Question

The numerator of a fraction is 2 less than its denominator. If 3 is subtracted from numerator and from the denominator, the new fraction obtained is 3/4. Find fraction

Answer 1

$\Large{\underline{\underline{\bf{GIVEN:-}}}}$

• The numerator of a fraction is 2 less than its denominator.
• If 3 is subtracted from numerator and from the denominator, the new fraction obtained is 3/4.

$\Large{\underline{\underline{\bf{TO \: FIND:-}}}}$

• What is the fraction ?

$\Large{\underline{\underline{\bf{SOLUTION:-}}}}$

Let the numerator of the fraction be 'x' and Denominator be 'y'

• FRACTION = x/y

CASE:- 1)

◆ The numerator of a fraction is 2 less than its denominator.

According to question:-

➜ x = y –2....❶

CASE:- 2)

◆ If 3 is subtracted from numerator and from the denominator, the new fraction obtained is 3/4.

According to question:-

➜ $\sf{\dfrac{x -3}{y-3} = \dfrac{3}{4}}$

Use cross product

➜ 4(x–3) = 3(y–3)

➜ 4x –12 = 3y –9

➜ 4x –3y = –9+12

➜ 4x –3y = 3....❷ 

Put the value of 'x' from equation 1) in equation 2)

➜ 4(y–2) –3y = 3

➜ 4y –8 –3y = 3

➜ y = 3+8

⠀⠀⠀⠀  ❛ y = 11 ❜

Put the value of 'y' in equation 1)

➜ x = 11 –2

⠀⠀⠀⠀  ❛ x = 9 ❜

• NUMERATOR = x = 9
• DENOMINATOR = y = 11
• FRACTION = 9/11

❝ Hence, the fraction formed is 9/11 ❞

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Answer 2

QUESTION:-

✯ᴛʜᴇ ɴᴜᴍᴇʀᴀᴛᴏʀ ᴏғ ᴀ ғʀᴀᴄᴛɪᴏɴ ɪs 2 ʟᴇss ᴛʜᴀɴ ɪᴛs ᴅᴇɴᴏᴍɪɴᴀᴛᴏʀ. ɪғ 3 ɪs sᴜʙᴛʀᴀᴄᴛᴇᴅ ғʀᴏᴍ ɴᴜᴍᴇʀᴀᴛᴏʀ ᴀɴᴅ ғʀᴏᴍ ᴛʜᴇ ᴅᴇɴᴏᴍɪɴᴀᴛᴏʀ, ᴛʜᴇ ɴᴇᴡ ғʀᴀᴄᴛɪᴏɴ ᴏʙᴛᴀɪɴᴇᴅ ɪs 3/4. ғɪɴᴅ ғʀᴀᴄᴛɪᴏɴ

ANSWER✔

$\Large\underline\bold{GIVEN,}$

$\sf\dashrightarrow numerator\:is\:less\:than\:the\:denominator\:by\:2$

$\sf\dashrightarrow the\:new\:fraction\:obtained\:is\: \dfrac{3}{4}\:when\:3\:is\: subtracted\:from\:numerator\:and\:denominator$

$\Large\underline\bold{TO\:FIND,}$

$\sf\large\dashrightarrow THE\:FRACTION$

$\Large\underline\bold{SOLUTION,}$

$\sf\therefore let\:the\:numerator\:be\:'a'\:and\:denominator\:be\:'b'$

$\sf\therefore \dfrac{numerator}{denominator} = \dfrac{a}{b}$

$\sf\therefore taking\:two\:cases,$

$\sf\therefore in\:case1\:we\:will\:find\:the\:value\:for\: substituting\:in\:equations\:$

$\sf\therefore in\:case2\:we\:will\:find\:the\:value\:of\:numerators\:and\:denominators\:with\:the\:help\:of\:1_{st}case$

$\Large {\fbox { CASE:-1}}$

$\sf\therefore numerator\:is\:less\:than\:the\:denominator\:by\:2$

ACCORDING TO THE QUESTION,

$\sf\therefore a=b-2.......eq^1$

$\sf{\boxed{\sf{a=b-2}}}$

$\Large {\fbox { CASE:-2}}$

$\sf\dashrightarrow the\:new\:fraction\:obtained\:is\: \dfrac{3}{4}\:when\:3\:is\:subtracted\:from\:numerator\:and\:denominator$

THEREFORE,

ACCORDING TO THE QUESTION,

$\sf\therefore \dfrac{a -3}{b-3} = \dfrac{3}{4}$

$\sf\implies \dfrac{a -3}{b-3} = \dfrac{3}{4}$

$\sf\implies 4(a-3) = 3(b-3)$

$\sf\implies 4a -12 = 3b -9$

$\sf\implies 4a -3b = (-9)+12$

$\sf\implies 4a -3b = 3......eq^2$

$\sf{\boxed{\sf{4a -3b = 3......eq^2}}}$

$\sf\large\therefore substituting\:the\:value\:of\:'a'\:in\:eq^2$

$\sf\therefore a=b-2$

$\sf\implies 4(b-2) -3b = 3$

$\sf\implies 4b -8 -3b = 3$

$\sf\implies 4b-3b=3+8$

$\sf\implies b=11$

$\large{\boxed{\sf{b=11}}}$

NOW,

$\sf\therefore substituting\:the\:value\:of\:b\:in\:eq^1$

$\sf\therefore a=b-2$

$\sf\implies a=11-2$

$\sf\therefore a=9$

$\large{\boxed{\sf{a=9}}}$

$\sf\large\therefore the\:required\:fraction\:is,$

$\sf\large\therefore \dfrac{numerator}{debominator} = \dfrac{a}{b} = \dfrac{9}{11}$

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