The difference of two numbers is \[1365\]. On dividing the larger number by the smaller, we get \[6\] as quotient and the 15 as remainder. What is the smaller number?
A) 240
B) 270
C) 295
D) 360
Answer
581.4k+ views
Hint:
Here, we have to find the smaller number. We will assign variables for the smaller and the larger number. Then we will form linear equations in two variables from the given conditions. We will equate both the equation and solve it further to find the solution of the variables which is the smaller and the larger number.
Complete step by step solution:
Let \[x\] and \[y\] be the smaller number and larger number respectively.
We are given that the difference of two numbers is \[1365\]. Therefore,
\[y - x = 1365\]
By rewriting the equation, we get
\[ \Rightarrow y = x + 1365\] ………………………………………………………………………………\[\left( 1 \right)\]
We are given that on dividing the larger number by the smaller, we get \[6\] as quotient and the 15 as remainder.
By division algorithm, we get
\[y = 6x + 15\] …………………………………………………………………………………\[\left( 2 \right)\]
By equating the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get
\[x + 1365 = 6x + 15\]
By rewriting the equations, we get
\[ \Rightarrow 6x - x = 1365 - 15\]
Subtracting the like terms, we get
\[ \Rightarrow 5x = 1350\]
Dividing by 5 on both the sides of the equation, we get
\[ \Rightarrow x = \dfrac{{1350}}{5}\]
\[ \Rightarrow x = 270\]
By substituting \[x = 270\] in the equation \[\left( 1 \right)\], we get
\[y = x + 1365\]
On adding the terms, we get
\[ \Rightarrow y = 270 + 1365\]
\[ \Rightarrow y = 1635\]
Thus the smaller number is 270 and the largest number is 1635.
Therefore, the smaller number is 270. Thus Option (B) is the correct answer.
Note:
We know that Division algorithm states that for any integer \[a\] and \[b\] be any positive integer such that there exists two unique integers, then we have \[a = qb + r\] where \[r\] is an integer equal to \[0\] or less than \[b\] i.e., \[0 \le r < b\]. We say \[a\] is the dividend, \[b\] is the divisor, \[q\] is the quotient, \[r\] is the remainder. Linear equation of two variables is an equation with the highest degree of variable as 1. Here, we have equated the two equations because both values are equal to \[y\].
Here, we have to find the smaller number. We will assign variables for the smaller and the larger number. Then we will form linear equations in two variables from the given conditions. We will equate both the equation and solve it further to find the solution of the variables which is the smaller and the larger number.
Complete step by step solution:
Let \[x\] and \[y\] be the smaller number and larger number respectively.
We are given that the difference of two numbers is \[1365\]. Therefore,
\[y - x = 1365\]
By rewriting the equation, we get
\[ \Rightarrow y = x + 1365\] ………………………………………………………………………………\[\left( 1 \right)\]
We are given that on dividing the larger number by the smaller, we get \[6\] as quotient and the 15 as remainder.
By division algorithm, we get
\[y = 6x + 15\] …………………………………………………………………………………\[\left( 2 \right)\]
By equating the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get
\[x + 1365 = 6x + 15\]
By rewriting the equations, we get
\[ \Rightarrow 6x - x = 1365 - 15\]
Subtracting the like terms, we get
\[ \Rightarrow 5x = 1350\]
Dividing by 5 on both the sides of the equation, we get
\[ \Rightarrow x = \dfrac{{1350}}{5}\]
\[ \Rightarrow x = 270\]
By substituting \[x = 270\] in the equation \[\left( 1 \right)\], we get
\[y = x + 1365\]
On adding the terms, we get
\[ \Rightarrow y = 270 + 1365\]
\[ \Rightarrow y = 1635\]
Thus the smaller number is 270 and the largest number is 1635.
Therefore, the smaller number is 270. Thus Option (B) is the correct answer.
Note:
We know that Division algorithm states that for any integer \[a\] and \[b\] be any positive integer such that there exists two unique integers, then we have \[a = qb + r\] where \[r\] is an integer equal to \[0\] or less than \[b\] i.e., \[0 \le r < b\]. We say \[a\] is the dividend, \[b\] is the divisor, \[q\] is the quotient, \[r\] is the remainder. Linear equation of two variables is an equation with the highest degree of variable as 1. Here, we have equated the two equations because both values are equal to \[y\].
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