)

(i) 13/15 x 5

= 13/15 x 5/1

= (13/3 x 5) x 5/1

= 13/3

(ii) 4/-5 x -5/4

= (4 x -5)/(-5 x 4)

= -20/-20

= 1

(iii) -2/5 x (-3/7)

= (-2) x (-3)/(5 x 7)

= 6/35

For a fraction,\(\frac {a}{b}\)

\(\frac {a}{b}=\frac{a\times n}{b\times n}\)

 Where, n ≠ 0 

(i) We have to express \(\frac{-3}{5}\) as a rational number with denominator 20. 

In order to make the denominator 20, multiply 5 by 4. 

Therefore, 

\(\frac{-3}{5}=\frac{-3\times4}{5 \times 4}\)

\(\frac{-3}{5}=\frac{-12}{20}\)

(ii) We have to express \(\frac{-3}{5}\) as a rational number with denominator -30. 

In order to make the denominator -30, multiply 5 by -6. 

Therefore,

\(\frac{-3}{5}=\frac{-3\times-6}{5\times-6}\)

 \(\Rightarrow\)  \(\frac{-3}{5}=\frac{18}{-30}\)

 (iii) We have to express \(\frac{-3}{5}\) as a rational number with denominator 35. 

In order to make the denominator 35, multiply 5 by 7. 

Therefore,

\(\frac{-3}{5}=\frac{-3\times7}{5\times7}\)

\(\Rightarrow\) \(\frac{-3}{5}=\frac{-21}{35}\)

 (iv) We have to express\(\frac{-3}{5}\) as a rational number with denominator -40. 

In order to make the denominator 20, multiply 5 by -8. 

Therefore, 

\(\frac{-3}{5}=\frac{-3\times-8}{5\times-8}\)

\(\Rightarrow\) \(\frac{-3}{5}=\frac{24}{-40}\)

-35

To get -35 in the denominator multiply by -5 for both numerator and denominator.

Then we get,

= [(4× (-5))/ (7× (-5))]

= (-20/-25)

(i) Given 1 by (1/2)

1 ÷ (1/2) 

= 1 × 2 

= 2

(ii) Given 5 by (-5/7)

5 ÷ (-5/7) 

= 5 × (-7/5)

= -7

(iii) Given (-3/4) by (9/-16)

(-3/4) ÷ (9/-16) 

= (-3/4) × (-16/9)

= (-4/-3)

= (4/3)

(iv) Given (-7/8) by (-21/16)

(-7/8) ÷ (-21/16) 

= (-7/8) × (16/-21)

= (-2/-3)

= (2/3)

(v) Given (7/-4) by (63/64)

(7/-4) ÷ (63/64) = (7/-4) × (64/63)

= (-16/9)

(vi) Given 0 by (-7/5)

0 ÷ (-7/5) = 0 × (5/7)

= 0

(vii) Given (-3/4) by -6

(-3/4) ÷ -6 = (-3/4) × (1/-6)

= (-1/-8)

= (1/8)

(viii) Given (2/3) by (-7/12)

(2/3) ÷ (-7/12) 

= (2/3) × (12/-7)

= (8/-7)

A rational number is in standard or simplest or lowest form when- 

1. Numerator and denominator have only 1 as its highest common factor.

 2. Denominator is a positive integer.

 (i) The HCF of 12 and 30 is 6 

Therefore,

\(\frac{-12}{30}=\frac{-12\div6}{30\div6}\)

\(\Rightarrow\)\(\frac{-12}{30}=\frac{-2}{5}\)

(ii) The HCF of 49 and 14 is 7

 Therefore,

\(\frac{-14}{49}=\frac{-14\div7}{49\div7}\)

\(\Rightarrow\)\(\frac{-14}{49}=\frac{-2}{7}\)

(iii) The HCF of 24 and 64 is 8 

Therefore,

\(\frac{24}{-64}=\frac{24\div8}{-64\div8}\)

\(\Rightarrow\)\(\frac{24}{-64}=\frac{3}{-8}\)

In order, to make the denominator positive, multiply both numerator and denominator by -1

\(\Rightarrow\)\(\frac{24}{-64}=\frac{3}{-8}=\frac{3\times-1}{-8\times-1}\)

\(\Rightarrow\)\(\frac{24}{-64}=\frac{-3}{8}\)

(iv) The HCF of 36 and 63 is 9 

Therefore,

\(\frac{-36}{-63}=\frac{-36\div9}{-63\div9}\)

\(\Rightarrow\)\(\frac{-36}{-63}=\frac{-4}{-7}\)

In order, to make the denominator positive, multiply both numerator and denominator by -1

 \(\Rightarrow\)\(\frac{-36}{-63}=\frac{-4}{-7}=\frac{-4\times-1}{-7\times-1}\)

\(\Rightarrow\)\(\frac{-36}{-63}=\frac{4}{7}\)

(i) \(\frac{\frac{2}{5}}{\frac{26}{15}}\)

= \(\frac{\frac{2}{1}}{\frac{26}{3}}\)

= \(\frac{2}{26}\times \frac{3}{1}\)

= \(\frac{3}{13}\)

 (ii) \(\frac{\frac{10}{3}}{\frac{-35}{12}}\)

= \(\frac{\frac{10}{1}}{\frac{-35}{4}}\)

= \(\frac{10}{1}\times \frac{-35}{4}\)

= \(\frac{-8}{7}\)

 (iii) \(\frac{-6}{\frac{-8}{17}}\)

= \(\frac{\frac{-6}{1}}{\frac{-8}{17}}\)

= \(-6\times \frac{17}{-8}\)

= \(\frac{51}{4}\)

 (iv) \(\frac{\frac{-40}{99}}{-20}\)

= \(\frac{\frac{2}{99}}{\frac{1}{1}}\)

= \(\frac{2}{99}\times \frac{1}{1}\)

= \(\frac{2}{99}\)

 (v) \(\frac{\frac{-22}{27}}{\frac{-110}{18}}\)

= \(\frac{\frac{22}{3}}{\frac{110}{2}}\)

= \(\frac{22}{3}\times \frac{2}{110}\)

= \(\frac{2}{15}\)

 (vi) \(\frac{\frac{-36}{125}}{\frac{-3}{75}}\)

= \(\frac{\frac{36}{5}}{\frac{3}{3}}\)

= \(\frac{36}{5}\times \frac{3}{3}\)

= \(\frac{36}{5}\)

(i) 1 by \(\frac{1}{2}\)

= \(\frac{1}{\frac{1}{2}}\)

= \(1\times \frac{1}{2}\)

= 2

(ii) 5 by \(\frac{-5}{7}\)

= \(5 \div \frac{-5}{7}\)

= \(\frac{5}{-5}\times 7\)

= -7

(iii) \(\frac{-3}{4}\) by \(\frac{9}{-16}\)

= \(\frac{3}{4} \div \frac{9}{16}\)

= \(\frac{3}{4}\times \frac{16}{9}\)

= \(\frac{4}{3}\)

(iv) \(\frac{-7}{8}\) by \(\frac{21}{-16}\)

= \(\frac{-7}{8} \div \frac{21}{16}\)

= \(\frac{-7}{8}\times \frac{16}{21}\)

= \(\frac{2}{3}\)

(v) \(\frac{7}{-4}\) by \(\frac{63}{864}\)

= \(\frac{7}{-4} \div \frac{63}{64}\)

= \(\frac{-7}{4}\times \frac{64}{63}\)

= \(\frac{-16}{9}\)

(vi) 0 by \(\frac{-7}{5}\)

= \(0 \div \frac{-7}{5}\)

= \(0\times \frac{-5}{7}\)

= 0

(vii) \(\frac{-3}{4}\) by -6

= \(\frac{3}{4} \div -6\)

= \(\frac{-3}{4}\times \frac{-1}{6}\)

= \(\frac{1}{8}\)

(viii) \(\frac{2}{3}\) by \(\frac{-7}{12}\)

= \(\frac{2}{3} \div \frac{-7}{12}\)

= \(\frac{2}{3}\times \frac{12}{-7}\)

= \(\frac{-24}{21}\)

= \(\frac{-8}{7}\)

(ix) -4 by \(\frac{-3}{5}\)

= \(-4 \div \frac{-3}{5}\)

= \(-4\times \frac{-5}{3}\)

= \(\frac{20}{3}\)

(x) \(\frac{-3}{13}\) by \(\frac{-4}{65}\)

= \(\frac{-3}{13} \div \frac{-4}{65}\)

= \(\frac{-3}{13}\times \frac{65}{-4}\)

= \(\frac{15}{4}\)

The additive inverse of -(1/3) = 1/3

Additive inverse (-5/-12) = -5/12

Correct option is (C)  63/4

\(\frac{1575}{100}\) \(=\frac{1575\div25}{100\div25}\) \(=\frac{63}{4}.\)

Correct option is   C) \(\frac{63}{4}\)

It is given that,

\(\frac{7\times 3+2}{3}\) = \(\frac{23}{3}\)metre of rope is Rs. \(\frac{12\times 4+3}{4}\)

Let cost of 1 metre be x

So,

\(x\times \frac{23}{3}\) = \(\frac{51}{4}\)

x = \(\frac{151}{92}\) = \(1\frac{61}{92}\)

Therefore,

Cost of rope is Rs. \(1\frac{61}{92}\) per metre

(D) all of these

Hint:

Additive inverse of 7 is -7

Additive inverse of −5/7 is −5/7

Additive inverse of 0 is 0.

Total Earnings = Rs 160 

Spend on tea and snacks = Rs \(26\frac{3}{5}\) 

Spend on food = Rs \(50\frac{1}{2}\) 

Spend on repairs = Rs \(16\frac{2}{5}\) 

Total Expenditure = Spend on tea and snacks + Spend on food + Spend on repairs

\(=Rs\,26\frac{3}{5}+\,Rs50\frac{1}{2}+\,RS16\frac{2}{5}\)

\(=Rs\,\frac{133}{5}+\,Rs\frac{101}{2}+\,Rs\frac{82}{5}\)

\(=Rs\,\frac{266+505+164}{10}\)

\(=Rs\,\frac{665}{10}\)

\(=Rs\,66\frac{1}{2}\)

Hence, 

Saving \(=Rs\,66\frac{1}{2}\)

Given (-5/-7), (12/-5), (7/4), (13/-9), 0, (-18/-7), (-95/116), (-1/-9)

A rational number is said to be positive if its numerator and denominator are either positive integers or both negative integers.

Therefore positive rational numbers are: (-5/-7), (-18/-7), (7/4), (-1/-9)

A rational number is said to be negative integers if its numerator and denominator are such that one of them is positive integer and another one is a negative integer.

Therefore negative rational numbers are: (12/-5), (13/-9), (-95/116)

(i) The product of two positive rational numbers is always positive.

(ii) The product of two negative rational numbers is always positive.

(iii) If two rational numbers have opposite signs then their product is always negative.

(iv) The reciprocal of a positive rational number is positive and the reciprocal of a negative raitonal number is negative.

(v) Rational number 0 has no reciprocal.

(vi) The product of a rational number and its reciprocal is 1. 

(vii) The numbers 1 and -1 are their own reciprocals. 

(viii)If m is reciprocal of n, then the reciprocal of n is m.

i. True

ii. True

iii. False

iv. True

v. False

True.

Example, let us take -½ is a negative rational number.

Then negative of negative rational number = – (-½)

= ½ (positive rational number)

\(\frac{-5}{16}+\frac{7}{12}\)

LCM of 12 and 16 = 48

\(=\frac{-5\times3+7\times4}{48}\)

\(=\frac{-15+28}{48}\)

\(=\frac{13}{48}\)