The denominator of a common fraction exceeds the numerator by $7$ . If $3$ is deducted from the numerator as well as the denominator, the function $\frac{2}{3}$ is obtained. Find the original common fraction.

Open in App

Solution

$Letthenumeratorbexthen,denomenator=x+7originalfraction=xx+7afterdeducting3frombothnumeatoranddenominator⇒x−3x+7−3=23⇒x−3x+4=23⇒3x−9=2x+8⇒x=17originalfraction=1717+7=1724$ .

Suggest Corrections

Similar questions

Q. If

3

is subtracted from the numerator as well as the denominator of a fraction, the reduced form of the new fraction so obtained is

\frac{4}{9}

. If

3

is added to the numerator as well as the denominator of the same fraction, the reduced form of the new fraction so obtained is

\frac{2}{3}

. Find the original fraction.

The denominator of a fraction is 3 more than the numerator. If the numerator as well as the denominator is increased by 4, the fraction becomes $\frac{4}{5}$ . What was the original fraction?

If 2 and 3 are respectively added to the numerator and the denominator of a fraction, the fraction becomes $\frac{7}{9}$ and if both the numerator as well as the denominator is decreased by 1, the fraction becomes $\frac{4}{5}$ . What is the original fraction?

Q. The numerator of a fraction is 3 less than its denominator. If the numerator is tripled and the denominator is increased by 2, the value of the fraction obtained is

\frac{1}{2}

. What was the original fraction?

Q. The denominator of a fraction is greater than its numerator by 3. If 3 is subtracted from the numerator and 2 added to the denominator, the value of the fraction becomes

\frac{1}{5}

. Find the original fraction.

Join BYJU'S Learning Program

The denominator of a common fraction exceeds the numerator by 7. If 3 is deducted from the numerator as well as the denominator, the function 23 is obtained. Find the original common fraction.

Letthenumeratorbexthen,denomenator=x+7originalfraction=xx+7afterdeducting3frombothnumeatoranddenominator⇒x−3x+7−3=23⇒x−3x+4=23⇒3x−9=2x+8⇒x=17originalfraction=1717+7=1724.

The denominator of a common fraction exceeds the numerator by $7$ . If $3$ is deducted from the numerator as well as the denominator, the function $\frac{2}{3}$ is obtained. Find the original common fraction.

$Letthenumeratorbexthen,denomenator=x+7originalfraction=xx+7afterdeducting3frombothnumeatoranddenominator⇒x−3x+7−3=23⇒x−3x+4=23⇒3x−9=2x+8⇒x=17originalfraction=1717+7=1724$ .