
Two numbers are in the ratio of 2:5. If they differ by 15, find the numbers.
$
{\text{A}}{\text{. 25 and 10}} \\
{\text{B}}{\text{. }} - {\text{25 and 10}} \\
{\text{C}}{\text{. }}\dfrac{1}{{25}}{\text{ and }}\dfrac{1}{{10}} \\
{\text{D}}{\text{. None of the above }} \\
$
Answer
602.1k+ views
Hint- Here, we will proceed by assuming the two numbers as x and y and then according to the problem statements we will obtain two equations which will be solved by using the substitution method.
“Complete step-by-step answer:”
Let the two numbers be x and y where x is the smaller number and y is the larger number.
Given the difference between the two numbers is 15 i.e., $
y - x = 15 \\
\Rightarrow y = 15 + x{\text{ }} \to {\text{(1)}} \\
$
Also, the ratio of these two numbers is 2:5 i.e., $\dfrac{x}{y} = \dfrac{2}{5}$
By cross multiplying the above equation, we get
$5x = 2y{\text{ }} \to {\text{(2)}}$
Putting the value of y from equation (1), equation (2) becomes
Put x=10 in equation (1), the value of y is given by
So, the assumed two numbers are 10 and 25.
Hence, option A is correct.
Note- In this particular problem, we have developed two equations in two variables which can be easily solved by substitution method (used in the above solution) or elimination method (in this method we make the coefficient of all the variables same except one variable whose value will be eventually calculated by performing algebra to the given equation).
“Complete step-by-step answer:”
Let the two numbers be x and y where x is the smaller number and y is the larger number.
Given the difference between the two numbers is 15 i.e., $
y - x = 15 \\
\Rightarrow y = 15 + x{\text{ }} \to {\text{(1)}} \\
$
Also, the ratio of these two numbers is 2:5 i.e., $\dfrac{x}{y} = \dfrac{2}{5}$
By cross multiplying the above equation, we get
$5x = 2y{\text{ }} \to {\text{(2)}}$
Putting the value of y from equation (1), equation (2) becomes
Put x=10 in equation (1), the value of y is given by
So, the assumed two numbers are 10 and 25.
Hence, option A is correct.
Note- In this particular problem, we have developed two equations in two variables which can be easily solved by substitution method (used in the above solution) or elimination method (in this method we make the coefficient of all the variables same except one variable whose value will be eventually calculated by performing algebra to the given equation).
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