Question

(1) If
b
then find the values of the following ratios.
3
a-6
7a+ 9b
7a-95
3a + 3b
Sa-36
2a +36
(iv)
(iii) 6
(ii) 2a - 36​

Answer 1

Question:

If $\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$, then find the values of the given ratios.

Answer:

( i ) $\displaystyle{\boxed{\red{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{22}{13}}}}$

( ii ) $\displaystyle{\boxed{\pink{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}$

( iii ) $\displaystyle{\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}$

( iv ) $\displaystyle{\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}$

Step-by-step-explanation:

( i )

We have given that,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}}$

Now,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

$\displaystyle{\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{5}{3}\:=\:\dfrac{7}{3}\:\times\:\dfrac{5}{3}\:\:\:\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{5}{3}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{5a}{3b}\:=\:\dfrac{35}{9}}$

$\displaystyle{\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{35\:+\:9}{35\:-\:9}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}$

$\displaystyle{\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\cancel{\dfrac{44}{26}}}$

$\displaystyle{\implies\boxed{\red{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{22}{13}}}}$

───────────────────────────

( ii )

We have given that,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

Now,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

$\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}$

$\displaystyle{\implies\sf\:a\:=\:7k\:\:\&\:\:b\:=\:3k}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}}$

Now,

$\displaystyle{\implies\sf\:\dfrac{2\:\times\:(\:7k\:)^2\:+\:3\:\times\:(\:3k\:)^2}{2\:\times\:(\:7k\:)^2\:-\:3\:\times\:(\:3\:k\:)^2}}$

$\displaystyle{\implies\sf\:\dfrac{2\:\times\:49k^2\:+\:3\:\times\:9k^2}{2\:\times\:49k^2\:-\:3\:\times\:9k^2}}$

$\displaystyle{\implies\sf\:\dfrac{98k^2\:+\:27k^2}{98k^2\:-\:27k^2}}$

$\displaystyle{\implies\sf\:\dfrac{125\:\cancel{k^2}}{71\:\cancel{k^2}}}$

$\displaystyle{\implies\boxed{\pink{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}$

───────────────────────────

( iii )

We have given that,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

Now,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

$\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}$

$\displaystyle{\implies\sf\:a\:=\:7k\:\:\&\:\:b\:=\:3k}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{a^3\:-\:b^3}{b^3}}$

Now,

$\displaystyle{\implies\sf\:\dfrac{(\:7k\:)^3\:-\:(\:3k\:)^3}{(\:3k\:)^3}}$

$\displaystyle{\implies\sf\:\dfrac{343\:k^3\:-\:27\:k^3}{27\:k^3}}$

$\displaystyle{\implies\sf\:\dfrac{316\:\cancel{k^3}}{27\:\cancel{k^3}}}$

$\displaystyle{\implies\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}$

───────────────────────────

( iv )

We have given that,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}}$

Now,

$\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$

$\displaystyle{\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{7}{9}\:=\:\dfrac{7}{3}\:\times\:\dfrac{7}{9}\:\:\:\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{7}{9}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{7a}{9b}\:=\:\dfrac{49}{27}}$

$\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{49\:+\:27}{49\:-\:27}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}$

$\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\cancel{\dfrac{76}{22}}}$

$\displaystyle{\implies\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}$

Answer 2

Answer:

If \displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

, then find the values of the given ratios.

Answer:

( i ) \displaystyle{\boxed{\red{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{22}{13}}}}

5a−3b

5a+3b

=

13

22

( ii ) \displaystyle{\boxed{\pink{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}

2a

2

−3b

2

2a

2

+3b

2

=

71

125

( iii ) \displaystyle{\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}

b

3

a

3

−b

3

=

27

316

( iv ) \displaystyle{\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}

7a−9b

7a+9b

=

11

38

Step-by-step-explanation:

( i )

We have given that,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

We have to find the value of

\displaystyle{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}}

5a−3b

5a+3b

Now,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

\displaystyle{\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{5}{3}\:=\:\dfrac{7}{3}\:\times\:\dfrac{5}{3}\:\:\:\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{5}{3}\:\right]}⟹

b

a

×

3

5

=

3

7

×

3

5

−−[Multiplyingbothsidesby

3

5

]

\displaystyle{\implies\sf\:\dfrac{5a}{3b}\:=\:\dfrac{35}{9}}⟹

3b

5a

=

9

35

\displaystyle{\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{35\:+\:9}{35\:-\:9}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}⟹

5a−3b

5a+3b

=

35−9

35+9

−−[ByComponendo−Dividendo]

\displaystyle{\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\cancel{\dfrac{44}{26}}}⟹

5a−3b

5a+3b

=

26

44

\displaystyle{\implies\boxed{\red{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{22}{13}}}}⟹

5a−3b

5a+3b

=

13

22

───────────────────────────

( ii )

We have given that,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

Now,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}⟹

7

a

=

3

b

=k−−−[Bycrossmultiplication]

\displaystyle{\implies\sf\:a\:=\:7k\:\:\&\:\:b\:=\:3k}⟹a=7k&b=3k

We have to find the value of

\displaystyle{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}}

2a

2

−3b

2

2a

2

+3b

2

Now,

\displaystyle{\implies\sf\:\dfrac{2\:\times\:(\:7k\:)^2\:+\:3\:\times\:(\:3k\:)^2}{2\:\times\:(\:7k\:)^2\:-\:3\:\times\:(\:3\:k\:)^2}}⟹

2×(7k)

2

−3×(3k)

2

2×(7k)

2

+3×(3k)

2

\displaystyle{\implies\sf\:\dfrac{2\:\times\:49k^2\:+\:3\:\times\:9k^2}{2\:\times\:49k^2\:-\:3\:\times\:9k^2}}⟹

2×49k

2

−3×9k

2

2×49k

2

+3×9k

2

\displaystyle{\implies\sf\:\dfrac{98k^2\:+\:27k^2}{98k^2\:-\:27k^2}}⟹

98k

2

−27k

2

98k

2

+27k

2

\displaystyle{\implies\sf\:\dfrac{125\:\cancel{k^2}}{71\:\cancel{k^2}}}⟹

71

k

2

125

k

2

\displaystyle{\implies\boxed{\pink{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}⟹

2a

2

−3b

2

2a

2

+3b

2

=

71

125

───────────────────────────

( iii )

We have given that,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

Now,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}⟹

7

a

=

3

b

=k−−−[Bycrossmultiplication]

\displaystyle{\implies\sf\:a\:=\:7k\:\:\&\:\:b\:=\:3k}⟹a=7k&b=3k

We have to find the value of

\displaystyle{\sf\:\dfrac{a^3\:-\:b^3}{b^3}}

b

3

a

3

−b

3

Now,

\displaystyle{\implies\sf\:\dfrac{(\:7k\:)^3\:-\:(\:3k\:)^3}{(\:3k\:)^3}}⟹

(3k)

3

(7k)

3

−(3k)

3

\displaystyle{\implies\sf\:\dfrac{343\:k^3\:-\:27\:k^3}{27\:k^3}}⟹

27k

3

343k

3

−27k

3

\displaystyle{\implies\sf\:\dfrac{316\:\cancel{k^3}}{27\:\cancel{k^3}}}⟹

27

k

3

316

k

3

\displaystyle{\implies\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}⟹

b

3

a

3

−b

3

=

27

316

───────────────────────────

( iv )

We have given that,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

We have to find the value of

\displaystyle{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}}

7a−9b

7a+9b

Now,

\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}

b

a

=

3

7

\displaystyle{\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{7}{9}\:=\:\dfrac{7}{3}\:\times\:\dfrac{7}{9}\:\:\:\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{7}{9}\:\right]}⟹

b

a

×

9

7

=

3

7

×

9

7

−−[Multiplyingbothsidesby

9

7

]

\displaystyle{\implies\sf\:\dfrac{7a}{9b}\:=\:\dfrac{49}{27}}⟹

9b

7a

=

27

49

\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{49\:+\:27}{49\:-\:27}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}⟹

7a−9b

7a+9b

=

49−27

49+27

−−[ByComponendo−Dividendo]

\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\cancel{\dfrac{76}{22}}}⟹

7a−9b

7a+9b

=

22

76

[tex]\displaystyle{\implies\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}⟹

7a−9b

7a+9b

=

11

38