In a restaurant there are two kinds of meal one costs 450 and the other costs 600. One evening the number of the people who ordered the 600 meal is one more than 3 times the number of people who ordered the 450 meal. If the receipt for that particular night was 50 100 how many people ate that evening?
Answers
Answer :-
___________________________
- Here the concept of Linear equations in Two Variables has been used. In this we make the value of one variable depend on other, so that we can find the value of both. The general form of Linear Equations in Two Variables is given by :-
• ax + by + c = 0
• px + qy + d = 0
___________________________
★ Solution :-
Given,
» There are two types of meals.
» Cost of one type of meal = ₹ 450
» Cost of second type of meal = ₹ 600
» Total payment receipt received = ₹ 50,100
☯️ Let the number of people who ordered ₹ 450 meal be 'x'.
☯️ And let the number of people who ordered ₹ 600 meal be 'y'.
___________________________
According to the question,
~ Case I :-
The number of the people who ordered the 600 meal is one more than 3 times the number of people who ordered the 450 meal.
So,
y = 3x + 1 ... (i)
~ Case II :-
✍ Total cost of ₹ 450 meal = 450x
✍ Total cost of ₹ 600 meal = 600y
So,
✍ 450x + 600y = 50100
Dividing both sides by 50 we get,
✍ 9x + 12y = 1002 ... (ii)
From equation (i) and equation (ii) , we get,
➠ 9x + 12(3x + 1) = 1002
➠ 9x + 36x + 12 = 1002
➠ 45x = 1002 - 12
➠ 45x = 990
➠ x = 990/45
➠ x = 22
Hence, we get, the value of x = 22.
By applying the value of x in equation (i), we get,
➠ y = 3x + 1
➠ y = 3(22) + 1 = 66 + 1
➠ y = 67
Hence we get the value of y = 67.
________________________________
• Hence the number of people who ate ₹450 meal = 22.
• Hence tue number of people who ate ₹600 meal = 67
Then,
➠ Total number of people who ate that evening
= x + y = 22 + 67 = 89
★ Hence 89 people ate that evening.
_____________________
- Linear Equations are the equations formed using constants and variables.
- On the basis of variables :-
- Linear Equation in One Variable
- Linear Equation in Two Variable
- Linear Equation in Poly Variable
- Methods to solve Linear Equations in Two Variables :-
- Cross Multiplication Method
- Substitution Method
- Elimination Method
- Reducing the pair method
Answer:
Here the concept of Linear equations in Two Variables has been used. In this we make the value of one variable depend on other, so that we can find the value of both. The general form of Linear Equations in Two Variables is given by :-
• ax + by + c = 0
• px + qy + d = 0
___________________________
★ Solution :-
Given,
» There are two types of meals.
» Cost of one type of meal = ₹ 450
» Cost of second type of meal = ₹ 600
» Total payment receipt received = ₹ 50,100
☯️ Let the number of people who ordered ₹ 450 meal be 'x'.
☯️ And let the number of people who ordered ₹ 600 meal be 'y'.
___________________________
According to the question,
~ Case I :-
The number of the people who ordered the 600 meal is one more than 3 times the number of people who ordered the 450 meal.
So,
y = 3x + 1 ... (i)
~ Case II :-
✍ Total cost of ₹ 450 meal = 450x
✍ Total cost of ₹ 600 meal = 600y
So,
✍ 450x + 600y = 50100
Dividing both sides by 50 we get,
✍ 9x + 12y = 1002 ... (ii)
From equation (i) and equation (ii) , we get,
➠ 9x + 12(3x + 1) = 1002
➠ 9x + 36x + 12 = 1002
➠ 45x = 1002 - 12
➠ 45x = 990
➠ x = 990/45
➠ x = 22
Hence, we get, the value of x = 22.
By applying the value of x in equation (i), we get,
➠ y = 3x + 1
➠ y = 3(22) + 1 = 66 + 1
➠ y = 67
Hence we get the value of y = 67.
________________________________
• Hence the number of people who ate ₹450 meal = 22.
• Hence tue number of people who ate ₹600 meal = 67
Then,
➠ Total number of people who ate that evening
= x + y = 22 + 67 = 89
★ Hence 89 people ate that evening.
_____________________
-★Moretoknow:−
Linear Equations are the equations formed using constants and variables.
On the basis of variables :-
Linear Equation in One Variable
Linear Equation in Two Variable
Linear Equation in Poly Variable
Methods to solve Linear Equations in Two Variables :-
Cross Multiplication Method
Substitution Method
Elimination Method
Reducing the pair method