Math, asked by moynal, 1 year ago

a fraction becomes 1 by 3 if 1 is subtracted from both its numerator and denominator if 1 is added to both numerator and denominator it becomes 1 by 2 find the fraction

Answers

Answered by rawatsukriti891
4

Answer:

Let the fraction be x/y

x-1/y-1=1/3

3x-3=y-1

3x-y=-1+3

3x-y=2..............eq 1

x+1/y+1=1/2

2x+2=y+1

2x-y=1-2

2x-y=-1............eq2

By subtracting eq2 from eq1

3x-y=2

2x-y=-1

-. +. +

x=1.........

Put value of x in eq1

3(1)-y=2

3-y=3

6=y..........

Therefore, the fraction is 1/6

Here's ur answer!!!!!!


moynal: Thank you
rawatsukriti891: Wlc
PRAJYOT2123X: your answer is wrong
Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
4

\huge\sf\pink{Answer}

\sf Original\: fraction=\dfrac{3}{7}

\rule{110}1

\huge\sf\blue{Given}

✭ A fraction becomes ⅓ if 1 is substracted from both its numerator and denominator

✭ If 1 is added to the both the numerator and denominator, it becomes ½

\rule{110}1

\huge\sf\gray{To \:Find}

◈ The original fraction?

\rule{110}1

\huge\sf\purple{Steps}

Let us Assume that,

Numerator = x

Denominator = y

\underline{\textsf{As Per the Question}}

Case 1 :

\sf{\dfrac{x-1}{y-1}=\dfrac{1}{3}}

\sf{3x-3=y-1}

\sf{3x=y+2}

\sf{y=3x-2\qquad -eq(1)}

Case 2 :

If 1 is added to the both the numerator and denominator , it becomes 1/2.

\sf{\dfrac{x+1}{y+1}=\dfrac{1}{2}}

\sf{2x+2=y+1}

\sf{2x+2=3x-2+1\quad\bigg\lgroup Put\:y=3x-2\: from\:eq(1)\bigg\rgroup}

\sf{2x-3x=-2+1-2}

\sf{-x=-3}

\orange{\sf{x=3}}

Now , put x = 3 in eq(1)

»» \sf{y=3x-2}

»» \sf{y=3\times\:3-2}

»» \orange{\sf{y=7}}

\rule{170}3

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