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प्रश्न
A fraction becomes `(1)/(3)` when 2 is subtracted from the numerator and it becomes `(1)/(2)` when 1 is subtracted from the denominator. Find the fraction.
उत्तर
Let's assume the fraction be `("x")/("y")`
1st condition:
`("x" -2)/("y") = (1)/(3)`
⇒ `3"x "- 6 = "y"`
⇒ `3"x" - "y" = 6` ...(1)
2nd condition:
`("x")/("y"-1) = (1)/(2)`
⇒ `2"x" - "y" - 1`
⇒ `2"x" - "y" = -1`
Using elimination method:
Multiplying (2) by -1 and then adding (1) and (2)
⇒ `3"x" - "y" = 6`
⇒ `-2"x" + "y" =1`
⇒ `"x" = 7`
Now, from (1)
⇒ `3"x" - "y" = 6`
⇒ `3 (7) - "y" = 6`
⇒ `21 - "y" = 6`
⇒ `"y" = 15`
Hence, the required fraction is `(7)/(15)`
संबंधित प्रश्न
Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:
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5x + 7y = 17 ; 3x - 2y = 4
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4x + y = 34 ; x + 4y = 16
The difference between an angle and its complement is 10° find measure of the larger angle.
If 52x + 65y = 183 and 65x + 52y = 168, then find x + y = ?
The sum of the two-digit number and the number obtained by interchanging the digits is 132. The digit in the ten’s place is 2 more than the digit in the unit’s place. Complete the activity to find the original number.
Activity: Let the digit in the unit’s place be y and the digit in the ten’s place be x.
∴ The number = 10x + y
∴ The number obtained by interchanging the digits = `square`
∴ The sum of the number and the number obtained by interchanging the digits = 132
∴ 10x + y + 10y + x = `square`
∴ x + y = `square` .....(i)
By second condition,
Digit in the ten’s place = digit in the unit’s place + 2
∴ x – y = 2 ......(ii)
Solving equations (i) and (ii)
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The ratio of two numbers is 2:3. If 5 is added in each numbers, then the ratio becomes 5:7 find the numbers.
The ratio of two numbers is 2:3.
So, let the first number be 2x and the second number be `square`.
From the given condition,
`((2x) + square)/(square + square) = square/square`
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