1) The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.
Answers & Comments
Answer:
Explanation:
Given that,
So,
Let numerator be 'x' and denominator be 'y'
Then,
→ y = x + 8 ....equation(1)
Also given that,
so,
→ (x + 17)/(y - 1) = 3/2
[cross multiplying]
→ 2x + 34 = 3y - 3
→ 2x - 3y = -37
[Using equation(1)]
→ 2x - 3(x + 8) = -37
→ 2x - 3x -24 = -37
→ -x = -13
→ x = 13
Putting value of x in equation (1)
→ y = x + 8
→ y = 13 + 8
→ y = 21
Therefore,
The required rational number will be, x/y = 13/21.
Answer:
Given :-
To Find :-
Solution :-
Let, the numerator be x
And, the denominator will be x + 8
Then, the number is![\sf\dfrac{x}{x + 8} \sf\dfrac{x}{x + 8}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%7D%7Bx%20%2B%208%7D)
According to the question,
⇒![\sf\dfrac{x + 17}{x + 8 - 1} =\: \dfrac{3}{2} \sf\dfrac{x + 17}{x + 8 - 1} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%20%2B%2017%7D%7Bx%20%2B%208%20-%201%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
⇒![\sf\dfrac{x + 17}{x + 7} =\: \dfrac{3}{2} \sf\dfrac{x + 17}{x + 7} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%20%2B%2017%7D%7Bx%20%2B%207%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
By doing cross multiplication we get,
⇒![\sf 3(x + 7) = 2(x + 17) \sf 3(x + 7) = 2(x + 17)](https://tex.z-dn.net/?f=%5Csf%203%28x%20%2B%207%29%20%3D%202%28x%20%2B%2017%29)
⇒![\sf 3x + 21 = 2x + 34 \sf 3x + 21 = 2x + 34](https://tex.z-dn.net/?f=%5Csf%203x%20%2B%2021%20%3D%202x%20%2B%2034)
⇒![\sf 3x - 2x = 34 - 21 \sf 3x - 2x = 34 - 21](https://tex.z-dn.net/?f=%5Csf%203x%20-%202x%20%3D%2034%20-%2021)
➠![\sf x =\: 13 \sf x =\: 13](https://tex.z-dn.net/?f=%5Csf%20x%20%3D%5C%3A%2013)
Hence, the required rational number are,
↦![\sf\dfrac{x}{x + 8} \sf\dfrac{x}{x + 8}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%7D%7Bx%20%2B%208%7D)
↦![\sf\dfrac{13}{13 + 8} \sf\dfrac{13}{13 + 8}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B13%7D%7B13%20%2B%208%7D)
➦![\sf\green{\dfrac{13}{21}} \sf\green{\dfrac{13}{21}}](https://tex.z-dn.net/?f=%5Csf%5Cgreen%7B%5Cdfrac%7B13%7D%7B21%7D%7D)
Let's Verify :-
⇒![\sf\dfrac{x + 17}{x + 8 - 1} =\: \dfrac{3}{2} \sf\dfrac{x + 17}{x + 8 - 1} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%20%2B%2017%7D%7Bx%20%2B%208%20-%201%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
⇒![\sf\dfrac{x + 17}{x + 7} =\: \dfrac{3}{2} \sf\dfrac{x + 17}{x + 7} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7Bx%20%2B%2017%7D%7Bx%20%2B%207%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
Put x = 13 we get,
⇒![\sf\dfrac{13 + 17}{13 + 7} =\: \dfrac{3}{2} \sf\dfrac{13 + 17}{13 + 7} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B13%20%2B%2017%7D%7B13%20%2B%207%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
⇒![\sf\dfrac{30}{20} =\: \dfrac{3}{2} \sf\dfrac{30}{20} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B30%7D%7B20%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
⇒![\sf\dfrac{\cancel{30}}{\cancel{20}} =\: \dfrac{3}{2} \sf\dfrac{\cancel{30}}{\cancel{20}} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B%5Ccancel%7B30%7D%7D%7B%5Ccancel%7B20%7D%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
⇒![\sf\dfrac{3}{2} =\: \dfrac{3}{2} \sf\dfrac{3}{2} =\: \dfrac{3}{2}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B3%7D%7B2%7D%20%3D%5C%3A%20%5Cdfrac%7B3%7D%7B2%7D)
➦ LHS= RHS
➲ Hence, Verified ✔