Question

A positive fraction is such that the denominator is five more than the numerator. Find the fraction
if the sum of squares of the numerator and denominator is 73.​

Answer 1

A n s w e r

$\:\:$

G i v e n

$\:\:$

• In a positive fraction denominator is five more than the numerator

• Sum of squares of the numerator and denominator is 73

$\:\:$

F i n d

$\:\:$

• The fraction

$\:\:$

S o l u t i o n

$\:\:$

• Let the numerator of the fraction be 'n'

• Let the denominator of the fraction be 'd'

$\:\:$

$\underline{ \underline{\bold{\texttt{Original fraction :}}}}$

$\:\:$

➠ $\sf \dfrac { n } { d }$ ⚊⚊⚊⚊ ⓵

$\:\:$

Given that , denominator is five more than the numerator

$\:\:$

Thus ,

$\:\:$

: ➜ d = n + 5 ⚊⚊⚊⚊ ⓶

$\:\:$

Also given that , Sum of squares of the numerator and denominator is 73

$\:\:$

: ➜ n² + d² = 73 ⚊⚊⚊⚊ ⓷

$\:\:$

⟮ Putting d = n + 5 from ⓶ to ⓷ ⟯

$\:\:$

: ➜ n² + d² = 73

$\:\:$

: ➜ n² + (n + 5)² = 73

$\:\:$

: ➜ n² + n² + 25 + 10n = 73

$\:\:$

: ➜ 2n² + 10n + 25 - 73 = 0

$\:\:$

: ➜ 2n² + 10n - 48 = 0

$\:\:$

⟮ Dividing the above equation by 2 ⟯

$\:\:$

: ➜ $\sf \dfrac { 2n² + 10n - 48 } { 2 } = \dfrac { 0 } { 2 }$

$\:\:$

: ➜ n² + 5n - 24 = 0

$\:\:$

: ➜ n² + 8n - 3n -24 = 0

$\:\:$

: ➜ n(n + 8) - 3(n + 8) = 0

$\:\:$

: ➜ (n - 3)(n + 8) = 0

$\:\:$

• n = 3
• n = -8

$\:\:$

Given that the fraction need to be positive

$\:\:$

Thus ,

$\:\:$

: ➜ n = 3 ⚊⚊⚊⚊ ⓸

$\:\:$

• Hence the numerator of fraction is 3

$\:\:$

⟮ Putting n = 3 from ⓸ to ⓶ ⟯

$\:\:$

: ➜ d = n + 5

$\:\:$

: ➜ d = 3 + 5

$\:\:$

: ➜ d = 8 ⚊⚊⚊⚊ ⓹

$\:\:$

• Hence the denominator of fraction is 8

$\:\:$

⟮ Putting n = 3 from ⓸ & d = 8 from ⓹ to ⓵ ⟯

$\:\:$

: ➜ $\sf \dfrac { n } { d }$

$\:\:$

: : ➨ $\sf \dfrac { 3} { 8}$

$\:\:$

• $\sf Hence \: the \: original \: fraction \: is \: \sf \dfrac { 3} { 8}$

Answer 2

S o l u t i o n

$\:\:$

• Let the numerator of the fraction be 'n'

• Let the denominator of the fraction be 'd'

$\:\:$

$\underline{ \underline{\bold{\texttt{Original fraction :}}}}$

$\:\:$

➠ $\sf \dfrac { n } { d }$ ⚊⚊⚊⚊ ⓵

$\:\:$

Given that , denominator is five more than the numerator

$\:\:$

Thus ,

$\:\:$

: ➜ d = n + 5 ⚊⚊⚊⚊ ⓶

$\:\:$

Also given that , Sum of squares of the numerator and denominator is 73

$\:\:$

: ➜ n² + d² = 73 ⚊⚊⚊⚊ ⓷

$\:\:$

⟮ Putting d = n + 5 from ⓶ to ⓷ ⟯

$\:\:$

: ➜ n² + d² = 73

$\:\:$

: ➜ n² + (n + 5)² = 73

$\:\:$

: ➜ n² + n² + 25 + 10n = 73

$\:\:$

: ➜ 2n² + 10n + 25 - 73 = 0

$\:\:$

: ➜ 2n² + 10n - 48 = 0

$\:\:$

⟮ Dividing the above equation by 2 ⟯

$\:\:$

: ➜ $\sf \dfrac { 2n² + 10n - 48 } { 2 } = \dfrac { 0 } { 2 }$

$\:\:$

: ➜ n² + 5n - 24 = 0

$\:\:$

: ➜ n² + 8n - 3n -24 = 0

$\:\:$

: ➜ n(n + 8) - 3(n + 8) = 0

$\:\:$

: ➜ (n - 3)(n + 8) = 0

$\:\:$

n = 3

n = -8

$\:\:$

Given that the fraction need to be positive

$\:\:$

Thus ,

$\:\:$

: ➜ n = 3 ⚊⚊⚊⚊ ⓸

$\:\:$

Hence the numerator of fraction is 3

$\:\:$

⟮ Putting n = 3 from ⓸ to ⓶ ⟯

$\:\:$

: ➜ d = n + 5

$\:\:$

: ➜ d = 3 + 5

$\:\:$

: ➜ d = 8 ⚊⚊⚊⚊ ⓹

$\:\:$

Hence the denominator of fraction is 8

$\:\:$

⟮ Putting n = 3 from ⓸ & d = 8 from ⓹ to ⓵ ⟯

$\:\:$

: ➜ $\sf \dfrac { n } { d }$

$\:\:$

: : ➨ $\sf \dfrac { 3} { 8}$

$\:\:$

$\sf Hence \: the \: original \: fraction \: is \: \sf \dfrac { 3} { 8}$