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प्रश्न
The sum of two numbers is 8. If their sum is four times their difference, find the numbers.
उत्तर
Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.
The sum of the two numbers is 8. Thus, we have ` x+y =8`
The sum of the two numbers is four times their difference. Thus, we have
` x + y = 4( x -y)`
` ⇒ x + y = 4x - 4y = 0`
`⇒ 4x - 4y - x - y =0`
` ⇒ 3 x -5 y = 0`
So, we have two equations
` x + y = 8`
`3x -5y = 0`
Here x and y are unknowns. We have to solve the above equations for x and y.
Multiplying the first equation by 5 and then adding with the second equation, we have
`5(x + y)+ (3x - 5y)= 5 xx 8 + 0 `
` ⇒ 8 x = 40 `
`⇒ x = 40/8`
`⇒ x = 5`
Substituting the value of x in the first equation, we have
` 5 + y = 8`
`⇒ y = 8 - 5 `
`⇒ y =3`
Hence, the numbers are 5 and 3.
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संबंधित प्रश्न
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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`
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
