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प्रश्न
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
उत्तर
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have
`x+y =2x+4`
`⇒ 2x +4 -x -y=0`
`⇒ x - y + 4=0`
If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have
` x+ 3 : y+ 3 = 2:3`
`⇒ (x+3)/(y+3)=2/3`
`⇒ 3(x+3)=2/3`
`⇒ 3x + 9 =2y+6`
`⇒ 3x -2y +3=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-1)xx3-(-2)xx4)=(-y)/(1xx3-3xx4)=(1)/(1xx(-2)-3xx(-1))`
`⇒ x/(-3+8)=(-y)/(3-12)=1/(-2+3)`
`⇒ x/5=(-5)/(-9)=1/1`
`⇒ x/5 = y/9=1`
`⇒ x=5,y=9`
Hence, the fraction is `5/9`
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Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes `6/5`. Thus, we have
`(2x)/(y-5)=6/5`
`⇒ 10x=6(y-5)`
`⇒ 10x=6y-30`
`⇒ 10x-6y+30 =0`
`⇒ 2(5x-3y+15)=0`
`⇒ 5x - 3y+15=0`
If the denominator is doubled and the numerator is increased by 8, the fraction becomes `2/5`. Thus, we have
`(x+8)/(2y)=2/5`
`⇒ 5(x+8)=4y`
`⇒ 5x+40=4y`
`⇒ 5x-4y+40=0`
So, we have two equations
`5x-3y+15=0`
`5x-4y+40=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-3)xx40-(-4)xx15)=-y/(5xx40-5xx15)=1/(5xx(-4)-5xx(-3))`
`⇒ x/(-120+60)=(-y)/(200-75)=1/(-20+15)`
`⇒x/(-60)=-y/125``=1/-5`
`⇒ x= 60/5,y=125/5`
`⇒ x=12,y=25`
Hence, the fraction is `12/25`
