The difference between two positive integers is 30. If the ratio of these integers is 2 : 5, find the integers.
Answer
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Hint: The concept of ratios and linear equations will be applied in such problems. We will assume two variables for the two integers. We will now try to form two linear equations to solve their values. The first equation will be formed using the fact that their difference is 30, and the second will be formed using their ratios.
Complete step-by-step answer:
Let the two positive integers be x and y, where x > y. We will try and form two equations in x and y according to the given information, and then determine their values. Firstly, we have been given that the difference of these integers is 30, so we can write that-
$\,x - y = 30$...(1)
Also, we have been given that the ratio of these two integers is 2 : 5. The ratio can be also be written in the form of a fraction as-
$2:5 = \dfrac{2}{5}$
We have assumed that x > y, so their ratio can be written as-
$\dfrac{y}{x} = \dfrac{2}{5}$
$Multiplying\;both\;sides\;by\;x,$
$y = \dfrac{{2x}}{5}...\left( 2 \right)$
Substituting the value of y from equation (2) in equation (1) we get-
$x - y = 30$
$x - \dfrac{{2x}}{5} = 30$
$Taking\;5\;as\;LCM\;in\;denominator,$
$\dfrac{{5x - 2x}}{5} = 30$
$3x = 30 \times 5 = 150$
$x = \dfrac{{150}}{3} = 50$
We have determined the value of x, so the corresponding value of y from equation (1) is-
$x - y = 30$
$y = x - 30 = 50 - 30 = 20$
The required values of x and y are 50 and 20 respectively. This is the answer.
Note: In this question, it is very important to make the assumption that x > y, and most of the students neglect this. If we do not assume this, we may end up getting the wrong answer. This is because in the second equation, we may assume that-
$\dfrac{x}{y} = \dfrac{2}{5}$
$x = \dfrac{2}{5}y$
On substituting this value in equation (1), we get-
$x - y = 30$
$\dfrac{2}{5}y - y = 30$
$ - \dfrac{{3y}}{5} = 30$
$y = - 50$
$x = 30 + y = - 20$
Here we obtained negative values of x and y, which also satisfy our equations. But in the question,we have been asked to find the positive integers, which makes this solution wrong. Hence, we should take care about such mistakes.
Complete step-by-step answer:
Let the two positive integers be x and y, where x > y. We will try and form two equations in x and y according to the given information, and then determine their values. Firstly, we have been given that the difference of these integers is 30, so we can write that-
$\,x - y = 30$...(1)
Also, we have been given that the ratio of these two integers is 2 : 5. The ratio can be also be written in the form of a fraction as-
$2:5 = \dfrac{2}{5}$
We have assumed that x > y, so their ratio can be written as-
$\dfrac{y}{x} = \dfrac{2}{5}$
$Multiplying\;both\;sides\;by\;x,$
$y = \dfrac{{2x}}{5}...\left( 2 \right)$
Substituting the value of y from equation (2) in equation (1) we get-
$x - y = 30$
$x - \dfrac{{2x}}{5} = 30$
$Taking\;5\;as\;LCM\;in\;denominator,$
$\dfrac{{5x - 2x}}{5} = 30$
$3x = 30 \times 5 = 150$
$x = \dfrac{{150}}{3} = 50$
We have determined the value of x, so the corresponding value of y from equation (1) is-
$x - y = 30$
$y = x - 30 = 50 - 30 = 20$
The required values of x and y are 50 and 20 respectively. This is the answer.
Note: In this question, it is very important to make the assumption that x > y, and most of the students neglect this. If we do not assume this, we may end up getting the wrong answer. This is because in the second equation, we may assume that-
$\dfrac{x}{y} = \dfrac{2}{5}$
$x = \dfrac{2}{5}y$
On substituting this value in equation (1), we get-
$x - y = 30$
$\dfrac{2}{5}y - y = 30$
$ - \dfrac{{3y}}{5} = 30$
$y = - 50$
$x = 30 + y = - 20$
Here we obtained negative values of x and y, which also satisfy our equations. But in the question,we have been asked to find the positive integers, which makes this solution wrong. Hence, we should take care about such mistakes.
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