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प्रश्न
Father's age is three times the sum of age of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
उत्तर
Let the present age of father be x years and the present ages of his two children’s be y and zyears.
The present age of father is three times the sum of the ages of the two children’s. Thus, we have
`x=3(y+2)`
`⇒ y+z=x/5`
After 5 years, father’s age will be (x+5) years and the children’s age will be (y+5) and (z+5) years. Thus using the given information, we have
`x+5 =2 {(y+5)+(z+5)}`
`⇒ x+5 =2 (y+5+z+5)`
`⇒ x = 2(y+z)+20-5`
`⇒ x = 2 (y+z)+15`
So, we have two equations
`y+z =x/3`
`x=2(y+z)+15`
Here x, y and z are unknowns. We have to find the value of x.
Substituting the value of (y+z) from the first equation in the second equation, we have
By using cross-multiplication, we have
`x = (2x)/3+15`
`⇒ x=(2x)/3=15`
`⇒ x(1-2/3)=15`
`⇒ x/3=15`
`⇒ x= 15xx3`
`⇒ x =45`
Hence, the present age of father is 45 years.
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संबंधित प्रश्न
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If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes `6/5`. Thus, we have
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`⇒ 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`
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